Most ambiguous and inconsistent phrases and notations in maths

139

What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

$f^{-1}(x)$ can be an inverse and a preimage and sometimes even $frac{1}{f(x)}$.

$f^2(x)$ can be $(fcirc f)(x)$ and $(f(x))^2$.

$f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

And for some functions the parentheses for the argument are omitted: $f:x = f(x)$.

So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $left(f^{(2)}circ f^{(-3)}right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

edited Nov 20 '14 at 18:48

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6 revs, 3 users 63%
fvel

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$f^{-1}$ can be $frac{1}{f}$. When you look at rings of (continuous, maybe) real or complex-valued functions on a space $X$, then it's natural to denote the multiplicative inverse of $f$ (if it exists, i.e. $f(x)neq 0$ for all $xin X$) by $f^{-1}$.
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– Daniel Fischer♦
Nov 16 '14 at 14:27

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I hate it when symbols $infty$, $omega$ and $aleph_0$ are misused.
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– user2345215
Nov 16 '14 at 14:31

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"nice" or "good" are used (depending on the context) to ensure that the setting is sufficient to prove theorems. Also I don't like "properly" to much. In german it is even worse.
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– Daniel Valenzuela
Nov 16 '14 at 14:44

27

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Technically there is no unambiguous way to write plain scalar multiplication. * is function convolution, $cdot$ is dot product, $times$ is cross product, and just putting the terms next to each other could be function application. I suppose you could divide by the reciprocal, but that is so evil we'll forget I said it.
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– hacatu
Nov 16 '14 at 18:17

25

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Most of the examples listed are perceived as problematic only because for some reason which I cannot even imagine people seem to think that a notation can have one meaning, or that there is anything that trumps convenience... The only possible sin with respect to notation is not being explicit about what one means with it. There is no sacrosant association between upper indices and exponentiation, say...
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– Mariano Suárez-Álvarez
Nov 17 '14 at 3:00

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show 19 more comments

139

What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

$f^{-1}(x)$ can be an inverse and a preimage and sometimes even $frac{1}{f(x)}$.

$f^2(x)$ can be $(fcirc f)(x)$ and $(f(x))^2$.

$f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

And for some functions the parentheses for the argument are omitted: $f:x = f(x)$.

So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $left(f^{(2)}circ f^{(-3)}right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

edited Nov 20 '14 at 18:48

community wiki

6 revs, 3 users 63%
fvel

12

$begingroup$
$f^{-1}$ can be $frac{1}{f}$. When you look at rings of (continuous, maybe) real or complex-valued functions on a space $X$, then it's natural to denote the multiplicative inverse of $f$ (if it exists, i.e. $f(x)neq 0$ for all $xin X$) by $f^{-1}$.
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– Daniel Fischer♦
Nov 16 '14 at 14:27

10

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I hate it when symbols $infty$, $omega$ and $aleph_0$ are misused.
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– user2345215
Nov 16 '14 at 14:31

3

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"nice" or "good" are used (depending on the context) to ensure that the setting is sufficient to prove theorems. Also I don't like "properly" to much. In german it is even worse.
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– Daniel Valenzuela
Nov 16 '14 at 14:44

27

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Technically there is no unambiguous way to write plain scalar multiplication. * is function convolution, $cdot$ is dot product, $times$ is cross product, and just putting the terms next to each other could be function application. I suppose you could divide by the reciprocal, but that is so evil we'll forget I said it.
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– hacatu
Nov 16 '14 at 18:17

25

$begingroup$
Most of the examples listed are perceived as problematic only because for some reason which I cannot even imagine people seem to think that a notation can have one meaning, or that there is anything that trumps convenience... The only possible sin with respect to notation is not being explicit about what one means with it. There is no sacrosant association between upper indices and exponentiation, say...
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 3:00

|
show 19 more comments

139

What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

$f^{-1}(x)$ can be an inverse and a preimage and sometimes even $frac{1}{f(x)}$.

$f^2(x)$ can be $(fcirc f)(x)$ and $(f(x))^2$.

$f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

And for some functions the parentheses for the argument are omitted: $f:x = f(x)$.

So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $left(f^{(2)}circ f^{(-3)}right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

edited Nov 20 '14 at 18:48

community wiki

6 revs, 3 users 63%
fvel

What are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts?

For instance, a function $f$:

$f^{-1}(x)$ can be an inverse and a preimage and sometimes even $frac{1}{f(x)}$.

$f^2(x)$ can be $(fcirc f)(x)$ and $(f(x))^2$.

$f^{(2)}(x)$ on the other hand, is the second derivative, even though adding parentheses to a number usually does nothing.

And for some functions the parentheses for the argument are omitted: $f:x = f(x)$.

So how should $f^{(2-3)}(x)$ be interpreted? $f^{(-1)}$, an integral of $f$? or a composition, $left(f^{(2)}circ f^{(-3)}right)(x)$? Or just $f^{2-3}(x) = f^{-1}(x)$?

Another example is mathematicians notorious use of the word normal to describe... normal things?

Using similar symbols and expressions for different things is unavoidable, but it can create some ambiguity when first introduced to their other uses.

soft-question notation big-list

edited Nov 20 '14 at 18:48

community wiki

6 revs, 3 users 63%
fvel

edited Nov 20 '14 at 18:48

community wiki

6 revs, 3 users 63%
fvel

edited Nov 20 '14 at 18:48

community wiki

6 revs, 3 users 63%
fvel

community wiki

6 revs, 3 users 63%
fvel

community wiki

6 revs, 3 users 63%
fvel

12

$begingroup$
$f^{-1}$ can be $frac{1}{f}$. When you look at rings of (continuous, maybe) real or complex-valued functions on a space $X$, then it's natural to denote the multiplicative inverse of $f$ (if it exists, i.e. $f(x)neq 0$ for all $xin X$) by $f^{-1}$.
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– Daniel Fischer♦
Nov 16 '14 at 14:27

10

$begingroup$
I hate it when symbols $infty$, $omega$ and $aleph_0$ are misused.
$endgroup$
– user2345215
Nov 16 '14 at 14:31

3

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"nice" or "good" are used (depending on the context) to ensure that the setting is sufficient to prove theorems. Also I don't like "properly" to much. In german it is even worse.
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– Daniel Valenzuela
Nov 16 '14 at 14:44

27

$begingroup$
Technically there is no unambiguous way to write plain scalar multiplication. * is function convolution, $cdot$ is dot product, $times$ is cross product, and just putting the terms next to each other could be function application. I suppose you could divide by the reciprocal, but that is so evil we'll forget I said it.
$endgroup$
– hacatu
Nov 16 '14 at 18:17

25

$begingroup$
Most of the examples listed are perceived as problematic only because for some reason which I cannot even imagine people seem to think that a notation can have one meaning, or that there is anything that trumps convenience... The only possible sin with respect to notation is not being explicit about what one means with it. There is no sacrosant association between upper indices and exponentiation, say...
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 3:00

|
show 19 more comments

12

$begingroup$
$f^{-1}$ can be $frac{1}{f}$. When you look at rings of (continuous, maybe) real or complex-valued functions on a space $X$, then it's natural to denote the multiplicative inverse of $f$ (if it exists, i.e. $f(x)neq 0$ for all $xin X$) by $f^{-1}$.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 14:27

10

$begingroup$
I hate it when symbols $infty$, $omega$ and $aleph_0$ are misused.
$endgroup$
– user2345215
Nov 16 '14 at 14:31

3

$begingroup$
"nice" or "good" are used (depending on the context) to ensure that the setting is sufficient to prove theorems. Also I don't like "properly" to much. In german it is even worse.
$endgroup$
– Daniel Valenzuela
Nov 16 '14 at 14:44

27

$begingroup$
Technically there is no unambiguous way to write plain scalar multiplication. * is function convolution, $cdot$ is dot product, $times$ is cross product, and just putting the terms next to each other could be function application. I suppose you could divide by the reciprocal, but that is so evil we'll forget I said it.
$endgroup$
– hacatu
Nov 16 '14 at 18:17

25

$begingroup$
Most of the examples listed are perceived as problematic only because for some reason which I cannot even imagine people seem to think that a notation can have one meaning, or that there is anything that trumps convenience... The only possible sin with respect to notation is not being explicit about what one means with it. There is no sacrosant association between upper indices and exponentiation, say...
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 3:00

$f^{-1}$ can be $frac{1}{f}$. When you look at rings of (continuous, maybe) real or complex-valued functions on a space $X$, then it's natural to denote the multiplicative inverse of $f$ (if it exists, i.e. $f(x)neq 0$ for all $xin X$) by $f^{-1}$.

– Daniel Fischer♦
Nov 16 '14 at 14:27

I hate it when symbols $infty$, $omega$ and $aleph_0$ are misused.

– user2345215
Nov 16 '14 at 14:31

"nice" or "good" are used (depending on the context) to ensure that the setting is sufficient to prove theorems. Also I don't like "properly" to much. In german it is even worse.

– Daniel Valenzuela
Nov 16 '14 at 14:44

Technically there is no unambiguous way to write plain scalar multiplication. * is function convolution, $cdot$ is dot product, $times$ is cross product, and just putting the terms next to each other could be function application. I suppose you could divide by the reciprocal, but that is so evil we'll forget I said it.

– hacatu
Nov 16 '14 at 18:17

Most of the examples listed are perceived as problematic only because for some reason which I cannot even imagine people seem to think that a notation can have one meaning, or that there is anything that trumps convenience... The only possible sin with respect to notation is not being explicit about what one means with it. There is no sacrosant association between upper indices and exponentiation, say...

– Mariano Suárez-Álvarez
Nov 17 '14 at 3:00

|
show 19 more comments

23 Answers
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'The function $f(x)$'. No, the function is $f$.

Let $f$ and $g$ be real differentiable functions defined in $mathbb R$. Some people denote $(fcirc g)'$ by $dfrac{mathrm df(g(x))}{mathrm dx}$. Contrast with the above. I discuss this in greater detail here.

The differential equation $y'=x^2y+y^3$. Just a minor variant of 1. Correct would be $y'=fy+y^3$ where $fcolon Ito mathbb R, xmapsto x^2$, for some interval $I$.

This is one I find particularly disgusting. "If $t(s)$ is a function of $s$ and it is invertible, then $s(t)$ is the inverse", lol what? The concept of 'function of a variable' isn't even definable in a satisfiable way in $sf ZFC$. Also $left(frac{mathrm dy}{mathrm dx}right)^{-1}=frac{mathrm dx}{mathrm dy}$. Contrast with 1.

In algebra it's common to denote the algebraic structure by the underlying set.

When $langle ,cdot,rangle$ is a function which takes sets as their inputs, it's common do abuse $langle{x} rangle$ as $langle xrangle$. More generally it's common to look at at finite set ${x_1, ldots ,x_n}$ as the finite sequence $x_1, ldots ,x_n$. This happens for instance in logic. Also in linear algebra and it's usual to go even further and talk about 'linearly independent vectors' instead of 'linearly independent set' - this only an abuse when linear (in)dependence is defined for sets instead of 'lists'.

'Consider the set $A={xin mathbb Rcolon P(x)}$'. I'm probably the only person who reads this as the set being the whole equality $A={xin mathbb Rcolon P(x)}$ instead of $A$ or ${xin mathbb Rcolon P(x)}$, in any case it is an abuse. Another example of this is 'multiply by $1=frac 2 2$'.

Denoting by $+$ both scalar addition and function addition.

Instead of $((varphiland psi)to rho)$ people first abandon the out parentheses and use $(varphiland psi)to rho$ and then $land$ is given precedence over $to$, yielding the much more common (though formally incorrect) $varphiland psito rho$.

Even ignoring the problem in 1., the symbol $int x,mathrm dx=frac {x^2}2$ is ambiguous as it can mean a number of things. Under one of the common interpretations the equal sign doesn't even denote an equality. I allude to that meaning here, (it is the same issue as with $f=O(g)$).

There's also the very common '$ldots$' mentioned by Lucian in the comments.

Lucian also mentions $mathbb C=mathbb R^2$ which is an abuse sometimes, but not all the time, depending on how you define things.

Given a linear map $L$ and $x$ on its domain, it's not unusual to write $Lx$ instead of $L(x)$. I'm not sure if this can even be considered an abuse of notation because $Lx$ is meaningless and we should be free to define $Lx:=L(x)$, there's no ambiguity. Unless, of course, you equate linear maps with matrices and this is an abuse. On the topic of matrices, it's common to look at $1times 1$ matrices as scalars.

Geometers like to say $mathbb Rsubseteq mathbb R^2subseteq mathbb R^3$.

Using $mathcal M_{mtimes n}(mathbb F)$ and $mathbb F^{mtimes n}$ interchangeably. On the same note, $A^{m+ n}=A^mtimes A^n$ and $left(A^mright)^n=A^{mtimes n}$.

I don't know how I forgot this one. The omission of quantifiers.

Calling 'well formed formulas' by 'formulas'.

Saying $forall x(P(x)to Q(x))$ is a conditional statement instead of a universal conditional statement.

Stuff like $exists yP(x,y)forall x$ instead of (most likely, but not certainly) $exists yforall xP(x,y)$.

The classic $u=x^2implies mathrm du=2xmathrm dx$.

This one disturbs me deeply. Sometimes people want to say "If $A$, then $B$" or "$Aimplies B$" and they say "If $Aimplies B$". "If $Aimplies B$" isn't even a statement, it's part of an incomplete conditional statement whose antecedent is $Aimplies B$. Again: mathematics is to be parsed with priority over natural language.

Saying that $x=yimplies f(x)=f(y)$ proves that $f$ is a function.

Using $f(A)$ to denote ${f(x)colon xin A}$. Why not stick to $f[A]$ which is so standard? Another possibility is $f^to(A)$ (or should it be square brackets?) which I learned from egreg in this comment.

edited Apr 13 '17 at 12:20

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16 revs
Git Gud

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Let $f colon mathbb{R} to C(mathbb{R},mathbb{R})$. The function $f(x)$ ...
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– Daniel Fischer♦
Nov 16 '14 at 15:00

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Please, Git Gud, don't treat my note as a suggestion, unless your goal is obfuscation; your "Consider the set $A$, where $A = ...$," is far more readable for the rest of us, and also works for your "peculiarly syntactic" reading.
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– John Hughes
Nov 16 '14 at 15:42

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I've always interpreted constructions like #7 to mean "Consider the set $A$, which equals ${ x in mathbb{R} : P(x) }$," where the equals sign itself introduces an appositive. In my mind, $``="$ is a word that takes on two grammatical roles, acting as a verb ($2+2$ equals $4$) or a subordinating conjunction (as above), depending on the context.
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– David Zhang
Nov 16 '14 at 17:20

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Wow, Git Gud, it must be really painful for you to read any mathematics at all! How do you stand it?
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– TonyK
Nov 16 '14 at 22:26

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This is getting slightly ridiculous. There are conventions and we're a mathematical society. We can interpret things. If everyone was pedantic like this, we would have already killed each other.
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– Pedro Tamaroff♦
Nov 20 '14 at 1:49

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The inconsistent treatment of raising trig functions to powers: $$ sin^n x ,.$$

Seriously, starting ab inito $$sin^2 x$$ could mean either $$sin( sin(x) )$$ if you are a quantum mechanic and like to see everything as an operator or as
$$(sin x)^2$$
which is the conventional meaning.

So why is $$sin^{-1} x$$ used for $$arcsin x$$ (which is vaguely consistent with the former) instead of $$(sin x)^{-1} $$ in keeping with the latter.

edited Nov 19 '14 at 5:33

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dmckee

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@Mariano It's pretty common to see $sin^2 x = (sin x)^2$ and $sin^{-1} x = arcsin x$ both used in precalculus or calculus textbooks. So, one or both of them (definitely including the second one) is an abuse.
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– Ryan Reich
Nov 17 '14 at 15:04

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@Mariano Even if it's just the superposition that's wrong, that superposition is itself a common convention. Also: $sin^{-1} x$ is definitely an abuse of notation. It is not the inverse function, but refers to the inverse of a specific restriction of $sin x$ to a domain on which it's injective. Note that $cos^{-1}$ and $tan^{-1}$ are also used, and their restrictions are different. This notation can only be defined in an extremely kludgy way. You seem to believe that notation can't be abused if it's well-defined: well, this notation can't be well-defined.
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– Ryan Reich
Nov 17 '14 at 18:47

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Maybe it's time to realize trig functions are functions. If we define a function f(x) = 3x, you would never write f40 = f(40), yet people will gladly write sin40 = sin(40). The parenthesis are (should be, at least) mandatory, which means anything outside them should be evaluated after the function. So sin(40)^2 = sin(40)*sin(40). I could be wrong, but isn't f(40)^2 commonly interpreted as f(40)*f(40)?
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– Dan Watkins
Nov 18 '14 at 4:28

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@DanWatkins, I don't understand why you think $f(x)$ is not very often written simply $fx$... That happens all the time! Why would anyone dream of making those parenthesis that you want mandatory? What possible sensible interpretations of $sin40$ can you think of apart from the intended one? Even if you can think of another one, just as with anything related to human communication what is mandatory is that you become accustumed with the usual traditions...
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– Mariano Suárez-Álvarez
Nov 19 '14 at 9:33

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@JackM The $sin^2x$ notation is historical; in trigonometric formulas it's common to have those powers and people devised that way for getting less clumsy formulas. The concept of composition of function as an operation came much later and the notation remained, which is surely more practical; how many times did you find in your life $sin(sin x)$ compared to $(sin x)^2$? What I find abusive is $sin^{-1}$ for the reasons well expressed by Ryan Reich.
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– egreg
Nov 20 '14 at 11:38

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Double factorial $n!!=n(n-2)(n-4)cdots$, where the product run through positive integers.

At the first time this notation confused me a lot because it looks the same as $(n!)!$ .

Similar argument about multifactorial.

edited Nov 19 '14 at 20:49

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Danylo Dubinin

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Not to mention an exclamation mark at the end of a sentence ending with a number: The answer is 10!
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– rabota
Nov 17 '14 at 20:13

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I wholeheartedly agree.
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– Jeel Shah
Nov 17 '14 at 20:14

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@barto I would say most writers should phrase it The answer is 10!. with the period at the end if they mean 10 factorial, and without it to mean 10 only, to be unambiguous.
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– Ryan
Nov 18 '14 at 14:04

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Your link says multinomial, but it points to multifactorial. Is this another abuse of (an)notation?
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– Marc van Leeuwen
Nov 19 '14 at 18:01

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@Ryan: What about the question? The answer is 10!!?
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– jxh
Nov 19 '14 at 19:38

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In the first year at college, I was really confused with the notion of a sequence

$${ x_n : n in mathbb N }$$

because this could also be a set! Then I discovered

$$(x_n)_{n in mathbb N}$$

And now I am fine with sequences.

answered Nov 17 '14 at 0:40

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ThePortakal

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I don't get it, ${ x_n : n in mathbb N }$ is unmistakably the image of the sequence. What's the abuse I'm missing here?
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– Git Gud
Nov 17 '14 at 0:50

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@GitGud ${ x_1, x_2, x_3, dots } = { x_2, x_1, x_3, dots }$ as sets, but not as sequences. (And I think I abused $dots$ here)
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– ThePortakal
Nov 17 '14 at 0:56

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This is a real mistake. It is more problematic in uses like «the set ${x_1,x_2}$ is linarly independent,» which is usually not intended to include the claim that when $x_1=x_2neq0$, the one-element set mentioned in linearly indepenent.
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– Mariano Suárez-Álvarez
Nov 17 '14 at 4:41

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A sequence in the set $X$ should be defined as a function $x:mathbb{N} rightarrow X$ where we denote the image $x(n)$ of $n in mathbb{N}$ by $x_{n}$.
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– Vincent Pfenninger
Nov 17 '14 at 17:29

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I see ${x_n:nin mathbb{N}}$ for more often than $(x_n)_{nin mathbb{N}}$, but I've never cared for it. Set notation to me implies unordered list without repeated elements, but a sequence is ordered and may have repeats. Of course, there is an implicit order inherited from the order on $mathbb{N},$ but using set notation still seems like an abuse.
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– Randy E
Nov 17 '14 at 20:02

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We often write $f(n) = O(g(n))$, when in fact $O(g(n))$ is a set, and should be written as $f(n) in O(g(n))$. Similarly for other asymptotic notation, such as $Theta$ and $Omega$.

answered Nov 17 '14 at 2:04

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nexolute

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Do you mean $f in O(g)$? From the typical usage of these symbols, I'd expect $f(n)$ and $g(n)$ to be numbers, and things like $4 in O(8)$ do not make sense.
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– Prateek
Nov 17 '14 at 10:39

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Yeah, and that major inconsistency that became second nature to us now.
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– nexolute
Nov 17 '14 at 11:01

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@Prateek Identifying a number with the constant function attaining that value (which is a useful abuse of notation often employed), the statement $4in O(8)$ makes perfect sense (and is correct).
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– Daniel Fischer♦
Nov 17 '14 at 11:37

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@Prateek: If you find that annoying, I bet you'll really hate things like $frac{rm d}{{rm d}x}x^2=2x$ (since, if you insisted on assigning a value to $x$ and evaluating from the inside out, you'd end up with something totally nonsensical like $frac{rm d}{{rm d}5}25=10$). It makes more sense if you allow "unevaluated expressions in one or more variables" as first-class objects in your mathematical framework (and not just when buried inside function definitions). If you don't, you'll have to go for something like $fin O(nmapsto n^2)$ to make asymptotical notation rigorous.
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– Ilmari Karonen
Nov 17 '14 at 11:42

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There is really no problem in writing a function as $x^2$. Define $x$ to be the identity function. Define the ring operations on functions as usual. Then $x^2$ is the function which takes $t$ to $t^2$, etc.
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– Martin Brandenburg
Nov 18 '14 at 9:02

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$mathbb N;;;;;;;;;;;;$

answered Nov 19 '14 at 18:35

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TonyK

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@DanielFischer I propose we let $mathbb{N}$ denote all natural numbers including $0$, and let $mathbb{C}$ denote all counting numbers.
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– Frank Vel
Nov 19 '14 at 19:48

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What the [censored] are "counting numbers"?
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– Daniel Fischer♦
Nov 19 '14 at 19:57

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@fvel: yeah, and $mathbb{K}$ all complex numbers, and $mathbb{F}$ some generic field, and...
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– leftaroundabout
Nov 19 '14 at 19:58

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@fvel Apart from the obvious, "You can have zero of a thing," it gives $mathbb{N}$ an identity under addition, which is nice. On the other hand, people (well, the ones who aren't programmers, anyway) seem to like to count from one, and you don't have to worry about dividing by zero in $mathbb{Z}^+$.
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– Stuart Olsen
Nov 20 '14 at 2:27

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Then $+$ would be an example too.
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– egreg
Nov 20 '14 at 11:32

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Einstein summation convention is a self-explanatory example.

Fourier transforms

I feel the majority of people (myself included) abuse notation when describing Fourier transforms.

For example, it's common to see:

$$mathcal{F}{{f(x)}} = F(omega)$$

to which my natural response is: uhm, no, I believe you mean

$$mathcal{F_x}{{f}} = F$$

or perhaps

$$mathcal{F}{f(,cdot,)} = F(,cdot,)$$

The original is clearly incorrect because $f(x)$ is the value of $f$ at some point $x$, and its Fourier transform $F$ is defined as $$F(omega) = f(x) delta(omega)$$
which is clearly not what is intended.

edited Nov 17 '14 at 15:37

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4 revs
Mehrdad

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In what possible way is Einstein summation inconsisten and or an abused notation?!
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– Mariano Suárez-Álvarez
Nov 17 '14 at 3:54

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Huh? Why would upper indices have one and only one meaning? If in some context you are never going to use exponents and you will use thousands of times indices of coordinates, why would you not use upper indices for the latter role? The notation, moreover, makes it impossible to write things which simply do not make sense it in the context it is used (for example: powers of coordinates, which make no sense in pretty much any situation...)
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– Mariano Suárez-Álvarez
Nov 17 '14 at 4:07

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Of course, if you ar not familiar with the notation, you will have to get used to it —just as if you are not familiar with English you are going to get some familiarity before trying to read Shakespeare.
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– Mariano Suárez-Álvarez
Nov 17 '14 at 4:09

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@MarianoSuárez-Alvarez: I don't follow your reasoning. Every widely-used abuse of notation happens for a reason; just because there's a reason or context behind it doesn't mean it's not an abuse of notation. When people write $mathcal{F}{f(x)} = F(omega)$ it's 100% clear from the context that they don't intend to take the FT of a single number, but that doesn't make it any less abusive of the notation.
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– Mehrdad
Nov 17 '14 at 4:09

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@MarianoSuárez-Alvarez: Could you give me a list of what you consider to be abuses of notation? I feel like I can't find any common ground with you to base my reasons off.
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– Mehrdad
Nov 17 '14 at 4:12

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Don't know whether these count as notational abuses as such, but a few common causes of confusion I have come across are

begin{align}
&log(x)text{ and }ln(x)\
&sin^2(x)text{ and }sin(x)^2\
&sin^{-1}(x)text{ and }arcsin(x)\
&log_2(x)text{ meaning }log(log(x)),text{ & }log_2(x)text{ meaning base }2\
&log(log(x))text{ and }loglog x text{ etc. }\
&mathbb{S}^ntext{ in topology & }mathbb{S}^ntext{ in geometry }\
end{align}

re last one, from MathWorld:

... A geometer would therefore regard the object described b $ x_1^2+x_2^2=R^2 $
as a $2$-sphere, while a topologist would consider it a $1$-sphere and denote it $mathbb{S}^1$. Similarly, a geometer would regard the object described by
$x_1^2+x_2^2+x_3^2=R^2 $ as a $3$-sphere, while a topologist would call it a $2$-sphere and denote it $mathbb{S}^2$. Extreme caution is therefore advised when consulting the literature. Following the literature, both conventions are used in this work, depending on context, which is stated explicitly wherever it might be ambiguous.

This last one was included as a curiosity, but generally though, subscript and superscript is often abused / ambiguous unless explicitly stated.

... more interesting anecdotes and opinions.

edited Apr 13 '17 at 12:58

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5 revs
martin

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I haven't seen the fourth one before. So $f_2(x)$ can be yet another way of writing $(fcirc f)(x)$...
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– Frank Vel
Nov 16 '14 at 19:01

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That somebody would call the unit circle a $2$-sphere. It beggars belief.
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– Daniel Fischer♦
Nov 16 '14 at 19:15

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I know. And I'm not in the habit of shooting the messenger, so you need not be afraid of me.
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– Daniel Fischer♦
Nov 16 '14 at 19:17

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I won't shoot the messenger, but I will shoot the message author. I don't know a single geometer that disagrees with a topologist on what the $n$-sphere is.
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– user98602
Nov 17 '14 at 3:17

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@MatthewLeingang I guess it's correct to say a geometer thinks the 2-sphere is a circle. ;)
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– user98602
Nov 18 '14 at 15:33

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In financial mathematics one encounters index numbers, such as consumer price indexes, where a base year (such as 1992) is commonly specified by truly ghastly expressions such as

1992=100.

And no, they're not working modulo a divisor of 1892: they're referring to the fact that the index number for the base year is 100. It feels wrong to even bring this up in polite company.

answered Nov 18 '14 at 23:27

community wiki

Shane O Rourke

add a comment |

Calculus I,II,III: The '$dx$'s in integrals and derivatives are just notation to help keep track of the important variables in a given problem, and they're otherwise meaningless in isolation.

Real Analysis: In fact, when the integration variable is unambiguous we may as well dispense with the differentials altogether and just denote the integral of $f$ over a region $Rsubset operatorname{dom}{f}$ as,

$$int_{R}f.$$

Differential Geometry: Wait nevermind, $dx$ is a differential 1-form, $d$ is itself an operator, and the placement of '$dx$'s and so forth in notation for integrals couldn't be less optional, case ~~and~~ in point being the general Stokes' theorem:

$$int_{R}domega=oint_{partial R}omega.$$

Sorry we lied.

Edit: correctified for grammars.

edited Mar 1 '15 at 22:43

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2 revs
David H

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By "case and point" I believe you mean "case in point".
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– Mehrdad
Nov 23 '14 at 10:00

1

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@Mehrdad heh, oops. Finally corrected. Apparently automatic notifications aren't sent for comments on community wikis.
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– David H
Mar 1 '15 at 22:45

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I haven't gotten to Differential Geometry yet, but I assume that is the final and correct answer?
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– Ovi
May 10 '17 at 23:19

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In measure theory books they really use $ dx = d mu (x) $ etc as just notation to help keep track of the important variables in a given problem and there is no manifold (or even a topology) involved. Examples found in Rudin's book of real and complex analysis in the section about Fubini's theorem.
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– Gustavo
Nov 6 '17 at 14:13

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The word "trivial." The many uses of this word include:

1.) The colloquial usage as a synonym for "easy."

2.) The trivial group which consists only of the identity element.

3.) The trivial ring which consists only of the multiplicative and additive identities.

4.) A trivial solution to an equation, often when a variable equals 0 (or constant in the cases of differential equations).

5.) Similar to #2 and #3, any object which satisfies the bare minimum of some particular definition but has no further structure. Often some sort of identity or null element.

6.) A trivial application of a theorem can refer to a special case where the truth of the theorem is more or less self-evident, e.g. a theorem in set theory which is obviously true when applied to the empty set.

This list could go on forever, but I'll stop here.

Note that the converse, of course, is the word "non-trivial," which is just as ambiguous.

I'll also note that although you may consider this a trivial answer, but I have had students get tripped up by my usage of the word "trivial" in lecture, so the ambiguity must be somewhat non-trivial.

answered Nov 19 '14 at 17:34

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josh314

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I feel like your answer needs a little more justification. Right now you've shown that uses of the term "trivial" are broad but not that they are ambiguous. To me they all fall under the umbrella of "nothing much to it": the "nothing much" in the colloquial usage being in terms of effort, while in the mathematical uses, it is "nothing much" in terms of structure or complexity.
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– Rahul
Nov 20 '14 at 18:04

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I agree with you, that is answer doesn't really emphasise its ambiguity, but I feel that I am not the only one who has spend some considerable time to proof "trivial" statement, that turned out to be not so trivial at all... Some mathematicians have the tendency to call virtually every result trivial, which they have seen a proof for.
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– Stefan Mesken
Nov 20 '14 at 20:13

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There can be ambiguity when there's more than one degenerate object of a certain kind. E.g. if $V$ is a vector space, certainly ${0}subset V$ is a trivial subspace. Is $V$ also a trivial subspace?
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– Julian Rosen
Nov 22 '14 at 16:08

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My pet peeve when a paper says things like "trivially easy" about a concept I don't have any grasp on...
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– Calmarius
Jan 16 '15 at 15:27

add a comment |

My favorite has always been with Fourier transforms.

Suppose a particle is in a potential $V$ given by $$V(x) = V_0cos(x/a).$$ Then $$V(k) = V_0sqrt{dfrac{pi}{2}} left( delta(k-a) + delta(k+a) right).$$

I know physicists do this a lot. I am not sure about mathematicians.

edited Nov 17 '14 at 19:43

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2 revs
NowIGetToLearnWhatAHeadIs

4

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The same goes for change of variables. E.g. you switch from $x$ to $xi$ in $psi(x)$, and then your equation suddenly is not for $psi(x)$ but for $psi(xi)$ where $psi$ is now a different function.
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– Ruslan
Nov 21 '14 at 6:42

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A well known mathematician, with a Field Medal, also did that.
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– Felix Marin
Dec 7 '14 at 5:58

add a comment |

Here is one which I think goes back to Euler:
$$i^i = e^{-frac{pi}{2}}$$

answered Nov 18 '14 at 22:22

community wiki

AD.

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What's wrong with this? Raising by an imaginary number??
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– Frank Vel
Nov 18 '14 at 22:50

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Ah, so the equality should be $i^i = expleft(-frac{tau}{4}+ktauright)$. Kinda like if you took $int{1}mathrm{d}x=x$ by assuming $C$ is $0$.
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– Frank Vel
Nov 18 '14 at 23:14

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@AD that was exactly what I said? $tau := 2pi$.
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– Frank Vel
Nov 19 '14 at 8:13

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@AD. Probably not the most known constant, but it exists! I thought it'd be worth mentioning since this thread is about ambiguous notations.
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– Frank Vel
Nov 19 '14 at 10:40

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@MarkHurd Now you did it again, the main reason I put the post in this list is that $i^i$ is not one number - it is a rather a set of real numbers (given in the comments above) - and yes! That is remarkable!
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– AD.
Feb 24 '15 at 7:05

|
show 4 more comments

Outer measures aren't measures.

answered Nov 18 '14 at 8:31

community wiki

Martín-Blas Pérez Pinilla

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Just like how the imaginary part $y$ is neither imaginary nor a part of $x+yi$ for $x,y$ real (which might explain why some people seem to call $yi$ the imaginary part instead...). I recall there's a linguistic term (likely something along the lines of "generalised noun/term/usage/adjective") describing situations where "an X Y" does not mean "a Y that is X" but can't put my finger on it -- someone like Tim Gowers mentioned this in connection with pedagogy.
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– Vandermonde
Nov 20 '14 at 6:37

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When we place an adjective in front a noun (or after it if you are French!) it may signify something less than, or more than, or a variant of the object that noun symbolises. (Virtual keyboard is not a keyboard.) So, IMO, outer measure is a perfectly valid choice.
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– P Vanchinathan
Nov 21 '14 at 1:27

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Another example is that a multivalued function is not a function. See the entry, radial concept in A Handbook of Mathematical Discourse by Charles Wells. Note that Tim Gowers discusses multivalued functions in gowers.wordpress.com/2009/06/08/….
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– J W
Nov 22 '14 at 15:54

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See also abstractmath.org/Word%20Press/?tag=multivalued-function in Wells' Gyre&Gimble blog.
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– J W
Nov 22 '14 at 16:04

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For my own future reference as much as anyone else's, the archetypal example I couldn't remember of a noun's meaning being changed by an adjective as opposed to being restricted is "almost foo".
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– Vandermonde
Dec 17 '14 at 21:06

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show 2 more comments

This may be a regional thing, but when I started studying at a British university, so many of the lecturers wrote multiplication as a single . (full-stop). This got really confusing when, after having studied in the states I was used to
$$ 0.5 + 0.5 = 1$$
whereas here it meant
$$0.5 + 0.5 = 0 + 0 = 0$$

answered Nov 22 '14 at 10:36

community wiki

kinbiko

3

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No. $$color{red}Huge cdot$$ (cdot) is used to represent multiplication. I've never seen a lower dot being used in this situation.
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– beep-boop
Nov 22 '14 at 15:29

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en.wikipedia.org/wiki/Interpunct#English - the interpunct used to have other uses in the UK, I assume this is why full stops were used. :)
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– kinbiko
Nov 22 '14 at 15:42

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@alexqwx It varies by country, but a lower dot is definitely used some places. Why are you presupposing that because you have never seen it, it isn't a real thing?
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– 6005
Nov 23 '14 at 0:38

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@Goos It's definitely not widely used in British universities (I have studied at 3 different ones and have watched online lectures from several more, and not one of them has ever had a lecturer using a lower dot to represent multiplication).
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– beep-boop
Nov 23 '14 at 10:57

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@alexqwx It seems to be dying out, fortunately, but e.g. in Hardy/Wright, the full stop is being used for multiplication. But with some space on both sides, so it'd be $0, .,5 + 0 , .,5 = 0$ (or perhaps wider spacing), which in print is distinguishable well enough from $0.5 + 0.5$. In handwriting, it may be indistinguishable.
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– Daniel Fischer♦
Nov 23 '14 at 12:41

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show 2 more comments

The inconsistency between the reading of "Negative" versus "Minus" has, in my opinion, been a thorn in the side of every teacher and student since their acceptance.

edited Apr 13 '17 at 12:21

community wiki

2 revs
skill patrol

add a comment |

$mathbb N subset mathbb Z subset mathbb Q subset mathbb R subset mathbb C subset mathbb H$

If $f colon X rightarrow Y$ is a map and $A subseteq X$, then people often write $f(A)$ to denote $f'' A := { y in Y mid exists x in X : f(x) = y }$, which can be quite confusing in cases where $A in X$.

"Canonical" ...

$(a,b,c) = ((a,b),c) = (a,(b,c)) = f,$ where $f colon {0,1,2} rightarrow {a,b,c}$ is such that $f(0) = a$, $f(1) = b$ and $f(2) = c$.

"regular", "dense", "dimension","rank", "computable", "recursive", "closed", "compatible", "compactification"... and other notions which, in a given context, may have several different meanings.

"$f colon X rightarrow Y$ smooth" or similar expressions, where "smooth" may refer to $f$, $X$ or $Y$.

"pictures" and "diagrams" can sometimes be ambiguous to an extent where they don't mean anything to anyone - or maybe that's just me.

the "constant" $c$.

$a < b < c in d$

$1 = left( frac 2 7 right) neq frac 2 7$

$prod_{i in I} (X_i, le_i) subseteq prod_{i in J} (Y_j, le_j)$

edited Nov 20 '14 at 19:11

community wiki

4 revs, 2 users 98%
Einer

2

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What's wrong with 1.?
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– Frank Vel
Nov 20 '14 at 10:25

3

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@fvel There are different ways to construct $mathbb Z, mathbb Q, ldots$ from $mathbb N$. Following the "standard" definitions in set theory, one has $mathbb N = omega$ such that each $n in mathbb N$ has cardinality $|n| = n < aleph_0$, while each element $z in mathbb Z$ is a subset of $mathbb N times mathbb N$ of size $|z| = aleph_0$. These identifications basically forget about the underlying sets and deal only with the "induced" structures.
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– Stefan Mesken
Nov 20 '14 at 10:46

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@fvel Maybe, I should give an example how this identification can cause trouble when one ignores that fact, that they typically are not subsets: In the "standard" construction of $mathbb R$, every element $x in mathbb R$ has size $|x| = 2^{aleph_0}$. If we had $mathbb N subseteq mathbb R$, then pick any $n in mathbb N$. As $n in mathbb N subseteq mathbb R$, $n$ itself has cardinality $2^{aleph_0}$. Now, $mathbb N = omega$ is a transitive set, so $n subseteq mathbb N$. This yields $2^{aleph_0} = |n| le |mathbb N| = aleph_0$, which contradicts Cantor's Theorem.
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– Stefan Mesken
Nov 20 '14 at 11:27

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That is an absurd objection, really. Anything coming out of a specific construction of the real numbers is absolutely irrelevant...
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– Mariano Suárez-Álvarez
Nov 20 '14 at 23:07

3

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Anf, to be honest, I don't even see what the standard construction of the real numbers is for you!
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:11

add a comment |

Lebesgue and (proper or improper) Riemann integrals written in the same way, ex.: $$int_a^b f(x)dx.$$
I have even found "mixed integrals" where $int_{Acup B}f(x)+g(x)dx$ means (by writing the measure in the Lebesgue integral and marking the Riemann integral with $mathscr{R}$) $int_Af+gdmu+int_Afdmu+mathscr{R}int_B g(x)dx$ (p. 422 here).

Lebesgue-Stieltjes and Riemann-Stieltjes integrals written both$$int_a^bf(x)dPhi(x).$$

Limit point can be found as meaning accumulation point, or adherent point, or point that is the limit of a subsequence of a sequence. I have found them used for all the three things even in the same textbook. I am not aware of other possible usages of the term.

Hilbert spaces sometimes intended to be separable, sometimes not necessarily separable.

Writing $A=B$ for isomorphisms $Asimeq B$ is something that can be confusing for less advanced students, especially if done to prove something without explaining why the isomorphism proves the desired result.

I find it particularly confusing when books do not specify the scalar field associated to a vector space which is beeing discussed, and I find it even more confusing when texts do not specify domains and codomains of maps. Such omissions are not too rare in engineering-oriented or older textbooks in general.

edited Apr 28 '16 at 16:00

community wiki

3 revs, 2 users 97%
Self-teaching worker

add a comment |

"X is dense in itself" is not equivalent to "X is dense in X".

An ellipse is not an elliptical curve.

I absolutely detest the "infinity is not a number, it's just a concept".

I believe it's already been addressed - natural numbers, whole numbers, counting numbers.

answered Mar 31 '15 at 19:12

community wiki

Paul Foster

2

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Do you mean an ellipse is not an elliptic curve?
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– A.P.
Mar 31 '15 at 19:30

add a comment |

"$subset$" is called proper subset (http://mathworld.wolfram.com/ProperSubset.html) and for example if $A = {1,2,3}$ then by definition, ${1,2,3}$ is not a proper subset of $A$ so we cannot write ${1,2,3} subset A$ or $A subset A$ simply. Instead we use $subseteq$ but for some of the people, use of $subset$ is ambiguous and it can include the equality of the sets. In other words, if $A subset B$ and $B subset A$ then $A = B$ rather than a contradiction.

Here is an example of an ambiguity from MSE: Proof verification: prove $Asubseteq B$ if and only if $Acap B=A$.

answered Jan 8 '18 at 5:00

community wiki

ArsenBerk

add a comment |

-1

When we list seasons it goes, summer, autumn, winter, spring, summer, autman, and one says seasons occur cyclically.

In the group denoted (by those without broken pieces of chalk) as $mathbf{Z}$ the elements are (half of them) go like this 1,2,3,4, etc without ever repeating and yet it is called the infinite cyclic group!

answered Nov 20 '14 at 14:44

community wiki

P Vanchinathan

4

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This has absolutely nothing to do with the question.
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– Mariano Suárez-Álvarez
Nov 20 '14 at 23:09

5

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We call crrtain graphs «trees» and they are not trees! :-/
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– Mariano Suárez-Álvarez
Nov 20 '14 at 23:10

2

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@ Mariano Su'arez-Alvarez: The title of the post says "inconsistent phrase" and the phrase "cyclic group" for a group where there is no periodicity definitely qualifies for it.
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– P Vanchinathan
Nov 21 '14 at 1:00

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About trees: Mathematicians try to name things by analogy. Trees of botanical kind, after shedding their leaves, resemble those of combinatorial kind and are perfectly valid. I think OP wanted to know if the analogy behind the name actually contradicts the concept it tries to illustrate.
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– P Vanchinathan
Nov 21 '14 at 2:33

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Well, infinite cyclic groups are quite analogous to the finite ones, and I would say the analogy is infinitely closer than between trees and trees.
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– Mariano Suárez-Álvarez
Nov 21 '14 at 2:49

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show 2 more comments

-2

$$bigcup { A ,B, C} = A cup B cup C $$ or $$bigcup mathcal A $$ where $mathcal A $ is a set of sets. It should be

$$bigcup_{X in mathcal A} X. $$ This can come up in topology sometimes (unioning over a collection of open sets), and can get rather confusing if you start doing things (edit:) like $left(bigcup mathcal Aright) cup U cup V$ where $U$ and $V$ are just sets (eg. when adding open sets to a cover).

Edit: It's like defining $sum S = sum_{xin S} x$ where $S subset mathbb R$ is finite (or ordered). Then usually you expect in something like $(sum S) + x + y$ for $S$, $x$, and, $y$ to be the same sort of thing.

edited Nov 20 '14 at 8:53

community wiki

3 revs
fhyve

7

$begingroup$
This is just wrong, and rather a misunderstanding on your part. What do you mean by "the union of a single set is the set"? This is false in general. For sure, $bigcup {x}=x$, but $bigcup x$ does not need to be $x$. Consider for example $x={emptyset}$. This is a singleton, so $xneemptyset=bigcup x$. You are probably more used to seeing $cup$ as a binary relation, written in between two sets, while $bigcup$ is unary, written preceding the set it acts on. Perhaps that's the source of the confusion. The notation is quite precise here, there is no ambiguity.
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– Andrés E. Caicedo
Nov 20 '14 at 8:16

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I edited to correct that out. I was mistakenly generalizing the case of the intersection of one $A$ set when considered as a subset of some ambient set $S$, in which case we can define it as a one object product as sets over $S$.
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– fhyve
Nov 20 '14 at 8:43

$begingroup$
I remember something that uses this sort of convention coming up in my topology class where it was definitely abusive, and where doing it rigorously would be a bit of annoying leg work, but I can't find it in my notes :(
$endgroup$
– fhyve
Nov 20 '14 at 8:56

add a comment |

-4

$sqrt{2}$ is usually read as "root two". The degree of root should be mentioned though (like "square root two").

$tan^{-1}$ is used for $arctan$ mostly in Physics and Electronics.

$j$ is used for $sqrt{-1}$ in Electronics, while $i$ is used for it in Mathematics.

Dots are used for time derivative in Pyhsics and Control Engineering (e.g.; $ddot{y}+4dot{y}+3y = 3dot{x} + 2x + 8$)

Vectors are sometimes denoted in bold ($mathbf{x}$), with bar on ($bar{x}$) or with arrow on ($oversetrightarrow x$).

Maybe the most frustrating things is when someone uses the word "integral" in the meaning of "anti-derivative" (e.g.; "Integral is the inverse of derivation.").

The unit scalers are used in the place of units themselves (e.g.; "I bought two kilos of potatoes.")

The $nabla$ (del) operator can be very confusing sometimes.

I have never understood the purpose of omitting the preceding zero in a decimal number (e.g.; writing $.5$ instead of $0.5$).

In C++ language, log() is used for ln() and log10() is used for log().

I still don't know what $2^{3^4}$ does equal to. $2^{(3^4)}=2^{81}$ or ${(2^3)}^4=8^4$?

The division operator $/$ is misused. Example: $N/Acdot m$ is written for $N/(Acdot m)$.

edited Sep 10 '16 at 17:51

community wiki

4 revs
hkBattousai

6

$begingroup$
Well, C++ isn't all bad. In mathematics, the prevailing convention is that $log$ denotes the (a local) inverse of $exp$, and other bases are explicitly indicated ($log_{text{base}}$). Only when speaking to/writing for physicists/engineers has $ln$ somewhat widespread use.
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– Daniel Fischer♦
Nov 23 '14 at 12:50

3

$begingroup$
-1 for the C++ comment. $log$ is the standard way mathematicians write the natural log, and mathematicians never write $log$ for the base-10 log.
$endgroup$
– Charles
Sep 10 '16 at 17:30

add a comment |

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23 Answers
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23 Answers
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'The function $f(x)$'. No, the function is $f$.

Let $f$ and $g$ be real differentiable functions defined in $mathbb R$. Some people denote $(fcirc g)'$ by $dfrac{mathrm df(g(x))}{mathrm dx}$. Contrast with the above. I discuss this in greater detail here.

The differential equation $y'=x^2y+y^3$. Just a minor variant of 1. Correct would be $y'=fy+y^3$ where $fcolon Ito mathbb R, xmapsto x^2$, for some interval $I$.

This is one I find particularly disgusting. "If $t(s)$ is a function of $s$ and it is invertible, then $s(t)$ is the inverse", lol what? The concept of 'function of a variable' isn't even definable in a satisfiable way in $sf ZFC$. Also $left(frac{mathrm dy}{mathrm dx}right)^{-1}=frac{mathrm dx}{mathrm dy}$. Contrast with 1.

In algebra it's common to denote the algebraic structure by the underlying set.

When $langle ,cdot,rangle$ is a function which takes sets as their inputs, it's common do abuse $langle{x} rangle$ as $langle xrangle$. More generally it's common to look at at finite set ${x_1, ldots ,x_n}$ as the finite sequence $x_1, ldots ,x_n$. This happens for instance in logic. Also in linear algebra and it's usual to go even further and talk about 'linearly independent vectors' instead of 'linearly independent set' - this only an abuse when linear (in)dependence is defined for sets instead of 'lists'.

'Consider the set $A={xin mathbb Rcolon P(x)}$'. I'm probably the only person who reads this as the set being the whole equality $A={xin mathbb Rcolon P(x)}$ instead of $A$ or ${xin mathbb Rcolon P(x)}$, in any case it is an abuse. Another example of this is 'multiply by $1=frac 2 2$'.

Denoting by $+$ both scalar addition and function addition.

Instead of $((varphiland psi)to rho)$ people first abandon the out parentheses and use $(varphiland psi)to rho$ and then $land$ is given precedence over $to$, yielding the much more common (though formally incorrect) $varphiland psito rho$.

Even ignoring the problem in 1., the symbol $int x,mathrm dx=frac {x^2}2$ is ambiguous as it can mean a number of things. Under one of the common interpretations the equal sign doesn't even denote an equality. I allude to that meaning here, (it is the same issue as with $f=O(g)$).

There's also the very common '$ldots$' mentioned by Lucian in the comments.

Lucian also mentions $mathbb C=mathbb R^2$ which is an abuse sometimes, but not all the time, depending on how you define things.

Given a linear map $L$ and $x$ on its domain, it's not unusual to write $Lx$ instead of $L(x)$. I'm not sure if this can even be considered an abuse of notation because $Lx$ is meaningless and we should be free to define $Lx:=L(x)$, there's no ambiguity. Unless, of course, you equate linear maps with matrices and this is an abuse. On the topic of matrices, it's common to look at $1times 1$ matrices as scalars.

Geometers like to say $mathbb Rsubseteq mathbb R^2subseteq mathbb R^3$.

Using $mathcal M_{mtimes n}(mathbb F)$ and $mathbb F^{mtimes n}$ interchangeably. On the same note, $A^{m+ n}=A^mtimes A^n$ and $left(A^mright)^n=A^{mtimes n}$.

I don't know how I forgot this one. The omission of quantifiers.

Calling 'well formed formulas' by 'formulas'.

Saying $forall x(P(x)to Q(x))$ is a conditional statement instead of a universal conditional statement.

Stuff like $exists yP(x,y)forall x$ instead of (most likely, but not certainly) $exists yforall xP(x,y)$.

The classic $u=x^2implies mathrm du=2xmathrm dx$.

This one disturbs me deeply. Sometimes people want to say "If $A$, then $B$" or "$Aimplies B$" and they say "If $Aimplies B$". "If $Aimplies B$" isn't even a statement, it's part of an incomplete conditional statement whose antecedent is $Aimplies B$. Again: mathematics is to be parsed with priority over natural language.

Saying that $x=yimplies f(x)=f(y)$ proves that $f$ is a function.

Using $f(A)$ to denote ${f(x)colon xin A}$. Why not stick to $f[A]$ which is so standard? Another possibility is $f^to(A)$ (or should it be square brackets?) which I learned from egreg in this comment.

edited Apr 13 '17 at 12:20

community wiki

16 revs
Git Gud

36

$begingroup$
Let $f colon mathbb{R} to C(mathbb{R},mathbb{R})$. The function $f(x)$ ...
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 15:00

21

$begingroup$
Please, Git Gud, don't treat my note as a suggestion, unless your goal is obfuscation; your "Consider the set $A$, where $A = ...$," is far more readable for the rest of us, and also works for your "peculiarly syntactic" reading.
$endgroup$
– John Hughes
Nov 16 '14 at 15:42

22

$begingroup$
I've always interpreted constructions like #7 to mean "Consider the set $A$, which equals ${ x in mathbb{R} : P(x) }$," where the equals sign itself introduces an appositive. In my mind, $``="$ is a word that takes on two grammatical roles, acting as a verb ($2+2$ equals $4$) or a subordinating conjunction (as above), depending on the context.
$endgroup$
– David Zhang
Nov 16 '14 at 17:20

63

$begingroup$
Wow, Git Gud, it must be really painful for you to read any mathematics at all! How do you stand it?
$endgroup$
– TonyK
Nov 16 '14 at 22:26

16

$begingroup$
This is getting slightly ridiculous. There are conventions and we're a mathematical society. We can interpret things. If everyone was pedantic like this, we would have already killed each other.
$endgroup$
– Pedro Tamaroff♦
Nov 20 '14 at 1:49

|
show 80 more comments

'The function $f(x)$'. No, the function is $f$.

Let $f$ and $g$ be real differentiable functions defined in $mathbb R$. Some people denote $(fcirc g)'$ by $dfrac{mathrm df(g(x))}{mathrm dx}$. Contrast with the above. I discuss this in greater detail here.

The differential equation $y'=x^2y+y^3$. Just a minor variant of 1. Correct would be $y'=fy+y^3$ where $fcolon Ito mathbb R, xmapsto x^2$, for some interval $I$.

This is one I find particularly disgusting. "If $t(s)$ is a function of $s$ and it is invertible, then $s(t)$ is the inverse", lol what? The concept of 'function of a variable' isn't even definable in a satisfiable way in $sf ZFC$. Also $left(frac{mathrm dy}{mathrm dx}right)^{-1}=frac{mathrm dx}{mathrm dy}$. Contrast with 1.

In algebra it's common to denote the algebraic structure by the underlying set.

When $langle ,cdot,rangle$ is a function which takes sets as their inputs, it's common do abuse $langle{x} rangle$ as $langle xrangle$. More generally it's common to look at at finite set ${x_1, ldots ,x_n}$ as the finite sequence $x_1, ldots ,x_n$. This happens for instance in logic. Also in linear algebra and it's usual to go even further and talk about 'linearly independent vectors' instead of 'linearly independent set' - this only an abuse when linear (in)dependence is defined for sets instead of 'lists'.

'Consider the set $A={xin mathbb Rcolon P(x)}$'. I'm probably the only person who reads this as the set being the whole equality $A={xin mathbb Rcolon P(x)}$ instead of $A$ or ${xin mathbb Rcolon P(x)}$, in any case it is an abuse. Another example of this is 'multiply by $1=frac 2 2$'.

Denoting by $+$ both scalar addition and function addition.

Instead of $((varphiland psi)to rho)$ people first abandon the out parentheses and use $(varphiland psi)to rho$ and then $land$ is given precedence over $to$, yielding the much more common (though formally incorrect) $varphiland psito rho$.

Even ignoring the problem in 1., the symbol $int x,mathrm dx=frac {x^2}2$ is ambiguous as it can mean a number of things. Under one of the common interpretations the equal sign doesn't even denote an equality. I allude to that meaning here, (it is the same issue as with $f=O(g)$).

There's also the very common '$ldots$' mentioned by Lucian in the comments.

Lucian also mentions $mathbb C=mathbb R^2$ which is an abuse sometimes, but not all the time, depending on how you define things.

Given a linear map $L$ and $x$ on its domain, it's not unusual to write $Lx$ instead of $L(x)$. I'm not sure if this can even be considered an abuse of notation because $Lx$ is meaningless and we should be free to define $Lx:=L(x)$, there's no ambiguity. Unless, of course, you equate linear maps with matrices and this is an abuse. On the topic of matrices, it's common to look at $1times 1$ matrices as scalars.

Geometers like to say $mathbb Rsubseteq mathbb R^2subseteq mathbb R^3$.

Using $mathcal M_{mtimes n}(mathbb F)$ and $mathbb F^{mtimes n}$ interchangeably. On the same note, $A^{m+ n}=A^mtimes A^n$ and $left(A^mright)^n=A^{mtimes n}$.

I don't know how I forgot this one. The omission of quantifiers.

Calling 'well formed formulas' by 'formulas'.

Saying $forall x(P(x)to Q(x))$ is a conditional statement instead of a universal conditional statement.

Stuff like $exists yP(x,y)forall x$ instead of (most likely, but not certainly) $exists yforall xP(x,y)$.

The classic $u=x^2implies mathrm du=2xmathrm dx$.

This one disturbs me deeply. Sometimes people want to say "If $A$, then $B$" or "$Aimplies B$" and they say "If $Aimplies B$". "If $Aimplies B$" isn't even a statement, it's part of an incomplete conditional statement whose antecedent is $Aimplies B$. Again: mathematics is to be parsed with priority over natural language.

Saying that $x=yimplies f(x)=f(y)$ proves that $f$ is a function.

Using $f(A)$ to denote ${f(x)colon xin A}$. Why not stick to $f[A]$ which is so standard? Another possibility is $f^to(A)$ (or should it be square brackets?) which I learned from egreg in this comment.

edited Apr 13 '17 at 12:20

community wiki

16 revs
Git Gud

36

$begingroup$
Let $f colon mathbb{R} to C(mathbb{R},mathbb{R})$. The function $f(x)$ ...
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 15:00

21

$begingroup$
Please, Git Gud, don't treat my note as a suggestion, unless your goal is obfuscation; your "Consider the set $A$, where $A = ...$," is far more readable for the rest of us, and also works for your "peculiarly syntactic" reading.
$endgroup$
– John Hughes
Nov 16 '14 at 15:42

22

$begingroup$
I've always interpreted constructions like #7 to mean "Consider the set $A$, which equals ${ x in mathbb{R} : P(x) }$," where the equals sign itself introduces an appositive. In my mind, $``="$ is a word that takes on two grammatical roles, acting as a verb ($2+2$ equals $4$) or a subordinating conjunction (as above), depending on the context.
$endgroup$
– David Zhang
Nov 16 '14 at 17:20

63

$begingroup$
Wow, Git Gud, it must be really painful for you to read any mathematics at all! How do you stand it?
$endgroup$
– TonyK
Nov 16 '14 at 22:26

16

$begingroup$
This is getting slightly ridiculous. There are conventions and we're a mathematical society. We can interpret things. If everyone was pedantic like this, we would have already killed each other.
$endgroup$
– Pedro Tamaroff♦
Nov 20 '14 at 1:49

|
show 80 more comments

'The function $f(x)$'. No, the function is $f$.

Let $f$ and $g$ be real differentiable functions defined in $mathbb R$. Some people denote $(fcirc g)'$ by $dfrac{mathrm df(g(x))}{mathrm dx}$. Contrast with the above. I discuss this in greater detail here.

The differential equation $y'=x^2y+y^3$. Just a minor variant of 1. Correct would be $y'=fy+y^3$ where $fcolon Ito mathbb R, xmapsto x^2$, for some interval $I$.

This is one I find particularly disgusting. "If $t(s)$ is a function of $s$ and it is invertible, then $s(t)$ is the inverse", lol what? The concept of 'function of a variable' isn't even definable in a satisfiable way in $sf ZFC$. Also $left(frac{mathrm dy}{mathrm dx}right)^{-1}=frac{mathrm dx}{mathrm dy}$. Contrast with 1.

In algebra it's common to denote the algebraic structure by the underlying set.

When $langle ,cdot,rangle$ is a function which takes sets as their inputs, it's common do abuse $langle{x} rangle$ as $langle xrangle$. More generally it's common to look at at finite set ${x_1, ldots ,x_n}$ as the finite sequence $x_1, ldots ,x_n$. This happens for instance in logic. Also in linear algebra and it's usual to go even further and talk about 'linearly independent vectors' instead of 'linearly independent set' - this only an abuse when linear (in)dependence is defined for sets instead of 'lists'.

'Consider the set $A={xin mathbb Rcolon P(x)}$'. I'm probably the only person who reads this as the set being the whole equality $A={xin mathbb Rcolon P(x)}$ instead of $A$ or ${xin mathbb Rcolon P(x)}$, in any case it is an abuse. Another example of this is 'multiply by $1=frac 2 2$'.

Denoting by $+$ both scalar addition and function addition.

Instead of $((varphiland psi)to rho)$ people first abandon the out parentheses and use $(varphiland psi)to rho$ and then $land$ is given precedence over $to$, yielding the much more common (though formally incorrect) $varphiland psito rho$.

Even ignoring the problem in 1., the symbol $int x,mathrm dx=frac {x^2}2$ is ambiguous as it can mean a number of things. Under one of the common interpretations the equal sign doesn't even denote an equality. I allude to that meaning here, (it is the same issue as with $f=O(g)$).

There's also the very common '$ldots$' mentioned by Lucian in the comments.

Lucian also mentions $mathbb C=mathbb R^2$ which is an abuse sometimes, but not all the time, depending on how you define things.

Given a linear map $L$ and $x$ on its domain, it's not unusual to write $Lx$ instead of $L(x)$. I'm not sure if this can even be considered an abuse of notation because $Lx$ is meaningless and we should be free to define $Lx:=L(x)$, there's no ambiguity. Unless, of course, you equate linear maps with matrices and this is an abuse. On the topic of matrices, it's common to look at $1times 1$ matrices as scalars.

Geometers like to say $mathbb Rsubseteq mathbb R^2subseteq mathbb R^3$.

Using $mathcal M_{mtimes n}(mathbb F)$ and $mathbb F^{mtimes n}$ interchangeably. On the same note, $A^{m+ n}=A^mtimes A^n$ and $left(A^mright)^n=A^{mtimes n}$.

I don't know how I forgot this one. The omission of quantifiers.

Calling 'well formed formulas' by 'formulas'.

Saying $forall x(P(x)to Q(x))$ is a conditional statement instead of a universal conditional statement.

Stuff like $exists yP(x,y)forall x$ instead of (most likely, but not certainly) $exists yforall xP(x,y)$.

The classic $u=x^2implies mathrm du=2xmathrm dx$.

This one disturbs me deeply. Sometimes people want to say "If $A$, then $B$" or "$Aimplies B$" and they say "If $Aimplies B$". "If $Aimplies B$" isn't even a statement, it's part of an incomplete conditional statement whose antecedent is $Aimplies B$. Again: mathematics is to be parsed with priority over natural language.

Saying that $x=yimplies f(x)=f(y)$ proves that $f$ is a function.

Using $f(A)$ to denote ${f(x)colon xin A}$. Why not stick to $f[A]$ which is so standard? Another possibility is $f^to(A)$ (or should it be square brackets?) which I learned from egreg in this comment.

edited Apr 13 '17 at 12:20

community wiki

16 revs
Git Gud

'The function $f(x)$'. No, the function is $f$.

Let $f$ and $g$ be real differentiable functions defined in $mathbb R$. Some people denote $(fcirc g)'$ by $dfrac{mathrm df(g(x))}{mathrm dx}$. Contrast with the above. I discuss this in greater detail here.

The differential equation $y'=x^2y+y^3$. Just a minor variant of 1. Correct would be $y'=fy+y^3$ where $fcolon Ito mathbb R, xmapsto x^2$, for some interval $I$.

This is one I find particularly disgusting. "If $t(s)$ is a function of $s$ and it is invertible, then $s(t)$ is the inverse", lol what? The concept of 'function of a variable' isn't even definable in a satisfiable way in $sf ZFC$. Also $left(frac{mathrm dy}{mathrm dx}right)^{-1}=frac{mathrm dx}{mathrm dy}$. Contrast with 1.

In algebra it's common to denote the algebraic structure by the underlying set.

When $langle ,cdot,rangle$ is a function which takes sets as their inputs, it's common do abuse $langle{x} rangle$ as $langle xrangle$. More generally it's common to look at at finite set ${x_1, ldots ,x_n}$ as the finite sequence $x_1, ldots ,x_n$. This happens for instance in logic. Also in linear algebra and it's usual to go even further and talk about 'linearly independent vectors' instead of 'linearly independent set' - this only an abuse when linear (in)dependence is defined for sets instead of 'lists'.

'Consider the set $A={xin mathbb Rcolon P(x)}$'. I'm probably the only person who reads this as the set being the whole equality $A={xin mathbb Rcolon P(x)}$ instead of $A$ or ${xin mathbb Rcolon P(x)}$, in any case it is an abuse. Another example of this is 'multiply by $1=frac 2 2$'.

Denoting by $+$ both scalar addition and function addition.

Instead of $((varphiland psi)to rho)$ people first abandon the out parentheses and use $(varphiland psi)to rho$ and then $land$ is given precedence over $to$, yielding the much more common (though formally incorrect) $varphiland psito rho$.

Even ignoring the problem in 1., the symbol $int x,mathrm dx=frac {x^2}2$ is ambiguous as it can mean a number of things. Under one of the common interpretations the equal sign doesn't even denote an equality. I allude to that meaning here, (it is the same issue as with $f=O(g)$).

There's also the very common '$ldots$' mentioned by Lucian in the comments.

Lucian also mentions $mathbb C=mathbb R^2$ which is an abuse sometimes, but not all the time, depending on how you define things.

Given a linear map $L$ and $x$ on its domain, it's not unusual to write $Lx$ instead of $L(x)$. I'm not sure if this can even be considered an abuse of notation because $Lx$ is meaningless and we should be free to define $Lx:=L(x)$, there's no ambiguity. Unless, of course, you equate linear maps with matrices and this is an abuse. On the topic of matrices, it's common to look at $1times 1$ matrices as scalars.

Geometers like to say $mathbb Rsubseteq mathbb R^2subseteq mathbb R^3$.

Using $mathcal M_{mtimes n}(mathbb F)$ and $mathbb F^{mtimes n}$ interchangeably. On the same note, $A^{m+ n}=A^mtimes A^n$ and $left(A^mright)^n=A^{mtimes n}$.

I don't know how I forgot this one. The omission of quantifiers.

Calling 'well formed formulas' by 'formulas'.

Saying $forall x(P(x)to Q(x))$ is a conditional statement instead of a universal conditional statement.

Stuff like $exists yP(x,y)forall x$ instead of (most likely, but not certainly) $exists yforall xP(x,y)$.

The classic $u=x^2implies mathrm du=2xmathrm dx$.

This one disturbs me deeply. Sometimes people want to say "If $A$, then $B$" or "$Aimplies B$" and they say "If $Aimplies B$". "If $Aimplies B$" isn't even a statement, it's part of an incomplete conditional statement whose antecedent is $Aimplies B$. Again: mathematics is to be parsed with priority over natural language.

Saying that $x=yimplies f(x)=f(y)$ proves that $f$ is a function.

Using $f(A)$ to denote ${f(x)colon xin A}$. Why not stick to $f[A]$ which is so standard? Another possibility is $f^to(A)$ (or should it be square brackets?) which I learned from egreg in this comment.

edited Apr 13 '17 at 12:20

community wiki

16 revs
Git Gud

edited Apr 13 '17 at 12:20

community wiki

16 revs
Git Gud

community wiki

16 revs
Git Gud

community wiki

16 revs
Git Gud

36

$begingroup$
Let $f colon mathbb{R} to C(mathbb{R},mathbb{R})$. The function $f(x)$ ...
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 15:00

21

$begingroup$
Please, Git Gud, don't treat my note as a suggestion, unless your goal is obfuscation; your "Consider the set $A$, where $A = ...$," is far more readable for the rest of us, and also works for your "peculiarly syntactic" reading.
$endgroup$
– John Hughes
Nov 16 '14 at 15:42

22

$begingroup$
I've always interpreted constructions like #7 to mean "Consider the set $A$, which equals ${ x in mathbb{R} : P(x) }$," where the equals sign itself introduces an appositive. In my mind, $``="$ is a word that takes on two grammatical roles, acting as a verb ($2+2$ equals $4$) or a subordinating conjunction (as above), depending on the context.
$endgroup$
– David Zhang
Nov 16 '14 at 17:20

63

$begingroup$
Wow, Git Gud, it must be really painful for you to read any mathematics at all! How do you stand it?
$endgroup$
– TonyK
Nov 16 '14 at 22:26

16

$begingroup$
This is getting slightly ridiculous. There are conventions and we're a mathematical society. We can interpret things. If everyone was pedantic like this, we would have already killed each other.
$endgroup$
– Pedro Tamaroff♦
Nov 20 '14 at 1:49

|
show 80 more comments

36

$begingroup$
Let $f colon mathbb{R} to C(mathbb{R},mathbb{R})$. The function $f(x)$ ...
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 15:00

21

$begingroup$
Please, Git Gud, don't treat my note as a suggestion, unless your goal is obfuscation; your "Consider the set $A$, where $A = ...$," is far more readable for the rest of us, and also works for your "peculiarly syntactic" reading.
$endgroup$
– John Hughes
Nov 16 '14 at 15:42

22

$begingroup$
I've always interpreted constructions like #7 to mean "Consider the set $A$, which equals ${ x in mathbb{R} : P(x) }$," where the equals sign itself introduces an appositive. In my mind, $``="$ is a word that takes on two grammatical roles, acting as a verb ($2+2$ equals $4$) or a subordinating conjunction (as above), depending on the context.
$endgroup$
– David Zhang
Nov 16 '14 at 17:20

63

$begingroup$
Wow, Git Gud, it must be really painful for you to read any mathematics at all! How do you stand it?
$endgroup$
– TonyK
Nov 16 '14 at 22:26

16

$begingroup$
This is getting slightly ridiculous. There are conventions and we're a mathematical society. We can interpret things. If everyone was pedantic like this, we would have already killed each other.
$endgroup$
– Pedro Tamaroff♦
Nov 20 '14 at 1:49

Let $f colon mathbb{R} to C(mathbb{R},mathbb{R})$. The function $f(x)$ ...

– Daniel Fischer♦
Nov 16 '14 at 15:00

Please, Git Gud, don't treat my note as a suggestion, unless your goal is obfuscation; your "Consider the set $A$, where $A = ...$," is far more readable for the rest of us, and also works for your "peculiarly syntactic" reading.

– John Hughes
Nov 16 '14 at 15:42

I've always interpreted constructions like #7 to mean "Consider the set $A$, which equals ${ x in mathbb{R} : P(x) }$," where the equals sign itself introduces an appositive. In my mind, $``="$ is a word that takes on two grammatical roles, acting as a verb ($2+2$ equals $4$) or a subordinating conjunction (as above), depending on the context.

– David Zhang
Nov 16 '14 at 17:20

Wow, Git Gud, it must be really painful for you to read any mathematics at all! How do you stand it?

– TonyK
Nov 16 '14 at 22:26

This is getting slightly ridiculous. There are conventions and we're a mathematical society. We can interpret things. If everyone was pedantic like this, we would have already killed each other.

– Pedro Tamaroff♦
Nov 20 '14 at 1:49

|
show 80 more comments

The inconsistent treatment of raising trig functions to powers: $$ sin^n x ,.$$

So why is $$sin^{-1} x$$ used for $$arcsin x$$ (which is vaguely consistent with the former) instead of $$(sin x)^{-1} $$ in keeping with the latter.

edited Nov 19 '14 at 5:33

community wiki

3 revs
dmckee

17

$begingroup$
@Mariano It's pretty common to see $sin^2 x = (sin x)^2$ and $sin^{-1} x = arcsin x$ both used in precalculus or calculus textbooks. So, one or both of them (definitely including the second one) is an abuse.
$endgroup$
– Ryan Reich
Nov 17 '14 at 15:04

8

$begingroup$
@Mariano Even if it's just the superposition that's wrong, that superposition is itself a common convention. Also: $sin^{-1} x$ is definitely an abuse of notation. It is not the inverse function, but refers to the inverse of a specific restriction of $sin x$ to a domain on which it's injective. Note that $cos^{-1}$ and $tan^{-1}$ are also used, and their restrictions are different. This notation can only be defined in an extremely kludgy way. You seem to believe that notation can't be abused if it's well-defined: well, this notation can't be well-defined.
$endgroup$
– Ryan Reich
Nov 17 '14 at 18:47

2

$begingroup$
Maybe it's time to realize trig functions are functions. If we define a function f(x) = 3x, you would never write f40 = f(40), yet people will gladly write sin40 = sin(40). The parenthesis are (should be, at least) mandatory, which means anything outside them should be evaluated after the function. So sin(40)^2 = sin(40)*sin(40). I could be wrong, but isn't f(40)^2 commonly interpreted as f(40)*f(40)?
$endgroup$
– Dan Watkins
Nov 18 '14 at 4:28

2

$begingroup$
@DanWatkins, I don't understand why you think $f(x)$ is not very often written simply $fx$... That happens all the time! Why would anyone dream of making those parenthesis that you want mandatory? What possible sensible interpretations of $sin40$ can you think of apart from the intended one? Even if you can think of another one, just as with anything related to human communication what is mandatory is that you become accustumed with the usual traditions...
$endgroup$
– Mariano Suárez-Álvarez
Nov 19 '14 at 9:33

2

$begingroup$
@JackM The $sin^2x$ notation is historical; in trigonometric formulas it's common to have those powers and people devised that way for getting less clumsy formulas. The concept of composition of function as an operation came much later and the notation remained, which is surely more practical; how many times did you find in your life $sin(sin x)$ compared to $(sin x)^2$? What I find abusive is $sin^{-1}$ for the reasons well expressed by Ryan Reich.
$endgroup$
– egreg
Nov 20 '14 at 11:38

|
show 5 more comments

The inconsistent treatment of raising trig functions to powers: $$ sin^n x ,.$$

So why is $$sin^{-1} x$$ used for $$arcsin x$$ (which is vaguely consistent with the former) instead of $$(sin x)^{-1} $$ in keeping with the latter.

edited Nov 19 '14 at 5:33

community wiki

3 revs
dmckee

17

$begingroup$
@Mariano It's pretty common to see $sin^2 x = (sin x)^2$ and $sin^{-1} x = arcsin x$ both used in precalculus or calculus textbooks. So, one or both of them (definitely including the second one) is an abuse.
$endgroup$
– Ryan Reich
Nov 17 '14 at 15:04

8

$begingroup$
@Mariano Even if it's just the superposition that's wrong, that superposition is itself a common convention. Also: $sin^{-1} x$ is definitely an abuse of notation. It is not the inverse function, but refers to the inverse of a specific restriction of $sin x$ to a domain on which it's injective. Note that $cos^{-1}$ and $tan^{-1}$ are also used, and their restrictions are different. This notation can only be defined in an extremely kludgy way. You seem to believe that notation can't be abused if it's well-defined: well, this notation can't be well-defined.
$endgroup$
– Ryan Reich
Nov 17 '14 at 18:47

2

$begingroup$
Maybe it's time to realize trig functions are functions. If we define a function f(x) = 3x, you would never write f40 = f(40), yet people will gladly write sin40 = sin(40). The parenthesis are (should be, at least) mandatory, which means anything outside them should be evaluated after the function. So sin(40)^2 = sin(40)*sin(40). I could be wrong, but isn't f(40)^2 commonly interpreted as f(40)*f(40)?
$endgroup$
– Dan Watkins
Nov 18 '14 at 4:28

2

$begingroup$
@DanWatkins, I don't understand why you think $f(x)$ is not very often written simply $fx$... That happens all the time! Why would anyone dream of making those parenthesis that you want mandatory? What possible sensible interpretations of $sin40$ can you think of apart from the intended one? Even if you can think of another one, just as with anything related to human communication what is mandatory is that you become accustumed with the usual traditions...
$endgroup$
– Mariano Suárez-Álvarez
Nov 19 '14 at 9:33

2

$begingroup$
@JackM The $sin^2x$ notation is historical; in trigonometric formulas it's common to have those powers and people devised that way for getting less clumsy formulas. The concept of composition of function as an operation came much later and the notation remained, which is surely more practical; how many times did you find in your life $sin(sin x)$ compared to $(sin x)^2$? What I find abusive is $sin^{-1}$ for the reasons well expressed by Ryan Reich.
$endgroup$
– egreg
Nov 20 '14 at 11:38

|
show 5 more comments

The inconsistent treatment of raising trig functions to powers: $$ sin^n x ,.$$

So why is $$sin^{-1} x$$ used for $$arcsin x$$ (which is vaguely consistent with the former) instead of $$(sin x)^{-1} $$ in keeping with the latter.

edited Nov 19 '14 at 5:33

community wiki

3 revs
dmckee

The inconsistent treatment of raising trig functions to powers: $$ sin^n x ,.$$

So why is $$sin^{-1} x$$ used for $$arcsin x$$ (which is vaguely consistent with the former) instead of $$(sin x)^{-1} $$ in keeping with the latter.

edited Nov 19 '14 at 5:33

community wiki

3 revs
dmckee

edited Nov 19 '14 at 5:33

community wiki

3 revs
dmckee

community wiki

3 revs
dmckee

community wiki

3 revs
dmckee

17

$begingroup$
@Mariano It's pretty common to see $sin^2 x = (sin x)^2$ and $sin^{-1} x = arcsin x$ both used in precalculus or calculus textbooks. So, one or both of them (definitely including the second one) is an abuse.
$endgroup$
– Ryan Reich
Nov 17 '14 at 15:04

8

$begingroup$
@Mariano Even if it's just the superposition that's wrong, that superposition is itself a common convention. Also: $sin^{-1} x$ is definitely an abuse of notation. It is not the inverse function, but refers to the inverse of a specific restriction of $sin x$ to a domain on which it's injective. Note that $cos^{-1}$ and $tan^{-1}$ are also used, and their restrictions are different. This notation can only be defined in an extremely kludgy way. You seem to believe that notation can't be abused if it's well-defined: well, this notation can't be well-defined.
$endgroup$
– Ryan Reich
Nov 17 '14 at 18:47

2

$begingroup$
Maybe it's time to realize trig functions are functions. If we define a function f(x) = 3x, you would never write f40 = f(40), yet people will gladly write sin40 = sin(40). The parenthesis are (should be, at least) mandatory, which means anything outside them should be evaluated after the function. So sin(40)^2 = sin(40)*sin(40). I could be wrong, but isn't f(40)^2 commonly interpreted as f(40)*f(40)?
$endgroup$
– Dan Watkins
Nov 18 '14 at 4:28

2

$begingroup$
@DanWatkins, I don't understand why you think $f(x)$ is not very often written simply $fx$... That happens all the time! Why would anyone dream of making those parenthesis that you want mandatory? What possible sensible interpretations of $sin40$ can you think of apart from the intended one? Even if you can think of another one, just as with anything related to human communication what is mandatory is that you become accustumed with the usual traditions...
$endgroup$
– Mariano Suárez-Álvarez
Nov 19 '14 at 9:33

2

$begingroup$
@JackM The $sin^2x$ notation is historical; in trigonometric formulas it's common to have those powers and people devised that way for getting less clumsy formulas. The concept of composition of function as an operation came much later and the notation remained, which is surely more practical; how many times did you find in your life $sin(sin x)$ compared to $(sin x)^2$? What I find abusive is $sin^{-1}$ for the reasons well expressed by Ryan Reich.
$endgroup$
– egreg
Nov 20 '14 at 11:38

|
show 5 more comments

17

$begingroup$
@Mariano It's pretty common to see $sin^2 x = (sin x)^2$ and $sin^{-1} x = arcsin x$ both used in precalculus or calculus textbooks. So, one or both of them (definitely including the second one) is an abuse.
$endgroup$
– Ryan Reich
Nov 17 '14 at 15:04

8

$begingroup$
@Mariano Even if it's just the superposition that's wrong, that superposition is itself a common convention. Also: $sin^{-1} x$ is definitely an abuse of notation. It is not the inverse function, but refers to the inverse of a specific restriction of $sin x$ to a domain on which it's injective. Note that $cos^{-1}$ and $tan^{-1}$ are also used, and their restrictions are different. This notation can only be defined in an extremely kludgy way. You seem to believe that notation can't be abused if it's well-defined: well, this notation can't be well-defined.
$endgroup$
– Ryan Reich
Nov 17 '14 at 18:47

2

$begingroup$
Maybe it's time to realize trig functions are functions. If we define a function f(x) = 3x, you would never write f40 = f(40), yet people will gladly write sin40 = sin(40). The parenthesis are (should be, at least) mandatory, which means anything outside them should be evaluated after the function. So sin(40)^2 = sin(40)*sin(40). I could be wrong, but isn't f(40)^2 commonly interpreted as f(40)*f(40)?
$endgroup$
– Dan Watkins
Nov 18 '14 at 4:28

2

$begingroup$
@DanWatkins, I don't understand why you think $f(x)$ is not very often written simply $fx$... That happens all the time! Why would anyone dream of making those parenthesis that you want mandatory? What possible sensible interpretations of $sin40$ can you think of apart from the intended one? Even if you can think of another one, just as with anything related to human communication what is mandatory is that you become accustumed with the usual traditions...
$endgroup$
– Mariano Suárez-Álvarez
Nov 19 '14 at 9:33

2

$begingroup$
@JackM The $sin^2x$ notation is historical; in trigonometric formulas it's common to have those powers and people devised that way for getting less clumsy formulas. The concept of composition of function as an operation came much later and the notation remained, which is surely more practical; how many times did you find in your life $sin(sin x)$ compared to $(sin x)^2$? What I find abusive is $sin^{-1}$ for the reasons well expressed by Ryan Reich.
$endgroup$
– egreg
Nov 20 '14 at 11:38

@Mariano It's pretty common to see $sin^2 x = (sin x)^2$ and $sin^{-1} x = arcsin x$ both used in precalculus or calculus textbooks. So, one or both of them (definitely including the second one) is an abuse.

– Ryan Reich
Nov 17 '14 at 15:04

@Mariano Even if it's just the superposition that's wrong, that superposition is itself a common convention. Also: $sin^{-1} x$ is definitely an abuse of notation. It is not the inverse function, but refers to the inverse of a specific restriction of $sin x$ to a domain on which it's injective. Note that $cos^{-1}$ and $tan^{-1}$ are also used, and their restrictions are different. This notation can only be defined in an extremely kludgy way. You seem to believe that notation can't be abused if it's well-defined: well, this notation can't be well-defined.

– Ryan Reich
Nov 17 '14 at 18:47

Maybe it's time to realize trig functions are functions. If we define a function f(x) = 3x, you would never write f40 = f(40), yet people will gladly write sin40 = sin(40). The parenthesis are (should be, at least) mandatory, which means anything outside them should be evaluated after the function. So sin(40)^2 = sin(40)*sin(40). I could be wrong, but isn't f(40)^2 commonly interpreted as f(40)*f(40)?

– Dan Watkins
Nov 18 '14 at 4:28

@DanWatkins, I don't understand why you think $f(x)$ is not very often written simply $fx$... That happens all the time! Why would anyone dream of making those parenthesis that you want mandatory? What possible sensible interpretations of $sin40$ can you think of apart from the intended one? Even if you can think of another one, just as with anything related to human communication what is mandatory is that you become accustumed with the usual traditions...

– Mariano Suárez-Álvarez
Nov 19 '14 at 9:33

@JackM The $sin^2x$ notation is historical; in trigonometric formulas it's common to have those powers and people devised that way for getting less clumsy formulas. The concept of composition of function as an operation came much later and the notation remained, which is surely more practical; how many times did you find in your life $sin(sin x)$ compared to $(sin x)^2$? What I find abusive is $sin^{-1}$ for the reasons well expressed by Ryan Reich.

– egreg
Nov 20 '14 at 11:38

|
show 5 more comments

edited Nov 19 '14 at 20:49

community wiki

2 revs
Danylo Dubinin

36

$begingroup$
Not to mention an exclamation mark at the end of a sentence ending with a number: The answer is 10!
$endgroup$
– rabota
Nov 17 '14 at 20:13

$begingroup$
I wholeheartedly agree.
$endgroup$
– Jeel Shah
Nov 17 '14 at 20:14

4

$begingroup$
@barto I would say most writers should phrase it The answer is 10!. with the period at the end if they mean 10 factorial, and without it to mean 10 only, to be unambiguous.
$endgroup$
– Ryan
Nov 18 '14 at 14:04

$begingroup$
Your link says multinomial, but it points to multifactorial. Is this another abuse of (an)notation?
$endgroup$
– Marc van Leeuwen
Nov 19 '14 at 18:01

8

$begingroup$
@Ryan: What about the question? The answer is 10!!?
$endgroup$
– jxh
Nov 19 '14 at 19:38

|
show 1 more comment

edited Nov 19 '14 at 20:49

community wiki

2 revs
Danylo Dubinin

36

$begingroup$
Not to mention an exclamation mark at the end of a sentence ending with a number: The answer is 10!
$endgroup$
– rabota
Nov 17 '14 at 20:13

$begingroup$
I wholeheartedly agree.
$endgroup$
– Jeel Shah
Nov 17 '14 at 20:14

4

$begingroup$
@barto I would say most writers should phrase it The answer is 10!. with the period at the end if they mean 10 factorial, and without it to mean 10 only, to be unambiguous.
$endgroup$
– Ryan
Nov 18 '14 at 14:04

$begingroup$
Your link says multinomial, but it points to multifactorial. Is this another abuse of (an)notation?
$endgroup$
– Marc van Leeuwen
Nov 19 '14 at 18:01

8

$begingroup$
@Ryan: What about the question? The answer is 10!!?
$endgroup$
– jxh
Nov 19 '14 at 19:38

|
show 1 more comment

edited Nov 19 '14 at 20:49

community wiki

2 revs
Danylo Dubinin

edited Nov 19 '14 at 20:49

community wiki

2 revs
Danylo Dubinin

edited Nov 19 '14 at 20:49

community wiki

2 revs
Danylo Dubinin

community wiki

2 revs
Danylo Dubinin

community wiki

2 revs
Danylo Dubinin

36

$begingroup$
Not to mention an exclamation mark at the end of a sentence ending with a number: The answer is 10!
$endgroup$
– rabota
Nov 17 '14 at 20:13

$begingroup$
I wholeheartedly agree.
$endgroup$
– Jeel Shah
Nov 17 '14 at 20:14

4

$begingroup$
@barto I would say most writers should phrase it The answer is 10!. with the period at the end if they mean 10 factorial, and without it to mean 10 only, to be unambiguous.
$endgroup$
– Ryan
Nov 18 '14 at 14:04

$begingroup$
Your link says multinomial, but it points to multifactorial. Is this another abuse of (an)notation?
$endgroup$
– Marc van Leeuwen
Nov 19 '14 at 18:01

8

$begingroup$
@Ryan: What about the question? The answer is 10!!?
$endgroup$
– jxh
Nov 19 '14 at 19:38

|
show 1 more comment

36

$begingroup$
Not to mention an exclamation mark at the end of a sentence ending with a number: The answer is 10!
$endgroup$
– rabota
Nov 17 '14 at 20:13

$begingroup$
I wholeheartedly agree.
$endgroup$
– Jeel Shah
Nov 17 '14 at 20:14

4

$begingroup$
@barto I would say most writers should phrase it The answer is 10!. with the period at the end if they mean 10 factorial, and without it to mean 10 only, to be unambiguous.
$endgroup$
– Ryan
Nov 18 '14 at 14:04

$begingroup$
Your link says multinomial, but it points to multifactorial. Is this another abuse of (an)notation?
$endgroup$
– Marc van Leeuwen
Nov 19 '14 at 18:01

8

$begingroup$
@Ryan: What about the question? The answer is 10!!?
$endgroup$
– jxh
Nov 19 '14 at 19:38

Not to mention an exclamation mark at the end of a sentence ending with a number: The answer is 10!

– rabota
Nov 17 '14 at 20:13

I wholeheartedly agree.

– Jeel Shah
Nov 17 '14 at 20:14

@barto I would say most writers should phrase it The answer is 10!. with the period at the end if they mean 10 factorial, and without it to mean 10 only, to be unambiguous.

– Ryan
Nov 18 '14 at 14:04

Your link says multinomial, but it points to multifactorial. Is this another abuse of (an)notation?

– Marc van Leeuwen
Nov 19 '14 at 18:01

@Ryan: What about the question? The answer is 10!!?

– jxh
Nov 19 '14 at 19:38

|
show 1 more comment

In the first year at college, I was really confused with the notion of a sequence

$${ x_n : n in mathbb N }$$

because this could also be a set! Then I discovered

$$(x_n)_{n in mathbb N}$$

And now I am fine with sequences.

answered Nov 17 '14 at 0:40

community wiki

ThePortakal

1

$begingroup$
I don't get it, ${ x_n : n in mathbb N }$ is unmistakably the image of the sequence. What's the abuse I'm missing here?
$endgroup$
– Git Gud
Nov 17 '14 at 0:50

21

$begingroup$
@GitGud ${ x_1, x_2, x_3, dots } = { x_2, x_1, x_3, dots }$ as sets, but not as sequences. (And I think I abused $dots$ here)
$endgroup$
– ThePortakal
Nov 17 '14 at 0:56

6

$begingroup$
This is a real mistake. It is more problematic in uses like «the set ${x_1,x_2}$ is linarly independent,» which is usually not intended to include the claim that when $x_1=x_2neq0$, the one-element set mentioned in linearly indepenent.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:41

2

$begingroup$
A sequence in the set $X$ should be defined as a function $x:mathbb{N} rightarrow X$ where we denote the image $x(n)$ of $n in mathbb{N}$ by $x_{n}$.
$endgroup$
– Vincent Pfenninger
Nov 17 '14 at 17:29

1

$begingroup$
I see ${x_n:nin mathbb{N}}$ for more often than $(x_n)_{nin mathbb{N}}$, but I've never cared for it. Set notation to me implies unordered list without repeated elements, but a sequence is ordered and may have repeats. Of course, there is an implicit order inherited from the order on $mathbb{N},$ but using set notation still seems like an abuse.
$endgroup$
– Randy E
Nov 17 '14 at 20:02

|
show 1 more comment

In the first year at college, I was really confused with the notion of a sequence

$${ x_n : n in mathbb N }$$

because this could also be a set! Then I discovered

$$(x_n)_{n in mathbb N}$$

And now I am fine with sequences.

answered Nov 17 '14 at 0:40

community wiki

ThePortakal

1

$begingroup$
I don't get it, ${ x_n : n in mathbb N }$ is unmistakably the image of the sequence. What's the abuse I'm missing here?
$endgroup$
– Git Gud
Nov 17 '14 at 0:50

21

$begingroup$
@GitGud ${ x_1, x_2, x_3, dots } = { x_2, x_1, x_3, dots }$ as sets, but not as sequences. (And I think I abused $dots$ here)
$endgroup$
– ThePortakal
Nov 17 '14 at 0:56

6

$begingroup$
This is a real mistake. It is more problematic in uses like «the set ${x_1,x_2}$ is linarly independent,» which is usually not intended to include the claim that when $x_1=x_2neq0$, the one-element set mentioned in linearly indepenent.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:41

2

$begingroup$
A sequence in the set $X$ should be defined as a function $x:mathbb{N} rightarrow X$ where we denote the image $x(n)$ of $n in mathbb{N}$ by $x_{n}$.
$endgroup$
– Vincent Pfenninger
Nov 17 '14 at 17:29

1

$begingroup$
I see ${x_n:nin mathbb{N}}$ for more often than $(x_n)_{nin mathbb{N}}$, but I've never cared for it. Set notation to me implies unordered list without repeated elements, but a sequence is ordered and may have repeats. Of course, there is an implicit order inherited from the order on $mathbb{N},$ but using set notation still seems like an abuse.
$endgroup$
– Randy E
Nov 17 '14 at 20:02

|
show 1 more comment

In the first year at college, I was really confused with the notion of a sequence

$${ x_n : n in mathbb N }$$

because this could also be a set! Then I discovered

$$(x_n)_{n in mathbb N}$$

And now I am fine with sequences.

answered Nov 17 '14 at 0:40

community wiki

ThePortakal

In the first year at college, I was really confused with the notion of a sequence

$${ x_n : n in mathbb N }$$

because this could also be a set! Then I discovered

$$(x_n)_{n in mathbb N}$$

And now I am fine with sequences.

answered Nov 17 '14 at 0:40

community wiki

ThePortakal

answered Nov 17 '14 at 0:40

community wiki

ThePortakal

community wiki

ThePortakal

community wiki

ThePortakal

1

$begingroup$
I don't get it, ${ x_n : n in mathbb N }$ is unmistakably the image of the sequence. What's the abuse I'm missing here?
$endgroup$
– Git Gud
Nov 17 '14 at 0:50

21

$begingroup$
@GitGud ${ x_1, x_2, x_3, dots } = { x_2, x_1, x_3, dots }$ as sets, but not as sequences. (And I think I abused $dots$ here)
$endgroup$
– ThePortakal
Nov 17 '14 at 0:56

6

$begingroup$
This is a real mistake. It is more problematic in uses like «the set ${x_1,x_2}$ is linarly independent,» which is usually not intended to include the claim that when $x_1=x_2neq0$, the one-element set mentioned in linearly indepenent.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:41

2

$begingroup$
A sequence in the set $X$ should be defined as a function $x:mathbb{N} rightarrow X$ where we denote the image $x(n)$ of $n in mathbb{N}$ by $x_{n}$.
$endgroup$
– Vincent Pfenninger
Nov 17 '14 at 17:29

1

$begingroup$
I see ${x_n:nin mathbb{N}}$ for more often than $(x_n)_{nin mathbb{N}}$, but I've never cared for it. Set notation to me implies unordered list without repeated elements, but a sequence is ordered and may have repeats. Of course, there is an implicit order inherited from the order on $mathbb{N},$ but using set notation still seems like an abuse.
$endgroup$
– Randy E
Nov 17 '14 at 20:02

|
show 1 more comment

1

$begingroup$
I don't get it, ${ x_n : n in mathbb N }$ is unmistakably the image of the sequence. What's the abuse I'm missing here?
$endgroup$
– Git Gud
Nov 17 '14 at 0:50

21

$begingroup$
@GitGud ${ x_1, x_2, x_3, dots } = { x_2, x_1, x_3, dots }$ as sets, but not as sequences. (And I think I abused $dots$ here)
$endgroup$
– ThePortakal
Nov 17 '14 at 0:56

6

$begingroup$
This is a real mistake. It is more problematic in uses like «the set ${x_1,x_2}$ is linarly independent,» which is usually not intended to include the claim that when $x_1=x_2neq0$, the one-element set mentioned in linearly indepenent.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:41

2

$begingroup$
A sequence in the set $X$ should be defined as a function $x:mathbb{N} rightarrow X$ where we denote the image $x(n)$ of $n in mathbb{N}$ by $x_{n}$.
$endgroup$
– Vincent Pfenninger
Nov 17 '14 at 17:29

1

$begingroup$
I see ${x_n:nin mathbb{N}}$ for more often than $(x_n)_{nin mathbb{N}}$, but I've never cared for it. Set notation to me implies unordered list without repeated elements, but a sequence is ordered and may have repeats. Of course, there is an implicit order inherited from the order on $mathbb{N},$ but using set notation still seems like an abuse.
$endgroup$
– Randy E
Nov 17 '14 at 20:02

I don't get it, ${ x_n : n in mathbb N }$ is unmistakably the image of the sequence. What's the abuse I'm missing here?

– Git Gud
Nov 17 '14 at 0:50

@GitGud ${ x_1, x_2, x_3, dots } = { x_2, x_1, x_3, dots }$ as sets, but not as sequences. (And I think I abused $dots$ here)

– ThePortakal
Nov 17 '14 at 0:56

This is a real mistake. It is more problematic in uses like «the set ${x_1,x_2}$ is linarly independent,» which is usually not intended to include the claim that when $x_1=x_2neq0$, the one-element set mentioned in linearly indepenent.

– Mariano Suárez-Álvarez
Nov 17 '14 at 4:41

A sequence in the set $X$ should be defined as a function $x:mathbb{N} rightarrow X$ where we denote the image $x(n)$ of $n in mathbb{N}$ by $x_{n}$.

– Vincent Pfenninger
Nov 17 '14 at 17:29

I see ${x_n:nin mathbb{N}}$ for more often than $(x_n)_{nin mathbb{N}}$, but I've never cared for it. Set notation to me implies unordered list without repeated elements, but a sequence is ordered and may have repeats. Of course, there is an implicit order inherited from the order on $mathbb{N},$ but using set notation still seems like an abuse.

– Randy E
Nov 17 '14 at 20:02

|
show 1 more comment

We often write $f(n) = O(g(n))$, when in fact $O(g(n))$ is a set, and should be written as $f(n) in O(g(n))$. Similarly for other asymptotic notation, such as $Theta$ and $Omega$.

answered Nov 17 '14 at 2:04

community wiki

nexolute

21

$begingroup$
Do you mean $f in O(g)$? From the typical usage of these symbols, I'd expect $f(n)$ and $g(n)$ to be numbers, and things like $4 in O(8)$ do not make sense.
$endgroup$
– Prateek
Nov 17 '14 at 10:39

1

$begingroup$
Yeah, and that major inconsistency that became second nature to us now.
$endgroup$
– nexolute
Nov 17 '14 at 11:01

2

$begingroup$
@Prateek Identifying a number with the constant function attaining that value (which is a useful abuse of notation often employed), the statement $4in O(8)$ makes perfect sense (and is correct).
$endgroup$
– Daniel Fischer♦
Nov 17 '14 at 11:37

12

$begingroup$
@Prateek: If you find that annoying, I bet you'll really hate things like $frac{rm d}{{rm d}x}x^2=2x$ (since, if you insisted on assigning a value to $x$ and evaluating from the inside out, you'd end up with something totally nonsensical like $frac{rm d}{{rm d}5}25=10$). It makes more sense if you allow "unevaluated expressions in one or more variables" as first-class objects in your mathematical framework (and not just when buried inside function definitions). If you don't, you'll have to go for something like $fin O(nmapsto n^2)$ to make asymptotical notation rigorous.
$endgroup$
– Ilmari Karonen
Nov 17 '14 at 11:42

5

$begingroup$
There is really no problem in writing a function as $x^2$. Define $x$ to be the identity function. Define the ring operations on functions as usual. Then $x^2$ is the function which takes $t$ to $t^2$, etc.
$endgroup$
– Martin Brandenburg
Nov 18 '14 at 9:02

|
show 1 more comment

We often write $f(n) = O(g(n))$, when in fact $O(g(n))$ is a set, and should be written as $f(n) in O(g(n))$. Similarly for other asymptotic notation, such as $Theta$ and $Omega$.

answered Nov 17 '14 at 2:04

community wiki

nexolute

21

$begingroup$
Do you mean $f in O(g)$? From the typical usage of these symbols, I'd expect $f(n)$ and $g(n)$ to be numbers, and things like $4 in O(8)$ do not make sense.
$endgroup$
– Prateek
Nov 17 '14 at 10:39

1

$begingroup$
Yeah, and that major inconsistency that became second nature to us now.
$endgroup$
– nexolute
Nov 17 '14 at 11:01

2

$begingroup$
@Prateek Identifying a number with the constant function attaining that value (which is a useful abuse of notation often employed), the statement $4in O(8)$ makes perfect sense (and is correct).
$endgroup$
– Daniel Fischer♦
Nov 17 '14 at 11:37

12

$begingroup$
@Prateek: If you find that annoying, I bet you'll really hate things like $frac{rm d}{{rm d}x}x^2=2x$ (since, if you insisted on assigning a value to $x$ and evaluating from the inside out, you'd end up with something totally nonsensical like $frac{rm d}{{rm d}5}25=10$). It makes more sense if you allow "unevaluated expressions in one or more variables" as first-class objects in your mathematical framework (and not just when buried inside function definitions). If you don't, you'll have to go for something like $fin O(nmapsto n^2)$ to make asymptotical notation rigorous.
$endgroup$
– Ilmari Karonen
Nov 17 '14 at 11:42

5

$begingroup$
There is really no problem in writing a function as $x^2$. Define $x$ to be the identity function. Define the ring operations on functions as usual. Then $x^2$ is the function which takes $t$ to $t^2$, etc.
$endgroup$
– Martin Brandenburg
Nov 18 '14 at 9:02

|
show 1 more comment

We often write $f(n) = O(g(n))$, when in fact $O(g(n))$ is a set, and should be written as $f(n) in O(g(n))$. Similarly for other asymptotic notation, such as $Theta$ and $Omega$.

answered Nov 17 '14 at 2:04

community wiki

nexolute

We often write $f(n) = O(g(n))$, when in fact $O(g(n))$ is a set, and should be written as $f(n) in O(g(n))$. Similarly for other asymptotic notation, such as $Theta$ and $Omega$.

answered Nov 17 '14 at 2:04

community wiki

nexolute

answered Nov 17 '14 at 2:04

community wiki

nexolute

community wiki

nexolute

community wiki

nexolute

21

$begingroup$
Do you mean $f in O(g)$? From the typical usage of these symbols, I'd expect $f(n)$ and $g(n)$ to be numbers, and things like $4 in O(8)$ do not make sense.
$endgroup$
– Prateek
Nov 17 '14 at 10:39

1

$begingroup$
Yeah, and that major inconsistency that became second nature to us now.
$endgroup$
– nexolute
Nov 17 '14 at 11:01

2

$begingroup$
@Prateek Identifying a number with the constant function attaining that value (which is a useful abuse of notation often employed), the statement $4in O(8)$ makes perfect sense (and is correct).
$endgroup$
– Daniel Fischer♦
Nov 17 '14 at 11:37

12

$begingroup$
@Prateek: If you find that annoying, I bet you'll really hate things like $frac{rm d}{{rm d}x}x^2=2x$ (since, if you insisted on assigning a value to $x$ and evaluating from the inside out, you'd end up with something totally nonsensical like $frac{rm d}{{rm d}5}25=10$). It makes more sense if you allow "unevaluated expressions in one or more variables" as first-class objects in your mathematical framework (and not just when buried inside function definitions). If you don't, you'll have to go for something like $fin O(nmapsto n^2)$ to make asymptotical notation rigorous.
$endgroup$
– Ilmari Karonen
Nov 17 '14 at 11:42

5

$begingroup$
There is really no problem in writing a function as $x^2$. Define $x$ to be the identity function. Define the ring operations on functions as usual. Then $x^2$ is the function which takes $t$ to $t^2$, etc.
$endgroup$
– Martin Brandenburg
Nov 18 '14 at 9:02

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show 1 more comment

21

$begingroup$
Do you mean $f in O(g)$? From the typical usage of these symbols, I'd expect $f(n)$ and $g(n)$ to be numbers, and things like $4 in O(8)$ do not make sense.
$endgroup$
– Prateek
Nov 17 '14 at 10:39

1

$begingroup$
Yeah, and that major inconsistency that became second nature to us now.
$endgroup$
– nexolute
Nov 17 '14 at 11:01

2

$begingroup$
@Prateek Identifying a number with the constant function attaining that value (which is a useful abuse of notation often employed), the statement $4in O(8)$ makes perfect sense (and is correct).
$endgroup$
– Daniel Fischer♦
Nov 17 '14 at 11:37

12

$begingroup$
@Prateek: If you find that annoying, I bet you'll really hate things like $frac{rm d}{{rm d}x}x^2=2x$ (since, if you insisted on assigning a value to $x$ and evaluating from the inside out, you'd end up with something totally nonsensical like $frac{rm d}{{rm d}5}25=10$). It makes more sense if you allow "unevaluated expressions in one or more variables" as first-class objects in your mathematical framework (and not just when buried inside function definitions). If you don't, you'll have to go for something like $fin O(nmapsto n^2)$ to make asymptotical notation rigorous.
$endgroup$
– Ilmari Karonen
Nov 17 '14 at 11:42

5

$begingroup$
There is really no problem in writing a function as $x^2$. Define $x$ to be the identity function. Define the ring operations on functions as usual. Then $x^2$ is the function which takes $t$ to $t^2$, etc.
$endgroup$
– Martin Brandenburg
Nov 18 '14 at 9:02

Do you mean $f in O(g)$? From the typical usage of these symbols, I'd expect $f(n)$ and $g(n)$ to be numbers, and things like $4 in O(8)$ do not make sense.

– Prateek
Nov 17 '14 at 10:39

Yeah, and that major inconsistency that became second nature to us now.

– nexolute
Nov 17 '14 at 11:01

@Prateek Identifying a number with the constant function attaining that value (which is a useful abuse of notation often employed), the statement $4in O(8)$ makes perfect sense (and is correct).

– Daniel Fischer♦
Nov 17 '14 at 11:37

@Prateek: If you find that annoying, I bet you'll really hate things like $frac{rm d}{{rm d}x}x^2=2x$ (since, if you insisted on assigning a value to $x$ and evaluating from the inside out, you'd end up with something totally nonsensical like $frac{rm d}{{rm d}5}25=10$). It makes more sense if you allow "unevaluated expressions in one or more variables" as first-class objects in your mathematical framework (and not just when buried inside function definitions). If you don't, you'll have to go for something like $fin O(nmapsto n^2)$ to make asymptotical notation rigorous.

– Ilmari Karonen
Nov 17 '14 at 11:42

There is really no problem in writing a function as $x^2$. Define $x$ to be the identity function. Define the ring operations on functions as usual. Then $x^2$ is the function which takes $t$ to $t^2$, etc.

– Martin Brandenburg
Nov 18 '14 at 9:02

|
show 1 more comment

$mathbb N;;;;;;;;;;;;$

answered Nov 19 '14 at 18:35

community wiki

TonyK

2

$begingroup$
@DanielFischer I propose we let $mathbb{N}$ denote all natural numbers including $0$, and let $mathbb{C}$ denote all counting numbers.
$endgroup$
– Frank Vel
Nov 19 '14 at 19:48

8

$begingroup$
What the [censored] are "counting numbers"?
$endgroup$
– Daniel Fischer♦
Nov 19 '14 at 19:57

3

$begingroup$
@fvel: yeah, and $mathbb{K}$ all complex numbers, and $mathbb{F}$ some generic field, and...
$endgroup$
– leftaroundabout
Nov 19 '14 at 19:58

3

$begingroup$
@fvel Apart from the obvious, "You can have zero of a thing," it gives $mathbb{N}$ an identity under addition, which is nice. On the other hand, people (well, the ones who aren't programmers, anyway) seem to like to count from one, and you don't have to worry about dividing by zero in $mathbb{Z}^+$.
$endgroup$
– Stuart Olsen
Nov 20 '14 at 2:27

4

$begingroup$
Then $+$ would be an example too.
$endgroup$
– egreg
Nov 20 '14 at 11:32

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show 3 more comments

$mathbb N;;;;;;;;;;;;$

answered Nov 19 '14 at 18:35

community wiki

TonyK

2

$begingroup$
@DanielFischer I propose we let $mathbb{N}$ denote all natural numbers including $0$, and let $mathbb{C}$ denote all counting numbers.
$endgroup$
– Frank Vel
Nov 19 '14 at 19:48

8

$begingroup$
What the [censored] are "counting numbers"?
$endgroup$
– Daniel Fischer♦
Nov 19 '14 at 19:57

3

$begingroup$
@fvel: yeah, and $mathbb{K}$ all complex numbers, and $mathbb{F}$ some generic field, and...
$endgroup$
– leftaroundabout
Nov 19 '14 at 19:58

3

$begingroup$
@fvel Apart from the obvious, "You can have zero of a thing," it gives $mathbb{N}$ an identity under addition, which is nice. On the other hand, people (well, the ones who aren't programmers, anyway) seem to like to count from one, and you don't have to worry about dividing by zero in $mathbb{Z}^+$.
$endgroup$
– Stuart Olsen
Nov 20 '14 at 2:27

4

$begingroup$
Then $+$ would be an example too.
$endgroup$
– egreg
Nov 20 '14 at 11:32

|
show 3 more comments

$mathbb N;;;;;;;;;;;;$

answered Nov 19 '14 at 18:35

community wiki

TonyK

$mathbb N;;;;;;;;;;;;$

answered Nov 19 '14 at 18:35

community wiki

TonyK

answered Nov 19 '14 at 18:35

community wiki

TonyK

community wiki

TonyK

community wiki

TonyK

2

$begingroup$
@DanielFischer I propose we let $mathbb{N}$ denote all natural numbers including $0$, and let $mathbb{C}$ denote all counting numbers.
$endgroup$
– Frank Vel
Nov 19 '14 at 19:48

8

$begingroup$
What the [censored] are "counting numbers"?
$endgroup$
– Daniel Fischer♦
Nov 19 '14 at 19:57

3

$begingroup$
@fvel: yeah, and $mathbb{K}$ all complex numbers, and $mathbb{F}$ some generic field, and...
$endgroup$
– leftaroundabout
Nov 19 '14 at 19:58

3

$begingroup$
@fvel Apart from the obvious, "You can have zero of a thing," it gives $mathbb{N}$ an identity under addition, which is nice. On the other hand, people (well, the ones who aren't programmers, anyway) seem to like to count from one, and you don't have to worry about dividing by zero in $mathbb{Z}^+$.
$endgroup$
– Stuart Olsen
Nov 20 '14 at 2:27

4

$begingroup$
Then $+$ would be an example too.
$endgroup$
– egreg
Nov 20 '14 at 11:32

|
show 3 more comments

2

$begingroup$
@DanielFischer I propose we let $mathbb{N}$ denote all natural numbers including $0$, and let $mathbb{C}$ denote all counting numbers.
$endgroup$
– Frank Vel
Nov 19 '14 at 19:48

8

$begingroup$
What the [censored] are "counting numbers"?
$endgroup$
– Daniel Fischer♦
Nov 19 '14 at 19:57

3

$begingroup$
@fvel: yeah, and $mathbb{K}$ all complex numbers, and $mathbb{F}$ some generic field, and...
$endgroup$
– leftaroundabout
Nov 19 '14 at 19:58

3

$begingroup$
@fvel Apart from the obvious, "You can have zero of a thing," it gives $mathbb{N}$ an identity under addition, which is nice. On the other hand, people (well, the ones who aren't programmers, anyway) seem to like to count from one, and you don't have to worry about dividing by zero in $mathbb{Z}^+$.
$endgroup$
– Stuart Olsen
Nov 20 '14 at 2:27

4

$begingroup$
Then $+$ would be an example too.
$endgroup$
– egreg
Nov 20 '14 at 11:32

@DanielFischer I propose we let $mathbb{N}$ denote all natural numbers including $0$, and let $mathbb{C}$ denote all counting numbers.

– Frank Vel
Nov 19 '14 at 19:48

What the [censored] are "counting numbers"?

– Daniel Fischer♦
Nov 19 '14 at 19:57

@fvel: yeah, and $mathbb{K}$ all complex numbers, and $mathbb{F}$ some generic field, and...

– leftaroundabout
Nov 19 '14 at 19:58

@fvel Apart from the obvious, "You can have zero of a thing," it gives $mathbb{N}$ an identity under addition, which is nice. On the other hand, people (well, the ones who aren't programmers, anyway) seem to like to count from one, and you don't have to worry about dividing by zero in $mathbb{Z}^+$.

– Stuart Olsen
Nov 20 '14 at 2:27

Then $+$ would be an example too.

– egreg
Nov 20 '14 at 11:32

|
show 3 more comments

Einstein summation convention is a self-explanatory example.

Fourier transforms

I feel the majority of people (myself included) abuse notation when describing Fourier transforms.

For example, it's common to see:

$$mathcal{F}{{f(x)}} = F(omega)$$

to which my natural response is: uhm, no, I believe you mean

$$mathcal{F_x}{{f}} = F$$

or perhaps

$$mathcal{F}{f(,cdot,)} = F(,cdot,)$$

edited Nov 17 '14 at 15:37

community wiki

4 revs
Mehrdad

12

$begingroup$
In what possible way is Einstein summation inconsisten and or an abused notation?!
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 3:54

11

$begingroup$
Huh? Why would upper indices have one and only one meaning? If in some context you are never going to use exponents and you will use thousands of times indices of coordinates, why would you not use upper indices for the latter role? The notation, moreover, makes it impossible to write things which simply do not make sense it in the context it is used (for example: powers of coordinates, which make no sense in pretty much any situation...)
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:07

7

$begingroup$
Of course, if you ar not familiar with the notation, you will have to get used to it —just as if you are not familiar with English you are going to get some familiarity before trying to read Shakespeare.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:09

6

$begingroup$
@MarianoSuárez-Alvarez: I don't follow your reasoning. Every widely-used abuse of notation happens for a reason; just because there's a reason or context behind it doesn't mean it's not an abuse of notation. When people write $mathcal{F}{f(x)} = F(omega)$ it's 100% clear from the context that they don't intend to take the FT of a single number, but that doesn't make it any less abusive of the notation.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:09

4

$begingroup$
@MarianoSuárez-Alvarez: Could you give me a list of what you consider to be abuses of notation? I feel like I can't find any common ground with you to base my reasons off.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:12

|
show 27 more comments

Einstein summation convention is a self-explanatory example.

Fourier transforms

I feel the majority of people (myself included) abuse notation when describing Fourier transforms.

For example, it's common to see:

$$mathcal{F}{{f(x)}} = F(omega)$$

to which my natural response is: uhm, no, I believe you mean

$$mathcal{F_x}{{f}} = F$$

or perhaps

$$mathcal{F}{f(,cdot,)} = F(,cdot,)$$

edited Nov 17 '14 at 15:37

community wiki

4 revs
Mehrdad

12

$begingroup$
In what possible way is Einstein summation inconsisten and or an abused notation?!
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 3:54

11

$begingroup$
Huh? Why would upper indices have one and only one meaning? If in some context you are never going to use exponents and you will use thousands of times indices of coordinates, why would you not use upper indices for the latter role? The notation, moreover, makes it impossible to write things which simply do not make sense it in the context it is used (for example: powers of coordinates, which make no sense in pretty much any situation...)
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:07

7

$begingroup$
Of course, if you ar not familiar with the notation, you will have to get used to it —just as if you are not familiar with English you are going to get some familiarity before trying to read Shakespeare.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:09

6

$begingroup$
@MarianoSuárez-Alvarez: I don't follow your reasoning. Every widely-used abuse of notation happens for a reason; just because there's a reason or context behind it doesn't mean it's not an abuse of notation. When people write $mathcal{F}{f(x)} = F(omega)$ it's 100% clear from the context that they don't intend to take the FT of a single number, but that doesn't make it any less abusive of the notation.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:09

4

$begingroup$
@MarianoSuárez-Alvarez: Could you give me a list of what you consider to be abuses of notation? I feel like I can't find any common ground with you to base my reasons off.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:12

|
show 27 more comments

Einstein summation convention is a self-explanatory example.

Fourier transforms

I feel the majority of people (myself included) abuse notation when describing Fourier transforms.

For example, it's common to see:

$$mathcal{F}{{f(x)}} = F(omega)$$

to which my natural response is: uhm, no, I believe you mean

$$mathcal{F_x}{{f}} = F$$

or perhaps

$$mathcal{F}{f(,cdot,)} = F(,cdot,)$$

edited Nov 17 '14 at 15:37

community wiki

4 revs
Mehrdad

Einstein summation convention is a self-explanatory example.

Fourier transforms

I feel the majority of people (myself included) abuse notation when describing Fourier transforms.

For example, it's common to see:

$$mathcal{F}{{f(x)}} = F(omega)$$

to which my natural response is: uhm, no, I believe you mean

$$mathcal{F_x}{{f}} = F$$

or perhaps

$$mathcal{F}{f(,cdot,)} = F(,cdot,)$$

edited Nov 17 '14 at 15:37

community wiki

4 revs
Mehrdad

edited Nov 17 '14 at 15:37

community wiki

4 revs
Mehrdad

community wiki

4 revs
Mehrdad

community wiki

4 revs
Mehrdad

12

$begingroup$
In what possible way is Einstein summation inconsisten and or an abused notation?!
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 3:54

11

$begingroup$
Huh? Why would upper indices have one and only one meaning? If in some context you are never going to use exponents and you will use thousands of times indices of coordinates, why would you not use upper indices for the latter role? The notation, moreover, makes it impossible to write things which simply do not make sense it in the context it is used (for example: powers of coordinates, which make no sense in pretty much any situation...)
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:07

7

$begingroup$
Of course, if you ar not familiar with the notation, you will have to get used to it —just as if you are not familiar with English you are going to get some familiarity before trying to read Shakespeare.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:09

6

$begingroup$
@MarianoSuárez-Alvarez: I don't follow your reasoning. Every widely-used abuse of notation happens for a reason; just because there's a reason or context behind it doesn't mean it's not an abuse of notation. When people write $mathcal{F}{f(x)} = F(omega)$ it's 100% clear from the context that they don't intend to take the FT of a single number, but that doesn't make it any less abusive of the notation.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:09

4

$begingroup$
@MarianoSuárez-Alvarez: Could you give me a list of what you consider to be abuses of notation? I feel like I can't find any common ground with you to base my reasons off.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:12

|
show 27 more comments

12

$begingroup$
In what possible way is Einstein summation inconsisten and or an abused notation?!
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 3:54

11

$begingroup$
Huh? Why would upper indices have one and only one meaning? If in some context you are never going to use exponents and you will use thousands of times indices of coordinates, why would you not use upper indices for the latter role? The notation, moreover, makes it impossible to write things which simply do not make sense it in the context it is used (for example: powers of coordinates, which make no sense in pretty much any situation...)
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:07

7

$begingroup$
Of course, if you ar not familiar with the notation, you will have to get used to it —just as if you are not familiar with English you are going to get some familiarity before trying to read Shakespeare.
$endgroup$
– Mariano Suárez-Álvarez
Nov 17 '14 at 4:09

6

$begingroup$
@MarianoSuárez-Alvarez: I don't follow your reasoning. Every widely-used abuse of notation happens for a reason; just because there's a reason or context behind it doesn't mean it's not an abuse of notation. When people write $mathcal{F}{f(x)} = F(omega)$ it's 100% clear from the context that they don't intend to take the FT of a single number, but that doesn't make it any less abusive of the notation.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:09

4

$begingroup$
@MarianoSuárez-Alvarez: Could you give me a list of what you consider to be abuses of notation? I feel like I can't find any common ground with you to base my reasons off.
$endgroup$
– Mehrdad
Nov 17 '14 at 4:12

In what possible way is Einstein summation inconsisten and or an abused notation?!

– Mariano Suárez-Álvarez
Nov 17 '14 at 3:54

Huh? Why would upper indices have one and only one meaning? If in some context you are never going to use exponents and you will use thousands of times indices of coordinates, why would you not use upper indices for the latter role? The notation, moreover, makes it impossible to write things which simply do not make sense it in the context it is used (for example: powers of coordinates, which make no sense in pretty much any situation...)

– Mariano Suárez-Álvarez
Nov 17 '14 at 4:07

Of course, if you ar not familiar with the notation, you will have to get used to it —just as if you are not familiar with English you are going to get some familiarity before trying to read Shakespeare.

– Mariano Suárez-Álvarez
Nov 17 '14 at 4:09

@MarianoSuárez-Alvarez: I don't follow your reasoning. Every widely-used abuse of notation happens for a reason; just because there's a reason or context behind it doesn't mean it's not an abuse of notation. When people write $mathcal{F}{f(x)} = F(omega)$ it's 100% clear from the context that they don't intend to take the FT of a single number, but that doesn't make it any less abusive of the notation.

– Mehrdad
Nov 17 '14 at 4:09

@MarianoSuárez-Alvarez: Could you give me a list of what you consider to be abuses of notation? I feel like I can't find any common ground with you to base my reasons off.

– Mehrdad
Nov 17 '14 at 4:12

|
show 27 more comments

Don't know whether these count as notational abuses as such, but a few common causes of confusion I have come across are

re last one, from MathWorld:

... A geometer would therefore regard the object described b $ x_1^2+x_2^2=R^2 $
as a $2$-sphere, while a topologist would consider it a $1$-sphere and denote it $mathbb{S}^1$. Similarly, a geometer would regard the object described by
$x_1^2+x_2^2+x_3^2=R^2 $ as a $3$-sphere, while a topologist would call it a $2$-sphere and denote it $mathbb{S}^2$. Extreme caution is therefore advised when consulting the literature. Following the literature, both conventions are used in this work, depending on context, which is stated explicitly wherever it might be ambiguous.

This last one was included as a curiosity, but generally though, subscript and superscript is often abused / ambiguous unless explicitly stated.

... more interesting anecdotes and opinions.

edited Apr 13 '17 at 12:58

community wiki

5 revs
martin

1

$begingroup$
I haven't seen the fourth one before. So $f_2(x)$ can be yet another way of writing $(fcirc f)(x)$...
$endgroup$
– Frank Vel
Nov 16 '14 at 19:01

11

$begingroup$
That somebody would call the unit circle a $2$-sphere. It beggars belief.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:15

1

$begingroup$
I know. And I'm not in the habit of shooting the messenger, so you need not be afraid of me.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:17

19

$begingroup$
I won't shoot the messenger, but I will shoot the message author. I don't know a single geometer that disagrees with a topologist on what the $n$-sphere is.
$endgroup$
– user98602
Nov 17 '14 at 3:17

2

$begingroup$
@MatthewLeingang I guess it's correct to say a geometer thinks the 2-sphere is a circle. ;)
$endgroup$
– user98602
Nov 18 '14 at 15:33

|
show 8 more comments

Don't know whether these count as notational abuses as such, but a few common causes of confusion I have come across are

re last one, from MathWorld:

... A geometer would therefore regard the object described b $ x_1^2+x_2^2=R^2 $
as a $2$-sphere, while a topologist would consider it a $1$-sphere and denote it $mathbb{S}^1$. Similarly, a geometer would regard the object described by
$x_1^2+x_2^2+x_3^2=R^2 $ as a $3$-sphere, while a topologist would call it a $2$-sphere and denote it $mathbb{S}^2$. Extreme caution is therefore advised when consulting the literature. Following the literature, both conventions are used in this work, depending on context, which is stated explicitly wherever it might be ambiguous.

This last one was included as a curiosity, but generally though, subscript and superscript is often abused / ambiguous unless explicitly stated.

... more interesting anecdotes and opinions.

edited Apr 13 '17 at 12:58

community wiki

5 revs
martin

1

$begingroup$
I haven't seen the fourth one before. So $f_2(x)$ can be yet another way of writing $(fcirc f)(x)$...
$endgroup$
– Frank Vel
Nov 16 '14 at 19:01

11

$begingroup$
That somebody would call the unit circle a $2$-sphere. It beggars belief.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:15

1

$begingroup$
I know. And I'm not in the habit of shooting the messenger, so you need not be afraid of me.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:17

19

$begingroup$
I won't shoot the messenger, but I will shoot the message author. I don't know a single geometer that disagrees with a topologist on what the $n$-sphere is.
$endgroup$
– user98602
Nov 17 '14 at 3:17

2

$begingroup$
@MatthewLeingang I guess it's correct to say a geometer thinks the 2-sphere is a circle. ;)
$endgroup$
– user98602
Nov 18 '14 at 15:33

|
show 8 more comments

Don't know whether these count as notational abuses as such, but a few common causes of confusion I have come across are

re last one, from MathWorld:

... A geometer would therefore regard the object described b $ x_1^2+x_2^2=R^2 $
as a $2$-sphere, while a topologist would consider it a $1$-sphere and denote it $mathbb{S}^1$. Similarly, a geometer would regard the object described by
$x_1^2+x_2^2+x_3^2=R^2 $ as a $3$-sphere, while a topologist would call it a $2$-sphere and denote it $mathbb{S}^2$. Extreme caution is therefore advised when consulting the literature. Following the literature, both conventions are used in this work, depending on context, which is stated explicitly wherever it might be ambiguous.

This last one was included as a curiosity, but generally though, subscript and superscript is often abused / ambiguous unless explicitly stated.

... more interesting anecdotes and opinions.

edited Apr 13 '17 at 12:58

community wiki

5 revs
martin

Don't know whether these count as notational abuses as such, but a few common causes of confusion I have come across are

re last one, from MathWorld:

... A geometer would therefore regard the object described b $ x_1^2+x_2^2=R^2 $
as a $2$-sphere, while a topologist would consider it a $1$-sphere and denote it $mathbb{S}^1$. Similarly, a geometer would regard the object described by
$x_1^2+x_2^2+x_3^2=R^2 $ as a $3$-sphere, while a topologist would call it a $2$-sphere and denote it $mathbb{S}^2$. Extreme caution is therefore advised when consulting the literature. Following the literature, both conventions are used in this work, depending on context, which is stated explicitly wherever it might be ambiguous.

This last one was included as a curiosity, but generally though, subscript and superscript is often abused / ambiguous unless explicitly stated.

... more interesting anecdotes and opinions.

edited Apr 13 '17 at 12:58

community wiki

5 revs
martin

edited Apr 13 '17 at 12:58

community wiki

5 revs
martin

community wiki

5 revs
martin

community wiki

5 revs
martin

1

$begingroup$
I haven't seen the fourth one before. So $f_2(x)$ can be yet another way of writing $(fcirc f)(x)$...
$endgroup$
– Frank Vel
Nov 16 '14 at 19:01

11

$begingroup$
That somebody would call the unit circle a $2$-sphere. It beggars belief.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:15

1

$begingroup$
I know. And I'm not in the habit of shooting the messenger, so you need not be afraid of me.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:17

19

$begingroup$
I won't shoot the messenger, but I will shoot the message author. I don't know a single geometer that disagrees with a topologist on what the $n$-sphere is.
$endgroup$
– user98602
Nov 17 '14 at 3:17

2

$begingroup$
@MatthewLeingang I guess it's correct to say a geometer thinks the 2-sphere is a circle. ;)
$endgroup$
– user98602
Nov 18 '14 at 15:33

|
show 8 more comments

1

$begingroup$
I haven't seen the fourth one before. So $f_2(x)$ can be yet another way of writing $(fcirc f)(x)$...
$endgroup$
– Frank Vel
Nov 16 '14 at 19:01

11

$begingroup$
That somebody would call the unit circle a $2$-sphere. It beggars belief.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:15

1

$begingroup$
I know. And I'm not in the habit of shooting the messenger, so you need not be afraid of me.
$endgroup$
– Daniel Fischer♦
Nov 16 '14 at 19:17

19

$begingroup$
I won't shoot the messenger, but I will shoot the message author. I don't know a single geometer that disagrees with a topologist on what the $n$-sphere is.
$endgroup$
– user98602
Nov 17 '14 at 3:17

2

$begingroup$
@MatthewLeingang I guess it's correct to say a geometer thinks the 2-sphere is a circle. ;)
$endgroup$
– user98602
Nov 18 '14 at 15:33

I haven't seen the fourth one before. So $f_2(x)$ can be yet another way of writing $(fcirc f)(x)$...

– Frank Vel
Nov 16 '14 at 19:01

That somebody would call the unit circle a $2$-sphere. It beggars belief.

– Daniel Fischer♦
Nov 16 '14 at 19:15

I know. And I'm not in the habit of shooting the messenger, so you need not be afraid of me.

– Daniel Fischer♦
Nov 16 '14 at 19:17

I won't shoot the messenger, but I will shoot the message author. I don't know a single geometer that disagrees with a topologist on what the $n$-sphere is.

– user98602
Nov 17 '14 at 3:17

@MatthewLeingang I guess it's correct to say a geometer thinks the 2-sphere is a circle. ;)

– user98602
Nov 18 '14 at 15:33

|
show 8 more comments

In financial mathematics one encounters index numbers, such as consumer price indexes, where a base year (such as 1992) is commonly specified by truly ghastly expressions such as

1992=100.

And no, they're not working modulo a divisor of 1892: they're referring to the fact that the index number for the base year is 100. It feels wrong to even bring this up in polite company.

answered Nov 18 '14 at 23:27

community wiki

Shane O Rourke

add a comment |

In financial mathematics one encounters index numbers, such as consumer price indexes, where a base year (such as 1992) is commonly specified by truly ghastly expressions such as

1992=100.

And no, they're not working modulo a divisor of 1892: they're referring to the fact that the index number for the base year is 100. It feels wrong to even bring this up in polite company.

answered Nov 18 '14 at 23:27

community wiki

Shane O Rourke

add a comment |

In financial mathematics one encounters index numbers, such as consumer price indexes, where a base year (such as 1992) is commonly specified by truly ghastly expressions such as

1992=100.

And no, they're not working modulo a divisor of 1892: they're referring to the fact that the index number for the base year is 100. It feels wrong to even bring this up in polite company.

answered Nov 18 '14 at 23:27

community wiki

Shane O Rourke

In financial mathematics one encounters index numbers, such as consumer price indexes, where a base year (such as 1992) is commonly specified by truly ghastly expressions such as

1992=100.

And no, they're not working modulo a divisor of 1892: they're referring to the fact that the index number for the base year is 100. It feels wrong to even bring this up in polite company.

answered Nov 18 '14 at 23:27

community wiki

Shane O Rourke

answered Nov 18 '14 at 23:27

community wiki

Shane O Rourke

community wiki

Shane O Rourke

community wiki

Shane O Rourke

add a comment |

Calculus I,II,III: The '$dx$'s in integrals and derivatives are just notation to help keep track of the important variables in a given problem, and they're otherwise meaningless in isolation.

$$int_{R}f.$$

$$int_{R}domega=oint_{partial R}omega.$$

Sorry we lied.

Edit: correctified for grammars.

edited Mar 1 '15 at 22:43

community wiki

2 revs
David H

3

$begingroup$
By "case and point" I believe you mean "case in point".
$endgroup$
– Mehrdad
Nov 23 '14 at 10:00

1

$begingroup$
@Mehrdad heh, oops. Finally corrected. Apparently automatic notifications aren't sent for comments on community wikis.
$endgroup$
– David H
Mar 1 '15 at 22:45

$begingroup$
I haven't gotten to Differential Geometry yet, but I assume that is the final and correct answer?
$endgroup$
– Ovi
May 10 '17 at 23:19

$begingroup$
In measure theory books they really use $ dx = d mu (x) $ etc as just notation to help keep track of the important variables in a given problem and there is no manifold (or even a topology) involved. Examples found in Rudin's book of real and complex analysis in the section about Fubini's theorem.
$endgroup$
– Gustavo
Nov 6 '17 at 14:13

add a comment |

Calculus I,II,III: The '$dx$'s in integrals and derivatives are just notation to help keep track of the important variables in a given problem, and they're otherwise meaningless in isolation.

$$int_{R}f.$$

$$int_{R}domega=oint_{partial R}omega.$$

Sorry we lied.

Edit: correctified for grammars.

edited Mar 1 '15 at 22:43

community wiki

2 revs
David H

3

$begingroup$
By "case and point" I believe you mean "case in point".
$endgroup$
– Mehrdad
Nov 23 '14 at 10:00

1

$begingroup$
@Mehrdad heh, oops. Finally corrected. Apparently automatic notifications aren't sent for comments on community wikis.
$endgroup$
– David H
Mar 1 '15 at 22:45

$begingroup$
I haven't gotten to Differential Geometry yet, but I assume that is the final and correct answer?
$endgroup$
– Ovi
May 10 '17 at 23:19

$begingroup$
In measure theory books they really use $ dx = d mu (x) $ etc as just notation to help keep track of the important variables in a given problem and there is no manifold (or even a topology) involved. Examples found in Rudin's book of real and complex analysis in the section about Fubini's theorem.
$endgroup$
– Gustavo
Nov 6 '17 at 14:13

add a comment |

Calculus I,II,III: The '$dx$'s in integrals and derivatives are just notation to help keep track of the important variables in a given problem, and they're otherwise meaningless in isolation.

$$int_{R}f.$$

$$int_{R}domega=oint_{partial R}omega.$$

Sorry we lied.

Edit: correctified for grammars.

edited Mar 1 '15 at 22:43

community wiki

2 revs
David H

Calculus I,II,III: The '$dx$'s in integrals and derivatives are just notation to help keep track of the important variables in a given problem, and they're otherwise meaningless in isolation.

$$int_{R}f.$$

$$int_{R}domega=oint_{partial R}omega.$$

Sorry we lied.

Edit: correctified for grammars.

edited Mar 1 '15 at 22:43

community wiki

2 revs
David H

edited Mar 1 '15 at 22:43

community wiki

2 revs
David H

community wiki

2 revs
David H

community wiki

2 revs
David H

3

$begingroup$
By "case and point" I believe you mean "case in point".
$endgroup$
– Mehrdad
Nov 23 '14 at 10:00

1

$begingroup$
@Mehrdad heh, oops. Finally corrected. Apparently automatic notifications aren't sent for comments on community wikis.
$endgroup$
– David H
Mar 1 '15 at 22:45

$begingroup$
I haven't gotten to Differential Geometry yet, but I assume that is the final and correct answer?
$endgroup$
– Ovi
May 10 '17 at 23:19

$begingroup$
In measure theory books they really use $ dx = d mu (x) $ etc as just notation to help keep track of the important variables in a given problem and there is no manifold (or even a topology) involved. Examples found in Rudin's book of real and complex analysis in the section about Fubini's theorem.
$endgroup$
– Gustavo
Nov 6 '17 at 14:13

add a comment |

3

$begingroup$
By "case and point" I believe you mean "case in point".
$endgroup$
– Mehrdad
Nov 23 '14 at 10:00

1

$begingroup$
@Mehrdad heh, oops. Finally corrected. Apparently automatic notifications aren't sent for comments on community wikis.
$endgroup$
– David H
Mar 1 '15 at 22:45

$begingroup$
I haven't gotten to Differential Geometry yet, but I assume that is the final and correct answer?
$endgroup$
– Ovi
May 10 '17 at 23:19

$begingroup$
In measure theory books they really use $ dx = d mu (x) $ etc as just notation to help keep track of the important variables in a given problem and there is no manifold (or even a topology) involved. Examples found in Rudin's book of real and complex analysis in the section about Fubini's theorem.
$endgroup$
– Gustavo
Nov 6 '17 at 14:13

By "case and point" I believe you mean "case in point".

– Mehrdad
Nov 23 '14 at 10:00

@Mehrdad heh, oops. Finally corrected. Apparently automatic notifications aren't sent for comments on community wikis.

– David H
Mar 1 '15 at 22:45

I haven't gotten to Differential Geometry yet, but I assume that is the final and correct answer?

– Ovi
May 10 '17 at 23:19

In measure theory books they really use $ dx = d mu (x) $ etc as just notation to help keep track of the important variables in a given problem and there is no manifold (or even a topology) involved. Examples found in Rudin's book of real and complex analysis in the section about Fubini's theorem.

– Gustavo
Nov 6 '17 at 14:13

add a comment |

The word "trivial." The many uses of this word include:

1.) The colloquial usage as a synonym for "easy."

2.) The trivial group which consists only of the identity element.

3.) The trivial ring which consists only of the multiplicative and additive identities.

4.) A trivial solution to an equation, often when a variable equals 0 (or constant in the cases of differential equations).

5.) Similar to #2 and #3, any object which satisfies the bare minimum of some particular definition but has no further structure. Often some sort of identity or null element.

This list could go on forever, but I'll stop here.

Note that the converse, of course, is the word "non-trivial," which is just as ambiguous.

answered Nov 19 '14 at 17:34

community wiki

josh314

1

$begingroup$
I feel like your answer needs a little more justification. Right now you've shown that uses of the term "trivial" are broad but not that they are ambiguous. To me they all fall under the umbrella of "nothing much to it": the "nothing much" in the colloquial usage being in terms of effort, while in the mathematical uses, it is "nothing much" in terms of structure or complexity.
$endgroup$
– Rahul
Nov 20 '14 at 18:04

1

$begingroup$
I agree with you, that is answer doesn't really emphasise its ambiguity, but I feel that I am not the only one who has spend some considerable time to proof "trivial" statement, that turned out to be not so trivial at all... Some mathematicians have the tendency to call virtually every result trivial, which they have seen a proof for.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 20:13

2

$begingroup$
There can be ambiguity when there's more than one degenerate object of a certain kind. E.g. if $V$ is a vector space, certainly ${0}subset V$ is a trivial subspace. Is $V$ also a trivial subspace?
$endgroup$
– Julian Rosen
Nov 22 '14 at 16:08

1

$begingroup$
My pet peeve when a paper says things like "trivially easy" about a concept I don't have any grasp on...
$endgroup$
– Calmarius
Jan 16 '15 at 15:27

add a comment |

The word "trivial." The many uses of this word include:

1.) The colloquial usage as a synonym for "easy."

2.) The trivial group which consists only of the identity element.

3.) The trivial ring which consists only of the multiplicative and additive identities.

4.) A trivial solution to an equation, often when a variable equals 0 (or constant in the cases of differential equations).

5.) Similar to #2 and #3, any object which satisfies the bare minimum of some particular definition but has no further structure. Often some sort of identity or null element.

This list could go on forever, but I'll stop here.

Note that the converse, of course, is the word "non-trivial," which is just as ambiguous.

answered Nov 19 '14 at 17:34

community wiki

josh314

1

$begingroup$
I feel like your answer needs a little more justification. Right now you've shown that uses of the term "trivial" are broad but not that they are ambiguous. To me they all fall under the umbrella of "nothing much to it": the "nothing much" in the colloquial usage being in terms of effort, while in the mathematical uses, it is "nothing much" in terms of structure or complexity.
$endgroup$
– Rahul
Nov 20 '14 at 18:04

1

$begingroup$
I agree with you, that is answer doesn't really emphasise its ambiguity, but I feel that I am not the only one who has spend some considerable time to proof "trivial" statement, that turned out to be not so trivial at all... Some mathematicians have the tendency to call virtually every result trivial, which they have seen a proof for.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 20:13

2

$begingroup$
There can be ambiguity when there's more than one degenerate object of a certain kind. E.g. if $V$ is a vector space, certainly ${0}subset V$ is a trivial subspace. Is $V$ also a trivial subspace?
$endgroup$
– Julian Rosen
Nov 22 '14 at 16:08

1

$begingroup$
My pet peeve when a paper says things like "trivially easy" about a concept I don't have any grasp on...
$endgroup$
– Calmarius
Jan 16 '15 at 15:27

add a comment |

The word "trivial." The many uses of this word include:

1.) The colloquial usage as a synonym for "easy."

2.) The trivial group which consists only of the identity element.

3.) The trivial ring which consists only of the multiplicative and additive identities.

4.) A trivial solution to an equation, often when a variable equals 0 (or constant in the cases of differential equations).

5.) Similar to #2 and #3, any object which satisfies the bare minimum of some particular definition but has no further structure. Often some sort of identity or null element.

This list could go on forever, but I'll stop here.

Note that the converse, of course, is the word "non-trivial," which is just as ambiguous.

answered Nov 19 '14 at 17:34

community wiki

josh314

The word "trivial." The many uses of this word include:

1.) The colloquial usage as a synonym for "easy."

2.) The trivial group which consists only of the identity element.

3.) The trivial ring which consists only of the multiplicative and additive identities.

4.) A trivial solution to an equation, often when a variable equals 0 (or constant in the cases of differential equations).

5.) Similar to #2 and #3, any object which satisfies the bare minimum of some particular definition but has no further structure. Often some sort of identity or null element.

This list could go on forever, but I'll stop here.

Note that the converse, of course, is the word "non-trivial," which is just as ambiguous.

answered Nov 19 '14 at 17:34

community wiki

josh314

answered Nov 19 '14 at 17:34

community wiki

josh314

community wiki

josh314

community wiki

josh314

1

$begingroup$
I feel like your answer needs a little more justification. Right now you've shown that uses of the term "trivial" are broad but not that they are ambiguous. To me they all fall under the umbrella of "nothing much to it": the "nothing much" in the colloquial usage being in terms of effort, while in the mathematical uses, it is "nothing much" in terms of structure or complexity.
$endgroup$
– Rahul
Nov 20 '14 at 18:04

1

$begingroup$
I agree with you, that is answer doesn't really emphasise its ambiguity, but I feel that I am not the only one who has spend some considerable time to proof "trivial" statement, that turned out to be not so trivial at all... Some mathematicians have the tendency to call virtually every result trivial, which they have seen a proof for.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 20:13

2

$begingroup$
There can be ambiguity when there's more than one degenerate object of a certain kind. E.g. if $V$ is a vector space, certainly ${0}subset V$ is a trivial subspace. Is $V$ also a trivial subspace?
$endgroup$
– Julian Rosen
Nov 22 '14 at 16:08

1

$begingroup$
My pet peeve when a paper says things like "trivially easy" about a concept I don't have any grasp on...
$endgroup$
– Calmarius
Jan 16 '15 at 15:27

add a comment |

1

$begingroup$
I feel like your answer needs a little more justification. Right now you've shown that uses of the term "trivial" are broad but not that they are ambiguous. To me they all fall under the umbrella of "nothing much to it": the "nothing much" in the colloquial usage being in terms of effort, while in the mathematical uses, it is "nothing much" in terms of structure or complexity.
$endgroup$
– Rahul
Nov 20 '14 at 18:04

1

$begingroup$
I agree with you, that is answer doesn't really emphasise its ambiguity, but I feel that I am not the only one who has spend some considerable time to proof "trivial" statement, that turned out to be not so trivial at all... Some mathematicians have the tendency to call virtually every result trivial, which they have seen a proof for.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 20:13

2

$begingroup$
There can be ambiguity when there's more than one degenerate object of a certain kind. E.g. if $V$ is a vector space, certainly ${0}subset V$ is a trivial subspace. Is $V$ also a trivial subspace?
$endgroup$
– Julian Rosen
Nov 22 '14 at 16:08

1

$begingroup$
My pet peeve when a paper says things like "trivially easy" about a concept I don't have any grasp on...
$endgroup$
– Calmarius
Jan 16 '15 at 15:27

I feel like your answer needs a little more justification. Right now you've shown that uses of the term "trivial" are broad but not that they are ambiguous. To me they all fall under the umbrella of "nothing much to it": the "nothing much" in the colloquial usage being in terms of effort, while in the mathematical uses, it is "nothing much" in terms of structure or complexity.

– Rahul
Nov 20 '14 at 18:04

I agree with you, that is answer doesn't really emphasise its ambiguity, but I feel that I am not the only one who has spend some considerable time to proof "trivial" statement, that turned out to be not so trivial at all... Some mathematicians have the tendency to call virtually every result trivial, which they have seen a proof for.

– Stefan Mesken
Nov 20 '14 at 20:13

There can be ambiguity when there's more than one degenerate object of a certain kind. E.g. if $V$ is a vector space, certainly ${0}subset V$ is a trivial subspace. Is $V$ also a trivial subspace?

– Julian Rosen
Nov 22 '14 at 16:08

My pet peeve when a paper says things like "trivially easy" about a concept I don't have any grasp on...

– Calmarius
Jan 16 '15 at 15:27

add a comment |

My favorite has always been with Fourier transforms.

Suppose a particle is in a potential $V$ given by $$V(x) = V_0cos(x/a).$$ Then $$V(k) = V_0sqrt{dfrac{pi}{2}} left( delta(k-a) + delta(k+a) right).$$

I know physicists do this a lot. I am not sure about mathematicians.

edited Nov 17 '14 at 19:43

community wiki

2 revs
NowIGetToLearnWhatAHeadIs

4

$begingroup$
The same goes for change of variables. E.g. you switch from $x$ to $xi$ in $psi(x)$, and then your equation suddenly is not for $psi(x)$ but for $psi(xi)$ where $psi$ is now a different function.
$endgroup$
– Ruslan
Nov 21 '14 at 6:42

$begingroup$
A well known mathematician, with a Field Medal, also did that.
$endgroup$
– Felix Marin
Dec 7 '14 at 5:58

add a comment |

My favorite has always been with Fourier transforms.

Suppose a particle is in a potential $V$ given by $$V(x) = V_0cos(x/a).$$ Then $$V(k) = V_0sqrt{dfrac{pi}{2}} left( delta(k-a) + delta(k+a) right).$$

I know physicists do this a lot. I am not sure about mathematicians.

edited Nov 17 '14 at 19:43

community wiki

2 revs
NowIGetToLearnWhatAHeadIs

4

$begingroup$
The same goes for change of variables. E.g. you switch from $x$ to $xi$ in $psi(x)$, and then your equation suddenly is not for $psi(x)$ but for $psi(xi)$ where $psi$ is now a different function.
$endgroup$
– Ruslan
Nov 21 '14 at 6:42

$begingroup$
A well known mathematician, with a Field Medal, also did that.
$endgroup$
– Felix Marin
Dec 7 '14 at 5:58

add a comment |

My favorite has always been with Fourier transforms.

Suppose a particle is in a potential $V$ given by $$V(x) = V_0cos(x/a).$$ Then $$V(k) = V_0sqrt{dfrac{pi}{2}} left( delta(k-a) + delta(k+a) right).$$

I know physicists do this a lot. I am not sure about mathematicians.

edited Nov 17 '14 at 19:43

community wiki

2 revs
NowIGetToLearnWhatAHeadIs

My favorite has always been with Fourier transforms.

Suppose a particle is in a potential $V$ given by $$V(x) = V_0cos(x/a).$$ Then $$V(k) = V_0sqrt{dfrac{pi}{2}} left( delta(k-a) + delta(k+a) right).$$

I know physicists do this a lot. I am not sure about mathematicians.

edited Nov 17 '14 at 19:43

community wiki

2 revs
NowIGetToLearnWhatAHeadIs

edited Nov 17 '14 at 19:43

community wiki

2 revs
NowIGetToLearnWhatAHeadIs

community wiki

2 revs
NowIGetToLearnWhatAHeadIs

community wiki

2 revs
NowIGetToLearnWhatAHeadIs

4

$begingroup$
The same goes for change of variables. E.g. you switch from $x$ to $xi$ in $psi(x)$, and then your equation suddenly is not for $psi(x)$ but for $psi(xi)$ where $psi$ is now a different function.
$endgroup$
– Ruslan
Nov 21 '14 at 6:42

$begingroup$
A well known mathematician, with a Field Medal, also did that.
$endgroup$
– Felix Marin
Dec 7 '14 at 5:58

add a comment |

4

$begingroup$
The same goes for change of variables. E.g. you switch from $x$ to $xi$ in $psi(x)$, and then your equation suddenly is not for $psi(x)$ but for $psi(xi)$ where $psi$ is now a different function.
$endgroup$
– Ruslan
Nov 21 '14 at 6:42

$begingroup$
A well known mathematician, with a Field Medal, also did that.
$endgroup$
– Felix Marin
Dec 7 '14 at 5:58

The same goes for change of variables. E.g. you switch from $x$ to $xi$ in $psi(x)$, and then your equation suddenly is not for $psi(x)$ but for $psi(xi)$ where $psi$ is now a different function.

– Ruslan
Nov 21 '14 at 6:42

A well known mathematician, with a Field Medal, also did that.

– Felix Marin
Dec 7 '14 at 5:58

add a comment |

Here is one which I think goes back to Euler:
$$i^i = e^{-frac{pi}{2}}$$

answered Nov 18 '14 at 22:22

community wiki

AD.

$begingroup$
What's wrong with this? Raising by an imaginary number??
$endgroup$
– Frank Vel
Nov 18 '14 at 22:50

4

$begingroup$
Ah, so the equality should be $i^i = expleft(-frac{tau}{4}+ktauright)$. Kinda like if you took $int{1}mathrm{d}x=x$ by assuming $C$ is $0$.
$endgroup$
– Frank Vel
Nov 18 '14 at 23:14

3

$begingroup$
@AD that was exactly what I said? $tau := 2pi$.
$endgroup$
– Frank Vel
Nov 19 '14 at 8:13

2

$begingroup$
@AD. Probably not the most known constant, but it exists! I thought it'd be worth mentioning since this thread is about ambiguous notations.
$endgroup$
– Frank Vel
Nov 19 '14 at 10:40

1

$begingroup$
@MarkHurd Now you did it again, the main reason I put the post in this list is that $i^i$ is not one number - it is a rather a set of real numbers (given in the comments above) - and yes! That is remarkable!
$endgroup$
– AD.
Feb 24 '15 at 7:05

|
show 4 more comments

Here is one which I think goes back to Euler:
$$i^i = e^{-frac{pi}{2}}$$

answered Nov 18 '14 at 22:22

community wiki

AD.

$begingroup$
What's wrong with this? Raising by an imaginary number??
$endgroup$
– Frank Vel
Nov 18 '14 at 22:50

4

$begingroup$
Ah, so the equality should be $i^i = expleft(-frac{tau}{4}+ktauright)$. Kinda like if you took $int{1}mathrm{d}x=x$ by assuming $C$ is $0$.
$endgroup$
– Frank Vel
Nov 18 '14 at 23:14

3

$begingroup$
@AD that was exactly what I said? $tau := 2pi$.
$endgroup$
– Frank Vel
Nov 19 '14 at 8:13

2

$begingroup$
@AD. Probably not the most known constant, but it exists! I thought it'd be worth mentioning since this thread is about ambiguous notations.
$endgroup$
– Frank Vel
Nov 19 '14 at 10:40

1

$begingroup$
@MarkHurd Now you did it again, the main reason I put the post in this list is that $i^i$ is not one number - it is a rather a set of real numbers (given in the comments above) - and yes! That is remarkable!
$endgroup$
– AD.
Feb 24 '15 at 7:05

|
show 4 more comments

Here is one which I think goes back to Euler:
$$i^i = e^{-frac{pi}{2}}$$

answered Nov 18 '14 at 22:22

community wiki

AD.

Here is one which I think goes back to Euler:
$$i^i = e^{-frac{pi}{2}}$$

answered Nov 18 '14 at 22:22

community wiki

AD.

answered Nov 18 '14 at 22:22

community wiki

AD.

community wiki

AD.

community wiki

AD.

$begingroup$
What's wrong with this? Raising by an imaginary number??
$endgroup$
– Frank Vel
Nov 18 '14 at 22:50

4

$begingroup$
Ah, so the equality should be $i^i = expleft(-frac{tau}{4}+ktauright)$. Kinda like if you took $int{1}mathrm{d}x=x$ by assuming $C$ is $0$.
$endgroup$
– Frank Vel
Nov 18 '14 at 23:14

3

$begingroup$
@AD that was exactly what I said? $tau := 2pi$.
$endgroup$
– Frank Vel
Nov 19 '14 at 8:13

2

$begingroup$
@AD. Probably not the most known constant, but it exists! I thought it'd be worth mentioning since this thread is about ambiguous notations.
$endgroup$
– Frank Vel
Nov 19 '14 at 10:40

1

$begingroup$
@MarkHurd Now you did it again, the main reason I put the post in this list is that $i^i$ is not one number - it is a rather a set of real numbers (given in the comments above) - and yes! That is remarkable!
$endgroup$
– AD.
Feb 24 '15 at 7:05

|
show 4 more comments

$begingroup$
What's wrong with this? Raising by an imaginary number??
$endgroup$
– Frank Vel
Nov 18 '14 at 22:50

4

$begingroup$
Ah, so the equality should be $i^i = expleft(-frac{tau}{4}+ktauright)$. Kinda like if you took $int{1}mathrm{d}x=x$ by assuming $C$ is $0$.
$endgroup$
– Frank Vel
Nov 18 '14 at 23:14

3

$begingroup$
@AD that was exactly what I said? $tau := 2pi$.
$endgroup$
– Frank Vel
Nov 19 '14 at 8:13

2

$begingroup$
@AD. Probably not the most known constant, but it exists! I thought it'd be worth mentioning since this thread is about ambiguous notations.
$endgroup$
– Frank Vel
Nov 19 '14 at 10:40

1

$begingroup$
@MarkHurd Now you did it again, the main reason I put the post in this list is that $i^i$ is not one number - it is a rather a set of real numbers (given in the comments above) - and yes! That is remarkable!
$endgroup$
– AD.
Feb 24 '15 at 7:05

What's wrong with this? Raising by an imaginary number??

– Frank Vel
Nov 18 '14 at 22:50

Ah, so the equality should be $i^i = expleft(-frac{tau}{4}+ktauright)$. Kinda like if you took $int{1}mathrm{d}x=x$ by assuming $C$ is $0$.

– Frank Vel
Nov 18 '14 at 23:14

@AD that was exactly what I said? $tau := 2pi$.

– Frank Vel
Nov 19 '14 at 8:13

@AD. Probably not the most known constant, but it exists! I thought it'd be worth mentioning since this thread is about ambiguous notations.

– Frank Vel
Nov 19 '14 at 10:40

@MarkHurd Now you did it again, the main reason I put the post in this list is that $i^i$ is not one number - it is a rather a set of real numbers (given in the comments above) - and yes! That is remarkable!

– AD.
Feb 24 '15 at 7:05

|
show 4 more comments

Outer measures aren't measures.

answered Nov 18 '14 at 8:31

community wiki

Martín-Blas Pérez Pinilla

$begingroup$
Just like how the imaginary part $y$ is neither imaginary nor a part of $x+yi$ for $x,y$ real (which might explain why some people seem to call $yi$ the imaginary part instead...). I recall there's a linguistic term (likely something along the lines of "generalised noun/term/usage/adjective") describing situations where "an X Y" does not mean "a Y that is X" but can't put my finger on it -- someone like Tim Gowers mentioned this in connection with pedagogy.
$endgroup$
– Vandermonde
Nov 20 '14 at 6:37

2

$begingroup$
When we place an adjective in front a noun (or after it if you are French!) it may signify something less than, or more than, or a variant of the object that noun symbolises. (Virtual keyboard is not a keyboard.) So, IMO, outer measure is a perfectly valid choice.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:27

$begingroup$
Another example is that a multivalued function is not a function. See the entry, radial concept in A Handbook of Mathematical Discourse by Charles Wells. Note that Tim Gowers discusses multivalued functions in gowers.wordpress.com/2009/06/08/….
$endgroup$
– J W
Nov 22 '14 at 15:54

$begingroup$
See also abstractmath.org/Word%20Press/?tag=multivalued-function in Wells' Gyre&Gimble blog.
$endgroup$
– J W
Nov 22 '14 at 16:04

1

$begingroup$
For my own future reference as much as anyone else's, the archetypal example I couldn't remember of a noun's meaning being changed by an adjective as opposed to being restricted is "almost foo".
$endgroup$
– Vandermonde
Dec 17 '14 at 21:06

|
show 2 more comments

Outer measures aren't measures.

answered Nov 18 '14 at 8:31

community wiki

Martín-Blas Pérez Pinilla

$begingroup$
Just like how the imaginary part $y$ is neither imaginary nor a part of $x+yi$ for $x,y$ real (which might explain why some people seem to call $yi$ the imaginary part instead...). I recall there's a linguistic term (likely something along the lines of "generalised noun/term/usage/adjective") describing situations where "an X Y" does not mean "a Y that is X" but can't put my finger on it -- someone like Tim Gowers mentioned this in connection with pedagogy.
$endgroup$
– Vandermonde
Nov 20 '14 at 6:37

2

$begingroup$
When we place an adjective in front a noun (or after it if you are French!) it may signify something less than, or more than, or a variant of the object that noun symbolises. (Virtual keyboard is not a keyboard.) So, IMO, outer measure is a perfectly valid choice.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:27

$begingroup$
Another example is that a multivalued function is not a function. See the entry, radial concept in A Handbook of Mathematical Discourse by Charles Wells. Note that Tim Gowers discusses multivalued functions in gowers.wordpress.com/2009/06/08/….
$endgroup$
– J W
Nov 22 '14 at 15:54

$begingroup$
See also abstractmath.org/Word%20Press/?tag=multivalued-function in Wells' Gyre&Gimble blog.
$endgroup$
– J W
Nov 22 '14 at 16:04

1

$begingroup$
For my own future reference as much as anyone else's, the archetypal example I couldn't remember of a noun's meaning being changed by an adjective as opposed to being restricted is "almost foo".
$endgroup$
– Vandermonde
Dec 17 '14 at 21:06

|
show 2 more comments

Outer measures aren't measures.

answered Nov 18 '14 at 8:31

community wiki

Martín-Blas Pérez Pinilla

Outer measures aren't measures.

answered Nov 18 '14 at 8:31

community wiki

Martín-Blas Pérez Pinilla

answered Nov 18 '14 at 8:31

community wiki

Martín-Blas Pérez Pinilla

community wiki

Martín-Blas Pérez Pinilla

community wiki

Martín-Blas Pérez Pinilla

$begingroup$
Just like how the imaginary part $y$ is neither imaginary nor a part of $x+yi$ for $x,y$ real (which might explain why some people seem to call $yi$ the imaginary part instead...). I recall there's a linguistic term (likely something along the lines of "generalised noun/term/usage/adjective") describing situations where "an X Y" does not mean "a Y that is X" but can't put my finger on it -- someone like Tim Gowers mentioned this in connection with pedagogy.
$endgroup$
– Vandermonde
Nov 20 '14 at 6:37

2

$begingroup$
When we place an adjective in front a noun (or after it if you are French!) it may signify something less than, or more than, or a variant of the object that noun symbolises. (Virtual keyboard is not a keyboard.) So, IMO, outer measure is a perfectly valid choice.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:27

$begingroup$
Another example is that a multivalued function is not a function. See the entry, radial concept in A Handbook of Mathematical Discourse by Charles Wells. Note that Tim Gowers discusses multivalued functions in gowers.wordpress.com/2009/06/08/….
$endgroup$
– J W
Nov 22 '14 at 15:54

$begingroup$
See also abstractmath.org/Word%20Press/?tag=multivalued-function in Wells' Gyre&Gimble blog.
$endgroup$
– J W
Nov 22 '14 at 16:04

1

$begingroup$
For my own future reference as much as anyone else's, the archetypal example I couldn't remember of a noun's meaning being changed by an adjective as opposed to being restricted is "almost foo".
$endgroup$
– Vandermonde
Dec 17 '14 at 21:06

|
show 2 more comments

$begingroup$
Just like how the imaginary part $y$ is neither imaginary nor a part of $x+yi$ for $x,y$ real (which might explain why some people seem to call $yi$ the imaginary part instead...). I recall there's a linguistic term (likely something along the lines of "generalised noun/term/usage/adjective") describing situations where "an X Y" does not mean "a Y that is X" but can't put my finger on it -- someone like Tim Gowers mentioned this in connection with pedagogy.
$endgroup$
– Vandermonde
Nov 20 '14 at 6:37

2

$begingroup$
When we place an adjective in front a noun (or after it if you are French!) it may signify something less than, or more than, or a variant of the object that noun symbolises. (Virtual keyboard is not a keyboard.) So, IMO, outer measure is a perfectly valid choice.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:27

$begingroup$
Another example is that a multivalued function is not a function. See the entry, radial concept in A Handbook of Mathematical Discourse by Charles Wells. Note that Tim Gowers discusses multivalued functions in gowers.wordpress.com/2009/06/08/….
$endgroup$
– J W
Nov 22 '14 at 15:54

$begingroup$
See also abstractmath.org/Word%20Press/?tag=multivalued-function in Wells' Gyre&Gimble blog.
$endgroup$
– J W
Nov 22 '14 at 16:04

1

$begingroup$
For my own future reference as much as anyone else's, the archetypal example I couldn't remember of a noun's meaning being changed by an adjective as opposed to being restricted is "almost foo".
$endgroup$
– Vandermonde
Dec 17 '14 at 21:06

Just like how the imaginary part $y$ is neither imaginary nor a part of $x+yi$ for $x,y$ real (which might explain why some people seem to call $yi$ the imaginary part instead...). I recall there's a linguistic term (likely something along the lines of "generalised noun/term/usage/adjective") describing situations where "an X Y" does not mean "a Y that is X" but can't put my finger on it -- someone like Tim Gowers mentioned this in connection with pedagogy.

– Vandermonde
Nov 20 '14 at 6:37

When we place an adjective in front a noun (or after it if you are French!) it may signify something less than, or more than, or a variant of the object that noun symbolises. (Virtual keyboard is not a keyboard.) So, IMO, outer measure is a perfectly valid choice.

– P Vanchinathan
Nov 21 '14 at 1:27

Another example is that a multivalued function is not a function. See the entry, radial concept in A Handbook of Mathematical Discourse by Charles Wells. Note that Tim Gowers discusses multivalued functions in gowers.wordpress.com/2009/06/08/….

– J W
Nov 22 '14 at 15:54

See also abstractmath.org/Word%20Press/?tag=multivalued-function in Wells' Gyre&Gimble blog.

– J W
Nov 22 '14 at 16:04

For my own future reference as much as anyone else's, the archetypal example I couldn't remember of a noun's meaning being changed by an adjective as opposed to being restricted is "almost foo".

– Vandermonde
Dec 17 '14 at 21:06

|
show 2 more comments

answered Nov 22 '14 at 10:36

community wiki

kinbiko

3

$begingroup$
No. $$color{red}Huge cdot$$ (cdot) is used to represent multiplication. I've never seen a lower dot being used in this situation.
$endgroup$
– beep-boop
Nov 22 '14 at 15:29

1

$begingroup$
en.wikipedia.org/wiki/Interpunct#English - the interpunct used to have other uses in the UK, I assume this is why full stops were used. :)
$endgroup$
– kinbiko
Nov 22 '14 at 15:42

4

$begingroup$
@alexqwx It varies by country, but a lower dot is definitely used some places. Why are you presupposing that because you have never seen it, it isn't a real thing?
$endgroup$
– 6005
Nov 23 '14 at 0:38

$begingroup$
@Goos It's definitely not widely used in British universities (I have studied at 3 different ones and have watched online lectures from several more, and not one of them has ever had a lecturer using a lower dot to represent multiplication).
$endgroup$
– beep-boop
Nov 23 '14 at 10:57

3

$begingroup$
@alexqwx It seems to be dying out, fortunately, but e.g. in Hardy/Wright, the full stop is being used for multiplication. But with some space on both sides, so it'd be $0, .,5 + 0 , .,5 = 0$ (or perhaps wider spacing), which in print is distinguishable well enough from $0.5 + 0.5$. In handwriting, it may be indistinguishable.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:41

|
show 2 more comments

answered Nov 22 '14 at 10:36

community wiki

kinbiko

3

$begingroup$
No. $$color{red}Huge cdot$$ (cdot) is used to represent multiplication. I've never seen a lower dot being used in this situation.
$endgroup$
– beep-boop
Nov 22 '14 at 15:29

1

$begingroup$
en.wikipedia.org/wiki/Interpunct#English - the interpunct used to have other uses in the UK, I assume this is why full stops were used. :)
$endgroup$
– kinbiko
Nov 22 '14 at 15:42

4

$begingroup$
@alexqwx It varies by country, but a lower dot is definitely used some places. Why are you presupposing that because you have never seen it, it isn't a real thing?
$endgroup$
– 6005
Nov 23 '14 at 0:38

$begingroup$
@Goos It's definitely not widely used in British universities (I have studied at 3 different ones and have watched online lectures from several more, and not one of them has ever had a lecturer using a lower dot to represent multiplication).
$endgroup$
– beep-boop
Nov 23 '14 at 10:57

3

$begingroup$
@alexqwx It seems to be dying out, fortunately, but e.g. in Hardy/Wright, the full stop is being used for multiplication. But with some space on both sides, so it'd be $0, .,5 + 0 , .,5 = 0$ (or perhaps wider spacing), which in print is distinguishable well enough from $0.5 + 0.5$. In handwriting, it may be indistinguishable.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:41

|
show 2 more comments

answered Nov 22 '14 at 10:36

community wiki

kinbiko

answered Nov 22 '14 at 10:36

community wiki

kinbiko

answered Nov 22 '14 at 10:36

community wiki

kinbiko

community wiki

kinbiko

community wiki

kinbiko

3

$begingroup$
No. $$color{red}Huge cdot$$ (cdot) is used to represent multiplication. I've never seen a lower dot being used in this situation.
$endgroup$
– beep-boop
Nov 22 '14 at 15:29

1

$begingroup$
en.wikipedia.org/wiki/Interpunct#English - the interpunct used to have other uses in the UK, I assume this is why full stops were used. :)
$endgroup$
– kinbiko
Nov 22 '14 at 15:42

4

$begingroup$
@alexqwx It varies by country, but a lower dot is definitely used some places. Why are you presupposing that because you have never seen it, it isn't a real thing?
$endgroup$
– 6005
Nov 23 '14 at 0:38

$begingroup$
@Goos It's definitely not widely used in British universities (I have studied at 3 different ones and have watched online lectures from several more, and not one of them has ever had a lecturer using a lower dot to represent multiplication).
$endgroup$
– beep-boop
Nov 23 '14 at 10:57

3

$begingroup$
@alexqwx It seems to be dying out, fortunately, but e.g. in Hardy/Wright, the full stop is being used for multiplication. But with some space on both sides, so it'd be $0, .,5 + 0 , .,5 = 0$ (or perhaps wider spacing), which in print is distinguishable well enough from $0.5 + 0.5$. In handwriting, it may be indistinguishable.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:41

|
show 2 more comments

3

$begingroup$
No. $$color{red}Huge cdot$$ (cdot) is used to represent multiplication. I've never seen a lower dot being used in this situation.
$endgroup$
– beep-boop
Nov 22 '14 at 15:29

1

$begingroup$
en.wikipedia.org/wiki/Interpunct#English - the interpunct used to have other uses in the UK, I assume this is why full stops were used. :)
$endgroup$
– kinbiko
Nov 22 '14 at 15:42

4

$begingroup$
@alexqwx It varies by country, but a lower dot is definitely used some places. Why are you presupposing that because you have never seen it, it isn't a real thing?
$endgroup$
– 6005
Nov 23 '14 at 0:38

$begingroup$
@Goos It's definitely not widely used in British universities (I have studied at 3 different ones and have watched online lectures from several more, and not one of them has ever had a lecturer using a lower dot to represent multiplication).
$endgroup$
– beep-boop
Nov 23 '14 at 10:57

3

$begingroup$
@alexqwx It seems to be dying out, fortunately, but e.g. in Hardy/Wright, the full stop is being used for multiplication. But with some space on both sides, so it'd be $0, .,5 + 0 , .,5 = 0$ (or perhaps wider spacing), which in print is distinguishable well enough from $0.5 + 0.5$. In handwriting, it may be indistinguishable.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:41

No. $$color{red}Huge cdot$$ (cdot) is used to represent multiplication. I've never seen a lower dot being used in this situation.

– beep-boop
Nov 22 '14 at 15:29

en.wikipedia.org/wiki/Interpunct#English - the interpunct used to have other uses in the UK, I assume this is why full stops were used. :)

– kinbiko
Nov 22 '14 at 15:42

@alexqwx It varies by country, but a lower dot is definitely used some places. Why are you presupposing that because you have never seen it, it isn't a real thing?

– 6005
Nov 23 '14 at 0:38

@Goos It's definitely not widely used in British universities (I have studied at 3 different ones and have watched online lectures from several more, and not one of them has ever had a lecturer using a lower dot to represent multiplication).

– beep-boop
Nov 23 '14 at 10:57

@alexqwx It seems to be dying out, fortunately, but e.g. in Hardy/Wright, the full stop is being used for multiplication. But with some space on both sides, so it'd be $0, .,5 + 0 , .,5 = 0$ (or perhaps wider spacing), which in print is distinguishable well enough from $0.5 + 0.5$. In handwriting, it may be indistinguishable.

– Daniel Fischer♦
Nov 23 '14 at 12:41

|
show 2 more comments

The inconsistency between the reading of "Negative" versus "Minus" has, in my opinion, been a thorn in the side of every teacher and student since their acceptance.

edited Apr 13 '17 at 12:21

community wiki

2 revs
skill patrol

add a comment |

The inconsistency between the reading of "Negative" versus "Minus" has, in my opinion, been a thorn in the side of every teacher and student since their acceptance.

edited Apr 13 '17 at 12:21

community wiki

2 revs
skill patrol

add a comment |

The inconsistency between the reading of "Negative" versus "Minus" has, in my opinion, been a thorn in the side of every teacher and student since their acceptance.

edited Apr 13 '17 at 12:21

community wiki

2 revs
skill patrol

The inconsistency between the reading of "Negative" versus "Minus" has, in my opinion, been a thorn in the side of every teacher and student since their acceptance.

edited Apr 13 '17 at 12:21

community wiki

2 revs
skill patrol

edited Apr 13 '17 at 12:21

community wiki

2 revs
skill patrol

community wiki

2 revs
skill patrol

community wiki

2 revs
skill patrol

add a comment |

$mathbb N subset mathbb Z subset mathbb Q subset mathbb R subset mathbb C subset mathbb H$

If $f colon X rightarrow Y$ is a map and $A subseteq X$, then people often write $f(A)$ to denote $f'' A := { y in Y mid exists x in X : f(x) = y }$, which can be quite confusing in cases where $A in X$.

"Canonical" ...

$(a,b,c) = ((a,b),c) = (a,(b,c)) = f,$ where $f colon {0,1,2} rightarrow {a,b,c}$ is such that $f(0) = a$, $f(1) = b$ and $f(2) = c$.

"regular", "dense", "dimension","rank", "computable", "recursive", "closed", "compatible", "compactification"... and other notions which, in a given context, may have several different meanings.

"$f colon X rightarrow Y$ smooth" or similar expressions, where "smooth" may refer to $f$, $X$ or $Y$.

"pictures" and "diagrams" can sometimes be ambiguous to an extent where they don't mean anything to anyone - or maybe that's just me.

the "constant" $c$.

$a < b < c in d$

$1 = left( frac 2 7 right) neq frac 2 7$

$prod_{i in I} (X_i, le_i) subseteq prod_{i in J} (Y_j, le_j)$

edited Nov 20 '14 at 19:11

community wiki

4 revs, 2 users 98%
Einer

2

$begingroup$
What's wrong with 1.?
$endgroup$
– Frank Vel
Nov 20 '14 at 10:25

3

$begingroup$
@fvel There are different ways to construct $mathbb Z, mathbb Q, ldots$ from $mathbb N$. Following the "standard" definitions in set theory, one has $mathbb N = omega$ such that each $n in mathbb N$ has cardinality $|n| = n < aleph_0$, while each element $z in mathbb Z$ is a subset of $mathbb N times mathbb N$ of size $|z| = aleph_0$. These identifications basically forget about the underlying sets and deal only with the "induced" structures.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 10:46

4

$begingroup$
@fvel Maybe, I should give an example how this identification can cause trouble when one ignores that fact, that they typically are not subsets: In the "standard" construction of $mathbb R$, every element $x in mathbb R$ has size $|x| = 2^{aleph_0}$. If we had $mathbb N subseteq mathbb R$, then pick any $n in mathbb N$. As $n in mathbb N subseteq mathbb R$, $n$ itself has cardinality $2^{aleph_0}$. Now, $mathbb N = omega$ is a transitive set, so $n subseteq mathbb N$. This yields $2^{aleph_0} = |n| le |mathbb N| = aleph_0$, which contradicts Cantor's Theorem.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 11:27

7

$begingroup$
That is an absurd objection, really. Anything coming out of a specific construction of the real numbers is absolutely irrelevant...
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:07

3

$begingroup$
Anf, to be honest, I don't even see what the standard construction of the real numbers is for you!
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:11

add a comment |

$mathbb N subset mathbb Z subset mathbb Q subset mathbb R subset mathbb C subset mathbb H$

If $f colon X rightarrow Y$ is a map and $A subseteq X$, then people often write $f(A)$ to denote $f'' A := { y in Y mid exists x in X : f(x) = y }$, which can be quite confusing in cases where $A in X$.

"Canonical" ...

$(a,b,c) = ((a,b),c) = (a,(b,c)) = f,$ where $f colon {0,1,2} rightarrow {a,b,c}$ is such that $f(0) = a$, $f(1) = b$ and $f(2) = c$.

"regular", "dense", "dimension","rank", "computable", "recursive", "closed", "compatible", "compactification"... and other notions which, in a given context, may have several different meanings.

"$f colon X rightarrow Y$ smooth" or similar expressions, where "smooth" may refer to $f$, $X$ or $Y$.

"pictures" and "diagrams" can sometimes be ambiguous to an extent where they don't mean anything to anyone - or maybe that's just me.

the "constant" $c$.

$a < b < c in d$

$1 = left( frac 2 7 right) neq frac 2 7$

$prod_{i in I} (X_i, le_i) subseteq prod_{i in J} (Y_j, le_j)$

edited Nov 20 '14 at 19:11

community wiki

4 revs, 2 users 98%
Einer

2

$begingroup$
What's wrong with 1.?
$endgroup$
– Frank Vel
Nov 20 '14 at 10:25

3

$begingroup$
@fvel There are different ways to construct $mathbb Z, mathbb Q, ldots$ from $mathbb N$. Following the "standard" definitions in set theory, one has $mathbb N = omega$ such that each $n in mathbb N$ has cardinality $|n| = n < aleph_0$, while each element $z in mathbb Z$ is a subset of $mathbb N times mathbb N$ of size $|z| = aleph_0$. These identifications basically forget about the underlying sets and deal only with the "induced" structures.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 10:46

4

$begingroup$
@fvel Maybe, I should give an example how this identification can cause trouble when one ignores that fact, that they typically are not subsets: In the "standard" construction of $mathbb R$, every element $x in mathbb R$ has size $|x| = 2^{aleph_0}$. If we had $mathbb N subseteq mathbb R$, then pick any $n in mathbb N$. As $n in mathbb N subseteq mathbb R$, $n$ itself has cardinality $2^{aleph_0}$. Now, $mathbb N = omega$ is a transitive set, so $n subseteq mathbb N$. This yields $2^{aleph_0} = |n| le |mathbb N| = aleph_0$, which contradicts Cantor's Theorem.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 11:27

7

$begingroup$
That is an absurd objection, really. Anything coming out of a specific construction of the real numbers is absolutely irrelevant...
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:07

3

$begingroup$
Anf, to be honest, I don't even see what the standard construction of the real numbers is for you!
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:11

add a comment |

$mathbb N subset mathbb Z subset mathbb Q subset mathbb R subset mathbb C subset mathbb H$

If $f colon X rightarrow Y$ is a map and $A subseteq X$, then people often write $f(A)$ to denote $f'' A := { y in Y mid exists x in X : f(x) = y }$, which can be quite confusing in cases where $A in X$.

"Canonical" ...

$(a,b,c) = ((a,b),c) = (a,(b,c)) = f,$ where $f colon {0,1,2} rightarrow {a,b,c}$ is such that $f(0) = a$, $f(1) = b$ and $f(2) = c$.

"regular", "dense", "dimension","rank", "computable", "recursive", "closed", "compatible", "compactification"... and other notions which, in a given context, may have several different meanings.

"$f colon X rightarrow Y$ smooth" or similar expressions, where "smooth" may refer to $f$, $X$ or $Y$.

"pictures" and "diagrams" can sometimes be ambiguous to an extent where they don't mean anything to anyone - or maybe that's just me.

the "constant" $c$.

$a < b < c in d$

$1 = left( frac 2 7 right) neq frac 2 7$

$prod_{i in I} (X_i, le_i) subseteq prod_{i in J} (Y_j, le_j)$

edited Nov 20 '14 at 19:11

community wiki

4 revs, 2 users 98%
Einer

$mathbb N subset mathbb Z subset mathbb Q subset mathbb R subset mathbb C subset mathbb H$

If $f colon X rightarrow Y$ is a map and $A subseteq X$, then people often write $f(A)$ to denote $f'' A := { y in Y mid exists x in X : f(x) = y }$, which can be quite confusing in cases where $A in X$.

"Canonical" ...

$(a,b,c) = ((a,b),c) = (a,(b,c)) = f,$ where $f colon {0,1,2} rightarrow {a,b,c}$ is such that $f(0) = a$, $f(1) = b$ and $f(2) = c$.

"regular", "dense", "dimension","rank", "computable", "recursive", "closed", "compatible", "compactification"... and other notions which, in a given context, may have several different meanings.

"$f colon X rightarrow Y$ smooth" or similar expressions, where "smooth" may refer to $f$, $X$ or $Y$.

"pictures" and "diagrams" can sometimes be ambiguous to an extent where they don't mean anything to anyone - or maybe that's just me.

the "constant" $c$.

$a < b < c in d$

$1 = left( frac 2 7 right) neq frac 2 7$

$prod_{i in I} (X_i, le_i) subseteq prod_{i in J} (Y_j, le_j)$

edited Nov 20 '14 at 19:11

community wiki

4 revs, 2 users 98%
Einer

edited Nov 20 '14 at 19:11

community wiki

4 revs, 2 users 98%
Einer

community wiki

4 revs, 2 users 98%
Einer

community wiki

4 revs, 2 users 98%
Einer

2

$begingroup$
What's wrong with 1.?
$endgroup$
– Frank Vel
Nov 20 '14 at 10:25

3

$begingroup$
@fvel There are different ways to construct $mathbb Z, mathbb Q, ldots$ from $mathbb N$. Following the "standard" definitions in set theory, one has $mathbb N = omega$ such that each $n in mathbb N$ has cardinality $|n| = n < aleph_0$, while each element $z in mathbb Z$ is a subset of $mathbb N times mathbb N$ of size $|z| = aleph_0$. These identifications basically forget about the underlying sets and deal only with the "induced" structures.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 10:46

4

$begingroup$
@fvel Maybe, I should give an example how this identification can cause trouble when one ignores that fact, that they typically are not subsets: In the "standard" construction of $mathbb R$, every element $x in mathbb R$ has size $|x| = 2^{aleph_0}$. If we had $mathbb N subseteq mathbb R$, then pick any $n in mathbb N$. As $n in mathbb N subseteq mathbb R$, $n$ itself has cardinality $2^{aleph_0}$. Now, $mathbb N = omega$ is a transitive set, so $n subseteq mathbb N$. This yields $2^{aleph_0} = |n| le |mathbb N| = aleph_0$, which contradicts Cantor's Theorem.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 11:27

7

$begingroup$
That is an absurd objection, really. Anything coming out of a specific construction of the real numbers is absolutely irrelevant...
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:07

3

$begingroup$
Anf, to be honest, I don't even see what the standard construction of the real numbers is for you!
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:11

add a comment |

2

$begingroup$
What's wrong with 1.?
$endgroup$
– Frank Vel
Nov 20 '14 at 10:25

3

$begingroup$
@fvel There are different ways to construct $mathbb Z, mathbb Q, ldots$ from $mathbb N$. Following the "standard" definitions in set theory, one has $mathbb N = omega$ such that each $n in mathbb N$ has cardinality $|n| = n < aleph_0$, while each element $z in mathbb Z$ is a subset of $mathbb N times mathbb N$ of size $|z| = aleph_0$. These identifications basically forget about the underlying sets and deal only with the "induced" structures.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 10:46

4

$begingroup$
@fvel Maybe, I should give an example how this identification can cause trouble when one ignores that fact, that they typically are not subsets: In the "standard" construction of $mathbb R$, every element $x in mathbb R$ has size $|x| = 2^{aleph_0}$. If we had $mathbb N subseteq mathbb R$, then pick any $n in mathbb N$. As $n in mathbb N subseteq mathbb R$, $n$ itself has cardinality $2^{aleph_0}$. Now, $mathbb N = omega$ is a transitive set, so $n subseteq mathbb N$. This yields $2^{aleph_0} = |n| le |mathbb N| = aleph_0$, which contradicts Cantor's Theorem.
$endgroup$
– Stefan Mesken
Nov 20 '14 at 11:27

7

$begingroup$
That is an absurd objection, really. Anything coming out of a specific construction of the real numbers is absolutely irrelevant...
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:07

3

$begingroup$
Anf, to be honest, I don't even see what the standard construction of the real numbers is for you!
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:11

What's wrong with 1.?

– Frank Vel
Nov 20 '14 at 10:25

@fvel There are different ways to construct $mathbb Z, mathbb Q, ldots$ from $mathbb N$. Following the "standard" definitions in set theory, one has $mathbb N = omega$ such that each $n in mathbb N$ has cardinality $|n| = n < aleph_0$, while each element $z in mathbb Z$ is a subset of $mathbb N times mathbb N$ of size $|z| = aleph_0$. These identifications basically forget about the underlying sets and deal only with the "induced" structures.

– Stefan Mesken
Nov 20 '14 at 10:46

@fvel Maybe, I should give an example how this identification can cause trouble when one ignores that fact, that they typically are not subsets: In the "standard" construction of $mathbb R$, every element $x in mathbb R$ has size $|x| = 2^{aleph_0}$. If we had $mathbb N subseteq mathbb R$, then pick any $n in mathbb N$. As $n in mathbb N subseteq mathbb R$, $n$ itself has cardinality $2^{aleph_0}$. Now, $mathbb N = omega$ is a transitive set, so $n subseteq mathbb N$. This yields $2^{aleph_0} = |n| le |mathbb N| = aleph_0$, which contradicts Cantor's Theorem.

– Stefan Mesken
Nov 20 '14 at 11:27

That is an absurd objection, really. Anything coming out of a specific construction of the real numbers is absolutely irrelevant...

– Mariano Suárez-Álvarez
Nov 20 '14 at 23:07

Anf, to be honest, I don't even see what the standard construction of the real numbers is for you!

– Mariano Suárez-Álvarez
Nov 20 '14 at 23:11

add a comment |

Lebesgue-Stieltjes and Riemann-Stieltjes integrals written both$$int_a^bf(x)dPhi(x).$$

Hilbert spaces sometimes intended to be separable, sometimes not necessarily separable.

edited Apr 28 '16 at 16:00

community wiki

3 revs, 2 users 97%
Self-teaching worker

add a comment |

Lebesgue-Stieltjes and Riemann-Stieltjes integrals written both$$int_a^bf(x)dPhi(x).$$

Hilbert spaces sometimes intended to be separable, sometimes not necessarily separable.

edited Apr 28 '16 at 16:00

community wiki

3 revs, 2 users 97%
Self-teaching worker

add a comment |

Lebesgue-Stieltjes and Riemann-Stieltjes integrals written both$$int_a^bf(x)dPhi(x).$$

Hilbert spaces sometimes intended to be separable, sometimes not necessarily separable.

edited Apr 28 '16 at 16:00

community wiki

3 revs, 2 users 97%
Self-teaching worker

Lebesgue-Stieltjes and Riemann-Stieltjes integrals written both$$int_a^bf(x)dPhi(x).$$

Hilbert spaces sometimes intended to be separable, sometimes not necessarily separable.

edited Apr 28 '16 at 16:00

community wiki

3 revs, 2 users 97%
Self-teaching worker

edited Apr 28 '16 at 16:00

community wiki

3 revs, 2 users 97%
Self-teaching worker

community wiki

3 revs, 2 users 97%
Self-teaching worker

community wiki

3 revs, 2 users 97%
Self-teaching worker

add a comment |

"X is dense in itself" is not equivalent to "X is dense in X".

An ellipse is not an elliptical curve.

I absolutely detest the "infinity is not a number, it's just a concept".

I believe it's already been addressed - natural numbers, whole numbers, counting numbers.

answered Mar 31 '15 at 19:12

community wiki

Paul Foster

2

$begingroup$
Do you mean an ellipse is not an elliptic curve?
$endgroup$
– A.P.
Mar 31 '15 at 19:30

add a comment |

"X is dense in itself" is not equivalent to "X is dense in X".

An ellipse is not an elliptical curve.

I absolutely detest the "infinity is not a number, it's just a concept".

I believe it's already been addressed - natural numbers, whole numbers, counting numbers.

answered Mar 31 '15 at 19:12

community wiki

Paul Foster

2

$begingroup$
Do you mean an ellipse is not an elliptic curve?
$endgroup$
– A.P.
Mar 31 '15 at 19:30

add a comment |

"X is dense in itself" is not equivalent to "X is dense in X".

An ellipse is not an elliptical curve.

I absolutely detest the "infinity is not a number, it's just a concept".

I believe it's already been addressed - natural numbers, whole numbers, counting numbers.

answered Mar 31 '15 at 19:12

community wiki

Paul Foster

"X is dense in itself" is not equivalent to "X is dense in X".

An ellipse is not an elliptical curve.

I absolutely detest the "infinity is not a number, it's just a concept".

I believe it's already been addressed - natural numbers, whole numbers, counting numbers.

answered Mar 31 '15 at 19:12

community wiki

Paul Foster

answered Mar 31 '15 at 19:12

community wiki

Paul Foster

community wiki

Paul Foster

community wiki

Paul Foster

2

$begingroup$
Do you mean an ellipse is not an elliptic curve?
$endgroup$
– A.P.
Mar 31 '15 at 19:30

add a comment |

2

$begingroup$
Do you mean an ellipse is not an elliptic curve?
$endgroup$
– A.P.
Mar 31 '15 at 19:30

Do you mean an ellipse is not an elliptic curve?

– A.P.
Mar 31 '15 at 19:30

add a comment |

Here is an example of an ambiguity from MSE: Proof verification: prove $Asubseteq B$ if and only if $Acap B=A$.

answered Jan 8 '18 at 5:00

community wiki

ArsenBerk

add a comment |

Here is an example of an ambiguity from MSE: Proof verification: prove $Asubseteq B$ if and only if $Acap B=A$.

answered Jan 8 '18 at 5:00

community wiki

ArsenBerk

add a comment |

Here is an example of an ambiguity from MSE: Proof verification: prove $Asubseteq B$ if and only if $Acap B=A$.

answered Jan 8 '18 at 5:00

community wiki

ArsenBerk

Here is an example of an ambiguity from MSE: Proof verification: prove $Asubseteq B$ if and only if $Acap B=A$.

answered Jan 8 '18 at 5:00

community wiki

ArsenBerk

answered Jan 8 '18 at 5:00

community wiki

ArsenBerk

community wiki

ArsenBerk

community wiki

ArsenBerk

add a comment |

-1

When we list seasons it goes, summer, autumn, winter, spring, summer, autman, and one says seasons occur cyclically.

answered Nov 20 '14 at 14:44

community wiki

P Vanchinathan

4

$begingroup$
This has absolutely nothing to do with the question.
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:09

5

$begingroup$
We call crrtain graphs «trees» and they are not trees! :-/
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:10

2

$begingroup$
@ Mariano Su'arez-Alvarez: The title of the post says "inconsistent phrase" and the phrase "cyclic group" for a group where there is no periodicity definitely qualifies for it.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:00

$begingroup$
About trees: Mathematicians try to name things by analogy. Trees of botanical kind, after shedding their leaves, resemble those of combinatorial kind and are perfectly valid. I think OP wanted to know if the analogy behind the name actually contradicts the concept it tries to illustrate.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 2:33

$begingroup$
Well, infinite cyclic groups are quite analogous to the finite ones, and I would say the analogy is infinitely closer than between trees and trees.
$endgroup$
– Mariano Suárez-Álvarez
Nov 21 '14 at 2:49

|
show 2 more comments

-1

When we list seasons it goes, summer, autumn, winter, spring, summer, autman, and one says seasons occur cyclically.

answered Nov 20 '14 at 14:44

community wiki

P Vanchinathan

4

$begingroup$
This has absolutely nothing to do with the question.
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:09

5

$begingroup$
We call crrtain graphs «trees» and they are not trees! :-/
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:10

2

$begingroup$
@ Mariano Su'arez-Alvarez: The title of the post says "inconsistent phrase" and the phrase "cyclic group" for a group where there is no periodicity definitely qualifies for it.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:00

$begingroup$
About trees: Mathematicians try to name things by analogy. Trees of botanical kind, after shedding their leaves, resemble those of combinatorial kind and are perfectly valid. I think OP wanted to know if the analogy behind the name actually contradicts the concept it tries to illustrate.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 2:33

$begingroup$
Well, infinite cyclic groups are quite analogous to the finite ones, and I would say the analogy is infinitely closer than between trees and trees.
$endgroup$
– Mariano Suárez-Álvarez
Nov 21 '14 at 2:49

|
show 2 more comments

-1

When we list seasons it goes, summer, autumn, winter, spring, summer, autman, and one says seasons occur cyclically.

answered Nov 20 '14 at 14:44

community wiki

P Vanchinathan

When we list seasons it goes, summer, autumn, winter, spring, summer, autman, and one says seasons occur cyclically.

answered Nov 20 '14 at 14:44

community wiki

P Vanchinathan

answered Nov 20 '14 at 14:44

community wiki

P Vanchinathan

community wiki

P Vanchinathan

community wiki

P Vanchinathan

4

$begingroup$
This has absolutely nothing to do with the question.
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:09

5

$begingroup$
We call crrtain graphs «trees» and they are not trees! :-/
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:10

2

$begingroup$
@ Mariano Su'arez-Alvarez: The title of the post says "inconsistent phrase" and the phrase "cyclic group" for a group where there is no periodicity definitely qualifies for it.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:00

$begingroup$
About trees: Mathematicians try to name things by analogy. Trees of botanical kind, after shedding their leaves, resemble those of combinatorial kind and are perfectly valid. I think OP wanted to know if the analogy behind the name actually contradicts the concept it tries to illustrate.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 2:33

$begingroup$
Well, infinite cyclic groups are quite analogous to the finite ones, and I would say the analogy is infinitely closer than between trees and trees.
$endgroup$
– Mariano Suárez-Álvarez
Nov 21 '14 at 2:49

|
show 2 more comments

4

$begingroup$
This has absolutely nothing to do with the question.
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:09

5

$begingroup$
We call crrtain graphs «trees» and they are not trees! :-/
$endgroup$
– Mariano Suárez-Álvarez
Nov 20 '14 at 23:10

2

$begingroup$
@ Mariano Su'arez-Alvarez: The title of the post says "inconsistent phrase" and the phrase "cyclic group" for a group where there is no periodicity definitely qualifies for it.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 1:00

$begingroup$
About trees: Mathematicians try to name things by analogy. Trees of botanical kind, after shedding their leaves, resemble those of combinatorial kind and are perfectly valid. I think OP wanted to know if the analogy behind the name actually contradicts the concept it tries to illustrate.
$endgroup$
– P Vanchinathan
Nov 21 '14 at 2:33

$begingroup$
Well, infinite cyclic groups are quite analogous to the finite ones, and I would say the analogy is infinitely closer than between trees and trees.
$endgroup$
– Mariano Suárez-Álvarez
Nov 21 '14 at 2:49

This has absolutely nothing to do with the question.

– Mariano Suárez-Álvarez
Nov 20 '14 at 23:09

We call crrtain graphs «trees» and they are not trees! :-/

– Mariano Suárez-Álvarez
Nov 20 '14 at 23:10

@ Mariano Su'arez-Alvarez: The title of the post says "inconsistent phrase" and the phrase "cyclic group" for a group where there is no periodicity definitely qualifies for it.

– P Vanchinathan
Nov 21 '14 at 1:00

About trees: Mathematicians try to name things by analogy. Trees of botanical kind, after shedding their leaves, resemble those of combinatorial kind and are perfectly valid. I think OP wanted to know if the analogy behind the name actually contradicts the concept it tries to illustrate.

– P Vanchinathan
Nov 21 '14 at 2:33

Well, infinite cyclic groups are quite analogous to the finite ones, and I would say the analogy is infinitely closer than between trees and trees.

– Mariano Suárez-Álvarez
Nov 21 '14 at 2:49

|
show 2 more comments

-2

$$bigcup { A ,B, C} = A cup B cup C $$ or $$bigcup mathcal A $$ where $mathcal A $ is a set of sets. It should be

edited Nov 20 '14 at 8:53

community wiki

3 revs
fhyve

7

$begingroup$
This is just wrong, and rather a misunderstanding on your part. What do you mean by "the union of a single set is the set"? This is false in general. For sure, $bigcup {x}=x$, but $bigcup x$ does not need to be $x$. Consider for example $x={emptyset}$. This is a singleton, so $xneemptyset=bigcup x$. You are probably more used to seeing $cup$ as a binary relation, written in between two sets, while $bigcup$ is unary, written preceding the set it acts on. Perhaps that's the source of the confusion. The notation is quite precise here, there is no ambiguity.
$endgroup$
– Andrés E. Caicedo
Nov 20 '14 at 8:16

$begingroup$
I edited to correct that out. I was mistakenly generalizing the case of the intersection of one $A$ set when considered as a subset of some ambient set $S$, in which case we can define it as a one object product as sets over $S$.
$endgroup$
– fhyve
Nov 20 '14 at 8:43

$begingroup$
I remember something that uses this sort of convention coming up in my topology class where it was definitely abusive, and where doing it rigorously would be a bit of annoying leg work, but I can't find it in my notes :(
$endgroup$
– fhyve
Nov 20 '14 at 8:56

add a comment |

-2

$$bigcup { A ,B, C} = A cup B cup C $$ or $$bigcup mathcal A $$ where $mathcal A $ is a set of sets. It should be

edited Nov 20 '14 at 8:53

community wiki

3 revs
fhyve

7

$begingroup$
This is just wrong, and rather a misunderstanding on your part. What do you mean by "the union of a single set is the set"? This is false in general. For sure, $bigcup {x}=x$, but $bigcup x$ does not need to be $x$. Consider for example $x={emptyset}$. This is a singleton, so $xneemptyset=bigcup x$. You are probably more used to seeing $cup$ as a binary relation, written in between two sets, while $bigcup$ is unary, written preceding the set it acts on. Perhaps that's the source of the confusion. The notation is quite precise here, there is no ambiguity.
$endgroup$
– Andrés E. Caicedo
Nov 20 '14 at 8:16

$begingroup$
I edited to correct that out. I was mistakenly generalizing the case of the intersection of one $A$ set when considered as a subset of some ambient set $S$, in which case we can define it as a one object product as sets over $S$.
$endgroup$
– fhyve
Nov 20 '14 at 8:43

$begingroup$
I remember something that uses this sort of convention coming up in my topology class where it was definitely abusive, and where doing it rigorously would be a bit of annoying leg work, but I can't find it in my notes :(
$endgroup$
– fhyve
Nov 20 '14 at 8:56

add a comment |

-2

$$bigcup { A ,B, C} = A cup B cup C $$ or $$bigcup mathcal A $$ where $mathcal A $ is a set of sets. It should be

edited Nov 20 '14 at 8:53

community wiki

3 revs
fhyve

$$bigcup { A ,B, C} = A cup B cup C $$ or $$bigcup mathcal A $$ where $mathcal A $ is a set of sets. It should be

edited Nov 20 '14 at 8:53

community wiki

3 revs
fhyve

edited Nov 20 '14 at 8:53

community wiki

3 revs
fhyve

community wiki

3 revs
fhyve

community wiki

3 revs
fhyve

7

$begingroup$
This is just wrong, and rather a misunderstanding on your part. What do you mean by "the union of a single set is the set"? This is false in general. For sure, $bigcup {x}=x$, but $bigcup x$ does not need to be $x$. Consider for example $x={emptyset}$. This is a singleton, so $xneemptyset=bigcup x$. You are probably more used to seeing $cup$ as a binary relation, written in between two sets, while $bigcup$ is unary, written preceding the set it acts on. Perhaps that's the source of the confusion. The notation is quite precise here, there is no ambiguity.
$endgroup$
– Andrés E. Caicedo
Nov 20 '14 at 8:16

$begingroup$
I edited to correct that out. I was mistakenly generalizing the case of the intersection of one $A$ set when considered as a subset of some ambient set $S$, in which case we can define it as a one object product as sets over $S$.
$endgroup$
– fhyve
Nov 20 '14 at 8:43

$begingroup$
I remember something that uses this sort of convention coming up in my topology class where it was definitely abusive, and where doing it rigorously would be a bit of annoying leg work, but I can't find it in my notes :(
$endgroup$
– fhyve
Nov 20 '14 at 8:56

add a comment |

7

$begingroup$
This is just wrong, and rather a misunderstanding on your part. What do you mean by "the union of a single set is the set"? This is false in general. For sure, $bigcup {x}=x$, but $bigcup x$ does not need to be $x$. Consider for example $x={emptyset}$. This is a singleton, so $xneemptyset=bigcup x$. You are probably more used to seeing $cup$ as a binary relation, written in between two sets, while $bigcup$ is unary, written preceding the set it acts on. Perhaps that's the source of the confusion. The notation is quite precise here, there is no ambiguity.
$endgroup$
– Andrés E. Caicedo
Nov 20 '14 at 8:16

$begingroup$
I edited to correct that out. I was mistakenly generalizing the case of the intersection of one $A$ set when considered as a subset of some ambient set $S$, in which case we can define it as a one object product as sets over $S$.
$endgroup$
– fhyve
Nov 20 '14 at 8:43

$begingroup$
I remember something that uses this sort of convention coming up in my topology class where it was definitely abusive, and where doing it rigorously would be a bit of annoying leg work, but I can't find it in my notes :(
$endgroup$
– fhyve
Nov 20 '14 at 8:56

This is just wrong, and rather a misunderstanding on your part. What do you mean by "the union of a single set is the set"? This is false in general. For sure, $bigcup {x}=x$, but $bigcup x$ does not need to be $x$. Consider for example $x={emptyset}$. This is a singleton, so $xneemptyset=bigcup x$. You are probably more used to seeing $cup$ as a binary relation, written in between two sets, while $bigcup$ is unary, written preceding the set it acts on. Perhaps that's the source of the confusion. The notation is quite precise here, there is no ambiguity.

– Andrés E. Caicedo
Nov 20 '14 at 8:16

I edited to correct that out. I was mistakenly generalizing the case of the intersection of one $A$ set when considered as a subset of some ambient set $S$, in which case we can define it as a one object product as sets over $S$.

– fhyve
Nov 20 '14 at 8:43

I remember something that uses this sort of convention coming up in my topology class where it was definitely abusive, and where doing it rigorously would be a bit of annoying leg work, but I can't find it in my notes :(

– fhyve
Nov 20 '14 at 8:56

add a comment |

-4

$sqrt{2}$ is usually read as "root two". The degree of root should be mentioned though (like "square root two").

$tan^{-1}$ is used for $arctan$ mostly in Physics and Electronics.

$j$ is used for $sqrt{-1}$ in Electronics, while $i$ is used for it in Mathematics.

Dots are used for time derivative in Pyhsics and Control Engineering (e.g.; $ddot{y}+4dot{y}+3y = 3dot{x} + 2x + 8$)

Vectors are sometimes denoted in bold ($mathbf{x}$), with bar on ($bar{x}$) or with arrow on ($oversetrightarrow x$).

Maybe the most frustrating things is when someone uses the word "integral" in the meaning of "anti-derivative" (e.g.; "Integral is the inverse of derivation.").

The unit scalers are used in the place of units themselves (e.g.; "I bought two kilos of potatoes.")

The $nabla$ (del) operator can be very confusing sometimes.

I have never understood the purpose of omitting the preceding zero in a decimal number (e.g.; writing $.5$ instead of $0.5$).

In C++ language, log() is used for ln() and log10() is used for log().

I still don't know what $2^{3^4}$ does equal to. $2^{(3^4)}=2^{81}$ or ${(2^3)}^4=8^4$?

The division operator $/$ is misused. Example: $N/Acdot m$ is written for $N/(Acdot m)$.

edited Sep 10 '16 at 17:51

community wiki

4 revs
hkBattousai

6

$begingroup$
Well, C++ isn't all bad. In mathematics, the prevailing convention is that $log$ denotes the (a local) inverse of $exp$, and other bases are explicitly indicated ($log_{text{base}}$). Only when speaking to/writing for physicists/engineers has $ln$ somewhat widespread use.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:50

3

$begingroup$
-1 for the C++ comment. $log$ is the standard way mathematicians write the natural log, and mathematicians never write $log$ for the base-10 log.
$endgroup$
– Charles
Sep 10 '16 at 17:30

add a comment |

-4

$sqrt{2}$ is usually read as "root two". The degree of root should be mentioned though (like "square root two").

$tan^{-1}$ is used for $arctan$ mostly in Physics and Electronics.

$j$ is used for $sqrt{-1}$ in Electronics, while $i$ is used for it in Mathematics.

Dots are used for time derivative in Pyhsics and Control Engineering (e.g.; $ddot{y}+4dot{y}+3y = 3dot{x} + 2x + 8$)

Vectors are sometimes denoted in bold ($mathbf{x}$), with bar on ($bar{x}$) or with arrow on ($oversetrightarrow x$).

Maybe the most frustrating things is when someone uses the word "integral" in the meaning of "anti-derivative" (e.g.; "Integral is the inverse of derivation.").

The unit scalers are used in the place of units themselves (e.g.; "I bought two kilos of potatoes.")

The $nabla$ (del) operator can be very confusing sometimes.

I have never understood the purpose of omitting the preceding zero in a decimal number (e.g.; writing $.5$ instead of $0.5$).

In C++ language, log() is used for ln() and log10() is used for log().

I still don't know what $2^{3^4}$ does equal to. $2^{(3^4)}=2^{81}$ or ${(2^3)}^4=8^4$?

The division operator $/$ is misused. Example: $N/Acdot m$ is written for $N/(Acdot m)$.

edited Sep 10 '16 at 17:51

community wiki

4 revs
hkBattousai

6

$begingroup$
Well, C++ isn't all bad. In mathematics, the prevailing convention is that $log$ denotes the (a local) inverse of $exp$, and other bases are explicitly indicated ($log_{text{base}}$). Only when speaking to/writing for physicists/engineers has $ln$ somewhat widespread use.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:50

3

$begingroup$
-1 for the C++ comment. $log$ is the standard way mathematicians write the natural log, and mathematicians never write $log$ for the base-10 log.
$endgroup$
– Charles
Sep 10 '16 at 17:30

add a comment |

-4

$sqrt{2}$ is usually read as "root two". The degree of root should be mentioned though (like "square root two").

$tan^{-1}$ is used for $arctan$ mostly in Physics and Electronics.

$j$ is used for $sqrt{-1}$ in Electronics, while $i$ is used for it in Mathematics.

Dots are used for time derivative in Pyhsics and Control Engineering (e.g.; $ddot{y}+4dot{y}+3y = 3dot{x} + 2x + 8$)

Vectors are sometimes denoted in bold ($mathbf{x}$), with bar on ($bar{x}$) or with arrow on ($oversetrightarrow x$).

Maybe the most frustrating things is when someone uses the word "integral" in the meaning of "anti-derivative" (e.g.; "Integral is the inverse of derivation.").

The unit scalers are used in the place of units themselves (e.g.; "I bought two kilos of potatoes.")

The $nabla$ (del) operator can be very confusing sometimes.

I have never understood the purpose of omitting the preceding zero in a decimal number (e.g.; writing $.5$ instead of $0.5$).

In C++ language, log() is used for ln() and log10() is used for log().

I still don't know what $2^{3^4}$ does equal to. $2^{(3^4)}=2^{81}$ or ${(2^3)}^4=8^4$?

The division operator $/$ is misused. Example: $N/Acdot m$ is written for $N/(Acdot m)$.

edited Sep 10 '16 at 17:51

community wiki

4 revs
hkBattousai

$sqrt{2}$ is usually read as "root two". The degree of root should be mentioned though (like "square root two").

$tan^{-1}$ is used for $arctan$ mostly in Physics and Electronics.

$j$ is used for $sqrt{-1}$ in Electronics, while $i$ is used for it in Mathematics.

Dots are used for time derivative in Pyhsics and Control Engineering (e.g.; $ddot{y}+4dot{y}+3y = 3dot{x} + 2x + 8$)

Vectors are sometimes denoted in bold ($mathbf{x}$), with bar on ($bar{x}$) or with arrow on ($oversetrightarrow x$).

Maybe the most frustrating things is when someone uses the word "integral" in the meaning of "anti-derivative" (e.g.; "Integral is the inverse of derivation.").

The unit scalers are used in the place of units themselves (e.g.; "I bought two kilos of potatoes.")

The $nabla$ (del) operator can be very confusing sometimes.

I have never understood the purpose of omitting the preceding zero in a decimal number (e.g.; writing $.5$ instead of $0.5$).

In C++ language, log() is used for ln() and log10() is used for log().

I still don't know what $2^{3^4}$ does equal to. $2^{(3^4)}=2^{81}$ or ${(2^3)}^4=8^4$?

The division operator $/$ is misused. Example: $N/Acdot m$ is written for $N/(Acdot m)$.

edited Sep 10 '16 at 17:51

community wiki

4 revs
hkBattousai

edited Sep 10 '16 at 17:51

community wiki

4 revs
hkBattousai

community wiki

4 revs
hkBattousai

community wiki

4 revs
hkBattousai

6

$begingroup$
Well, C++ isn't all bad. In mathematics, the prevailing convention is that $log$ denotes the (a local) inverse of $exp$, and other bases are explicitly indicated ($log_{text{base}}$). Only when speaking to/writing for physicists/engineers has $ln$ somewhat widespread use.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:50

3

$begingroup$
-1 for the C++ comment. $log$ is the standard way mathematicians write the natural log, and mathematicians never write $log$ for the base-10 log.
$endgroup$
– Charles
Sep 10 '16 at 17:30

add a comment |

6

$begingroup$
Well, C++ isn't all bad. In mathematics, the prevailing convention is that $log$ denotes the (a local) inverse of $exp$, and other bases are explicitly indicated ($log_{text{base}}$). Only when speaking to/writing for physicists/engineers has $ln$ somewhat widespread use.
$endgroup$
– Daniel Fischer♦
Nov 23 '14 at 12:50

3

$begingroup$
-1 for the C++ comment. $log$ is the standard way mathematicians write the natural log, and mathematicians never write $log$ for the base-10 log.
$endgroup$
– Charles
Sep 10 '16 at 17:30

Well, C++ isn't all bad. In mathematics, the prevailing convention is that $log$ denotes the (a local) inverse of $exp$, and other bases are explicitly indicated ($log_{text{base}}$). Only when speaking to/writing for physicists/engineers has $ln$ somewhat widespread use.

– Daniel Fischer♦
Nov 23 '14 at 12:50

-1 for the C++ comment. $log$ is the standard way mathematicians write the natural log, and mathematicians never write $log$ for the base-10 log.

– Charles
Sep 10 '16 at 17:30

add a comment |

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