Mixing
Where is the mistake in this proof that “If the square of a number is even, then the number is even”?
$begingroup$
The proof given in This question is incorrect (the proof will be posted at the end for convenience). However, the question seem to address the fact that the statement to be proven is not stated correctly, and not the fact the the proof is incorrect.
Stated different, what I am asking is...
What step in the following (see below) proof of "If a number $n^2$
is even, then $n$ is even" is incorrect?
(Some step must be incorrect since If I change $n^2$ to $x$ and $n$ to $sqrt{x}$, we would have a proof saying "if you give me any even number $x$, its square root is even")
The proof (by contrapositive) given is as follows:
If a number $n^2$ is even, then $n$ is even. The contrapositive is that is that if $n$ is not even (odd), then $n^2$ must also be not be even (be odd).
We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(p^2+2) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.
Is the problem that, in general, the negation of "$n$ is even", but rather "$n$ is odd OR $n$ is not an integer$?
proof-explanation
asked Jan 19 at 3:13
user106860user106860
390214
$endgroup$
add a comment |
$begingroup$
The proof given in This question is incorrect (the proof will be posted at the end for convenience). However, the question seem to address the fact that the statement to be proven is not stated correctly, and not the fact the the proof is incorrect.
Stated different, what I am asking is...
What step in the following (see below) proof of "If a number $n^2$
is even, then $n$ is even" is incorrect?
(Some step must be incorrect since If I change $n^2$ to $x$ and $n$ to $sqrt{x}$, we would have a proof saying "if you give me any even number $x$, its square root is even")
The proof (by contrapositive) given is as follows:
If a number $n^2$ is even, then $n$ is even. The contrapositive is that is that if $n$ is not even (odd), then $n^2$ must also be not be even (be odd).
We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(p^2+2) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.
Is the problem that, in general, the negation of "$n$ is even", but rather "$n$ is odd OR $n$ is not an integer$?
proof-explanation
asked Jan 19 at 3:13
user106860user106860
390214
$endgroup$
$begingroup$
Your claim that it is incorrect is itself incorrect; the issue here is that the statement elides the fact that “number” here is to be understood to mean “integer”, as those are the only types of real numbers for which “odd” and “even” apply. So the statement really says “For all integers $n$, if $n^2$ is even then $n$ is even”. The contrapositive says “For all integers $n$, if $n$ is not even then $n^2$ is not even”, and since “odd” literally means “not even” (for integers), this is the same as “For all integers $n$, if $n$ is odd then $n^2$ is odd”. Nothing wrong in the argument given then.
$endgroup$
– Arturo Magidin
Jan 19 at 3:17
$begingroup$
Questions about even/odd numbers are almost exclusively asked for the scenario where the domain of numbers to be considered are integers or natural numbers. We tacitly make the assumption then that when we are asked to prove that "if the square of a number is even, then the number is even" that the "number" is an integer. Could it have been explicitly written out to make it clearer that this is what we mean? Yes, of course. Is it necessary? I would argue not, especially given the context in which the problem is generally presented.
$endgroup$
– JMoravitz
Jan 19 at 3:18
$begingroup$
@ArturoMagidin But in fact there do exist other subrings of $,Bbb R,$ besides $,Bbb Z,$ where "odd and even apply", i.e. have $,Bbb Z/2,$ as an image so partiy arguments work, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 5:06
$begingroup$
I added an answer in the original thread which explains what occurs if we do in fact consider the argument in a quadratic number ring
$endgroup$
– Bill Dubuque
Jan 19 at 15:55
add a comment |
$begingroup$
The proof given in This question is incorrect (the proof will be posted at the end for convenience). However, the question seem to address the fact that the statement to be proven is not stated correctly, and not the fact the the proof is incorrect.
Stated different, what I am asking is...
What step in the following (see below) proof of "If a number $n^2$
is even, then $n$ is even" is incorrect?
(Some step must be incorrect since If I change $n^2$ to $x$ and $n$ to $sqrt{x}$, we would have a proof saying "if you give me any even number $x$, its square root is even")
The proof (by contrapositive) given is as follows:
If a number $n^2$ is even, then $n$ is even. The contrapositive is that is that if $n$ is not even (odd), then $n^2$ must also be not be even (be odd).
We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(p^2+2) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.
Is the problem that, in general, the negation of "$n$ is even", but rather "$n$ is odd OR $n$ is not an integer$?
proof-explanation
asked Jan 19 at 3:13
user106860user106860
390214
$endgroup$
The proof given in This question is incorrect (the proof will be posted at the end for convenience). However, the question seem to address the fact that the statement to be proven is not stated correctly, and not the fact the the proof is incorrect.
Stated different, what I am asking is...
What step in the following (see below) proof of "If a number $n^2$
is even, then $n$ is even" is incorrect?
(Some step must be incorrect since If I change $n^2$ to $x$ and $n$ to $sqrt{x}$, we would have a proof saying "if you give me any even number $x$, its square root is even")
The proof (by contrapositive) given is as follows:
If a number $n^2$ is even, then $n$ is even. The contrapositive is that is that if $n$ is not even (odd), then $n^2$ must also be not be even (be odd).
We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(p^2+2) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.
Is the problem that, in general, the negation of "$n$ is even", but rather "$n$ is odd OR $n$ is not an integer$?
proof-explanation
proof-explanation
asked Jan 19 at 3:13
user106860user106860
390214
asked Jan 19 at 3:13
user106860user106860
390214
asked Jan 19 at 3:13
user106860user106860
390214
asked Jan 19 at 3:13
user106860user106860
390214
asked Jan 19 at 3:13
user106860user106860
390214
390214
$begingroup$
Your claim that it is incorrect is itself incorrect; the issue here is that the statement elides the fact that “number” here is to be understood to mean “integer”, as those are the only types of real numbers for which “odd” and “even” apply. So the statement really says “For all integers $n$, if $n^2$ is even then $n$ is even”. The contrapositive says “For all integers $n$, if $n$ is not even then $n^2$ is not even”, and since “odd” literally means “not even” (for integers), this is the same as “For all integers $n$, if $n$ is odd then $n^2$ is odd”. Nothing wrong in the argument given then.
$endgroup$
– Arturo Magidin
Jan 19 at 3:17
$begingroup$
Questions about even/odd numbers are almost exclusively asked for the scenario where the domain of numbers to be considered are integers or natural numbers. We tacitly make the assumption then that when we are asked to prove that "if the square of a number is even, then the number is even" that the "number" is an integer. Could it have been explicitly written out to make it clearer that this is what we mean? Yes, of course. Is it necessary? I would argue not, especially given the context in which the problem is generally presented.
$endgroup$
– JMoravitz
Jan 19 at 3:18
$begingroup$
@ArturoMagidin But in fact there do exist other subrings of $,Bbb R,$ besides $,Bbb Z,$ where "odd and even apply", i.e. have $,Bbb Z/2,$ as an image so partiy arguments work, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 5:06
$begingroup$
I added an answer in the original thread which explains what occurs if we do in fact consider the argument in a quadratic number ring
$endgroup$
– Bill Dubuque
Jan 19 at 15:55
add a comment |
$begingroup$
Your claim that it is incorrect is itself incorrect; the issue here is that the statement elides the fact that “number” here is to be understood to mean “integer”, as those are the only types of real numbers for which “odd” and “even” apply. So the statement really says “For all integers $n$, if $n^2$ is even then $n$ is even”. The contrapositive says “For all integers $n$, if $n$ is not even then $n^2$ is not even”, and since “odd” literally means “not even” (for integers), this is the same as “For all integers $n$, if $n$ is odd then $n^2$ is odd”. Nothing wrong in the argument given then.
$endgroup$
– Arturo Magidin
Jan 19 at 3:17
$begingroup$
Questions about even/odd numbers are almost exclusively asked for the scenario where the domain of numbers to be considered are integers or natural numbers. We tacitly make the assumption then that when we are asked to prove that "if the square of a number is even, then the number is even" that the "number" is an integer. Could it have been explicitly written out to make it clearer that this is what we mean? Yes, of course. Is it necessary? I would argue not, especially given the context in which the problem is generally presented.
$endgroup$
– JMoravitz
Jan 19 at 3:18
$begingroup$
@ArturoMagidin But in fact there do exist other subrings of $,Bbb R,$ besides $,Bbb Z,$ where "odd and even apply", i.e. have $,Bbb Z/2,$ as an image so partiy arguments work, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 5:06
$begingroup$
I added an answer in the original thread which explains what occurs if we do in fact consider the argument in a quadratic number ring
$endgroup$
– Bill Dubuque
Jan 19 at 15:55
$begingroup$
Your claim that it is incorrect is itself incorrect; the issue here is that the statement elides the fact that “number” here is to be understood to mean “integer”, as those are the only types of real numbers for which “odd” and “even” apply. So the statement really says “For all integers $n$, if $n^2$ is even then $n$ is even”. The contrapositive says “For all integers $n$, if $n$ is not even then $n^2$ is not even”, and since “odd” literally means “not even” (for integers), this is the same as “For all integers $n$, if $n$ is odd then $n^2$ is odd”. Nothing wrong in the argument given then.
$endgroup$
– Arturo Magidin
Jan 19 at 3:17
$begingroup$
Your claim that it is incorrect is itself incorrect; the issue here is that the statement elides the fact that “number” here is to be understood to mean “integer”, as those are the only types of real numbers for which “odd” and “even” apply. So the statement really says “For all integers $n$, if $n^2$ is even then $n$ is even”. The contrapositive says “For all integers $n$, if $n$ is not even then $n^2$ is not even”, and since “odd” literally means “not even” (for integers), this is the same as “For all integers $n$, if $n$ is odd then $n^2$ is odd”. Nothing wrong in the argument given then.
$endgroup$
– Arturo Magidin
Jan 19 at 3:17
$begingroup$
Questions about even/odd numbers are almost exclusively asked for the scenario where the domain of numbers to be considered are integers or natural numbers. We tacitly make the assumption then that when we are asked to prove that "if the square of a number is even, then the number is even" that the "number" is an integer. Could it have been explicitly written out to make it clearer that this is what we mean? Yes, of course. Is it necessary? I would argue not, especially given the context in which the problem is generally presented.
$endgroup$
– JMoravitz
Jan 19 at 3:18
$begingroup$
Questions about even/odd numbers are almost exclusively asked for the scenario where the domain of numbers to be considered are integers or natural numbers. We tacitly make the assumption then that when we are asked to prove that "if the square of a number is even, then the number is even" that the "number" is an integer. Could it have been explicitly written out to make it clearer that this is what we mean? Yes, of course. Is it necessary? I would argue not, especially given the context in which the problem is generally presented.
$endgroup$
– JMoravitz
Jan 19 at 3:18
$begingroup$
@ArturoMagidin But in fact there do exist other subrings of $,Bbb R,$ besides $,Bbb Z,$ where "odd and even apply", i.e. have $,Bbb Z/2,$ as an image so partiy arguments work, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 5:06
$begingroup$
@ArturoMagidin But in fact there do exist other subrings of $,Bbb R,$ besides $,Bbb Z,$ where "odd and even apply", i.e. have $,Bbb Z/2,$ as an image so partiy arguments work, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 5:06
$begingroup$
I added an answer in the original thread which explains what occurs if we do in fact consider the argument in a quadratic number ring
$endgroup$
– Bill Dubuque
Jan 19 at 15:55
$begingroup$
I added an answer in the original thread which explains what occurs if we do in fact consider the argument in a quadratic number ring
$endgroup$
– Bill Dubuque
Jan 19 at 15:55
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We need to be clear about what values $n$ is allowed to take. For example; $n^2 = 2$ is even, but $n$ is not, because $n$ is not an integer.
If $n$ is an integer and $n^2$ is even, then $n$ is even. This is a correct statement.
If $n$ is a real number, and $n^2$ is even, then $n$ is not necessarily even; $n = sqrt{2}$ is a counterexample as mentioned above.
The flaw in the proof when $n$ is not assumed to be an integer is that if $n$ is not even, it does not necessarily follow that $n$ is odd, because $n$ may not be an integer. When $n$ is an integer, then either $n$ is divisible by $2$ (we call these 'even') or $n$ is not divisible by $2$ (we call these 'odd'). Therefore, when $n$ is not restricted to integers, the negation of the statement "$n$ is even" is not "$n$ is odd," but rather, "$n$ is not an even integer."
answered Jan 19 at 3:36
heropupheropup
64k762102
$endgroup$
$begingroup$
Thank you. "The flaw in the proof when $n$ is not assumed to be an integer" was exactly what I was looking for -- I just could figure out how to ask it so clearly.
$endgroup$
– user106860
Jan 21 at 15:32
add a comment |
$begingroup$
The proof you've cited is correct. It is a good proof by contrapositive. In fact, it is often the "role model" proof by contrapositive that I first introduce in a proofs course. (And let's say "integer" instead of "number" for clarity.)
Your mistake is in thinking that you can "change... $n$ to $sqrt{x}$" and get something meaningful. After all, $sqrt{x}$ is rarely an integer when $x$ is a natural number, so "even" and "odd" wouldn't even make sense. For example, if $x=3$, what would it mean for $sqrt{3}$ to be "even"? It doesn't make any sense at all.
answered Jan 19 at 3:17
RandallRandall
10.1k11230
$endgroup$
$begingroup$
Actually it does make sense.There are many rings of algebraic integers where parity arguments work fine, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 14:31
$begingroup$
Good lord that is certainly not the context of OP’s question.
$endgroup$
– Randall
Jan 19 at 14:56
$begingroup$
My point was not about the OP's context, Rather, it was only to emphasize that such arithmetic does in fact "make sense". I added an answer in the original thread on that topic.
$endgroup$
– Bill Dubuque
Jan 19 at 15:35
$begingroup$
I understand that all sorts of generalizations exist, but I don’t find such things helpful to OP. in fact, it may only confuse them. I mean, I know some things about smooth manifolds but I don’t bring it up in Calc I.
$endgroup$
– Randall
Jan 19 at 15:39
$begingroup$
Please pardon my ranting. It is growing tiresome with the amount of nitpicking on this site when you are only trying to help out relative novices or newcomers. This frustration need not be directed at you.
$endgroup$
– Randall
Jan 19 at 15:42
|
show 2 more comments
$begingroup$
This is a classic example of proof by contrapositive. It is completely correct, and I'll use truth tables to show you why:
$$begin{array}{c|c|c} P & to & Q \ hline T & T & T \ hline T & F & F \ hline F & T & T \ hline F & T & F end{array}$$
and, where $P'$ means not $P$:
$$begin{array}{c|c|c} P' & leftarrow & Q' \ hline F & T & F \ hline F & F & T \ hline T & T & F \ hline T & T & Tend{array}$$
The two middle rows match, so the tables' meanings are the same, or have the same truth value. That is to say:
"If P then Q" is the same as "If not Q, then not P".
This is what is used in the proof method.
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
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add a comment |
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3 Answers
3
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3 Answers
3
active
oldest
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$begingroup$
We need to be clear about what values $n$ is allowed to take. For example; $n^2 = 2$ is even, but $n$ is not, because $n$ is not an integer.
If $n$ is an integer and $n^2$ is even, then $n$ is even. This is a correct statement.
If $n$ is a real number, and $n^2$ is even, then $n$ is not necessarily even; $n = sqrt{2}$ is a counterexample as mentioned above.
The flaw in the proof when $n$ is not assumed to be an integer is that if $n$ is not even, it does not necessarily follow that $n$ is odd, because $n$ may not be an integer. When $n$ is an integer, then either $n$ is divisible by $2$ (we call these 'even') or $n$ is not divisible by $2$ (we call these 'odd'). Therefore, when $n$ is not restricted to integers, the negation of the statement "$n$ is even" is not "$n$ is odd," but rather, "$n$ is not an even integer."
answered Jan 19 at 3:36
heropupheropup
64k762102
$endgroup$
$begingroup$
Thank you. "The flaw in the proof when $n$ is not assumed to be an integer" was exactly what I was looking for -- I just could figure out how to ask it so clearly.
$endgroup$
– user106860
Jan 21 at 15:32
add a comment |
$begingroup$
We need to be clear about what values $n$ is allowed to take. For example; $n^2 = 2$ is even, but $n$ is not, because $n$ is not an integer.
If $n$ is an integer and $n^2$ is even, then $n$ is even. This is a correct statement.
If $n$ is a real number, and $n^2$ is even, then $n$ is not necessarily even; $n = sqrt{2}$ is a counterexample as mentioned above.
The flaw in the proof when $n$ is not assumed to be an integer is that if $n$ is not even, it does not necessarily follow that $n$ is odd, because $n$ may not be an integer. When $n$ is an integer, then either $n$ is divisible by $2$ (we call these 'even') or $n$ is not divisible by $2$ (we call these 'odd'). Therefore, when $n$ is not restricted to integers, the negation of the statement "$n$ is even" is not "$n$ is odd," but rather, "$n$ is not an even integer."
answered Jan 19 at 3:36
heropupheropup
64k762102
$endgroup$
$begingroup$
Thank you. "The flaw in the proof when $n$ is not assumed to be an integer" was exactly what I was looking for -- I just could figure out how to ask it so clearly.
$endgroup$
– user106860
Jan 21 at 15:32
add a comment |
$begingroup$
We need to be clear about what values $n$ is allowed to take. For example; $n^2 = 2$ is even, but $n$ is not, because $n$ is not an integer.
If $n$ is an integer and $n^2$ is even, then $n$ is even. This is a correct statement.
If $n$ is a real number, and $n^2$ is even, then $n$ is not necessarily even; $n = sqrt{2}$ is a counterexample as mentioned above.
The flaw in the proof when $n$ is not assumed to be an integer is that if $n$ is not even, it does not necessarily follow that $n$ is odd, because $n$ may not be an integer. When $n$ is an integer, then either $n$ is divisible by $2$ (we call these 'even') or $n$ is not divisible by $2$ (we call these 'odd'). Therefore, when $n$ is not restricted to integers, the negation of the statement "$n$ is even" is not "$n$ is odd," but rather, "$n$ is not an even integer."
answered Jan 19 at 3:36
heropupheropup
64k762102
$endgroup$
We need to be clear about what values $n$ is allowed to take. For example; $n^2 = 2$ is even, but $n$ is not, because $n$ is not an integer.
If $n$ is an integer and $n^2$ is even, then $n$ is even. This is a correct statement.
If $n$ is a real number, and $n^2$ is even, then $n$ is not necessarily even; $n = sqrt{2}$ is a counterexample as mentioned above.
The flaw in the proof when $n$ is not assumed to be an integer is that if $n$ is not even, it does not necessarily follow that $n$ is odd, because $n$ may not be an integer. When $n$ is an integer, then either $n$ is divisible by $2$ (we call these 'even') or $n$ is not divisible by $2$ (we call these 'odd'). Therefore, when $n$ is not restricted to integers, the negation of the statement "$n$ is even" is not "$n$ is odd," but rather, "$n$ is not an even integer."
answered Jan 19 at 3:36
heropupheropup
64k762102
answered Jan 19 at 3:36
heropupheropup
64k762102
answered Jan 19 at 3:36
heropupheropup
64k762102
answered Jan 19 at 3:36
heropupheropup
64k762102
64k762102
$begingroup$
Thank you. "The flaw in the proof when $n$ is not assumed to be an integer" was exactly what I was looking for -- I just could figure out how to ask it so clearly.
$endgroup$
– user106860
Jan 21 at 15:32
add a comment |
$begingroup$
Thank you. "The flaw in the proof when $n$ is not assumed to be an integer" was exactly what I was looking for -- I just could figure out how to ask it so clearly.
$endgroup$
– user106860
Jan 21 at 15:32
$begingroup$
Thank you. "The flaw in the proof when $n$ is not assumed to be an integer" was exactly what I was looking for -- I just could figure out how to ask it so clearly.
$endgroup$
– user106860
Jan 21 at 15:32
$begingroup$
Thank you. "The flaw in the proof when $n$ is not assumed to be an integer" was exactly what I was looking for -- I just could figure out how to ask it so clearly.
$endgroup$
– user106860
Jan 21 at 15:32
add a comment |
$begingroup$
The proof you've cited is correct. It is a good proof by contrapositive. In fact, it is often the "role model" proof by contrapositive that I first introduce in a proofs course. (And let's say "integer" instead of "number" for clarity.)
Your mistake is in thinking that you can "change... $n$ to $sqrt{x}$" and get something meaningful. After all, $sqrt{x}$ is rarely an integer when $x$ is a natural number, so "even" and "odd" wouldn't even make sense. For example, if $x=3$, what would it mean for $sqrt{3}$ to be "even"? It doesn't make any sense at all.
answered Jan 19 at 3:17
RandallRandall
10.1k11230
$endgroup$
$begingroup$
Actually it does make sense.There are many rings of algebraic integers where parity arguments work fine, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 14:31
$begingroup$
Good lord that is certainly not the context of OP’s question.
$endgroup$
– Randall
Jan 19 at 14:56
$begingroup$
My point was not about the OP's context, Rather, it was only to emphasize that such arithmetic does in fact "make sense". I added an answer in the original thread on that topic.
$endgroup$
– Bill Dubuque
Jan 19 at 15:35
$begingroup$
I understand that all sorts of generalizations exist, but I don’t find such things helpful to OP. in fact, it may only confuse them. I mean, I know some things about smooth manifolds but I don’t bring it up in Calc I.
$endgroup$
– Randall
Jan 19 at 15:39
$begingroup$
Please pardon my ranting. It is growing tiresome with the amount of nitpicking on this site when you are only trying to help out relative novices or newcomers. This frustration need not be directed at you.
$endgroup$
– Randall
Jan 19 at 15:42
|
show 2 more comments
$begingroup$
The proof you've cited is correct. It is a good proof by contrapositive. In fact, it is often the "role model" proof by contrapositive that I first introduce in a proofs course. (And let's say "integer" instead of "number" for clarity.)
Your mistake is in thinking that you can "change... $n$ to $sqrt{x}$" and get something meaningful. After all, $sqrt{x}$ is rarely an integer when $x$ is a natural number, so "even" and "odd" wouldn't even make sense. For example, if $x=3$, what would it mean for $sqrt{3}$ to be "even"? It doesn't make any sense at all.
answered Jan 19 at 3:17
RandallRandall
10.1k11230
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$begingroup$
Actually it does make sense.There are many rings of algebraic integers where parity arguments work fine, e.g. see this answer.
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– Bill Dubuque
Jan 19 at 14:31
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Good lord that is certainly not the context of OP’s question.
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– Randall
Jan 19 at 14:56
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My point was not about the OP's context, Rather, it was only to emphasize that such arithmetic does in fact "make sense". I added an answer in the original thread on that topic.
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– Bill Dubuque
Jan 19 at 15:35
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I understand that all sorts of generalizations exist, but I don’t find such things helpful to OP. in fact, it may only confuse them. I mean, I know some things about smooth manifolds but I don’t bring it up in Calc I.
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– Randall
Jan 19 at 15:39
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Please pardon my ranting. It is growing tiresome with the amount of nitpicking on this site when you are only trying to help out relative novices or newcomers. This frustration need not be directed at you.
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– Randall
Jan 19 at 15:42
|
show 2 more comments
$begingroup$
The proof you've cited is correct. It is a good proof by contrapositive. In fact, it is often the "role model" proof by contrapositive that I first introduce in a proofs course. (And let's say "integer" instead of "number" for clarity.)
Your mistake is in thinking that you can "change... $n$ to $sqrt{x}$" and get something meaningful. After all, $sqrt{x}$ is rarely an integer when $x$ is a natural number, so "even" and "odd" wouldn't even make sense. For example, if $x=3$, what would it mean for $sqrt{3}$ to be "even"? It doesn't make any sense at all.
answered Jan 19 at 3:17
RandallRandall
10.1k11230
$endgroup$
The proof you've cited is correct. It is a good proof by contrapositive. In fact, it is often the "role model" proof by contrapositive that I first introduce in a proofs course. (And let's say "integer" instead of "number" for clarity.)
Your mistake is in thinking that you can "change... $n$ to $sqrt{x}$" and get something meaningful. After all, $sqrt{x}$ is rarely an integer when $x$ is a natural number, so "even" and "odd" wouldn't even make sense. For example, if $x=3$, what would it mean for $sqrt{3}$ to be "even"? It doesn't make any sense at all.
answered Jan 19 at 3:17
RandallRandall
10.1k11230
answered Jan 19 at 3:17
RandallRandall
10.1k11230
answered Jan 19 at 3:17
RandallRandall
10.1k11230
answered Jan 19 at 3:17
RandallRandall
10.1k11230
10.1k11230
$begingroup$
Actually it does make sense.There are many rings of algebraic integers where parity arguments work fine, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 14:31
$begingroup$
Good lord that is certainly not the context of OP’s question.
$endgroup$
– Randall
Jan 19 at 14:56
$begingroup$
My point was not about the OP's context, Rather, it was only to emphasize that such arithmetic does in fact "make sense". I added an answer in the original thread on that topic.
$endgroup$
– Bill Dubuque
Jan 19 at 15:35
$begingroup$
I understand that all sorts of generalizations exist, but I don’t find such things helpful to OP. in fact, it may only confuse them. I mean, I know some things about smooth manifolds but I don’t bring it up in Calc I.
$endgroup$
– Randall
Jan 19 at 15:39
$begingroup$
Please pardon my ranting. It is growing tiresome with the amount of nitpicking on this site when you are only trying to help out relative novices or newcomers. This frustration need not be directed at you.
$endgroup$
– Randall
Jan 19 at 15:42
|
show 2 more comments
$begingroup$
Actually it does make sense.There are many rings of algebraic integers where parity arguments work fine, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 14:31
$begingroup$
Good lord that is certainly not the context of OP’s question.
$endgroup$
– Randall
Jan 19 at 14:56
$begingroup$
My point was not about the OP's context, Rather, it was only to emphasize that such arithmetic does in fact "make sense". I added an answer in the original thread on that topic.
$endgroup$
– Bill Dubuque
Jan 19 at 15:35
$begingroup$
I understand that all sorts of generalizations exist, but I don’t find such things helpful to OP. in fact, it may only confuse them. I mean, I know some things about smooth manifolds but I don’t bring it up in Calc I.
$endgroup$
– Randall
Jan 19 at 15:39
$begingroup$
Please pardon my ranting. It is growing tiresome with the amount of nitpicking on this site when you are only trying to help out relative novices or newcomers. This frustration need not be directed at you.
$endgroup$
– Randall
Jan 19 at 15:42
$begingroup$
Actually it does make sense.There are many rings of algebraic integers where parity arguments work fine, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 14:31
$begingroup$
Actually it does make sense.There are many rings of algebraic integers where parity arguments work fine, e.g. see this answer.
$endgroup$
– Bill Dubuque
Jan 19 at 14:31
$begingroup$
Good lord that is certainly not the context of OP’s question.
$endgroup$
– Randall
Jan 19 at 14:56
$begingroup$
Good lord that is certainly not the context of OP’s question.
$endgroup$
– Randall
Jan 19 at 14:56
$begingroup$
My point was not about the OP's context, Rather, it was only to emphasize that such arithmetic does in fact "make sense". I added an answer in the original thread on that topic.
$endgroup$
– Bill Dubuque
Jan 19 at 15:35
$begingroup$
My point was not about the OP's context, Rather, it was only to emphasize that such arithmetic does in fact "make sense". I added an answer in the original thread on that topic.
$endgroup$
– Bill Dubuque
Jan 19 at 15:35
$begingroup$
I understand that all sorts of generalizations exist, but I don’t find such things helpful to OP. in fact, it may only confuse them. I mean, I know some things about smooth manifolds but I don’t bring it up in Calc I.
$endgroup$
– Randall
Jan 19 at 15:39
$begingroup$
I understand that all sorts of generalizations exist, but I don’t find such things helpful to OP. in fact, it may only confuse them. I mean, I know some things about smooth manifolds but I don’t bring it up in Calc I.
$endgroup$
– Randall
Jan 19 at 15:39
$begingroup$
Please pardon my ranting. It is growing tiresome with the amount of nitpicking on this site when you are only trying to help out relative novices or newcomers. This frustration need not be directed at you.
$endgroup$
– Randall
Jan 19 at 15:42
$begingroup$
Please pardon my ranting. It is growing tiresome with the amount of nitpicking on this site when you are only trying to help out relative novices or newcomers. This frustration need not be directed at you.
$endgroup$
– Randall
Jan 19 at 15:42
|
show 2 more comments
$begingroup$
This is a classic example of proof by contrapositive. It is completely correct, and I'll use truth tables to show you why:
$$begin{array}{c|c|c} P & to & Q \ hline T & T & T \ hline T & F & F \ hline F & T & T \ hline F & T & F end{array}$$
and, where $P'$ means not $P$:
$$begin{array}{c|c|c} P' & leftarrow & Q' \ hline F & T & F \ hline F & F & T \ hline T & T & F \ hline T & T & Tend{array}$$
The two middle rows match, so the tables' meanings are the same, or have the same truth value. That is to say:
"If P then Q" is the same as "If not Q, then not P".
This is what is used in the proof method.
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
$endgroup$
add a comment |
$begingroup$
This is a classic example of proof by contrapositive. It is completely correct, and I'll use truth tables to show you why:
$$begin{array}{c|c|c} P & to & Q \ hline T & T & T \ hline T & F & F \ hline F & T & T \ hline F & T & F end{array}$$
and, where $P'$ means not $P$:
$$begin{array}{c|c|c} P' & leftarrow & Q' \ hline F & T & F \ hline F & F & T \ hline T & T & F \ hline T & T & Tend{array}$$
The two middle rows match, so the tables' meanings are the same, or have the same truth value. That is to say:
"If P then Q" is the same as "If not Q, then not P".
This is what is used in the proof method.
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
$endgroup$
add a comment |
$begingroup$
This is a classic example of proof by contrapositive. It is completely correct, and I'll use truth tables to show you why:
$$begin{array}{c|c|c} P & to & Q \ hline T & T & T \ hline T & F & F \ hline F & T & T \ hline F & T & F end{array}$$
and, where $P'$ means not $P$:
$$begin{array}{c|c|c} P' & leftarrow & Q' \ hline F & T & F \ hline F & F & T \ hline T & T & F \ hline T & T & Tend{array}$$
The two middle rows match, so the tables' meanings are the same, or have the same truth value. That is to say:
"If P then Q" is the same as "If not Q, then not P".
This is what is used in the proof method.
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
$endgroup$
This is a classic example of proof by contrapositive. It is completely correct, and I'll use truth tables to show you why:
$$begin{array}{c|c|c} P & to & Q \ hline T & T & T \ hline T & F & F \ hline F & T & T \ hline F & T & F end{array}$$
and, where $P'$ means not $P$:
$$begin{array}{c|c|c} P' & leftarrow & Q' \ hline F & T & F \ hline F & F & T \ hline T & T & F \ hline T & T & Tend{array}$$
The two middle rows match, so the tables' meanings are the same, or have the same truth value. That is to say:
"If P then Q" is the same as "If not Q, then not P".
This is what is used in the proof method.
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 4:26
Rhys HughesRhys Hughes
6,9441530
6,9441530
add a comment |
add a comment |
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$begingroup$
Your claim that it is incorrect is itself incorrect; the issue here is that the statement elides the fact that “number” here is to be understood to mean “integer”, as those are the only types of real numbers for which “odd” and “even” apply. So the statement really says “For all integers $n$, if $n^2$ is even then $n$ is even”. The contrapositive says “For all integers $n$, if $n$ is not even then $n^2$ is not even”, and since “odd” literally means “not even” (for integers), this is the same as “For all integers $n$, if $n$ is odd then $n^2$ is odd”. Nothing wrong in the argument given then.
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– Arturo Magidin
Jan 19 at 3:17
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Questions about even/odd numbers are almost exclusively asked for the scenario where the domain of numbers to be considered are integers or natural numbers. We tacitly make the assumption then that when we are asked to prove that "if the square of a number is even, then the number is even" that the "number" is an integer. Could it have been explicitly written out to make it clearer that this is what we mean? Yes, of course. Is it necessary? I would argue not, especially given the context in which the problem is generally presented.
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– JMoravitz
Jan 19 at 3:18
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@ArturoMagidin But in fact there do exist other subrings of $,Bbb R,$ besides $,Bbb Z,$ where "odd and even apply", i.e. have $,Bbb Z/2,$ as an image so partiy arguments work, e.g. see this answer.
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– Bill Dubuque
Jan 19 at 5:06
$begingroup$
I added an answer in the original thread which explains what occurs if we do in fact consider the argument in a quadratic number ring
$endgroup$
– Bill Dubuque
Jan 19 at 15:55