Mixing
Archimedean property concept
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I want to know what the "big deal" about the Archimedean property is. Abbott states it is an important fact about how $Bbb Q$ fits inside $Bbb R.$
First, I want to know if the following statements are true:
The Archimedean property states that $Bbb N$ isn't bounded above--some natural number can be found such that it is greater than some specified real number.
The Archimedean property also states that there is some rational $frac1n, n inBbb N$ such that it is less than some specified real number.
Secondly, what do the above statements imply about the connection between $Bbb Q$ and $Bbb R?$ Does it imply that $Bbb R$ fills the gaps of $Bbb Q$ and $Bbb N;$ is it the proof for $Bbb R$ completing $Bbb Q?$
Lastly, what insights have you obtained from the Archimedean property?
I am sorry if some of my questions are unclear.
real-analysis
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
$endgroup$
|
show 2 more comments
$begingroup$
I want to know what the "big deal" about the Archimedean property is. Abbott states it is an important fact about how $Bbb Q$ fits inside $Bbb R.$
First, I want to know if the following statements are true:
The Archimedean property states that $Bbb N$ isn't bounded above--some natural number can be found such that it is greater than some specified real number.
The Archimedean property also states that there is some rational $frac1n, n inBbb N$ such that it is less than some specified real number.
Secondly, what do the above statements imply about the connection between $Bbb Q$ and $Bbb R?$ Does it imply that $Bbb R$ fills the gaps of $Bbb Q$ and $Bbb N;$ is it the proof for $Bbb R$ completing $Bbb Q?$
Lastly, what insights have you obtained from the Archimedean property?
I am sorry if some of my questions are unclear.
real-analysis
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
$endgroup$
$begingroup$
@GitGud: I don't think the additional tags were needed.
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– Asaf Karagila♦
Aug 19 '13 at 23:38
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@AsafKaragila I was doubtful about topology, not about set theory, though. But you're the expert.
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– Git Gud
Aug 19 '13 at 23:41
1
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It is quite important, for it means that (in the reals) there are no "infinitesimals" of the type that were used in the early years of calculus.
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– André Nicolas
Aug 19 '13 at 23:43
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@AsafKaragila I changed my mind, I realise now this has little to do with the construction of the real numbers and concur with your decision.
$endgroup$
– Git Gud
Aug 19 '13 at 23:57
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@GitGud: You're right. Now, anyway. ;-)
$endgroup$
– Asaf Karagila♦
Aug 20 '13 at 0:02
|
show 2 more comments
$begingroup$
I want to know what the "big deal" about the Archimedean property is. Abbott states it is an important fact about how $Bbb Q$ fits inside $Bbb R.$
First, I want to know if the following statements are true:
The Archimedean property states that $Bbb N$ isn't bounded above--some natural number can be found such that it is greater than some specified real number.
The Archimedean property also states that there is some rational $frac1n, n inBbb N$ such that it is less than some specified real number.
Secondly, what do the above statements imply about the connection between $Bbb Q$ and $Bbb R?$ Does it imply that $Bbb R$ fills the gaps of $Bbb Q$ and $Bbb N;$ is it the proof for $Bbb R$ completing $Bbb Q?$
Lastly, what insights have you obtained from the Archimedean property?
I am sorry if some of my questions are unclear.
real-analysis
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
$endgroup$
I want to know what the "big deal" about the Archimedean property is. Abbott states it is an important fact about how $Bbb Q$ fits inside $Bbb R.$
First, I want to know if the following statements are true:
The Archimedean property states that $Bbb N$ isn't bounded above--some natural number can be found such that it is greater than some specified real number.
The Archimedean property also states that there is some rational $frac1n, n inBbb N$ such that it is less than some specified real number.
Secondly, what do the above statements imply about the connection between $Bbb Q$ and $Bbb R?$ Does it imply that $Bbb R$ fills the gaps of $Bbb Q$ and $Bbb N;$ is it the proof for $Bbb R$ completing $Bbb Q?$
Lastly, what insights have you obtained from the Archimedean property?
I am sorry if some of my questions are unclear.
real-analysis
real-analysis
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
edited Aug 20 '13 at 0:26
MathApprentice
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
asked Aug 19 '13 at 23:34
MathApprenticeMathApprentice
681112
681112
$begingroup$
@GitGud: I don't think the additional tags were needed.
$endgroup$
– Asaf Karagila♦
Aug 19 '13 at 23:38
$begingroup$
@AsafKaragila I was doubtful about topology, not about set theory, though. But you're the expert.
$endgroup$
– Git Gud
Aug 19 '13 at 23:41
1
$begingroup$
It is quite important, for it means that (in the reals) there are no "infinitesimals" of the type that were used in the early years of calculus.
$endgroup$
– André Nicolas
Aug 19 '13 at 23:43
$begingroup$
@AsafKaragila I changed my mind, I realise now this has little to do with the construction of the real numbers and concur with your decision.
$endgroup$
– Git Gud
Aug 19 '13 at 23:57
$begingroup$
@GitGud: You're right. Now, anyway. ;-)
$endgroup$
– Asaf Karagila♦
Aug 20 '13 at 0:02
|
show 2 more comments
$begingroup$
@GitGud: I don't think the additional tags were needed.
$endgroup$
– Asaf Karagila♦
Aug 19 '13 at 23:38
$begingroup$
@AsafKaragila I was doubtful about topology, not about set theory, though. But you're the expert.
$endgroup$
– Git Gud
Aug 19 '13 at 23:41
1
$begingroup$
It is quite important, for it means that (in the reals) there are no "infinitesimals" of the type that were used in the early years of calculus.
$endgroup$
– André Nicolas
Aug 19 '13 at 23:43
$begingroup$
@AsafKaragila I changed my mind, I realise now this has little to do with the construction of the real numbers and concur with your decision.
$endgroup$
– Git Gud
Aug 19 '13 at 23:57
$begingroup$
@GitGud: You're right. Now, anyway. ;-)
$endgroup$
– Asaf Karagila♦
Aug 20 '13 at 0:02
$begingroup$
@GitGud: I don't think the additional tags were needed.
$endgroup$
– Asaf Karagila♦
Aug 19 '13 at 23:38
$begingroup$
@GitGud: I don't think the additional tags were needed.
$endgroup$
– Asaf Karagila♦
Aug 19 '13 at 23:38
$begingroup$
@AsafKaragila I was doubtful about topology, not about set theory, though. But you're the expert.
$endgroup$
– Git Gud
Aug 19 '13 at 23:41
$begingroup$
@AsafKaragila I was doubtful about topology, not about set theory, though. But you're the expert.
$endgroup$
– Git Gud
Aug 19 '13 at 23:41
1
1
$begingroup$
It is quite important, for it means that (in the reals) there are no "infinitesimals" of the type that were used in the early years of calculus.
$endgroup$
– André Nicolas
Aug 19 '13 at 23:43
$begingroup$
It is quite important, for it means that (in the reals) there are no "infinitesimals" of the type that were used in the early years of calculus.
$endgroup$
– André Nicolas
Aug 19 '13 at 23:43
$begingroup$
@AsafKaragila I changed my mind, I realise now this has little to do with the construction of the real numbers and concur with your decision.
$endgroup$
– Git Gud
Aug 19 '13 at 23:57
$begingroup$
@AsafKaragila I changed my mind, I realise now this has little to do with the construction of the real numbers and concur with your decision.
$endgroup$
– Git Gud
Aug 19 '13 at 23:57
$begingroup$
@GitGud: You're right. Now, anyway. ;-)
$endgroup$
– Asaf Karagila♦
Aug 20 '13 at 0:02
$begingroup$
@GitGud: You're right. Now, anyway. ;-)
$endgroup$
– Asaf Karagila♦
Aug 20 '13 at 0:02
|
show 2 more comments
6 Answers
6
active
oldest
votes
$begingroup$
It has been pointed out above that your two versions of the Archimedean property should be equivalent--$Bbb N$ has no upper bound if and only if $left{frac1n:ninBbb Nright}$ has no positive lower bound. Your presentations of these two versions were misleading, however (and still are). More precisely:
- Given any real number $x$, there is a natural number $n$ greater than $x$. This natural number $n$ will of course depend on which $x$ we're looking at in the first place, but we will always be able to find some (infinitely-many, in fact) natural number greater than $x$, regardless of the $x$ we started with.
- Given any positive real number $y$, there is a natural number $n$ such that $frac1n$ is less than $y$. Again, the natural number $n$ will depend on which $y$ we're looking at.
It's a good exercise to show that these are equivalent statements.
Now, let me try to explain what Abbott meant by "how $Bbb Q$ fits inside $Bbb R$." Specifically, I will show that the following are equivalent:
- The Archimedean property holds.
- Given any $a,binBbb R$ with $a<b,$ there is some $rinBbb Q$ such that $a<r<b$. (More briefly, $Bbb Q$ is dense in $Bbb R$.)
- Given any $a,binBbb R$ with $a<b,$ there is some $pinBbb Z$ and some $ninBbb N$ such that $a<frac{p}{2^n}<b$.
It is clear that 3 $implies$ 2. To see that 2 $implies$ 1, take any $x>0,$ and find $rinBbb Q$ such that $0<r<x$. We know that $r=frac mn$ for some positive integers $m,n,$ so $frac1nlefrac mn=r<x,$ and so the Archimedean property holds.
To see that 1 $implies$ 3, we proceed by contrapositive, so suppose that there exist $a,binBbb R$ with $a<b$ such that for all $pinBbb Z$ and all $ninBbb N,$ we have $frac{p}{2^n}le a$ or $frac{p}{2^n}ge b$. For each $ninBbb N,$ let $p_n$ the greatest integer $p$ such that $frac{p}{2^n}le a.$ (By the Well-ordering property of $Bbb N,$ such a $p_n$ exists.) From this, it follows by hypothesis that $$frac{p_n}{2^n}le a<blefrac{p_n+1}{2^n}$$ for all $ninBbb N$. But then $$0<b-alefrac{p_n+1}{2^n}-frac{p_n}{2^n}=frac1{2^n}$$ for all $ninBbb N.$ Since $n<2^n$ for all $ninBbb N,$ then it follows that $0<b-a<frac1n$ for all $ninBbb N,$ and so the Archimedean property fails. Thus, by contrapositive, 1 $implies$ 3.
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
85.6k772160
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In a group with a linear order, if $a$ and $b$ are two positive elements, $a$ is said to be infinitesimal with respect to $b$, if no positive integer multiple of $a$ is greater than $b$. Then the linear order is called Archimedean if there are no infinitesimal elements. The first two statements you mentioned are easy consequences of $mathbb{R}$ being Archimedean as a linear order:
If $r$ is any positive real number, since $1$ is not infinitesimal with respect to $r$ by the Archimedean property, there exists a natural number $n$ such that $n.1=n> r$. This shows $mathbb{N}$ is not bounded above.
The second statement is also true for positive real numbers: Suppose we're given a positive real number $r$. Then by the previous property, there exists a natural number $n$ such that $1/r < n$. Therefore $1/n<r$.
But to make sense of the rest of your question, we have to use the notion of Archimedean normed fields. If $F$ is a field with an absolute value $|cdot|$, then $F$ is called Archimedean if, given any element $xin F$, some positive integer multiple of $x$ has absolute value greater than 1: $exists nin mathbb{N}, |nx|>1$.
This definition of Archimedean implies the earlier version. Indeed, suppose the normed Archimedean field $F$ is also a linear order, with $|x|=x$ for positive elements. Let $a,bin mathbb{F}$ be positive. Then letting $x=a/b$, there must exist an $n$ such that $|nx|>1$. Then by the triangle inequality $|x|+|x|+cdots +|x| text{ (n times)} geq |nx| > 1$. Therefore $1<n|x|=n|a/b|=na/b$, since $a/b$ is positive. Then we have $na>b$, so $a$ is not infinitesimal with respect to $b$. This shows $F$ is Archimedean in the previous sense, so somewhat more indirectly, $mathbb{R}$ being Archimedean as a normed field also implies the statements you gave.
This property alone does not prove the completeness of $mathbb{R}$. For instance, the same properties hold for $mathbb{Q}$ in place of $mathbb{R}$, but $mathbb{Q}$ is not complete. Completeness is altogether another property of $mathbb{R}$, which would take too long to explain here from scratch. The main point is that $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to the ordinary absolute value. It turns out that there are other absolute value functions one can define on $mathbb{Q}$, and if we repeat the same procedure that we use to make $mathbb{R}$ out of $mathbb{Q}$ ,using these absolute values in place of the usual one, we get new completions of $mathbb{Q}$, the so called $p$-adic fields, which are not Archimedean.
Let's firmly fix a prime number $p$. Given a rational number $x$, we can factor $x$ into primes (with negative exponents possibly). If $r$ is the exponent of $p$ that occurs in the factorization of $x$, we define the $p$-adic absolute value of $x$ to be $|x|_p=p^{-r}$. It turns out that this function has the same basic properties as ordinary absolute value. In fact in place of the triangle inequality, something even stronger holds:
$$|x+y|_p leq max{|x|_p,|y|_p}.$$
This is the so called ultra-metric inequality, and ultimately what makes this absolute value non-Archimedean.
The $p$-adic absolute value may look strange at first, but just like the ordinary absolute value, it measures something about numbers. Whereas the ordinary absolute value $|x|$ quantifies the overall size of a rational number $x$, the $p$-adic absolute value $|x|_p$ measures how far $x$ is divisible by $p$. The higher the power of $p$ that $x$ is divisible by, the closer $x$ is to zero in the $p$-adic absolute value. So for example $250=2times 5^3$ is much closer to $0$ in $5$-adic absolute value than $2$ is: $|250|_5 = 1/125, |2|_5 = 1$.
If we form the completion of $mathbb{Q}$ with respect to this new absolute value, similar to how we form $mathbb{R}$ from $mathbb{Q}$ (taking equivalence classes of Cauchy sequences), we get a new way to "fill the gaps" of $mathbb{Q}$. This is called the field $mathbb{Q}_p$ of $p$-adic numbers.
Unlike $mathbb{R}$, the field $mathbb{Q}_p$ is not Archimedean. This is a consequence of the ultra-metric inequality. To see this, suppose $xin mathbb{Q}_p$ is any $p$-adic number such that $|x|_p leq 1$, then we have:
$$|2x|_p =|x+x|_p leq max{|x|_p,|x|_p}=|x|_p leq 1.$$
Using $|2x|_pleq 1$ and $|x|_pleq 1,$
$$|3x|_p=|2x+x|_p leq max{|x|_p,|2x|_p} leq 1,$$
and similarly $|4x|_p=|3x+x|_pleq max{|3x|_p,|x|_p} leq 1$, and so by induction $|nx|_pleq 1$ for all $n$. Thus $mathbb{Q}_p$ is non-Archimedean.
We can also see that in $mathbb{Q}_p$, the natural numbers $mathbb{N}$ are bounded above: for any $nin mathbb{N}$, $|n|_p leq 1$, since the exponent $r$ of $p$ appearing in the prime factorization of $n$ is non-negative: $|n|_p = 5^{-r} leq 5^0 =1$.
This also shows that for $nin mathbb{N}$, $|1/n|_p geq 1$, so we can not find $1/n$ that are very small in $p$-adic absolute value. So both of the facts that you mentioned in your question are false for $mathbb{Q}_p$.
Now there is a surprising theorem of Ostrowski which says there are no other completions of $mathbb{Q}$! Any other absolute values that you may define on $mathbb{Q}$ will be equivalent either to the ordinary absolute value, or to one of the $p$-adics, and after carrying out the completion process, you will get either $mathbb{R}$, or one of the $p$-adic fields $mathbb{Q}_p$. This is the sense in which $mathbb{R}$ is the Archimedean completion of $mathbb{Q}$.
Finally, let me say that although the field $mathbb{Q}_p$ is at first a very strange place to play in, it's used by number theorists all the time. Arithmetic in $mathbb{Q}_p$ is sometimes interpreted as doing congruence arithmetic modulo all powers of p at the same time.
Hopefully, the notion of $mathbb{R}$ being Archimedean is clearer once we are aware of the existence of these non-Archimedean completions. The $p$-adic numbers have many more strangely beautiful properties than can't be mentioned all in one place.
answered Aug 20 '13 at 1:52
PrometheusPrometheus
5,2511324
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Your statements, strictly interpreted,
are not true.
You need to change the order of quantifiers.
You say
"The Archimedean property states that $Bbb{N}$ isn't bounded above--some natural number can be found such that it is greater than any real number."
There is no natural number "such that
it is greater than any real number."
Rather,
for any positive real number $x$,
there is a natural number $n$
such that $n > x$.
Similarly,
you say
"The Archimedean property also states that there is some rational $1/n,n∈ℕ$ such that it is less than any real number."
Rather,
for any positive real number $x$,
there is a positive integer $n$
such that
$1/n < x$.
Be careful out there.
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
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However, I am confused. If there can be any 1/n found less than some specified real, wouldn't the statement be true anyway for all? Why wouldn't it?
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– MathApprentice
Aug 20 '13 at 0:28
1
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@MathApprentice: The way you originally put it, you actually contradicted the Archimedean property! Suppose there is some natural number $n$ greater than any real number. Now, all natural numbers are real, so $n$ is a real number greater than all real numbers, and so greater than all natural numbers, meaning it is an upper bound of $Bbb N$ (and is also greater than itself). Now, I suspect that you meant that for any real number $x$ we can find a larger natural number (which is true), but your phrasing suggested otherwise.
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– Cameron Buie
Aug 20 '13 at 0:40
add a comment |
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Both statements you mention are true, and are two different ways of expressing the Archimedean property. For instance, if we have that $n in mathbb{N}$ and $M in mathbb{R}$ with $M$ positive, then $$ n >M Leftrightarrow frac{1}{n} < frac{1}{M}$$
Regarding your second question, the Archimedean property is not the proof that $mathbb{R}$ completes $mathbb{Q}$. That fact is usually proved (though the two ideas are equivalent) from the least-upper-bound axiom of $mathbb{R}$. The Archimedean property of $mathbb{N}$ is a consequence of the least-upper-bound axiom of the real numbers.
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
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Is there any more significance to the Archimedan Property?
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– MathApprentice
Aug 20 '13 at 0:21
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I'm not sure what you're looking for; there are plenty of other conclusions you can derive from the Archimedean property, but these seem to be the most general. I would add that the property shows how dangerous it can be to rely on infinitesimals as a basis for calculus.
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– Matt R.
Aug 20 '13 at 1:12
add a comment |
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Let's look at some concrete consequences for ordinary calculus. Consider the following limit:
$$lim_{ntoinfty}frac1n=0$$
I bet you want that equation to be true, right? Well, it requires the Archimedean property! Otherwise, the sequence doesn't converge.* That would be really inconvenient. Similarly, definitions like
$$exp(x)=sum_{n=0}^inftyfrac{x^n}{n!}$$
rely on the Archimedean property, since that sum doesn't converge otherwise.*
*Statements about non-convergence refer to the order topology of a non-Archimedean ordered field. I make no claims about non-ordered fields, such as the $p$-adic numbers.
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
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I think you are asking if Archimedean helps in deriving R from Q. Then you are wrong because Q itself has Archimedean but is not ordered field with least upper bound property.
I also mention a fact which will help you clearing idea about it(if thsi is question):
An ordered field satisfy Least upper bound property iff it satisfy Archimedean and cauchy completeness property. And R is only ordered field with least upper bound property. This is what we take as definition of R
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
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6 Answers
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It has been pointed out above that your two versions of the Archimedean property should be equivalent--$Bbb N$ has no upper bound if and only if $left{frac1n:ninBbb Nright}$ has no positive lower bound. Your presentations of these two versions were misleading, however (and still are). More precisely:
- Given any real number $x$, there is a natural number $n$ greater than $x$. This natural number $n$ will of course depend on which $x$ we're looking at in the first place, but we will always be able to find some (infinitely-many, in fact) natural number greater than $x$, regardless of the $x$ we started with.
- Given any positive real number $y$, there is a natural number $n$ such that $frac1n$ is less than $y$. Again, the natural number $n$ will depend on which $y$ we're looking at.
It's a good exercise to show that these are equivalent statements.
Now, let me try to explain what Abbott meant by "how $Bbb Q$ fits inside $Bbb R$." Specifically, I will show that the following are equivalent:
- The Archimedean property holds.
- Given any $a,binBbb R$ with $a<b,$ there is some $rinBbb Q$ such that $a<r<b$. (More briefly, $Bbb Q$ is dense in $Bbb R$.)
- Given any $a,binBbb R$ with $a<b,$ there is some $pinBbb Z$ and some $ninBbb N$ such that $a<frac{p}{2^n}<b$.
It is clear that 3 $implies$ 2. To see that 2 $implies$ 1, take any $x>0,$ and find $rinBbb Q$ such that $0<r<x$. We know that $r=frac mn$ for some positive integers $m,n,$ so $frac1nlefrac mn=r<x,$ and so the Archimedean property holds.
To see that 1 $implies$ 3, we proceed by contrapositive, so suppose that there exist $a,binBbb R$ with $a<b$ such that for all $pinBbb Z$ and all $ninBbb N,$ we have $frac{p}{2^n}le a$ or $frac{p}{2^n}ge b$. For each $ninBbb N,$ let $p_n$ the greatest integer $p$ such that $frac{p}{2^n}le a.$ (By the Well-ordering property of $Bbb N,$ such a $p_n$ exists.) From this, it follows by hypothesis that $$frac{p_n}{2^n}le a<blefrac{p_n+1}{2^n}$$ for all $ninBbb N$. But then $$0<b-alefrac{p_n+1}{2^n}-frac{p_n}{2^n}=frac1{2^n}$$ for all $ninBbb N.$ Since $n<2^n$ for all $ninBbb N,$ then it follows that $0<b-a<frac1n$ for all $ninBbb N,$ and so the Archimedean property fails. Thus, by contrapositive, 1 $implies$ 3.
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
85.6k772160
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It has been pointed out above that your two versions of the Archimedean property should be equivalent--$Bbb N$ has no upper bound if and only if $left{frac1n:ninBbb Nright}$ has no positive lower bound. Your presentations of these two versions were misleading, however (and still are). More precisely:
- Given any real number $x$, there is a natural number $n$ greater than $x$. This natural number $n$ will of course depend on which $x$ we're looking at in the first place, but we will always be able to find some (infinitely-many, in fact) natural number greater than $x$, regardless of the $x$ we started with.
- Given any positive real number $y$, there is a natural number $n$ such that $frac1n$ is less than $y$. Again, the natural number $n$ will depend on which $y$ we're looking at.
It's a good exercise to show that these are equivalent statements.
Now, let me try to explain what Abbott meant by "how $Bbb Q$ fits inside $Bbb R$." Specifically, I will show that the following are equivalent:
- The Archimedean property holds.
- Given any $a,binBbb R$ with $a<b,$ there is some $rinBbb Q$ such that $a<r<b$. (More briefly, $Bbb Q$ is dense in $Bbb R$.)
- Given any $a,binBbb R$ with $a<b,$ there is some $pinBbb Z$ and some $ninBbb N$ such that $a<frac{p}{2^n}<b$.
It is clear that 3 $implies$ 2. To see that 2 $implies$ 1, take any $x>0,$ and find $rinBbb Q$ such that $0<r<x$. We know that $r=frac mn$ for some positive integers $m,n,$ so $frac1nlefrac mn=r<x,$ and so the Archimedean property holds.
To see that 1 $implies$ 3, we proceed by contrapositive, so suppose that there exist $a,binBbb R$ with $a<b$ such that for all $pinBbb Z$ and all $ninBbb N,$ we have $frac{p}{2^n}le a$ or $frac{p}{2^n}ge b$. For each $ninBbb N,$ let $p_n$ the greatest integer $p$ such that $frac{p}{2^n}le a.$ (By the Well-ordering property of $Bbb N,$ such a $p_n$ exists.) From this, it follows by hypothesis that $$frac{p_n}{2^n}le a<blefrac{p_n+1}{2^n}$$ for all $ninBbb N$. But then $$0<b-alefrac{p_n+1}{2^n}-frac{p_n}{2^n}=frac1{2^n}$$ for all $ninBbb N.$ Since $n<2^n$ for all $ninBbb N,$ then it follows that $0<b-a<frac1n$ for all $ninBbb N,$ and so the Archimedean property fails. Thus, by contrapositive, 1 $implies$ 3.
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
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It has been pointed out above that your two versions of the Archimedean property should be equivalent--$Bbb N$ has no upper bound if and only if $left{frac1n:ninBbb Nright}$ has no positive lower bound. Your presentations of these two versions were misleading, however (and still are). More precisely:
- Given any real number $x$, there is a natural number $n$ greater than $x$. This natural number $n$ will of course depend on which $x$ we're looking at in the first place, but we will always be able to find some (infinitely-many, in fact) natural number greater than $x$, regardless of the $x$ we started with.
- Given any positive real number $y$, there is a natural number $n$ such that $frac1n$ is less than $y$. Again, the natural number $n$ will depend on which $y$ we're looking at.
It's a good exercise to show that these are equivalent statements.
Now, let me try to explain what Abbott meant by "how $Bbb Q$ fits inside $Bbb R$." Specifically, I will show that the following are equivalent:
- The Archimedean property holds.
- Given any $a,binBbb R$ with $a<b,$ there is some $rinBbb Q$ such that $a<r<b$. (More briefly, $Bbb Q$ is dense in $Bbb R$.)
- Given any $a,binBbb R$ with $a<b,$ there is some $pinBbb Z$ and some $ninBbb N$ such that $a<frac{p}{2^n}<b$.
It is clear that 3 $implies$ 2. To see that 2 $implies$ 1, take any $x>0,$ and find $rinBbb Q$ such that $0<r<x$. We know that $r=frac mn$ for some positive integers $m,n,$ so $frac1nlefrac mn=r<x,$ and so the Archimedean property holds.
To see that 1 $implies$ 3, we proceed by contrapositive, so suppose that there exist $a,binBbb R$ with $a<b$ such that for all $pinBbb Z$ and all $ninBbb N,$ we have $frac{p}{2^n}le a$ or $frac{p}{2^n}ge b$. For each $ninBbb N,$ let $p_n$ the greatest integer $p$ such that $frac{p}{2^n}le a.$ (By the Well-ordering property of $Bbb N,$ such a $p_n$ exists.) From this, it follows by hypothesis that $$frac{p_n}{2^n}le a<blefrac{p_n+1}{2^n}$$ for all $ninBbb N$. But then $$0<b-alefrac{p_n+1}{2^n}-frac{p_n}{2^n}=frac1{2^n}$$ for all $ninBbb N.$ Since $n<2^n$ for all $ninBbb N,$ then it follows that $0<b-a<frac1n$ for all $ninBbb N,$ and so the Archimedean property fails. Thus, by contrapositive, 1 $implies$ 3.
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
85.6k772160
$endgroup$
It has been pointed out above that your two versions of the Archimedean property should be equivalent--$Bbb N$ has no upper bound if and only if $left{frac1n:ninBbb Nright}$ has no positive lower bound. Your presentations of these two versions were misleading, however (and still are). More precisely:
- Given any real number $x$, there is a natural number $n$ greater than $x$. This natural number $n$ will of course depend on which $x$ we're looking at in the first place, but we will always be able to find some (infinitely-many, in fact) natural number greater than $x$, regardless of the $x$ we started with.
- Given any positive real number $y$, there is a natural number $n$ such that $frac1n$ is less than $y$. Again, the natural number $n$ will depend on which $y$ we're looking at.
It's a good exercise to show that these are equivalent statements.
Now, let me try to explain what Abbott meant by "how $Bbb Q$ fits inside $Bbb R$." Specifically, I will show that the following are equivalent:
- The Archimedean property holds.
- Given any $a,binBbb R$ with $a<b,$ there is some $rinBbb Q$ such that $a<r<b$. (More briefly, $Bbb Q$ is dense in $Bbb R$.)
- Given any $a,binBbb R$ with $a<b,$ there is some $pinBbb Z$ and some $ninBbb N$ such that $a<frac{p}{2^n}<b$.
It is clear that 3 $implies$ 2. To see that 2 $implies$ 1, take any $x>0,$ and find $rinBbb Q$ such that $0<r<x$. We know that $r=frac mn$ for some positive integers $m,n,$ so $frac1nlefrac mn=r<x,$ and so the Archimedean property holds.
To see that 1 $implies$ 3, we proceed by contrapositive, so suppose that there exist $a,binBbb R$ with $a<b$ such that for all $pinBbb Z$ and all $ninBbb N,$ we have $frac{p}{2^n}le a$ or $frac{p}{2^n}ge b$. For each $ninBbb N,$ let $p_n$ the greatest integer $p$ such that $frac{p}{2^n}le a.$ (By the Well-ordering property of $Bbb N,$ such a $p_n$ exists.) From this, it follows by hypothesis that $$frac{p_n}{2^n}le a<blefrac{p_n+1}{2^n}$$ for all $ninBbb N$. But then $$0<b-alefrac{p_n+1}{2^n}-frac{p_n}{2^n}=frac1{2^n}$$ for all $ninBbb N.$ Since $n<2^n$ for all $ninBbb N,$ then it follows that $0<b-a<frac1n$ for all $ninBbb N,$ and so the Archimedean property fails. Thus, by contrapositive, 1 $implies$ 3.
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
85.6k772160
edited Aug 20 '13 at 2:02
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
85.6k772160
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
85.6k772160
answered Aug 20 '13 at 1:21
Cameron BuieCameron Buie
85.6k772160
85.6k772160
add a comment |
add a comment |
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In a group with a linear order, if $a$ and $b$ are two positive elements, $a$ is said to be infinitesimal with respect to $b$, if no positive integer multiple of $a$ is greater than $b$. Then the linear order is called Archimedean if there are no infinitesimal elements. The first two statements you mentioned are easy consequences of $mathbb{R}$ being Archimedean as a linear order:
If $r$ is any positive real number, since $1$ is not infinitesimal with respect to $r$ by the Archimedean property, there exists a natural number $n$ such that $n.1=n> r$. This shows $mathbb{N}$ is not bounded above.
The second statement is also true for positive real numbers: Suppose we're given a positive real number $r$. Then by the previous property, there exists a natural number $n$ such that $1/r < n$. Therefore $1/n<r$.
But to make sense of the rest of your question, we have to use the notion of Archimedean normed fields. If $F$ is a field with an absolute value $|cdot|$, then $F$ is called Archimedean if, given any element $xin F$, some positive integer multiple of $x$ has absolute value greater than 1: $exists nin mathbb{N}, |nx|>1$.
This definition of Archimedean implies the earlier version. Indeed, suppose the normed Archimedean field $F$ is also a linear order, with $|x|=x$ for positive elements. Let $a,bin mathbb{F}$ be positive. Then letting $x=a/b$, there must exist an $n$ such that $|nx|>1$. Then by the triangle inequality $|x|+|x|+cdots +|x| text{ (n times)} geq |nx| > 1$. Therefore $1<n|x|=n|a/b|=na/b$, since $a/b$ is positive. Then we have $na>b$, so $a$ is not infinitesimal with respect to $b$. This shows $F$ is Archimedean in the previous sense, so somewhat more indirectly, $mathbb{R}$ being Archimedean as a normed field also implies the statements you gave.
This property alone does not prove the completeness of $mathbb{R}$. For instance, the same properties hold for $mathbb{Q}$ in place of $mathbb{R}$, but $mathbb{Q}$ is not complete. Completeness is altogether another property of $mathbb{R}$, which would take too long to explain here from scratch. The main point is that $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to the ordinary absolute value. It turns out that there are other absolute value functions one can define on $mathbb{Q}$, and if we repeat the same procedure that we use to make $mathbb{R}$ out of $mathbb{Q}$ ,using these absolute values in place of the usual one, we get new completions of $mathbb{Q}$, the so called $p$-adic fields, which are not Archimedean.
Let's firmly fix a prime number $p$. Given a rational number $x$, we can factor $x$ into primes (with negative exponents possibly). If $r$ is the exponent of $p$ that occurs in the factorization of $x$, we define the $p$-adic absolute value of $x$ to be $|x|_p=p^{-r}$. It turns out that this function has the same basic properties as ordinary absolute value. In fact in place of the triangle inequality, something even stronger holds:
$$|x+y|_p leq max{|x|_p,|y|_p}.$$
This is the so called ultra-metric inequality, and ultimately what makes this absolute value non-Archimedean.
The $p$-adic absolute value may look strange at first, but just like the ordinary absolute value, it measures something about numbers. Whereas the ordinary absolute value $|x|$ quantifies the overall size of a rational number $x$, the $p$-adic absolute value $|x|_p$ measures how far $x$ is divisible by $p$. The higher the power of $p$ that $x$ is divisible by, the closer $x$ is to zero in the $p$-adic absolute value. So for example $250=2times 5^3$ is much closer to $0$ in $5$-adic absolute value than $2$ is: $|250|_5 = 1/125, |2|_5 = 1$.
If we form the completion of $mathbb{Q}$ with respect to this new absolute value, similar to how we form $mathbb{R}$ from $mathbb{Q}$ (taking equivalence classes of Cauchy sequences), we get a new way to "fill the gaps" of $mathbb{Q}$. This is called the field $mathbb{Q}_p$ of $p$-adic numbers.
Unlike $mathbb{R}$, the field $mathbb{Q}_p$ is not Archimedean. This is a consequence of the ultra-metric inequality. To see this, suppose $xin mathbb{Q}_p$ is any $p$-adic number such that $|x|_p leq 1$, then we have:
$$|2x|_p =|x+x|_p leq max{|x|_p,|x|_p}=|x|_p leq 1.$$
Using $|2x|_pleq 1$ and $|x|_pleq 1,$
$$|3x|_p=|2x+x|_p leq max{|x|_p,|2x|_p} leq 1,$$
and similarly $|4x|_p=|3x+x|_pleq max{|3x|_p,|x|_p} leq 1$, and so by induction $|nx|_pleq 1$ for all $n$. Thus $mathbb{Q}_p$ is non-Archimedean.
We can also see that in $mathbb{Q}_p$, the natural numbers $mathbb{N}$ are bounded above: for any $nin mathbb{N}$, $|n|_p leq 1$, since the exponent $r$ of $p$ appearing in the prime factorization of $n$ is non-negative: $|n|_p = 5^{-r} leq 5^0 =1$.
This also shows that for $nin mathbb{N}$, $|1/n|_p geq 1$, so we can not find $1/n$ that are very small in $p$-adic absolute value. So both of the facts that you mentioned in your question are false for $mathbb{Q}_p$.
Now there is a surprising theorem of Ostrowski which says there are no other completions of $mathbb{Q}$! Any other absolute values that you may define on $mathbb{Q}$ will be equivalent either to the ordinary absolute value, or to one of the $p$-adics, and after carrying out the completion process, you will get either $mathbb{R}$, or one of the $p$-adic fields $mathbb{Q}_p$. This is the sense in which $mathbb{R}$ is the Archimedean completion of $mathbb{Q}$.
Finally, let me say that although the field $mathbb{Q}_p$ is at first a very strange place to play in, it's used by number theorists all the time. Arithmetic in $mathbb{Q}_p$ is sometimes interpreted as doing congruence arithmetic modulo all powers of p at the same time.
Hopefully, the notion of $mathbb{R}$ being Archimedean is clearer once we are aware of the existence of these non-Archimedean completions. The $p$-adic numbers have many more strangely beautiful properties than can't be mentioned all in one place.
answered Aug 20 '13 at 1:52
PrometheusPrometheus
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add a comment |
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In a group with a linear order, if $a$ and $b$ are two positive elements, $a$ is said to be infinitesimal with respect to $b$, if no positive integer multiple of $a$ is greater than $b$. Then the linear order is called Archimedean if there are no infinitesimal elements. The first two statements you mentioned are easy consequences of $mathbb{R}$ being Archimedean as a linear order:
If $r$ is any positive real number, since $1$ is not infinitesimal with respect to $r$ by the Archimedean property, there exists a natural number $n$ such that $n.1=n> r$. This shows $mathbb{N}$ is not bounded above.
The second statement is also true for positive real numbers: Suppose we're given a positive real number $r$. Then by the previous property, there exists a natural number $n$ such that $1/r < n$. Therefore $1/n<r$.
But to make sense of the rest of your question, we have to use the notion of Archimedean normed fields. If $F$ is a field with an absolute value $|cdot|$, then $F$ is called Archimedean if, given any element $xin F$, some positive integer multiple of $x$ has absolute value greater than 1: $exists nin mathbb{N}, |nx|>1$.
This definition of Archimedean implies the earlier version. Indeed, suppose the normed Archimedean field $F$ is also a linear order, with $|x|=x$ for positive elements. Let $a,bin mathbb{F}$ be positive. Then letting $x=a/b$, there must exist an $n$ such that $|nx|>1$. Then by the triangle inequality $|x|+|x|+cdots +|x| text{ (n times)} geq |nx| > 1$. Therefore $1<n|x|=n|a/b|=na/b$, since $a/b$ is positive. Then we have $na>b$, so $a$ is not infinitesimal with respect to $b$. This shows $F$ is Archimedean in the previous sense, so somewhat more indirectly, $mathbb{R}$ being Archimedean as a normed field also implies the statements you gave.
This property alone does not prove the completeness of $mathbb{R}$. For instance, the same properties hold for $mathbb{Q}$ in place of $mathbb{R}$, but $mathbb{Q}$ is not complete. Completeness is altogether another property of $mathbb{R}$, which would take too long to explain here from scratch. The main point is that $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to the ordinary absolute value. It turns out that there are other absolute value functions one can define on $mathbb{Q}$, and if we repeat the same procedure that we use to make $mathbb{R}$ out of $mathbb{Q}$ ,using these absolute values in place of the usual one, we get new completions of $mathbb{Q}$, the so called $p$-adic fields, which are not Archimedean.
Let's firmly fix a prime number $p$. Given a rational number $x$, we can factor $x$ into primes (with negative exponents possibly). If $r$ is the exponent of $p$ that occurs in the factorization of $x$, we define the $p$-adic absolute value of $x$ to be $|x|_p=p^{-r}$. It turns out that this function has the same basic properties as ordinary absolute value. In fact in place of the triangle inequality, something even stronger holds:
$$|x+y|_p leq max{|x|_p,|y|_p}.$$
This is the so called ultra-metric inequality, and ultimately what makes this absolute value non-Archimedean.
The $p$-adic absolute value may look strange at first, but just like the ordinary absolute value, it measures something about numbers. Whereas the ordinary absolute value $|x|$ quantifies the overall size of a rational number $x$, the $p$-adic absolute value $|x|_p$ measures how far $x$ is divisible by $p$. The higher the power of $p$ that $x$ is divisible by, the closer $x$ is to zero in the $p$-adic absolute value. So for example $250=2times 5^3$ is much closer to $0$ in $5$-adic absolute value than $2$ is: $|250|_5 = 1/125, |2|_5 = 1$.
If we form the completion of $mathbb{Q}$ with respect to this new absolute value, similar to how we form $mathbb{R}$ from $mathbb{Q}$ (taking equivalence classes of Cauchy sequences), we get a new way to "fill the gaps" of $mathbb{Q}$. This is called the field $mathbb{Q}_p$ of $p$-adic numbers.
Unlike $mathbb{R}$, the field $mathbb{Q}_p$ is not Archimedean. This is a consequence of the ultra-metric inequality. To see this, suppose $xin mathbb{Q}_p$ is any $p$-adic number such that $|x|_p leq 1$, then we have:
$$|2x|_p =|x+x|_p leq max{|x|_p,|x|_p}=|x|_p leq 1.$$
Using $|2x|_pleq 1$ and $|x|_pleq 1,$
$$|3x|_p=|2x+x|_p leq max{|x|_p,|2x|_p} leq 1,$$
and similarly $|4x|_p=|3x+x|_pleq max{|3x|_p,|x|_p} leq 1$, and so by induction $|nx|_pleq 1$ for all $n$. Thus $mathbb{Q}_p$ is non-Archimedean.
We can also see that in $mathbb{Q}_p$, the natural numbers $mathbb{N}$ are bounded above: for any $nin mathbb{N}$, $|n|_p leq 1$, since the exponent $r$ of $p$ appearing in the prime factorization of $n$ is non-negative: $|n|_p = 5^{-r} leq 5^0 =1$.
This also shows that for $nin mathbb{N}$, $|1/n|_p geq 1$, so we can not find $1/n$ that are very small in $p$-adic absolute value. So both of the facts that you mentioned in your question are false for $mathbb{Q}_p$.
Now there is a surprising theorem of Ostrowski which says there are no other completions of $mathbb{Q}$! Any other absolute values that you may define on $mathbb{Q}$ will be equivalent either to the ordinary absolute value, or to one of the $p$-adics, and after carrying out the completion process, you will get either $mathbb{R}$, or one of the $p$-adic fields $mathbb{Q}_p$. This is the sense in which $mathbb{R}$ is the Archimedean completion of $mathbb{Q}$.
Finally, let me say that although the field $mathbb{Q}_p$ is at first a very strange place to play in, it's used by number theorists all the time. Arithmetic in $mathbb{Q}_p$ is sometimes interpreted as doing congruence arithmetic modulo all powers of p at the same time.
Hopefully, the notion of $mathbb{R}$ being Archimedean is clearer once we are aware of the existence of these non-Archimedean completions. The $p$-adic numbers have many more strangely beautiful properties than can't be mentioned all in one place.
answered Aug 20 '13 at 1:52
PrometheusPrometheus
5,2511324
$endgroup$
add a comment |
$begingroup$
In a group with a linear order, if $a$ and $b$ are two positive elements, $a$ is said to be infinitesimal with respect to $b$, if no positive integer multiple of $a$ is greater than $b$. Then the linear order is called Archimedean if there are no infinitesimal elements. The first two statements you mentioned are easy consequences of $mathbb{R}$ being Archimedean as a linear order:
If $r$ is any positive real number, since $1$ is not infinitesimal with respect to $r$ by the Archimedean property, there exists a natural number $n$ such that $n.1=n> r$. This shows $mathbb{N}$ is not bounded above.
The second statement is also true for positive real numbers: Suppose we're given a positive real number $r$. Then by the previous property, there exists a natural number $n$ such that $1/r < n$. Therefore $1/n<r$.
But to make sense of the rest of your question, we have to use the notion of Archimedean normed fields. If $F$ is a field with an absolute value $|cdot|$, then $F$ is called Archimedean if, given any element $xin F$, some positive integer multiple of $x$ has absolute value greater than 1: $exists nin mathbb{N}, |nx|>1$.
This definition of Archimedean implies the earlier version. Indeed, suppose the normed Archimedean field $F$ is also a linear order, with $|x|=x$ for positive elements. Let $a,bin mathbb{F}$ be positive. Then letting $x=a/b$, there must exist an $n$ such that $|nx|>1$. Then by the triangle inequality $|x|+|x|+cdots +|x| text{ (n times)} geq |nx| > 1$. Therefore $1<n|x|=n|a/b|=na/b$, since $a/b$ is positive. Then we have $na>b$, so $a$ is not infinitesimal with respect to $b$. This shows $F$ is Archimedean in the previous sense, so somewhat more indirectly, $mathbb{R}$ being Archimedean as a normed field also implies the statements you gave.
This property alone does not prove the completeness of $mathbb{R}$. For instance, the same properties hold for $mathbb{Q}$ in place of $mathbb{R}$, but $mathbb{Q}$ is not complete. Completeness is altogether another property of $mathbb{R}$, which would take too long to explain here from scratch. The main point is that $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to the ordinary absolute value. It turns out that there are other absolute value functions one can define on $mathbb{Q}$, and if we repeat the same procedure that we use to make $mathbb{R}$ out of $mathbb{Q}$ ,using these absolute values in place of the usual one, we get new completions of $mathbb{Q}$, the so called $p$-adic fields, which are not Archimedean.
Let's firmly fix a prime number $p$. Given a rational number $x$, we can factor $x$ into primes (with negative exponents possibly). If $r$ is the exponent of $p$ that occurs in the factorization of $x$, we define the $p$-adic absolute value of $x$ to be $|x|_p=p^{-r}$. It turns out that this function has the same basic properties as ordinary absolute value. In fact in place of the triangle inequality, something even stronger holds:
$$|x+y|_p leq max{|x|_p,|y|_p}.$$
This is the so called ultra-metric inequality, and ultimately what makes this absolute value non-Archimedean.
The $p$-adic absolute value may look strange at first, but just like the ordinary absolute value, it measures something about numbers. Whereas the ordinary absolute value $|x|$ quantifies the overall size of a rational number $x$, the $p$-adic absolute value $|x|_p$ measures how far $x$ is divisible by $p$. The higher the power of $p$ that $x$ is divisible by, the closer $x$ is to zero in the $p$-adic absolute value. So for example $250=2times 5^3$ is much closer to $0$ in $5$-adic absolute value than $2$ is: $|250|_5 = 1/125, |2|_5 = 1$.
If we form the completion of $mathbb{Q}$ with respect to this new absolute value, similar to how we form $mathbb{R}$ from $mathbb{Q}$ (taking equivalence classes of Cauchy sequences), we get a new way to "fill the gaps" of $mathbb{Q}$. This is called the field $mathbb{Q}_p$ of $p$-adic numbers.
Unlike $mathbb{R}$, the field $mathbb{Q}_p$ is not Archimedean. This is a consequence of the ultra-metric inequality. To see this, suppose $xin mathbb{Q}_p$ is any $p$-adic number such that $|x|_p leq 1$, then we have:
$$|2x|_p =|x+x|_p leq max{|x|_p,|x|_p}=|x|_p leq 1.$$
Using $|2x|_pleq 1$ and $|x|_pleq 1,$
$$|3x|_p=|2x+x|_p leq max{|x|_p,|2x|_p} leq 1,$$
and similarly $|4x|_p=|3x+x|_pleq max{|3x|_p,|x|_p} leq 1$, and so by induction $|nx|_pleq 1$ for all $n$. Thus $mathbb{Q}_p$ is non-Archimedean.
We can also see that in $mathbb{Q}_p$, the natural numbers $mathbb{N}$ are bounded above: for any $nin mathbb{N}$, $|n|_p leq 1$, since the exponent $r$ of $p$ appearing in the prime factorization of $n$ is non-negative: $|n|_p = 5^{-r} leq 5^0 =1$.
This also shows that for $nin mathbb{N}$, $|1/n|_p geq 1$, so we can not find $1/n$ that are very small in $p$-adic absolute value. So both of the facts that you mentioned in your question are false for $mathbb{Q}_p$.
Now there is a surprising theorem of Ostrowski which says there are no other completions of $mathbb{Q}$! Any other absolute values that you may define on $mathbb{Q}$ will be equivalent either to the ordinary absolute value, or to one of the $p$-adics, and after carrying out the completion process, you will get either $mathbb{R}$, or one of the $p$-adic fields $mathbb{Q}_p$. This is the sense in which $mathbb{R}$ is the Archimedean completion of $mathbb{Q}$.
Finally, let me say that although the field $mathbb{Q}_p$ is at first a very strange place to play in, it's used by number theorists all the time. Arithmetic in $mathbb{Q}_p$ is sometimes interpreted as doing congruence arithmetic modulo all powers of p at the same time.
Hopefully, the notion of $mathbb{R}$ being Archimedean is clearer once we are aware of the existence of these non-Archimedean completions. The $p$-adic numbers have many more strangely beautiful properties than can't be mentioned all in one place.
answered Aug 20 '13 at 1:52
PrometheusPrometheus
5,2511324
$endgroup$
In a group with a linear order, if $a$ and $b$ are two positive elements, $a$ is said to be infinitesimal with respect to $b$, if no positive integer multiple of $a$ is greater than $b$. Then the linear order is called Archimedean if there are no infinitesimal elements. The first two statements you mentioned are easy consequences of $mathbb{R}$ being Archimedean as a linear order:
If $r$ is any positive real number, since $1$ is not infinitesimal with respect to $r$ by the Archimedean property, there exists a natural number $n$ such that $n.1=n> r$. This shows $mathbb{N}$ is not bounded above.
The second statement is also true for positive real numbers: Suppose we're given a positive real number $r$. Then by the previous property, there exists a natural number $n$ such that $1/r < n$. Therefore $1/n<r$.
But to make sense of the rest of your question, we have to use the notion of Archimedean normed fields. If $F$ is a field with an absolute value $|cdot|$, then $F$ is called Archimedean if, given any element $xin F$, some positive integer multiple of $x$ has absolute value greater than 1: $exists nin mathbb{N}, |nx|>1$.
This definition of Archimedean implies the earlier version. Indeed, suppose the normed Archimedean field $F$ is also a linear order, with $|x|=x$ for positive elements. Let $a,bin mathbb{F}$ be positive. Then letting $x=a/b$, there must exist an $n$ such that $|nx|>1$. Then by the triangle inequality $|x|+|x|+cdots +|x| text{ (n times)} geq |nx| > 1$. Therefore $1<n|x|=n|a/b|=na/b$, since $a/b$ is positive. Then we have $na>b$, so $a$ is not infinitesimal with respect to $b$. This shows $F$ is Archimedean in the previous sense, so somewhat more indirectly, $mathbb{R}$ being Archimedean as a normed field also implies the statements you gave.
This property alone does not prove the completeness of $mathbb{R}$. For instance, the same properties hold for $mathbb{Q}$ in place of $mathbb{R}$, but $mathbb{Q}$ is not complete. Completeness is altogether another property of $mathbb{R}$, which would take too long to explain here from scratch. The main point is that $mathbb{R}$ is the completion of $mathbb{Q}$ with respect to the ordinary absolute value. It turns out that there are other absolute value functions one can define on $mathbb{Q}$, and if we repeat the same procedure that we use to make $mathbb{R}$ out of $mathbb{Q}$ ,using these absolute values in place of the usual one, we get new completions of $mathbb{Q}$, the so called $p$-adic fields, which are not Archimedean.
Let's firmly fix a prime number $p$. Given a rational number $x$, we can factor $x$ into primes (with negative exponents possibly). If $r$ is the exponent of $p$ that occurs in the factorization of $x$, we define the $p$-adic absolute value of $x$ to be $|x|_p=p^{-r}$. It turns out that this function has the same basic properties as ordinary absolute value. In fact in place of the triangle inequality, something even stronger holds:
$$|x+y|_p leq max{|x|_p,|y|_p}.$$
This is the so called ultra-metric inequality, and ultimately what makes this absolute value non-Archimedean.
The $p$-adic absolute value may look strange at first, but just like the ordinary absolute value, it measures something about numbers. Whereas the ordinary absolute value $|x|$ quantifies the overall size of a rational number $x$, the $p$-adic absolute value $|x|_p$ measures how far $x$ is divisible by $p$. The higher the power of $p$ that $x$ is divisible by, the closer $x$ is to zero in the $p$-adic absolute value. So for example $250=2times 5^3$ is much closer to $0$ in $5$-adic absolute value than $2$ is: $|250|_5 = 1/125, |2|_5 = 1$.
If we form the completion of $mathbb{Q}$ with respect to this new absolute value, similar to how we form $mathbb{R}$ from $mathbb{Q}$ (taking equivalence classes of Cauchy sequences), we get a new way to "fill the gaps" of $mathbb{Q}$. This is called the field $mathbb{Q}_p$ of $p$-adic numbers.
Unlike $mathbb{R}$, the field $mathbb{Q}_p$ is not Archimedean. This is a consequence of the ultra-metric inequality. To see this, suppose $xin mathbb{Q}_p$ is any $p$-adic number such that $|x|_p leq 1$, then we have:
$$|2x|_p =|x+x|_p leq max{|x|_p,|x|_p}=|x|_p leq 1.$$
Using $|2x|_pleq 1$ and $|x|_pleq 1,$
$$|3x|_p=|2x+x|_p leq max{|x|_p,|2x|_p} leq 1,$$
and similarly $|4x|_p=|3x+x|_pleq max{|3x|_p,|x|_p} leq 1$, and so by induction $|nx|_pleq 1$ for all $n$. Thus $mathbb{Q}_p$ is non-Archimedean.
We can also see that in $mathbb{Q}_p$, the natural numbers $mathbb{N}$ are bounded above: for any $nin mathbb{N}$, $|n|_p leq 1$, since the exponent $r$ of $p$ appearing in the prime factorization of $n$ is non-negative: $|n|_p = 5^{-r} leq 5^0 =1$.
This also shows that for $nin mathbb{N}$, $|1/n|_p geq 1$, so we can not find $1/n$ that are very small in $p$-adic absolute value. So both of the facts that you mentioned in your question are false for $mathbb{Q}_p$.
Now there is a surprising theorem of Ostrowski which says there are no other completions of $mathbb{Q}$! Any other absolute values that you may define on $mathbb{Q}$ will be equivalent either to the ordinary absolute value, or to one of the $p$-adics, and after carrying out the completion process, you will get either $mathbb{R}$, or one of the $p$-adic fields $mathbb{Q}_p$. This is the sense in which $mathbb{R}$ is the Archimedean completion of $mathbb{Q}$.
Finally, let me say that although the field $mathbb{Q}_p$ is at first a very strange place to play in, it's used by number theorists all the time. Arithmetic in $mathbb{Q}_p$ is sometimes interpreted as doing congruence arithmetic modulo all powers of p at the same time.
Hopefully, the notion of $mathbb{R}$ being Archimedean is clearer once we are aware of the existence of these non-Archimedean completions. The $p$-adic numbers have many more strangely beautiful properties than can't be mentioned all in one place.
answered Aug 20 '13 at 1:52
PrometheusPrometheus
5,2511324
edited Sep 8 '14 at 2:16
answered Aug 20 '13 at 1:52
PrometheusPrometheus
5,2511324
answered Aug 20 '13 at 1:52
PrometheusPrometheus
5,2511324
answered Aug 20 '13 at 1:52
PrometheusPrometheus
5,2511324
5,2511324
add a comment |
add a comment |
$begingroup$
Your statements, strictly interpreted,
are not true.
You need to change the order of quantifiers.
You say
"The Archimedean property states that $Bbb{N}$ isn't bounded above--some natural number can be found such that it is greater than any real number."
There is no natural number "such that
it is greater than any real number."
Rather,
for any positive real number $x$,
there is a natural number $n$
such that $n > x$.
Similarly,
you say
"The Archimedean property also states that there is some rational $1/n,n∈ℕ$ such that it is less than any real number."
Rather,
for any positive real number $x$,
there is a positive integer $n$
such that
$1/n < x$.
Be careful out there.
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
$endgroup$
$begingroup$
However, I am confused. If there can be any 1/n found less than some specified real, wouldn't the statement be true anyway for all? Why wouldn't it?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:28
1
$begingroup$
@MathApprentice: The way you originally put it, you actually contradicted the Archimedean property! Suppose there is some natural number $n$ greater than any real number. Now, all natural numbers are real, so $n$ is a real number greater than all real numbers, and so greater than all natural numbers, meaning it is an upper bound of $Bbb N$ (and is also greater than itself). Now, I suspect that you meant that for any real number $x$ we can find a larger natural number (which is true), but your phrasing suggested otherwise.
$endgroup$
– Cameron Buie
Aug 20 '13 at 0:40
add a comment |
$begingroup$
Your statements, strictly interpreted,
are not true.
You need to change the order of quantifiers.
You say
"The Archimedean property states that $Bbb{N}$ isn't bounded above--some natural number can be found such that it is greater than any real number."
There is no natural number "such that
it is greater than any real number."
Rather,
for any positive real number $x$,
there is a natural number $n$
such that $n > x$.
Similarly,
you say
"The Archimedean property also states that there is some rational $1/n,n∈ℕ$ such that it is less than any real number."
Rather,
for any positive real number $x$,
there is a positive integer $n$
such that
$1/n < x$.
Be careful out there.
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
$endgroup$
$begingroup$
However, I am confused. If there can be any 1/n found less than some specified real, wouldn't the statement be true anyway for all? Why wouldn't it?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:28
1
$begingroup$
@MathApprentice: The way you originally put it, you actually contradicted the Archimedean property! Suppose there is some natural number $n$ greater than any real number. Now, all natural numbers are real, so $n$ is a real number greater than all real numbers, and so greater than all natural numbers, meaning it is an upper bound of $Bbb N$ (and is also greater than itself). Now, I suspect that you meant that for any real number $x$ we can find a larger natural number (which is true), but your phrasing suggested otherwise.
$endgroup$
– Cameron Buie
Aug 20 '13 at 0:40
add a comment |
$begingroup$
Your statements, strictly interpreted,
are not true.
You need to change the order of quantifiers.
You say
"The Archimedean property states that $Bbb{N}$ isn't bounded above--some natural number can be found such that it is greater than any real number."
There is no natural number "such that
it is greater than any real number."
Rather,
for any positive real number $x$,
there is a natural number $n$
such that $n > x$.
Similarly,
you say
"The Archimedean property also states that there is some rational $1/n,n∈ℕ$ such that it is less than any real number."
Rather,
for any positive real number $x$,
there is a positive integer $n$
such that
$1/n < x$.
Be careful out there.
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
$endgroup$
Your statements, strictly interpreted,
are not true.
You need to change the order of quantifiers.
You say
"The Archimedean property states that $Bbb{N}$ isn't bounded above--some natural number can be found such that it is greater than any real number."
There is no natural number "such that
it is greater than any real number."
Rather,
for any positive real number $x$,
there is a natural number $n$
such that $n > x$.
Similarly,
you say
"The Archimedean property also states that there is some rational $1/n,n∈ℕ$ such that it is less than any real number."
Rather,
for any positive real number $x$,
there is a positive integer $n$
such that
$1/n < x$.
Be careful out there.
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
edited Aug 20 '13 at 0:34
Cameron Buie
85.6k772160
85.6k772160
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
answered Aug 20 '13 at 0:23
marty cohenmarty cohen
74.3k549128
74.3k549128
$begingroup$
However, I am confused. If there can be any 1/n found less than some specified real, wouldn't the statement be true anyway for all? Why wouldn't it?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:28
1
$begingroup$
@MathApprentice: The way you originally put it, you actually contradicted the Archimedean property! Suppose there is some natural number $n$ greater than any real number. Now, all natural numbers are real, so $n$ is a real number greater than all real numbers, and so greater than all natural numbers, meaning it is an upper bound of $Bbb N$ (and is also greater than itself). Now, I suspect that you meant that for any real number $x$ we can find a larger natural number (which is true), but your phrasing suggested otherwise.
$endgroup$
– Cameron Buie
Aug 20 '13 at 0:40
add a comment |
$begingroup$
However, I am confused. If there can be any 1/n found less than some specified real, wouldn't the statement be true anyway for all? Why wouldn't it?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:28
1
$begingroup$
@MathApprentice: The way you originally put it, you actually contradicted the Archimedean property! Suppose there is some natural number $n$ greater than any real number. Now, all natural numbers are real, so $n$ is a real number greater than all real numbers, and so greater than all natural numbers, meaning it is an upper bound of $Bbb N$ (and is also greater than itself). Now, I suspect that you meant that for any real number $x$ we can find a larger natural number (which is true), but your phrasing suggested otherwise.
$endgroup$
– Cameron Buie
Aug 20 '13 at 0:40
$begingroup$
However, I am confused. If there can be any 1/n found less than some specified real, wouldn't the statement be true anyway for all? Why wouldn't it?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:28
$begingroup$
However, I am confused. If there can be any 1/n found less than some specified real, wouldn't the statement be true anyway for all? Why wouldn't it?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:28
1
1
$begingroup$
@MathApprentice: The way you originally put it, you actually contradicted the Archimedean property! Suppose there is some natural number $n$ greater than any real number. Now, all natural numbers are real, so $n$ is a real number greater than all real numbers, and so greater than all natural numbers, meaning it is an upper bound of $Bbb N$ (and is also greater than itself). Now, I suspect that you meant that for any real number $x$ we can find a larger natural number (which is true), but your phrasing suggested otherwise.
$endgroup$
– Cameron Buie
Aug 20 '13 at 0:40
$begingroup$
@MathApprentice: The way you originally put it, you actually contradicted the Archimedean property! Suppose there is some natural number $n$ greater than any real number. Now, all natural numbers are real, so $n$ is a real number greater than all real numbers, and so greater than all natural numbers, meaning it is an upper bound of $Bbb N$ (and is also greater than itself). Now, I suspect that you meant that for any real number $x$ we can find a larger natural number (which is true), but your phrasing suggested otherwise.
$endgroup$
– Cameron Buie
Aug 20 '13 at 0:40
add a comment |
$begingroup$
Both statements you mention are true, and are two different ways of expressing the Archimedean property. For instance, if we have that $n in mathbb{N}$ and $M in mathbb{R}$ with $M$ positive, then $$ n >M Leftrightarrow frac{1}{n} < frac{1}{M}$$
Regarding your second question, the Archimedean property is not the proof that $mathbb{R}$ completes $mathbb{Q}$. That fact is usually proved (though the two ideas are equivalent) from the least-upper-bound axiom of $mathbb{R}$. The Archimedean property of $mathbb{N}$ is a consequence of the least-upper-bound axiom of the real numbers.
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
$endgroup$
$begingroup$
Is there any more significance to the Archimedan Property?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:21
$begingroup$
I'm not sure what you're looking for; there are plenty of other conclusions you can derive from the Archimedean property, but these seem to be the most general. I would add that the property shows how dangerous it can be to rely on infinitesimals as a basis for calculus.
$endgroup$
– Matt R.
Aug 20 '13 at 1:12
add a comment |
$begingroup$
Both statements you mention are true, and are two different ways of expressing the Archimedean property. For instance, if we have that $n in mathbb{N}$ and $M in mathbb{R}$ with $M$ positive, then $$ n >M Leftrightarrow frac{1}{n} < frac{1}{M}$$
Regarding your second question, the Archimedean property is not the proof that $mathbb{R}$ completes $mathbb{Q}$. That fact is usually proved (though the two ideas are equivalent) from the least-upper-bound axiom of $mathbb{R}$. The Archimedean property of $mathbb{N}$ is a consequence of the least-upper-bound axiom of the real numbers.
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
$endgroup$
$begingroup$
Is there any more significance to the Archimedan Property?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:21
$begingroup$
I'm not sure what you're looking for; there are plenty of other conclusions you can derive from the Archimedean property, but these seem to be the most general. I would add that the property shows how dangerous it can be to rely on infinitesimals as a basis for calculus.
$endgroup$
– Matt R.
Aug 20 '13 at 1:12
add a comment |
$begingroup$
Both statements you mention are true, and are two different ways of expressing the Archimedean property. For instance, if we have that $n in mathbb{N}$ and $M in mathbb{R}$ with $M$ positive, then $$ n >M Leftrightarrow frac{1}{n} < frac{1}{M}$$
Regarding your second question, the Archimedean property is not the proof that $mathbb{R}$ completes $mathbb{Q}$. That fact is usually proved (though the two ideas are equivalent) from the least-upper-bound axiom of $mathbb{R}$. The Archimedean property of $mathbb{N}$ is a consequence of the least-upper-bound axiom of the real numbers.
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
$endgroup$
Both statements you mention are true, and are two different ways of expressing the Archimedean property. For instance, if we have that $n in mathbb{N}$ and $M in mathbb{R}$ with $M$ positive, then $$ n >M Leftrightarrow frac{1}{n} < frac{1}{M}$$
Regarding your second question, the Archimedean property is not the proof that $mathbb{R}$ completes $mathbb{Q}$. That fact is usually proved (though the two ideas are equivalent) from the least-upper-bound axiom of $mathbb{R}$. The Archimedean property of $mathbb{N}$ is a consequence of the least-upper-bound axiom of the real numbers.
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
edited Jan 23 at 14:40
Cameron Buie
85.6k772160
85.6k772160
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
answered Aug 19 '13 at 23:57
Matt R.Matt R.
298313
298313
$begingroup$
Is there any more significance to the Archimedan Property?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:21
$begingroup$
I'm not sure what you're looking for; there are plenty of other conclusions you can derive from the Archimedean property, but these seem to be the most general. I would add that the property shows how dangerous it can be to rely on infinitesimals as a basis for calculus.
$endgroup$
– Matt R.
Aug 20 '13 at 1:12
add a comment |
$begingroup$
Is there any more significance to the Archimedan Property?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:21
$begingroup$
I'm not sure what you're looking for; there are plenty of other conclusions you can derive from the Archimedean property, but these seem to be the most general. I would add that the property shows how dangerous it can be to rely on infinitesimals as a basis for calculus.
$endgroup$
– Matt R.
Aug 20 '13 at 1:12
$begingroup$
Is there any more significance to the Archimedan Property?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:21
$begingroup$
Is there any more significance to the Archimedan Property?
$endgroup$
– MathApprentice
Aug 20 '13 at 0:21
$begingroup$
I'm not sure what you're looking for; there are plenty of other conclusions you can derive from the Archimedean property, but these seem to be the most general. I would add that the property shows how dangerous it can be to rely on infinitesimals as a basis for calculus.
$endgroup$
– Matt R.
Aug 20 '13 at 1:12
$begingroup$
I'm not sure what you're looking for; there are plenty of other conclusions you can derive from the Archimedean property, but these seem to be the most general. I would add that the property shows how dangerous it can be to rely on infinitesimals as a basis for calculus.
$endgroup$
– Matt R.
Aug 20 '13 at 1:12
add a comment |
$begingroup$
Let's look at some concrete consequences for ordinary calculus. Consider the following limit:
$$lim_{ntoinfty}frac1n=0$$
I bet you want that equation to be true, right? Well, it requires the Archimedean property! Otherwise, the sequence doesn't converge.* That would be really inconvenient. Similarly, definitions like
$$exp(x)=sum_{n=0}^inftyfrac{x^n}{n!}$$
rely on the Archimedean property, since that sum doesn't converge otherwise.*
*Statements about non-convergence refer to the order topology of a non-Archimedean ordered field. I make no claims about non-ordered fields, such as the $p$-adic numbers.
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
$endgroup$
add a comment |
$begingroup$
Let's look at some concrete consequences for ordinary calculus. Consider the following limit:
$$lim_{ntoinfty}frac1n=0$$
I bet you want that equation to be true, right? Well, it requires the Archimedean property! Otherwise, the sequence doesn't converge.* That would be really inconvenient. Similarly, definitions like
$$exp(x)=sum_{n=0}^inftyfrac{x^n}{n!}$$
rely on the Archimedean property, since that sum doesn't converge otherwise.*
*Statements about non-convergence refer to the order topology of a non-Archimedean ordered field. I make no claims about non-ordered fields, such as the $p$-adic numbers.
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
$endgroup$
add a comment |
$begingroup$
Let's look at some concrete consequences for ordinary calculus. Consider the following limit:
$$lim_{ntoinfty}frac1n=0$$
I bet you want that equation to be true, right? Well, it requires the Archimedean property! Otherwise, the sequence doesn't converge.* That would be really inconvenient. Similarly, definitions like
$$exp(x)=sum_{n=0}^inftyfrac{x^n}{n!}$$
rely on the Archimedean property, since that sum doesn't converge otherwise.*
*Statements about non-convergence refer to the order topology of a non-Archimedean ordered field. I make no claims about non-ordered fields, such as the $p$-adic numbers.
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
$endgroup$
Let's look at some concrete consequences for ordinary calculus. Consider the following limit:
$$lim_{ntoinfty}frac1n=0$$
I bet you want that equation to be true, right? Well, it requires the Archimedean property! Otherwise, the sequence doesn't converge.* That would be really inconvenient. Similarly, definitions like
$$exp(x)=sum_{n=0}^inftyfrac{x^n}{n!}$$
rely on the Archimedean property, since that sum doesn't converge otherwise.*
*Statements about non-convergence refer to the order topology of a non-Archimedean ordered field. I make no claims about non-ordered fields, such as the $p$-adic numbers.
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
answered Aug 20 '13 at 1:58
Chris CulterChris Culter
21.4k43888
21.4k43888
add a comment |
add a comment |
$begingroup$
I think you are asking if Archimedean helps in deriving R from Q. Then you are wrong because Q itself has Archimedean but is not ordered field with least upper bound property.
I also mention a fact which will help you clearing idea about it(if thsi is question):
An ordered field satisfy Least upper bound property iff it satisfy Archimedean and cauchy completeness property. And R is only ordered field with least upper bound property. This is what we take as definition of R
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
$endgroup$
add a comment |
$begingroup$
I think you are asking if Archimedean helps in deriving R from Q. Then you are wrong because Q itself has Archimedean but is not ordered field with least upper bound property.
I also mention a fact which will help you clearing idea about it(if thsi is question):
An ordered field satisfy Least upper bound property iff it satisfy Archimedean and cauchy completeness property. And R is only ordered field with least upper bound property. This is what we take as definition of R
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
$endgroup$
add a comment |
$begingroup$
I think you are asking if Archimedean helps in deriving R from Q. Then you are wrong because Q itself has Archimedean but is not ordered field with least upper bound property.
I also mention a fact which will help you clearing idea about it(if thsi is question):
An ordered field satisfy Least upper bound property iff it satisfy Archimedean and cauchy completeness property. And R is only ordered field with least upper bound property. This is what we take as definition of R
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
$endgroup$
I think you are asking if Archimedean helps in deriving R from Q. Then you are wrong because Q itself has Archimedean but is not ordered field with least upper bound property.
I also mention a fact which will help you clearing idea about it(if thsi is question):
An ordered field satisfy Least upper bound property iff it satisfy Archimedean and cauchy completeness property. And R is only ordered field with least upper bound property. This is what we take as definition of R
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
answered Aug 10 '14 at 13:48
SushilSushil
1,035927
1,035927
add a comment |
add a comment |
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@GitGud: I don't think the additional tags were needed.
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– Asaf Karagila♦
Aug 19 '13 at 23:38
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@AsafKaragila I was doubtful about topology, not about set theory, though. But you're the expert.
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– Git Gud
Aug 19 '13 at 23:41
1
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It is quite important, for it means that (in the reals) there are no "infinitesimals" of the type that were used in the early years of calculus.
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– André Nicolas
Aug 19 '13 at 23:43
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@AsafKaragila I changed my mind, I realise now this has little to do with the construction of the real numbers and concur with your decision.
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– Git Gud
Aug 19 '13 at 23:57
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@GitGud: You're right. Now, anyway. ;-)
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– Asaf Karagila♦
Aug 20 '13 at 0:02