I know this is a common proof. I'm following Rudin's proof and I'm following everything except for one step.

Suppose $x, y \in \Bbb R$ and $x < y$. Then there exists an $n \in \Bbb N$ such that $n(y-x) > 1$.

Again by the Archimedean property, there exist $m_{1}, m_{2} \in \Bbb N$ such that $m_{1} > nx$ and $m_{2} > -nx$, i.e. $$ -m_{2} < nx < m_{1} $$

From here, Rudin says there must be an $m \in \Bbb Z$ with $-m_{2} \le m \le m_{1}$ and that $$ m-1 \le nx < m $$

I'm confused about these two steps. If $-m_{2} < nx < m_{1}$, then isn't $-m_{2} < m_{1}$?

edit: to be clear, I follow everything up until the introduction of $m$.

  • 1
    $\begingroup$ He seems to be using the Well-Ordering principle $\endgroup$ – Prahlad Vaidyanathan Sep 28 '13 at 15:07
  • $\begingroup$ What if $x=0$ ...? $\endgroup$ – Michael Hoppe Sep 28 '13 at 15:57
  • $\begingroup$ @PrahladVaidyanathan I see, thanks. But to me, the well-ordering principle would suggest the following: by Archimedean property, the set $\left\{m_{1} \in \Bbb N: m_{1} > nx\right\}$ is not empty. And by the well-ordering property, there is an $m \in \Bbb N$ such that $m > nx$ and $nx > m-1$, so $m-1 < nx < m$. We know that $ny > 1 + nx$, so combining the inequalities gives $nx < m < ny$. Is this not enough? $\endgroup$ – asdfghjkl Sep 28 '13 at 16:54
  • $\begingroup$ I guess I'm just confused what the $m_{2}$ is for $\endgroup$ – asdfghjkl Sep 28 '13 at 16:57
  • $\begingroup$ not sure if you have this same issue (even after accepting an answer), but the whole proof Rudin presents seems rather random and as if it comes out of no where. I have no intuition on why he introduces $m_1, m_2$. Sure $m_1 > nx$, $m_2 > -nx$, it feels as if I have no conceptual idea of why one would do this. Why is he doing this beyond the "it works". I am nearly 100% this formal proof comes out from conveying some conceptual idea and just making it formal. If you understand where it comes from, please provide your own answer and clarify it! It still a mystery for me. $\endgroup$ – Pinocchio Dec 25 '16 at 19:35

Yes, $-m_2<m_1$, but we knew this anyway: $m_1$ and $m_2$ are positive integers, so $-m_2$ is negative and $m_1$ is positive.

Rudin introduces both $m_1$ and $m_2$ in order to avoid having to split the argument into cases depending on whether $x>0$, $x=0$, or $x<0$. If you simply take $m_1$ to be the minimal such that $m_1>nx$, you’re in trouble if $x$ is negative: $m_1=1$, which doesn’t do what you want.

Once you have $-m_2<nx<m_1$, you can use the well-ordering principle to set

$$k_0=\min\{k\in\Bbb N:-m_2+k>nx\}\;;$$

$\{k\in\Bbb N:-m_2+k>nx\}$ is non-empty, because it contains $m_1-(-m_2)$ so the well-ordering principle ensures that $k_0$ exists. Now let $m=-m_2+k_0$, and you can easily check that $m-1\le nx<m$.

  • $\begingroup$ sorry if this is really obvious, but what is the trouble if x is negative? if x is negative, say $x=-1$, and $m_1=1$, then $m_1 > nx \iff 1 > -n$ which is clearly true if $n \in mathbb N$. No? Or am I missing something? I think I don't understand why he wants to avoid cases and how cases even arise in the proof. $\endgroup$ – Pinocchio Dec 25 '16 at 18:45
  • 1
    $\begingroup$ @Pinocchio: In order for the argument to work, we need to find the smallest integer $m$ such that $m>nx$. If $nx$ is positive, the well-ordering principle takes care of this. If $nx$ is negative, however, using the well-ordering principle to get the smallest member of $\Bbb N$ greater than $nx$ merely gives us $1$, and $1$ is not the smallest integer greater than $nx$. If $nx=-\pi$, for instance, we need $m$ to be $-3$, not $1$, or the rest of the argument won’t work. $\endgroup$ – Brian M. Scott Dec 25 '16 at 19:03
  • 1
    $\begingroup$ @Pinocchio: The basic idea is the one that works when $x>0$: use the well-ordering principle to find the smallest $m\in\Bbb N$ such that $nx<m$, note that $m-1\le nx<m$, and proceed from there. There are then two choices: split the argument into cases, or find some way to apply this idea even when $x<0$. Rudin chose the latter approach. He doesn’t have $0<nx$, so he finds a $-m_2<nx$. Then he uses the fact that $\{k\in\Bbb Z:k>-m_2x\}$ ‘looks like’ $\Bbb N$, just shifted left by $m_2$. This lets him find least elements of non-empty subsets of $\{k\in\Bbb Z:k>-m_2x\}$ by shifting the sets ... $\endgroup$ – Brian M. Scott Dec 25 '16 at 19:38
  • $\begingroup$ ... $m_2$ units to the right into $\Bbb N$, finding a least element there, and then shifting it back down $m_2$ units to the original set. $\endgroup$ – Brian M. Scott Dec 25 '16 at 19:38
  • $\begingroup$ @BrianM.Scott I see that makes much more sense to me (thanks!). Its clear to me now that what Rudin is trying to do (without telling us) is to show that the smallest integer larger than nx is indeed between nx and ny, that way he can just form the fraction/rational we are looking for by dividing the interger $m$ by n and havinv $nx < m < ny \implies x< \frac{m}{n} < y$ which is what we want. What I still need to figure out is what the issue is when $x < 0$ i.e $x$ is negative (btw, I am aware that having an $n$ s.t. $n(y-x)>1$ is important is not crucial). $\endgroup$ – Pinocchio Dec 25 '16 at 20:07

The whole proof seems to hinge on (1) Archimedian property, and (2) well ordering principle. While Rudin states and proves Archimedian property, just prior to the proof that we are discussing, well ordering principle is not mentioned by Rudin anywhere in his book, neither before nor after the current proof. Assuming something that is not stated anywhere in the book, is not a good proof strategy, especially in subjects like real analysis.


The proof seems rather oddly written. I find it more intuitive to note that $nx-ny>1$ implies there must be an integer between $ny$ and $nx$. Look at $m=\lfloor ny\rfloor+1$.

  • $\begingroup$ I’m pretty sure that that argument isn’t available to Rudin at that point in the book, because he hasn’t shown that $nx-ny>1$ implies that there is an integer between $ny$ and $nx$ or justified the existence of $\lfloor x\rfloor$ for an arbitrary $x\in\Bbb R$. He’s trying to be extremely rigorous. $\endgroup$ – Brian M. Scott Sep 28 '13 at 19:29
  • $\begingroup$ @BrianM.Scott We can be rigorous by defining $\lfloor x\rfloor$ using WOP, cannot we? $\endgroup$ – Pedro Tamaroff Sep 28 '13 at 19:35
  • $\begingroup$ That’s essentially what Rudin is doing in this argument, though he actually gets $\lfloor x\rfloor+1$. $\endgroup$ – Brian M. Scott Sep 28 '13 at 19:37
  • $\begingroup$ @BrianM.Scott Aha. That's why I said it is oddly written. I don't feel like correcting Rudin or anything, just saying it can be made clearer. $\endgroup$ – Pedro Tamaroff Sep 28 '13 at 19:40
  • $\begingroup$ I also find the proof to be extremely oddly written and super confusing. Having that if $ny -nx > 1$ implies there is some integer between $nx$ and $ny$ and just finding it (say $m$) and dividing by $n$ finds our rational $x<\frac{m}{n}<y$ is more obvious. Isn't it super simple to prove that a $m$ does exist if $ny -nx > 1$? $\endgroup$ – Pinocchio Dec 25 '16 at 19:24

I agree with Pedro above.

Choose $n >1/(b-a)$ so that $an > bn +1$.

Then we can prove there is a k in Z with an < k < bn because: Let $k = \min\{m \in Z : an \le m\}$ This is nonempty by Archimedian thingy (and we can easily prove that nonempty bounded below sets of integers have a min element using the fact that nonempty sets of $N$ have a min element)

Then $an \le k < bn$ (first $an\le k$ by def and $k < bn$ because if $k \ge bn$ then $k\ge bn > an +1$ implies $k-1 >an$ which would contradict minimality)

Then $a\le k/n < b$ ($n \ge 1$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.