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I am looking for some advice/help in regard to the proof that Q is dense in R, given in Walter Rudin's book "Principles of Mathematical Analysis". Mostly, I want to see if my reasoning is correct for some parts, and what I am not picking up as well.

I will post below the specific theorem and proof that I am referring to,

(Also I will include the archimedian property that I am referring to for convince to anyone who may be reading)

(archimedian property): "If $x \in R$ , $y \in R$ and $x \gt 0$ then there is a positive integer n such that $nx \gt y$"

$\mathbf{Theorem:}$

If $x \in R$ , $y \in R$ and $x \lt y$ then there exists a $p \in Q$ such that $x \lt p \lt y$.

$\mathbf{Proof:}$

We have $$x \lt y$$ so , $$y-x \gt 0$$ thus we can use the archimedean property of R and say that there must exist a positive integer n such that

$$n(y-x) \gt 1$$ ( I believe we just choose 1 for convince, as the property will hold for any R?).

Now here is where I get confused, the author then writes " apply the archimedean property again, to obtain positive integers $m_{1}$ and $m_{2}$ such that $m_{1} \gt nx$ and $m_{2} \gt -nx$, then $-m_{2} \lt nx \lt m_{1}..$

My question is, what are applying the property to now? I am just not seeing where this comes from, also because the property required us that $x \gt 0$ and here we only have $y-x \gt 0$, so how does this all tie in together?

I appreciate any help/advice, thanks all!

Quality
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  • You can apply the property, because $1>0$, so $m_1\cdot 1>nx$ for some positive integer $m_1$. – egreg Jun 24 '15 at 23:08
  • No, you can’t just set $n=1$: what if, for instance, $y=\sqrt2$ and $x=1$? Then you need $n\ge 3$. At the next step you’re applying the Archimedean property to the pair $1$ and $nx$ to get $m_1\cdot1>nx$ and to $1$ and $-nx$ to get $m_2\cdot1>-nx$. For a fuller explanation of why Rudin does this, see this answer. – Brian M. Scott Jun 24 '15 at 23:09
  • @muaddib: Whether one uses equivalence classes of Cauchy sequences in $\Bbb Q$ or Dedekind cuts in $\Bbb Q$ to construct $\Bbb R$, it is still a theorem that $\Bbb Q$ is dense in $\Bbb R$, one that has to be proved. – Brian M. Scott Jun 24 '15 at 23:13
  • @muaddib: No. You’re sliding over a great many details there, starting with the embedding of $\Bbb Q$ in the constructed $\Bbb R$ and defining the linear order on the constructed $\Bbb R$, which is not entirely trivial. Your ‘proof’ is only slightly better than hand-waving. – Brian M. Scott Jun 24 '15 at 23:23
  • @BrianM.Scott I did not mean to say I set n=1 , I mean the RHS because doesn't it hold for any y in R , of course including 1 ? – Quality Jun 24 '15 at 23:24
  • @Quality: Ah, okay: I misunderstood. The righthand side has to be $\ge 1$, because we want to be sure that there is at least one integer between $nx$ and $ny$. We don’t have to use $1$, but we can’t use anything smaller. – Brian M. Scott Jun 24 '15 at 23:28
  • @muaddib: No, I’m not wrong; you are. I can’t judge what’s in Strichartz, since I don’t have the book, but I can certainly judge what you wrote (and hope that it doesn’t accurately reflect Strichartz). – Brian M. Scott Jun 24 '15 at 23:30

1 Answers1

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You can obtain $m_1$ and $m_2$ by applying the Archimedean property as cited with $x=1 > 0$ and $y= nx$ or $y = -nx$.

Only $x$ in the Archimedean property needs to be positive. You can make a little sketch to see it.

hHhh
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    Thanks, I think this makes sense. So basically, we are using the 1 as our x and using commutative field properties to say that we know y of the form nx and -nx exist, and also since $x \gt 0$ so to is $nx$ and that is why we can eventually write$ - m_{2} \gt nx because m_{2} \gt -(nx)$ Does that reasoning sound okay? – Quality Jun 24 '15 at 23:32
  • Yes, that's right – hHhh Jun 24 '15 at 23:44
  • Thank you, and lastly if you can, I am wondering how having $-m_{2} \lt nx \lt m_{1}$ , implies that there is an integer m with $-m_{2} \le m \le m_{1}$ ? – Quality Jun 25 '15 at 00:05
  • because $m_2$ and $m_1$ are two different integers :) – hHhh Jun 25 '15 at 00:07