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Prove if $x$ and $y$ are real numbers with $x \lt y$, then there are infinitely many rational numbers in the interval $[x,y]$.

What I got so far:

Let $x,y \in \Bbb R$ with $x \lt y$

Let $S = [x,y]$

By the density of $\Bbb Q$ in $\Bbb R$, $\exists r \in \Bbb Q$ such that $x \lt r \lt y$ where $r \in S$.


This is where I got stuck.

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  • $\begingroup$ do the same with the interval $[x,r]$... and repeat and repeat infinitely many times (use induction if you prefer) $\endgroup$ – Tito Eliatron Mar 12 at 19:20
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    $\begingroup$ @dmtri This is not the correct way to do it. Example : $x=0$ and $y=2$. You will pick a rational in every $[1/n, 2-1/n]$, but how can you know that you are not picking the rational $1$ at each step ? Therefore, this does not tell you that $[0,2]$ contains an infinity of rationals. $\endgroup$ – TheSilverDoe Mar 12 at 19:24
  • $\begingroup$ @TheSivlerDoe, you are right, $\endgroup$ – dmtri Mar 12 at 19:34
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If you have only $n$ finitely many rational numbers between $x$ and $y$, put them in order, $x, r_1, r_2, \ldots r_n, y$. Can you now prove that there were more than $n$ rational numbers between $x$ and $y$? If so, you have contradicted your original assumption that there were only $n$ such numbers, which means no such $n$ can exists and there must be infinitely many rational numbers between $x$ and $y$.

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  • $\begingroup$ I wrote out a new proof on this thread. Care to critique? $\endgroup$ – Ash Mar 12 at 20:35
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Just find two distinct rational numbers $p,q\in[x,y]$. Now the rational number $p+(q-p)/n$ is in $[x,y]$ for all $n\in \Bbb N_{>0}$.

(And you can find such $p,q$ in the intervals $[x,x+(y-x)/3]$ and $[x+2(y-x)/3,y]$.)

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Okay, so I'll give it another shot given the feedback.

Proof:

Let $x,y \in \Bbb R$ with $x \lt y$ and $S = [x,y]$

Suppose there are only $n$ rational numbers between $x$ and $y$ such that:$$x \lt r_1 \lt \cdot \cdot \cdot \lt r_n \lt y$$

But since $\Bbb Q$ is dense in $\Bbb R$, there exists $r_{n+1} \in \Bbb Q$ such that: $$ x \lt r_{n+1} \lt r_1 \lt \cdot \cdot \cdot \lt r_n \lt y$$

which contradicts our assumption that there are only $n$ rational numbers in $[x,y]$.

Therefore, there must be infinitely many rational numbers in the interval $[x,y]$.

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  • $\begingroup$ @Robert Shore, how's this? $\endgroup$ – Ash Mar 12 at 20:34
  • $\begingroup$ Slightly sloppy but fundamentally correct. As framed, it only works if $n \geq 2$, because you need that in order to know that there are $r_k$ and $r_{k+1}$ to stuff the new rational between. Better to simply put the new rational between $x$ and $r_1$. Finally, I think you know that $x \lt r_1$, not just that $x \leq r_1$. $\endgroup$ – Robert Shore Mar 12 at 20:39
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    $\begingroup$ @RobertShore thank you for the feedback $\endgroup$ – Ash Mar 12 at 20:45
  • $\begingroup$ With your newest edits, it's now correct. As a matter of exposition, I'd probably elaborate a little (a phrase or a sentence) on the contradiction that you've demonstrated but your math and your logic are both correct. $\endgroup$ – Robert Shore Mar 12 at 20:48
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Consider the interval $[x,y]$. Find two rational numbers $q_{min}$ and $q_{max}$ such that $x < q_{min} < q_{max} < y$. Let $r=q_{min}-q_{max}$. Now make a new number: $q_1 = q_{min} + r/2$. Let $r_1=q_{max}-q_1$. Now make a new number: $q_2=q_1+r_1/2$. Let $r_2=q_{max}-q_2$. Follow the pattern. Each of these numbers is a rational number; each number is unique. The set constructed is countably infinite in size.

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    $\begingroup$ We are not given that $x$ and $y$ are rational. So your numbers will not in general be rational. $\endgroup$ – TonyK Mar 12 at 19:30
  • $\begingroup$ @TonyK Did I fix it? $\endgroup$ – NicNic8 Mar 12 at 21:12
  • $\begingroup$ Yes, I think so. But that $q^{(1)}$ notation is hard on the eyes! $\endgroup$ – TonyK Mar 12 at 21:24
  • $\begingroup$ @TonyK Tough customer. :) $\endgroup$ – NicNic8 Mar 13 at 15:37
  • $\begingroup$ And now you have two different $q_1$'s... $\endgroup$ – TonyK Mar 13 at 15:57

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