# Proving there are infinitely many rational numbers in $[x,y]$

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.

• do the same with the interval $[x,r]$... and repeat and repeat infinitely many times (use induction if you prefer) – Tito Eliatron Mar 12 at 19:20
• @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. – TheSilverDoe Mar 12 at 19:24
• @TheSivlerDoe, you are right, – dmtri Mar 12 at 19:34

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 $$y$$.

• I wrote out a new proof on this thread. Care to critique? – Ash Mar 12 at 20:35

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 $$[x+2(y-x)/3,y]$$.)

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 $$[x,y]$$.

• @Robert Shore, how's this? – Ash Mar 12 at 20:34
• 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 $x \leq r_1$. – Robert Shore Mar 12 at 20:39
• @RobertShore thank you for the feedback – Ash Mar 12 at 20:45
• 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. – Robert Shore Mar 12 at 20:48

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.

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