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Let $a$ and $b$ be reals with $a<b$. Show that there are infinitely many rationals $x$ such that $a<x<b$.

My plan of action was to assume that $x$ is the smallest such rational and find another rational in the interval $(a, x)$, but I am struggling to make it work. A hint will be much preferred to a full solution.

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Just take the average of $a$ and $x$. It's not hard to show that's greater than $a$ and smaller than $x$. –  crf Sep 20 '12 at 0:30
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@crf The average of $a$ and $x$ need not be rational. –  Jon Sep 20 '12 at 0:32
    
! sorry. Somehow I missed the whole thing about them being rationals. –  crf Sep 20 '12 at 0:35
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4 Answers

This PDF may help; howbeit, I amended the picture :

enter image description here

Whence we perceive by inspection : Because $\color{#1FB4BF}{1/n} < \color{#E431D2}{(y - x)}$ for all $x, y \in \mathbb{R}$,
thus $ \; k/n < x \iff k/n + \color{#1FB4BF}{1/n}< x + \color{#E431D2}{(y - x)}$.
In summary, $k/n + \color{#1FB4BF}{1/n} = \dfrac{k+1}{n}$ $\in (x,y)$ is the rational number desired.

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Note that the real numbers are an Archimedean field, so for any real number $r$ we have some integer $n>r$. This means that for any real number $\epsilon>0$, we have some $n>1/\epsilon$, so $1/n<\epsilon$. Furthermore, the rationals are dense in the reals, so we can find some rational $x$ such that $a<x<b$.

Let $n$ be such that $1/n<b-x$. Then $x+\frac{1}{n},x+\frac{1}{n+1},\ldots$ is an infinite set of rationals between $a$ and $b$.

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There is no smallest such rational. But your basic strategy will work if we can show there is at least one such rational. We sketch a proof of the fact that there are at least two. There is some detail that needs to be filled in.

Let $\epsilon=\dfrac{1}{b-a}$. Then by something that has undoubtedly already been proved, there is a positive integer $N$ such that $\dfrac{1}{N}\lt \epsilon/2$.

There is a largest integer $m$ such that $\dfrac{m}{N}\lt a$. Argue that $$a\lt \frac{m+1}{N}\lt \frac{m+2}{N}\lt b.$$

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Say $a\ne b$ and we want to show that infinitely many rationals are between $a$ and $b$. Then $|a-b|>0$. Is there an integer $n$ so big that $1/n < |a-b|$? If not, then $|a-b|>0$ is a lower bound of the set $\{1/n:n\in\{1,2,3,\ldots\}\}$, which therefore has an infimum $c$ that is positive and therefore has a reciprocal $1/c>0$, and $c=\sup\{1,2,3,\ldots\}$. Since $c>0$ is the smallest number greater than every positive integer, $c/2$ is smaller than sum positive integer $n$, so $c$ is smaller than $2n$, but $2n$ is a positive integer, so we have a contradiction. Conclusion: for some positive integer $n$, we have $1/n<|a-b|$. From there it's not hard to show that for every denominator $m\ge n$, some rationals with denominator $m$ are between $a$ and $b$.

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