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How do you prove this question? I was thinking proving contrapositive. But I was stuck..Thanks guys.

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  • $\begingroup$ If $n$ is a natural number greater than $(b-a)^{-1}$, then $\frac{1}{n}<b-a$, so one of the numbers in the sequence $1/n,2/n,\ldots$ should work for you. $\endgroup$ – G Tony Jacobs Dec 22 '14 at 2:20
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Lemma 1. For every real number $x$ there is exactly one integer $N$ such that $N \le x < N + 1$. (This integer $N$ is called the integer part of $x$, and is sometimes denoted $N = \lfloor x \rfloor$.)

Lemma 2. For any positive real number $x > 0$ there exists a positive integer $N$ such that $0 < 1/N < x$.

We now show

Proposition 3. Given any two real numbers $x < y$, we can find a rational number $q$ such that $x < q < y$.

Proof. By hypothesis, we have $y -x$ is positive. So exists a positive integer $N$ such that $0 < 1/N < y - x$, by Lemma 2. Since $xN$ is a real number, by Lemma 1, there exists a integer $n$ such that $n - 1 \le xN < n$, i.e., $n/N - 1/N \le x$ and $x < n/N$. Thus $x < n/N \le x + 1/N$. Since $1/N \le y - x$, i.e., $x + 1/N < y$, we have $x < n/N < y$. Thus $n/N$ is rational, the claim follows.

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Hint: Let $\epsilon = b-a$. There is an $n\in\Bbb N$ such that $\frac{1}{n}<\epsilon$.

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We have that $b-a>0$. Let $c$ be an integer such that $c(b-a)>1$. Then strictly between $ca$ and $cb$ there is an integer $d$, so $ca<d<cb$, hence $a<\frac{d}{c}< b$, and $\frac{d}{c}$ is rational.

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If $\mathbb R$ is defined in terms of Dedekind cuts, then $a<b$ means that $a\subsetneq b$, meaning there's a rational $r$ in $b$ but not in $a$. Then the set of rationals less than $r$ forms a cut $r'$ which is a strict subset of $b$. Also, since $r\notin a$, we have $a\leq r'$. If the inequality is strict, we're done. If $a=r'$, then if there is no rational strictly between $a=r'$ and $b$, then $r$ is the maximal rational in $b$, which contradicts being a cut. Thus, there is a rational strictly between $a$ and $b$.

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In fact this rational can be dyadic. Choose $x\in \mathbb{R}$. You can easily find $n, k$ so that $${{k-1}\over 2^n} \le x < {k\over 2^n}.$$ That puts $x$ within $1/2^{n+1}$ of a dyadic rational.

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