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How would I go about proving that for any two real numbers $a,b\in\mathbb{R}$ we may find a both a rational and irrational number between them?

It can be proven by using infinite decimals but how would one prove this directly from the completeness of $\mathbb{R}$.

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    $\begingroup$ You know, David, it really is polite to read the answers, upvote the helpful ones, accept the one that helped the most, and at the very least provide some commentary if you found them unhelpful. $\endgroup$
    – The Count
    Commented Apr 27, 2017 at 19:51

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You can use the Archimedean property of $\Bbb R$. Take $b \gt a$. There is some $n$ such that $\frac 2{n} \lt b-a$. There are at least two rationals with denominator $n$ between $a$ and $b$. Then the lower one plus $\frac {\sqrt 2-1}{n}$ is an irrational between $a$ and $b$

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You are trying to prove the density property.

It sometimes helps to know the name of something to look it up.

There is a proof here ( it seems fairly pointless/disingenuous just copying that proof in to this answer.)

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If $a$ and $b$ are both rational, then for the rational's existence, consider $\dfrac{a+b}{2}$. If either are irrational, this works as well to prove the irrational's existence. In either case, you need to be a little more clever for the other part, but as a hint: try proof by contradiction. Then what can be said about the density/completeness of the irrationals or rationals?

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  • $\begingroup$ We weren't given that $a,b$ are rational. $\endgroup$ Commented Apr 26, 2017 at 16:40
  • $\begingroup$ Oh, goodess. Fair point. Edited. $\endgroup$
    – The Count
    Commented Apr 26, 2017 at 16:41

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