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Proof that for $a,b \in \mathbb{R}$ there is an irrational number $r$ so that $a < r < b$. Basically, proof, that between any two irrationals, there is another irrational r.

I'm sure there are already many ways out there how to do it, however I have troubles proving it in the following way:

(1) For every $x,y \in \mathbb{R}$ there is a bijective function between $[0,1]$ and $[x,y]$ (already proven)

(2) $\frac{\sqrt{2}}{2} \in ]0,1[$

(3) Now when mapping $[0,1]$ onto $[x,y]$ $\frac{\sqrt{2}}{2}$ will also be mapped into the new intervall, therefore there has to be an irrational number in $[x,y]$

Now the problem I see is, that for example $\frac{\sqrt{2}}{2}$ could be mapped onto a rational number and therefore I'd have to proof, that there is a different irrational in $[x,y]$. It'd be nice if you could help me complete the proof.

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marked as duplicate by JMoravitz, carmichael561, dantopa, Claude Leibovici, user91500 May 25 '17 at 8:42

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  • $\begingroup$ do you have a description of the bijective function? or would like to only use the existencE? $\endgroup$ – clark May 25 '17 at 1:31
  • $\begingroup$ @clark only use the existence $\endgroup$ – WhatAMesh May 25 '17 at 1:33
  • $\begingroup$ your proof does not mention $a$ and $b$ at all $\endgroup$ – Mirko May 25 '17 at 1:35
  • $\begingroup$ What do you mean when you say you need to prove that there is a different irrational in [x,y]? Doesn't every non-empty open interval in $\mathbb{R}$ have an irrational? $\endgroup$ – Blitzkrieg May 25 '17 at 1:35
  • $\begingroup$ @JMoravitz It's not about the result it is about the way $\endgroup$ – WhatAMesh May 25 '17 at 1:35
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If there are only rational numbers in $[x, y]$, then $[x, y]$ is countable and is in bijection with $[0, 1]$ which is uncountable, which is impossible.

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If you want to continue your proof, then define $m = \frac{a + b}{2}$, $n = \frac{(a + b)\sqrt{2}}{2}$. You can show that $a < m < n < b$. Then either:

  1. $m$ is irrational, and you're done.
  2. $m$ is rational, then $n = \sqrt{2}m$ is the product of a rational and an irrational, and so is irrational, and you're done.
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  • 1
    $\begingroup$ why should n be less than b? If a=2, b=3, then n=3.53... Unfortunately your arguments does not work $\endgroup$ – fonfonx May 25 '17 at 2:38
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    $\begingroup$ You're right, I think my brain was off. $\endgroup$ – ConMan May 25 '17 at 22:57

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