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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.

WhatAMesh
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  • do you have a description of the bijective function? or would like to only use the existencE? – clark May 25 '17 at 01:31
  • @clark only use the existence – WhatAMesh May 25 '17 at 01:33
  • your proof does not mention $a$ and $b$ at all – Mirko May 25 '17 at 01:35
  • 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? –  May 25 '17 at 01:35
  • @JMoravitz It's not about the result it is about the way – WhatAMesh May 25 '17 at 01:35
  • @RajivKaipa Well that is bascially what has to be proven, or rather, that between any to irrationals that is another irraitonal – WhatAMesh May 25 '17 at 01:37
  • If the statement that has to be proved is irrationals are dense in $\mathbb{R}$, then the link provided by JMoravitz should answer that part right? This should answer that as well. Another way you could prove it is by stating that fact that group with elements of the form $n - m\xi$ where $\xi$ is irrational and $n$ and $m$ are integers is an additive group which is not not cyclic, which implies that it is dense in $\mathbb{R}$. –  May 25 '17 at 01:48

2 Answers2

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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.

fonfonx
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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.
ConMan
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