Proof there is an irrational number $r$ in every intervall $a < r < b$ Proof that for $a,b \in \mathbb{R}$ there is an irrational number $r$ so that
  $a &lt; r &lt; b$. 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.
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.
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 $a &lt; m &lt; n &lt; b$. Then either:


$m$ is irrational, and you're done.
$m$ is rational, then $n = \sqrt{2}m$ is the product of a rational and an irrational, and so is irrational, and you're done.