Let $a
Let $a<b$ and $a,b\in\Bbb R$. Then there is $c\in\Bbb R\setminus\Bbb Q$ such that $a<c<b$.

My attempt:


*

*$a+b$ is irrational


Let $c:=\dfrac{a+b}{2}$


*$a+b$ is rational


Let $x:=\dfrac{a+b}{\sqrt 2}$. Then $x$ is irrational.


*

*$a<x<b$
Let $c:=x$


*

*$b\le x$
Then $x-b<x-a$. Take $x'\in (x-b,x-a)$ such that $x'$ is rational.
Then $a<x-x'<b$ where $x-x'$ is irrational.
Let $c:=x-x'$.


*

*$x\le a$
Then $a-x<b-x$. Take $x'\in (a-x,b-x)$ such that $x'$ is rational.
Then $a<x+x'<b$ where $x+x'$ is irrational.
Let $c:=x+x'$.

My proof is quite short. I'm worried if it's sloppy and contains mistakes. Please help me verify it!

If $a, b$ both are rationals then, $\frac{b+a}{2}$ is rational. Then, lets pick $c= \frac{b+a}{2}+\frac{b-a}{2\sqrt{2}}$
If $a$ is rational, and $b$ is irrational, we take $c=\frac{b+a}{2}$
If $a, b$ are irrational:
$\frac{a+b}{2}$ is irrational then pick it as $c$.
Else we pick, $c=\frac{b+a}{2}+ (b-\frac{b+a}{2})/2$. $(b-\frac{b+a}{2})$ is irrational.
