I wanna prove this statement: $\forall a, b \in \mathbb{R}, a < b \exists x \in \mathbb{R} \setminus \mathbb{Q} : a < x < b$
So basically I want to prove, that there is a irrational number between two real numbers.
Its already given that $\sqrt{2}$ is a irrational number, so I thought about proving the statement with using $\sqrt{2}$:
$ a < b \Rightarrow a < m < b $
So its assumed that there are (infinitely) many rational numbers between two rational numbers. In this case, m is a rational number that is between a and b.
$ a < m < b \Rightarrow a + \sqrt{2} < n < b + \sqrt{2} $
Now, if u add $\sqrt{2}$ to a and b (the beginning and end of the interval), there is still a rational number n that is in the interval between a and b.
$a + \sqrt{2} < n < b + \sqrt{2}$ | $-\sqrt{2}$
$= a < n - \sqrt{2} < b $
If we now subtract $\sqrt{2}$, then n - $\sqrt{2}$ should be an irrational number. But somehow this makes no sense... damn