Between two Rational Numbers exists Irrational Number/Proof 2

Theorem

Let $a, b \in \Q$ where $a < b$.

Then:

$\exists \xi \in \R \setminus \Q: a < \xi < b$

Lemma

Let $\alpha \in \Q$ and $\beta \in \R \setminus \Q$.

Then:

$\alpha + \beta \in \R \setminus \Q$

Proof

From Between two Real Numbers exists Rational Number, there exists $x \in \Q$ such that:

$a - \sqrt2 < x < b - \sqrt2$

Since Square Root of 2 is Irrational, by the Lemma:

$x + \sqrt 2$ is irrational.

But:

$a < x + \sqrt 2 < b$

which is what we wanted to show.

$\blacksquare$