Between two Rational Numbers exists Irrational Number

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

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

Proof 1
Let $d = b - a$.

As $a, b \in \Q: a < b$ it follows from Rational Numbers form Ordered Integral Domain that $d \in \Q: d > 0$.

From Square Root of 2 is Irrational, $\sqrt 2$ is not a rational number, so it is an element of $\R \setminus \Q$.

From Square Number Less than One, for any given real number $x$, we have $x^2 < 1 \implies x \in \left({-1 \,.\,.\, 1}\right)$.

Let $k = \dfrac {\sqrt 2} 2$.

Then from Lemma 1:
 * $k \in \R \setminus \Q$

As $k^2 = \dfrac 1 2$, it follows that $0 < k < 1$.

Let $\xi = a + k d$.

Then, since $a, d \in \Q$ and $k \in \R \setminus \Q$, it follows from Lemma 1 and Lemma 2 that $\xi \in \R \setminus \Q$.

$d > 0$ and $k > 0$, so $\xi = a + k d > a + 0 \cdot 0 = a$.

$k < 1$, so $\xi = a + k d < a + 1 \cdot d < a + \left({b-a}\right) = b$.

We thus have:
 * $\xi \in \R \setminus \Q: \xi \in \left({a \,.\,.\, b}\right)$

Proof 2
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 Lemma 2
 * $ x+\sqrt 2$ is irrational.

But
 * $ a < x+\sqrt 2 < b$

which is what we wanted to show.