Between two Rational Numbers exists Irrational Number/Proof 1

Proof
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 $x$:
 * $x^2 < 1 \implies x \in \openint {-1} 1$

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:
 * $\xi = a + k d < a + 1 \cdot d < a + \paren {b - a} = b$

We thus have:
 * $\xi \in \R \setminus \Q: \xi \in \openint a b$

Hence the result.