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
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 $\R \setminus \Q$.

Furthermore, for any given real number $x$, we know $x^2 < 1 \iff x \in \left({-1 .. 1}\right)$.

Thus, if we set $k = \dfrac {\sqrt 2} 2$, we know that $k \in \R \setminus \Q$ and that $0 < k < 1$, since $k^2 = \dfrac 1 2$.

Let $\xi = a + k d$.

Then, since $a, d \in \Q$ and $k \in \R \setminus \Q$, it follows 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:
 * $\xi \in \R \setminus \Q: \xi \in \left({a ..b}\right)$