Between Every Two Rationals Exists an Irrational

Theorem

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

Then:

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

Proof

Lemma 1

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

Then:

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

Proof of Lemma 1

Suppose by contradiction that $\alpha \cdot \beta \in \Q$.

By the definition of rational numbers:

$\exists n, m, p, q \in \Z: \alpha = \dfrac n m$

$\alpha \cdot \beta = \dfrac p q$

Thus:

$\beta = \dfrac p q \cdot \dfrac 1 \alpha = \dfrac p q \cdot \dfrac m n$

By Rational Multiplication is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$.

Therefore $\alpha \cdot \beta \in \R \setminus \Q$.

$\Box$

Lemma 2

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

Then:

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

Proof of Lemma 2

Suppose by contradiction that $\alpha + \beta \in \Q$.

By the definition of rational numbers:

$\exists p, q \in \Z: \alpha + \beta = \dfrac p q$

Thus:

$\beta = \dfrac p q - \alpha$

By Rational Subtraction is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$.

Therefore:

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

$\Box$

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 $x^2 < 1 \implies x \in \left({-1 .. 1}\right)$.

Thus, if we set $k = \dfrac {\sqrt 2} 2$, we know from Lemma 1 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 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:

$\xi \in \R \setminus \Q: \xi \in \left({a ..b}\right)$

$\blacksquare$