Between two Rational Numbers exists Irrational Number
Theorem
Let $a, b \in \Q$ where $a < b$.
Then:
- $\exists \xi \in \R \setminus \Q: a < \xi < b$
Lemma 1
Let $\alpha \in \Q \setminus \set 0$ and $\beta \in \R \setminus \Q$.
Then:
- $\alpha \cdot \beta \in \R \setminus \Q$
Lemma 2
Let $\alpha \in \Q$ and $\beta \in \R \setminus \Q$.
Then:
- $\alpha + \beta \in \R \setminus \Q$
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 $x$:
- $x^2 < 1 \implies x \in \openint {-1} 1$
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.
$\blacksquare$
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 the Lemma:
- $x + \sqrt 2$ is irrational.
But:
- $a < x + \sqrt 2 < b$
which is what we wanted to show.
$\blacksquare$
Sources
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.14$: Exercise $18 \ (2)$