Between two Rational Numbers exists Irrational Number

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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

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