Irrational Number divided by Rational Number is Irrational

Theorem

Let $x$ be a irrational number.

Let $y$ be a non-zero rational number.

Then:

$\dfrac x y$

is irrational.

Proof

Aiming for a contradiction, suppose $\dfrac x y$ is rational number.

Then there exists an integer $p_1$ and a natural number $q_1$ such that:

$\dfrac x y = \dfrac {p_1} {q_1}$

That is:

$x = \dfrac {p_1} {q_1} y$

From the fact that $y$ is rational, we similarly have that there exists an integer $p_2$ and a natural number $q_2$ such that:

$y = \dfrac {p_2} {q_2}$

Then:

$x = \dfrac {p_1 p_2} {q_1 q_2}$

From Integer Multiplication is Closed, we have that $p_1 p_2$ is an integer.

From Natural Number Multiplication is Closed, we have that $q_1 q_2$ is a natural number.

Let $p_3 = p_1 p_2$ and $q_3 = q_1 q_2$.

Then $x$ is expressible in the form:

$x = \dfrac {p_3} {q_3}$

where $p_3$ is an integer and $q_3$ is a natural number.

This, however, implies that $x$ is rational, which is a contradiction.

By Proof by Contradiction, we conclude that $\dfrac x y$ is irrational.

$\blacksquare$