# Rational Number plus Irrational Number is Irrational

## Theorem

That is, let $x \in \Q$, $y \in \R \setminus \Q$ and $x + y = z$.

Then $z \in \R \setminus \Q$.

## Proof

Aiming for a contradiction, suppose $z \in \Q$.

By definition of rational numbers:

$\exists a, b \in \Z, b \ne 0: x = \dfrac a b$
$\exists c, d \in \Z, d \ne 0: z = \dfrac c d$

Then:

 $\displaystyle x + y$ $=$ $\displaystyle z$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac a b + y$ $=$ $\displaystyle \dfrac c d$ $\displaystyle \leadsto \ \$ $\displaystyle y$ $=$ $\displaystyle \dfrac {b c - a d} {b d}$ by rearrangement

This shows that $y$ is rational, which is a contradiction.

By Proof by Contradiction, $z$ is irrational.

$\blacksquare$