# Question #c9e1c

May 22, 2015

OK, because the problem statement is ambiguous, I'd rather try to prove a more general and--presumably--more useful result.

I shall prove that, if $\frac{a}{b}$ is rational and $x$ is irrational, then their sum $\frac{a}{b} + x$ is irrational. ($a$ and $b$ are integers and $b \ne 0$)

I'll use a proof by contradiction: let's assume that the sum of the two numbers is rational. This means that the sum can be written as a ratio of two integers $c$ and $d$, $d \ne 0$:

$\frac{a}{b} + x = \frac{c}{d} \iff x = \frac{c}{d} - \frac{a}{b} \iff x = \frac{b c - a d}{b d}$

Note that, because $a , b , c$ and $d$ are integers, then $b d$ is an integer and $b c - a d$ is also an integer. Therefore, $x$ is the ratio of two integers, i.e. $x$ is rational, which contradicts the hypothesis that $x$ is irrational.

Thus, the assumption that $\frac{a}{b} + x$ is a rational number led us to a contradiction. Hence, its opposite must be true, that is $\frac{a}{b} + x$ is an irrational number.

This kind of reasoning can be applied to any particular cases.