# Proof verification: Let $a$ be an irrational number and $r$ be a nonzero rational number. If $s$ is a rational number then $ar$ + $s$ is irrational

I have to prove the following:

Prove: Let $a$ be an irrational number and $r$ be a nonzero rational number. If $s$ is a rational number then $ar$ + $s$ is irrational

So, I decided to do a proof by contradiction and I was wondering if someone can check it for me?

Suppose $ar + s$ is rational then it can be expressed a ratio of integers, $\frac{p}{q}$. This implies

\begin{align} ar + s &= \frac{p}{q} \\ ar &= \frac{p}{q} - s \\ ar &= \frac{p-s}{q} \quad (*) \\ \end{align}

Contradiction $(*)$. We know that an irrational times a rational is irrational and therefore, it can't be expressed as a ratio of integers but here we are claiming that it can be expressed as such. Therefore, our original statement must be false. Therefore, $ar+s$ must be irrational. $QED$

Does this make sense?

Thanks!

Below is the proof rewritten for completeness.

Proof by contradiction Suppose $ar + s$ is rational then it can be expressed a ratio of integers, $\frac{p}{q}$. This implies

\begin{align} ar + s &= \frac{p}{q} \\ ar &= \frac{p}{q} - s \\ ar &= \frac{p-sq}{q} \\ a &= \frac{p-sq}{rq} \quad (*) \end{align}

Contradiction $(*)$. We know that $a$ is irrational which implies that it cannot be written as a ratio of integers, in other words $a = \frac{p-sq}{rq} \in \mathbb{Q}$ which is false. Therefore, our original statement must be false also. Therefore, $ar + s$ is irrational. $QED$.

• Why not add the last and really punch line?: $$ar=\frac{p-s}q\implies a=\frac{p-s}{rq}\in\Bbb Q\;\ldots\text{contradiction}$$ – DonAntonio Apr 9 '14 at 13:00
• You should include the condition : where $p$ and $q$ have no common factors other than 1 – Shubham Apr 9 '14 at 13:03
• @Shubham: Why bother? – TonyK Apr 9 '14 at 13:33

Looks pretty good. You have an error on the starred line ($s$ should be multiplied by $q$) and you will also want to divide the equation by $r$ to make the conclusion explicit.
• Ha! What a stupid/silly mistake on $(*)$. Thanks for your quick answer. – Jeel Shah Apr 9 '14 at 13:04