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? 
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-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. $QED$.

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.  
