" $$ a + br =...= (pn + qmr) / qn $$
...
Since r is irrational, we know that both the numerator and the denominator cannot be rational numbers.
"
I think your conclusion is illogical / not deductive.
Your assumption was:
"Assume that if a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is rational."
The numerator, $pn + qmr, $ is in the form $c+dr$ where $c$ and $d$ are rational and $r$ is irrational. How did you deduce that the numerator, $pn + qmr, $ is irrational, without assuming the thing you're trying to prove?
In fact, your assumption implies that the numerator $is$ rational.
You were on the right track in your proof until this part:
$a + br = p/q + (m/n)r/1$
I think you got "caught up in the maths" and forgot about the logical reasoning of the proof.
I would slightly modify the first line of your proof, which I assume you intended to be a proof by contradiction:
Proof: Assume that if a and b are rational numbers, b ≠ 0, and r is an irrational number, $and \ that \ $ a + br is rational. (Now your goal is to prove that some sort of contradiction will arise.)
By the definition of rational, we can substitute a and b with fractions where p, q, m, n are particular but arbitrary integers. (this bit is fine)
Your assumption assumed $a+br \ $ is rational, so now you should write:
$a + br = p/q$
and take it from there. Remember your goal now is to get a contradiction based on the fact that r is irrational and the rest of the numbers are rational.
Best Answer
There is very little left to do. You almost finished the problem.
Assume without loss of generality that $n$ is nonnegative. $n$ can't be $0$ because you can't divide by zero, so $n>0$. Therefore $q^n$ is an integer multiple of $q$. Now $p^m$ is an integer multiple of $q$, which is impossible.