Is this a valid proof of "a, b are rational, b ≠ 0, r is irrational, then a + br is irrational" Theorem: If a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is irrational.
Proof: Assume that if a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is rational.  
By the definition of rational, we can substitute a and b with fractions where p, q, m, n are particular but arbitrary integers.
    a = p/q   b = m/n

    a + br = p/q + (m/n)r/1

           = p/q + mr/n

           = (pn + qmr) / qn

Since r is irrational, we know that both the numerator and the denominator cannot be rational numbers, which implies a + br is irrational, which contradicts the fact that a + br is rational.  This contradiction shows the supposition is false, therefore the theorem is true.
 A: " $$ a + br =...= (pn + qmr) / qn $$
...
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.
A: One possibility is,since $b\neq 0$ and $b$ is rational we show
first $br$ must be irrational.
This is done by Contradiction
Suppose $br$ is rational ,then since $\frac{1}{b}$ is rational, the product $\frac{1}{b}(br)=r$ is rational.(Products of rational numbers are rational)
But this is a contradiction because by
Hypothesis $r$  is irrational.
Hence $br$ has to be irrational
(First step proved)
Remains to show that since br is irrational as we have shown already that also the sum
a+br is irrational.
If a is zero we have nothing more to show.
Otherwise we argue again by Contradiction and assume $a+br$ to be rational
Since $-a$ is rational(because a is rational) it follows that $-a+(a+br)=br$ is rational
(sums of rational numbers are rational),which contradicts the irrationality of $br$ shown in the first step.
Hence $a+br$ must be irrational
(Second step proved)
q.e.d
