I want to prove this constructively (ie, without using contradiction), and I tried to prove the contrapositive, that if a + b is rational then it can not be the case that WOLOG a is rational and b is irrational, but that proof also seems to lend itself to contradiction.
For an example of what I am looking for, consider this constructive proof that $\sqrt{2}$ is irrational under the header "Constructive Proof" in this wikipedia article: $\sqrt{2}$ is irrational
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I also thought of another proof. We want to prove the contrapositive, that if $a + b$ is rational then $a$ is irrational or $b$ is rational. This is equivalent to the statement that $a$ is rational implies $b$ is rational, which is true since $b = (a + b) - a$, and by assumption both $a + b$ and $a$ are rational and since the rationals are closed under subtraction we have that $b$ is rational.
There is a simple proof adapted from the example you gave of the irrationality of $\sqrt 2$.
Assume $a$ is a rational and $b$ an irrational. Then for any integers $p,q$ with $q\neq 0$ you have $$|(a+b)-\frac pq|=|b-\left(\frac pq-a\right)|$$ which is positive since $\frac pq-a$ is a rational and $b$ is not.
Since the distance between any rational and $a+b$ is positive, $a+b$ is not a rational.