Prove that if $a+b$ is an irrational number, then at least one of $a$ or $b$ is irrational. I came across this question in a book. I tried proving the condition as the following: Suppose that a and b are rational. Clearly the sum of $a$ and $b$ is rational, which contradicts the condition, which is that $a+b$ is irrational. Therefore at least one of a or b is irrational.
I have a feeling that something is amiss there. I feel like using contrapositive proof here could be better but I'm not sure because I'm new to the world of proofs.
 A: I want to address your "I don't see how the contrapositive works here" comment.
Let $\mathbb{I} = \mathbb{R} \setminus \mathbb{Q}$ (the set of irrational numbers).
You want to show that
$$ a+b \in \mathbb{I} \implies a \in \mathbb{I} \vee b \in \mathbb{I}$$
Before switching to the contrapositive, note that for $a \in \mathbb{R}$
$$ \lnot (a \in \mathbb{I}) \Leftrightarrow a \in (\mathbb{R} \setminus \mathbb{I}) \Leftrightarrow a \in \mathbb{Q}$$
Now, the contrapositive becomes
$$ \lnot (a \in \mathbb{I} \vee b \in \mathbb{I}) \implies \lnot (a+b \in \mathbb{I})$$
which, in light of the observation above, is
$$ a \in \mathbb{Q} \land b \in \mathbb{Q} \implies a+b \in \mathbb{Q}$$
which is a defining property of $\mathbb{Q}$.
Remember also that $\lnot (P \vee Q) = (\lnot P) \land (\lnot Q)$.
A: The statement you're trying to prove is $\forall a,b\, (a+b\notin \Bbb{Q} \implies a\notin \Bbb{Q} \text{ or } b \notin \Bbb{Q})$. This is simply the symbolic translation of the statement "for every $a,b$, if $a+b$ is irrational then atleast one of $a$ or $b$ is irrational".
Here, the statement $X$ is "$a+b\notin \Bbb{Q}$", and the statement $Y$ is "$a\notin \Bbb{Q} \text{ or } b \notin \Bbb{Q}$". So, the contrapositive of "for every $a,b$ ($X \implies Y$)" is "for every $a,b$ $(\neg Y \implies \neg X)$", which in this case is:

For every $a,b$ we have ($a\in \Bbb{Q}$ and $b\in \Bbb{Q} \implies a+b \in \Bbb{Q}$)

and this is what you argued.
