The problem defines a relation on the integers as follows:
For integers $a,b$, $a \sim b$ if $ab$ is even. In other words, $ab$ is even $ \Rightarrow a\sim b$.
To prove $\sim$ is symmetric, I need to show $a\sim b \Rightarrow b\sim a$. My original approach was to assume $a\sim b$, and from this deduce $ab$ is even. By the commutative property of multiplication, $ba$ is even as well, so $b\sim a$.
However, if I assume $a\sim b$, I'm not sure if I can deduce $ab$ is even. According to the statement, $a\sim b$ being true does not guarantee $ab$ is even, as $ab$ can be odd and the statement can still hold. Does this concern warrant me changing my original approach, or is there a way to still deduce $ab$ is even from the fact $a\sim b$?
Your approach is correct. The condition for the relation should be read as $$ a\sim b \Leftrightarrow ab \text{ is even}.$$ When defining relations, the forward implication is implicitly assumed to be there. This also happens with definitions in general. For example, a number $n\in \mathbb{Z}$ is said to be even if $n=2k$ for some $k\in \mathbb{Z}$. There is no need to use “if and only if” here, as it is assumed.