Vacuous proof of $a+b+c=0$ _"Prove that if $a$, $b$, and $c$ are odd integers such that $a+b+c=0$, then $abc < 0$"_ My attempt: I figured that since you can't make an odd by adding two other odd numbers there was no way to get to $0$ the way it shows so I wrote: $a+b \ne$ odd, therefore $a+b+c \ne 0$, so the implication is true (since the condition is always false) I'm not entirely sure how to do this, the textbook didn't give that many examples for trivial and vacuous proofs. What is the strategy for questions regarding odd/even proofs?

The standard way to phrase an argument like this is to use modular arithmetic - as used in the comments, if something is even, we write $a\equiv 0 \mod 2$. Because these statements are so 'obvious' e.g. the sum of three odd numbers is odd, you typically wouldn't bother proving it, you'd just take it as given.

However, if you're working at a very basic level learning some formal logic, it's worth seeing a clear proof. The general method is to _define_ an even integer $a$ as an integer that can can be written as $a=2k$ for some $k\in\mathbb{Z}$. Similarly, an odd number is any $b$ that can be written as $b=2l+1$ for some $l\in\mathbb{Z}$.

So then you can show if $a, b, c$ are odd, then there are three integers $k, m, n$ such that $a=2k+1$, etc. Then $a+b+c=2k+1+2m+1+2n+1=2(k+m+n)+3=2(k+m+n+1)+1$, so this is odd, while $0$ is even.

Then your conclusion works perfectly; the condition is always false, so the implication is true.