Discrete math proof-verification of divisibility. Case with both truth and a counterexample

Question

If you need me to expand or write the proof more formally please let me know.

I am confused by the following question, where it seems I have found both a universal value that makes the question true, but then I have also found a counterexample. Where did I go wrong in my proof? Is the statement true or false?

Answer 1

Your negation of the statement is not correct. The original statement is

$$\forall a \ \exists b\ (3 \mid a+b)$$

Its negation is

$$\exists a\ \forall b (3 \not\mid a+b)\tag{1}$$

What you showed is

$$\forall b\ \exists a\ (3 \not\mid a+b)\tag{2}$$

Interchanging those two quantifiers is a big deal. In (2), $a$ can depend on $b$. Indeed, this is what you did to show the statement is true. But in (1), $a$ comes first and $b$ is arbitrary.

Your proof of truth of the original statement is valid. To make it clear that you are using the quantifiers correctly, you should add words to your proof. “Given $a$, let $b=-a$. Then $b+a = 0$, and $3 \mid 0$.” The use of “given” refers to the “for all” quantifier, and the use of “let” refers to “there exists.”