Proof by contradiction : if n or m are not divisible by 3 then the sum or the difference is also not divisible by 3

Question

I'm just trying to prove this by contradiction: $$3\nmid n \vee 3\nmid m \implies 3\nmid (n+m) \vee 3\nmid(n-m)$$ Things I know: $$n,m \in \mathbb{Z} $$ $$n|m \Leftrightarrow \exists x\in \mathbb{N}^+ : n*x=m$$ I negate the assumption: $$3|n \wedge 3|m \implies 3|(n+m) \wedge 3|(n-m)$$ That implies: $$\exists x \in \mathbb{N}^+ : 3x=(n+m)$$ I tried to divide this equation by $3$: $$x=\frac{n+m}{3}$$ I wanted to show that $x\not\in \mathbb{N}^+$ and therefore is this a contradiction to the definition. But I know this is not correct. What I am missing here?

Answer 1

Suppose, by contradiction, that $3$ divides both $n+m$ and $n-m$ (that is, negate the thesis).

Then $3$ also divides $(n+m)+(n-m)=2n$. Since $3\nmid2$, we get $3\mid n$.

Can you show that $3$ divides also $m$?

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Actually this is not a proof by contradiction, which is not necessary. The statements “$A$ implies $B$” and “not $B$ implies not $A$” are completely equivalent. In formal terms, the propositional formula $$ (A\to B)\leftrightarrow(\lnot B\to\lnot A) $$ is a tautology.