Proof: sum of even and odd integer is odd

Question

Statement: Sum of even and odd integer is odd

$$ \forall(a,b) \in \mathbb{Z} : a \text{ mod } 2 = 0 \wedge b \text{ mod } 2 \neq 0 \implies a + b \text{ mod } 2 \neq 0 $$

Proof:

$$ a \text{ mod } 2 = 0 \implies \exists n \in \mathbb{Z}: a = 2n $$ $$ b \text{ mod } 2 \neq 0 \implies \exists k \in \mathbb{Z}: b = 2k +1 $$

$$ \implies (2n+2k+1) \text{ mod } 2 \neq 0 \implies \exists q \in \mathbb{Z}: 2q+1 = (2n+2k+1)$$

$$ \implies q = n + k $$

Not quite sure if this is correct or not. To me it would seem it's correct but could i have comment on this?

Answer 1

No, it is not correct, because of those $\implies$ signs. You wish to prove that$$a+b\not\equiv0\pmod2,\tag1$$which means that $2n+2k+1\not\equiv0\pmod2$. Therefore, you should have used a $\iff$ sign here. And you should also use that sign after that, since you wish to prove $(1)$, not to extract conclusions from it.

Answer 2

It is not correct for the use of the implication mark. You have to write in a different way the last part. Below is my suggestion.

---

We have:

• $a=0\pmod{2} \Rightarrow \exists n\in\mathbb{Z}\,:\, a=2n$
• $b\neq 0 \pmod{2} \Rightarrow \exists k\in\mathbb{Z} \,:\, b=2k+1$

Both the above imply $a+b = 2n+2k+1 = 2(n+k)+1$ and this is $\neq 0\pmod{2}$ because $\exists q\in\mathbb{Z}\,:\,a+b=2q+1$, and in particular $q=n+k$.