0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $q=n+k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.