0

If $m$ and $n$ are odd, then $m+n$ is even.

For direct proof I said $m=2a+1$ and $n=2b+1$ so $m+n=(2a+1)(2b+1) =2a+2b+2 =2(a+b+1)$ which is even I think.

For contradiction, I think it begins as "If $m$ and $n$ are odd, then $m+n$ is odd." but I don't know what to do after that. I ended up basically doing the same as the direct and said that the contradiction is false therefore the original statement is true.

For contrapositive, I think it might start as "If $m+n$ is odd, then $m$ and $n$ are even." but again I'm unsure if it's correct.

I apologise if I've gotten everything completely wrong but online school is a b****.

Souza
  • 889
Belle
  • 1

1 Answers1

1

Note that:

Statements: If $m$ and $n$ are odd, so $m+n$ is even. Now, you have the form $$H_{1} \wedge H_{2} \implies C $$ where:
$H_{1}:$ $m$ is odd.
$H_{2}:$ $n$ is odd.
$C:$ $m+n$ is even.

Note also that the contrapositive is of the form $$[\neg C \implies \neg(H_{1} \wedge H_{2})] \equiv [ \neg C \implies \neg H_{1} \vee \neg H_{2}]$$

In words this is: "If $m+n$ is even, so $m$ o $n$ are even".

Now, I think you have a typo in you direct proof, note that:
If $m$ is odd, so $m=2b+1$, $b \in \mathbb{Z}$ and if $n$ is odd, so $n=2c+1$, $c\in \mathbb{Z}$. Now, $$m+n=(2b+1)+(2c+1)=2(b+c)+2=2(b+c+1)$$ and since that $k:=b+c+1\in \mathbb{Z}$, so $m+n=2k, k \in \mathbb{Z}$ and therefore $m+n$ is even.