1. ## proofs question

Could someone tell me if I proved this correctly?

prove that for all integers m and n, if m and n are odd, then m + n is even.

p: m and n is odd
q: m+n is even(conclusion)

There exists an integer, k[1], such that m = 2 k[1] + 1
There exist an integer, k[2], such that n = 2 k[2] + 1

m + n is even

m + n = k[1]+1 + 2 k[2]+1 = 2(k[1]+k[2])+2

2. Roughly speaking, yes.

for all integers m and n, if m and n are odd, then m + n is even.
A proof of a statement that starts with "for all", unless it is a proof by induction, must start with words like "fix arbitrary", "consider", "let," etc. In this case, you can say, "Fix arbitrary m and n" or "Let m and n be some integers." Otherwise, when you are talking about m and n later, it is not clear what those are.

Further, a proof of a statement that starts with "if" must start with "assume." In this case, you say, "... and assume that m and n are odd."

p: m and n is odd
q: m+n is even(conclusion)
This is not necessary since you are not using the letters p and q later. In general, one should not just write a proposition as a part of a proof but rather explain its status: whether the proposition is assumed, whether it's the next goal to prove, etc.

m + n is even
To show that m + n is even, consider...

= 2(k[1]+k[2])+2
= 2(k[1] + k[2] + 1); therefore, this number is even.