Proof by contradition

[UCdavisEcon]

Not sure where to put this question...but here goes.

Prove by contradiction. If k is an even number, then k+2 is an even number. I know from definitions that k is an even number if k=2n, and k is an odd number if k=2n+1 for any integer n.

So, the layout of the proof is to be: Definition, Claim, and proof.

Is the correct definition to use "k is an even number if k=2n" or the odd definition?

Claim: If k is an even number, then k+2 is an even number
Proof: Assume if k is an even number, then k+2 is an odd number (is this correct?)
k=2m
k+2=2m+2
k+2=2(m+1)
k+2 is not an odd number. Therefore by contradiction, if k is an even number k+2 is an even number.

Please critique my work and give me an advice.

 

[stapel]

Prove by contradiction. If k is an even number, then k+2 is an even number. I know from definitions that k is an even number if k=2n, and k is an odd number if k=2n+1 for any integer n.

So, the layout of the proof is to be: Definition, Claim, and proof.

Is the correct definition to use "k is an even number if k=2n" or the odd definition?

Claim: If k is an even number, then k+2 is an even number
Proof: Assume if k is an even number, then k+2 is an odd number (is this correct?)

Yes, this is likely how you're expected to start this contradiction proof. You're supposed to prove that, given k even, then k + 2 is even. To prove by contradiction, you assume the given (that k is even), assume the contrary of what you want (that k + 2 is odd), and derive a contradition.

k=2m
k+2=2m+2
k+2=2(m+1)
k+2 is not an odd number. Therefore by contradiction, if k is an even number k+2 is an even number.

Please critique my work and give me an advice.

Looks fine to me.