# Thread: Proof involving even integers...

1. ## Proof involving even integers...

Question: Prove than an n ∈ ℤ is even if and only if n + 6 is even. Show the if part and the only if part.

My attempt...

Only if part

Prove that an integer n is even only if n + 6 is even.

Assume n is even; n = 2k.

Now, n + 6 is equivalent to 2k + 6.

2(k + 3) where k + 3 is obviously an integer.

∴ Integer n is even only if n + 6 is even.

If part

Prove that if n + 6 is even; n is an even integer.

Assume n + 6 is even; n + 6 = 2k.

n = 2k - 6.

n = 2(k - 3) where k - 3 is obviously an integer.

∴ If n + 6 is even; integer n is even.

Have I made any mistakes? I have properly read the "if" and "only if" parts out of the question? I want to be sure as I'm new to proofs.

2. ## Re: Proof involving even integers...

Originally Posted by Induction
[FONT="]Question: Prove than an n ∈ ℤ is even if and only if n + 6 is even. Show the if part and the only if part.

My attempt...

Only if part

Prove that an integer n is even only if n + 6 is even.

Assume n is even; n = 2k.

Now, n + 6 is equivalent to 2k + 6.

2(k + 3) where k + 3 is obviously an integer.

[/FONT]
∴ Integer n is even only if n + 6 i[FONT="]s even.

If part

Prove that if n + 6 is even; n is an even integer.

Assume n + 6 is even; n + 6 = 2k.

n = 2k - 6.

n = 2(k - 3) where k - 3 is obviously an integer.

[/FONT]
∴ If n + 6 is even; integer n is even.

Have I made any mistakes? I have properly read the "if" and "only if" parts out of the question? I want to be sure as I'm new to proofs.

Those are both correct. Well done!