Proof: If n + 5 is odd, then 3n + 2 is even | Simple Direct Proof

[mohabitar]

Let n be an integer. Prove that if n + 5 is odd, then 3n + 2 is even.

So the instructions say to use a direct proof. I couldn't figure that method out, so I used a controposition proof and that seemed to work ok. Here are my contraposition steps:

Assume 3n+2 is odd
Def of odd: n=2k+1
n+5=2k+1+5 = 2k+6 = 2(k+3)
n+5 is even (multiple of 2)
since negation of conclusion implies hypothesis is false, original statement is true.

Im pretty sure that's correct, but how could this be done using a direct proof?

 

[Office_Shredder]

3n+2 is odd, but you assumed that n itself was odd when you plugged n=2k+1 into n+5

 

[JonF]

Office Shredder is right, your proof if flawed. 3n+2 is odd means for some natural number m, 3n + 2 = 2m + 1

With that in mind, let’s try a direct proof.
By definition what does n + 5 being odd mean? It means that: n + 5 = 2m + 1 for some natural m.
Can we solve for n? If we do, can we plug n into 3n + 2? Can we simplify and pull out a factor of 2? If we can pull out a factor of two, what does that mean?

 

[mohabitar]

Ahh ok I see! So n=2k-4 and 3n+2=2(k-5) which means 3n+2 is even...thanks a lot!