Prove that 2n^2 + n is odd if and only if cos(npi/2) is even

Let n be an element of **Z**. Prove that is odd if and only if is even.

Since it's a biconditional, I'm certain there will be two parts: 1) Proving , and 2) proving

It's in the proof by contrapositive section, so I'll try to prove it that way.

1) If is odd, then is even.

Contrapositive: If is odd, then is even.

Assuming is odd, then

for some integer m.

Therefore, is even and the implication is true.

My professor says that my proof is wrong, but I don't see how.

Re: Prove that 2n^2 + n is odd if and only if cos(npi/2) is even

I would suggest you go back and reread the problem. is **never** odd.

Your title, however, says , not . Perhaps if you tried proving **that** you will do better.

Re: Prove that 2n^2 + n is odd if and only if cos(npi/2) is even

Quote:

Originally Posted by

**HallsofIvy** I would suggest you go back and reread the problem.

is

**never** odd.

Your title, however, says

, not

. Perhaps if you tried proving

**that** you will do better.

Sorry, I have no idea why I typed '2' instead of 'n'

Re: Prove that 2n^2 + n is odd if and only if cos(npi/2) is even

Correction: Let n be an element of Z. Prove that is odd if and only if is even.

1) If is odd, then is even.

Contrapositive: If is odd, then is even.

Assuming is odd, then

for some integer m.

Therefore, is even and the implication is true.