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 $\displaystyle 2n^2 + 2$ is odd if and only if $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is even.

Since it's a biconditional, I'm certain there will be two parts: 1) Proving $\displaystyle p \Rightarrow q$ , and 2) proving $\displaystyle q \Rightarrow p$

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

1) If $\displaystyle 2n^2 + 2$ is odd, then $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is even.

Contrapositive: If $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is odd, then $\displaystyle 2n^2 + 2$ is even.

Assuming $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is odd, then $\displaystyle n=2k$

$\displaystyle 2n^2 + 2=2(2k)^2+2$

$\displaystyle 8k^2+2=2(4k^2+1)=2m$ for some integer m.

Therefore, $\displaystyle 2n^2 + 2$ 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. $\displaystyle 2n^2+ 2= 2(n^2+ 1)$ is **never** odd.

Your title, however, says $\displaystyle 2n^2+ n$, not $\displaystyle 2n^2+ 2$. 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. $\displaystyle 2n^2+ 2= 2(n^2+ 1)$ is **never** odd.

Your title, however, says $\displaystyle 2n^2+ n$, not $\displaystyle 2n^2+ 2$. 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 $\displaystyle 2n^2 + n$ is odd if and only if $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is even.

1) If $\displaystyle 2n^2 + n$ is odd, then $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is even.

Contrapositive: If $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is odd, then $\displaystyle 2n^2 + n$ is even.

Assuming $\displaystyle cos\left(\frac{\pi n }{2}\right)$ is odd, then $\displaystyle n=2k$

$\displaystyle 2n^2 + n=2(2k)^2+2k$

$\displaystyle 8k^2+2k=2(4k^2+k)=2m$ for some integer m.

Therefore, $\displaystyle 2n^2 + n$ is even and the implication is true.