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

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

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

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

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

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

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

    2n^2 + 2=2(2k)^2+2

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

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

    My professor says that my proof is wrong, but I don't see how.
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  2. #2
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    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. 2n^2+ 2= 2(n^2+ 1) is never odd.

    Your title, however, says 2n^2+ n, not 2n^2+ 2. Perhaps if you tried proving that you will do better.
    Last edited by HallsofIvy; October 2nd 2013 at 05:13 PM.
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    Re: Prove that 2n^2 + n is odd if and only if cos(npi/2) is even

    Quote Originally Posted by HallsofIvy View Post
    I would suggest you go back and reread the problem. 2n^2+ 2= 2(n^2+ 1) is never odd.

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

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

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

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

    2n^2 + n=2(2k)^2+2k

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

    Therefore, 2n^2 + n is even and the implication is true.
    Last edited by MadSoulz; October 2nd 2013 at 05:31 PM.
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