Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By HallsofIvy

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

  1. #1
    Member
    Joined
    Sep 2013
    From
    United States
    Posts
    85
    Thanks
    3

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003

    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.
    Last edited by HallsofIvy; Oct 2nd 2013 at 04:13 PM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2013
    From
    United States
    Posts
    85
    Thanks
    3

    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. $\displaystyle 2n^2+ 2= 2(n^2+ 1)$ 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'
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2013
    From
    United States
    Posts
    85
    Thanks
    3

    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.
    Last edited by MadSoulz; Oct 2nd 2013 at 04:31 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove?
    Posted in the Geometry Forum
    Replies: 3
    Last Post: Feb 26th 2010, 09:02 AM
  2. Prove 1+1=2
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Feb 25th 2010, 02:23 PM
  3. Replies: 2
    Last Post: Aug 28th 2009, 02:59 AM
  4. Prove this!!!!
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Dec 19th 2008, 09:16 AM
  5. help me to prove this..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Nov 26th 2008, 12:05 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum