Results 1 to 9 of 9

Math Help - Proof n^2 + n is even

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    4

    Proof n^2 + n is even

    Hi!

    So Im not allowed to use induction on this one.
    How do i prove that n^2 + n is even when n is natural number?

    Not a clue on this one. Would you mind to help a bit?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,947
    Thanks
    6
    Quote Originally Posted by veturi View Post
    Hi!

    So Im not allowed to use induction on this one.
    How do i prove that n^2 + n is even when n is natural number?

    Not a clue on this one. Would you mind to help a bit?
    n^2 + n = n(n+1). Therefore one of the factors of n^2 + n is certain to be even. Therefore ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    4
    oh my. so easy. Thanks a lot! I'm a bit ashamed tho...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2009
    Posts
    36
    A number can be represent in two forms:

    n=2k with k\in\mathbb{Z}
    or
    n=2k+1 with k\in\mathbb{Z}

    First case we have:

    (2k)^2+2k=4k^2+2k=2(2k^2+k)

    so, 2|(n^2+n) for all even n.

    Second case:

    (2k+1)^2+(2k+1)=4k^2+4k+2=2(2k^2+2k+1)

    so, 2|(n^2+n) for all odd n.

    thus, 2|(n^2+n) \ \forall n\in\mathbb{Z}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    553
    Thanks
    1
    Quote Originally Posted by veturi View Post
    Hi!

    So Im not allowed to use induction on this one.
    How do i prove that n^2 + n is even when n is natural number?

    Not a clue on this one. Would you mind to help a bit?
    Hello : There is the solution
    I'have two cases:
    1 case : n \equiv 0\left[ {\bmod 2} \right]

    \left\{ \begin{array}{l}<br />
n \equiv 0\left[ {\bmod 2} \right] \\ <br />
n^2 \equiv 0\left[ {\bmod 2} \right] \\ <br />
\end{array} \right. \Rightarrow n^2 + n \equiv 0\left[ {\bmod 2} \right]
    2 case: n \equiv 1\left[ {\bmod 2} \right]<br />

    \left\{ \begin{array}{l}<br />
n \equiv 1\left[ {\bmod 2} \right] \\ <br />
n^2 \equiv 1\left[ {\bmod 2} \right] \\ <br />
\end{array} \right. \Rightarrow n^2 + n \equiv 0\left[ {\bmod 2} \right]<br />
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by veturi View Post
    Hi!

    So Im not allowed to use induction on this one.
    How do i prove that n^2 + n is even when n is natural number?

    Not a clue on this one. Would you mind to help a bit?
    By Contradiction:

    If n is even there is nothing to prove, so assume n is odd and n^2+n is also odd.

    As n is odd it can be written n=2k+1 for some natural k, then:

    n^2+n=4k^2+4k+1+2k+1= ...

    Hence ... a contradiction.

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    553
    Thanks
    1
    Quote Originally Posted by CaptainBlack View Post
    By Contradiction:

    If n is even there is nothing to prove, so assume n is odd and n^2+n is also odd.

    As n is odd it can be written n=2k+1 for some natural k, then:

    n^2+n=4k^2+4k+1+2k+1= ...

    Hence ... a contradiction.

    CB
    Hello : you have any contradiction , because Iícan written :
    \begin{array}{l}<br />
n^2 + n = 4k^2 + 4k + 1 + 2k + 1 \\ <br />
n^2 + n = 2\left( {2k^2 + 2k + 1 + k} \right) \\ <br />
n^2 + n = 2\alpha \\ <br />
\alpha = 2k^2 + 2k + 1 + k \\ <br />
\end{array}<br />
    Conclusion : even number
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dhiab View Post
    Hello : you have any contradiction , because Iícan written :
    \begin{array}{l}<br />
n^2 + n = 4k^2 + 4k + 1 + 2k + 1 \\ <br />
n^2 + n = 2\left( {2k^2 + 2k + 1 + k} \right) \\ <br />
n^2 + n = 2\alpha \\ <br />
\alpha = 2k^2 + 2k + 1 + k \\ <br />
\end{array}<br />
    Conclusion : even number
    It is a contradiction, I assume it odd and find it even hence the assumption that n odd and n^2+n is also odd fails, so if n is odd n^2+n is not odd.

    CB.

    CB
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member Renji Rodrigo's Avatar
    Joined
    Sep 2009
    From
    Rio de janeiro
    Posts
    38
    other proof
    we know that
    1+2+3+...+n= n(n+1)/2

    so 2|n(n+1), n(n+1) is even
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 19th 2010, 10:50 AM
  2. Replies: 0
    Last Post: June 29th 2010, 08:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: April 14th 2008, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum