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
    16,948
    Thanks
    5
    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
    541
    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
    541
    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, 11:50 AM
  2. Replies: 0
    Last Post: June 29th 2010, 09:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 11:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: June 8th 2008, 02:20 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, 05:07 PM

Search Tags


/mathhelpforum @mathhelpforum