Results 1 to 2 of 2

Math Help - [SOLVED] Proof question 3

  1. #1
    Super Member
    Joined
    Feb 2008
    Posts
    535

    [SOLVED] Proof question 3

    Suppose n,B e N, where n > 0. SHow that if B^n = 0, then B = 0.

    Im trying to do a proof by contradicion by assuming that B != 0, but I don't know where to go from there. Any advice...Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,974
    Thanks
    1121
    Quote Originally Posted by jzellt View Post
    Suppose n,B e N, where n > 0. SHow that if B^n = 0, then B = 0.

    Im trying to do a proof by contradicion by assuming that B != 0, but I don't know where to go from there. Any advice...Thanks.
    I would recommend a proof by induction on n combined with contradiction to prove the induction step.

    If n= 1 B^1= 0 gives immediately B= 0. Suppose that, for some k, B^k= 0 implies that B= 0, then if B^(k+1)= 0 and B is NOT 0, then we can divide both sides by B to get B^k= 0, which implies B= 0, a contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 30th 2010, 03:53 PM
  2. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  3. [SOLVED] Proof question for eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 25th 2010, 11:41 AM
  4. [SOLVED] Please help with proof
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 2nd 2010, 11:11 AM
  5. [SOLVED] set proof help
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: July 18th 2009, 08:48 AM

Search Tags


/mathhelpforum @mathhelpforum