Results 1 to 4 of 4

Math Help - Understanding proof by contradiction

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    5

    Post Understanding proof by contradiction

    Hello guys,
    I have problems understanding proofs by contradiction.

    This is one of my exercises :

    Give a proof by contradiction that if 2n^9 + 5n^7 - 6n^4 - n^2 - 27 = 0 where n is an integer, then n is a positive integer (n > 0)

    Obviously, I have to assume that n is negative integer. What I don't know is how to proceed with the contradiction. I try to substitute n in the equation with different positive and negative integers, but it is never equal to zero

    Thank you in advance for your answers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by cyb3r View Post
    Give a proof by contradiction that if 2n^9 + 5n^7 - 6n^4 - n^2 - 27 = 0 where n is an integer, then n is a positive integer (n > 0)

    Obviously, I have to assume that n is negative integer. What I don't know is how to proceed with the contradiction. I try to substitute n in the equation with different positive and negative integers, but it is never equal to zero
    Let's assume n\leq 0.

    • Is 2n^9 positive ( >0) or non positive ?
    • Is 5n^7 positive or non positive ?
    • Is -6n^4 positive or non positive ?
    • Is -n^2 positive or non positive ?
    • Is 2n^9 + 5n^7 - 6n^4 - n^2 positive or non positive ?


    Do you see why 2n^9 + 5n^7 - 6n^4 - n^2-27=0 isn't possible if n\leq0 ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    5
    So 2n^9, 5n^7, -6n^4, -n^2 are going to be all non-positive, and the whole equation is going to be non-positive, and Zero is not a non-positive number and we make the conclusion that n is positive ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by cyb3r View Post
    So 2n^9, 5n^7, -6n^4, -n^2 are going to be all non-positive
    Yes.
    and the whole equation is going to be non-positive
    Yes, 2n^9 + 5n^7 - 6n^4 - n^2 \leq  0 so 2n^9 + 5n^7 - 6n^4 - n^2 {\color{blue}- 27} < 0. In other words, 2n^9 + 5n^7 - 6n^4 - n^2 {\color{blue}- 27} is negative.
    and Zero is not a non-positive number
    Yes but what's interesting here is that 0 is non negative. As 2n^9 + 5n^7 - 6n^4 - n^2- 27 is negative, we've found the contradiction we were looking for.
    and we make the conclusion that n is positive ?
    Yes we do.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Proof by contradiction
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: February 28th 2011, 10:06 AM
  2. Help with Proof by Contradiction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 27th 2010, 10:33 PM
  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 by Contradiction
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 8th 2007, 05:51 AM
  5. proof by contradiction :)
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 26th 2007, 11:47 AM

Search Tags


/mathhelpforum @mathhelpforum