Results 1 to 6 of 6

Math Help - Proof of mathematical induction inequality

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    21

    Proof of mathematical induction inequality

    Hi,

    Im having trouble trying to prove the following:

    n^2 > 2n; n 3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2010
    Posts
    9
    Make first an equation out of it and try to understand how these numbers relate
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    180

    Or you can, do this

    In order to prove it by induction u have to prove that it is true for n=3
    and then assume that it is true for some n and that it then follows for n+1
    So
    Step one:
    Is it true for 3?
    3^2>2*3......True!
    Good not we assume that it is true for n and prove that it then follows for n+1.

    n^2>2*n
    (n+1)^2>2*(n+1)
    n^2+2n+1>2n+1
    and almost done now =).
    n^2>2n by our first assumtion.
    so we can subtract that from the inequality.
    then we are left with 2n+1>1
    and since n=positive number clearly this is true...
    Hope I was clear enough can you take it from here
    ?
    Regards Henry
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    hi
    let S=\left \{ n\in \mathbb{N}\mid n\geq 3,\text{and},n^2> 2n \right \}
    Clearly 3\in S since 9> 6,now assume m\in S to prove m+1\in S we must prove (m+1)^2> 2(m+1)...
    i'll let you do that .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2010
    Posts
    21
    Thanks for the replies.

    when I tried to do it I got as far as:

    <br />
k^2 + 2k + 1 > 2k + 2<br />
    I wasn't aware I could subtract k^2 > 2k from the inequality.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by blackhug View Post
    Thanks for the replies.

    when I tried to do it I got as far as:

    <br />
k^2 + 2k + 1 > 2k + 2<br />
    I wasn't aware I could subtract k^2 > 2k from the inequality.
    In that case you can simply write

    k^2+(2k+1)>(2k+1)+1 ?

    2k+1 is common to both sides so it is redundant,
    hence we can eliminate it.

    k^2>1 ?

    This is true for k>1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof by mathematical induction.
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 23rd 2011, 09:12 PM
  2. Mathematical Induction Proof
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: August 28th 2010, 07:10 AM
  3. Mathematical Induction with an inequality
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: March 17th 2010, 02:12 PM
  4. mathematical induction for an inequality
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: February 4th 2010, 02:28 PM
  5. Proof by Mathematical induction!
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: October 30th 2008, 09:45 AM

Search Tags


/mathhelpforum @mathhelpforum