1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof by induction

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Using mathematical induction, prove the inequality
    (1+a)^n >= 1+na for all a>-1 and all n

    2. Relevant equations



    3. The attempt at a solution
    Base case
    1+a >= 1+a. the inequality holds.

    I am struggling with the inductive step.
    by working backward, i multiply both sides by (1+a) and this gives me

    (1+a)^(k+1) >= (1+ka)(1+a)

    all i can think of at this point is that 1+a >= 0, but i dont think this helps me.

    working the other way, starting with n=k+1, i find

    (1+a)(1+a)^k>= (1+ka)+a

    would is be enough to state that (1+a)> a and thus the inequality holds?

    thanks in advance
     
  2. jcsd
  3. Jun 22, 2009 #2
    (1+ka)+a=1+(k+1)a,thus the inequality holds for n=k+1
     
  4. Jun 22, 2009 #3

    lanedance

    User Avatar
    Homework Helper

    Hey Easty, i think you were close with your first step, not to sure about your 2nd,

    how about this, so as usual start by assuming the case for n is true and trying to show n+1 is also true based on this:

    (1+a)^(n+1) = (1+a)*(1+a)^n >= (1+a)*(1+na) = (1 + a + na + na^2) = 1 + (n+1)a + na^2

    So now you have
    (1+a)^(n+1) >= 1 + (n+1)a + na^2

    any ideas from here?
     
  5. Jun 22, 2009 #4
    i had gotten to that stage, but thought it led no where.
    The only term that isnt part of the original equality is na^2, since a is >-1, a^2 >= 0, so couldnt the extra term add some arbitary large postive number such that the inequality wasnt true?
     
  6. Jun 22, 2009 #5

    lanedance

    User Avatar
    Homework Helper

    so if na^2 is always >=0 can you think of another inequality to write that will help get you close to the required form?
     
  7. Jun 23, 2009 #6
    I think i've got it
    na^2 >= 0

    add 1+(n+1)a to both sides

    1+(n+1)a+na^2 >= 1+(n+1)a

    therefore

    (1+a)^(n+1) >= 1 + (n+1)a + na^2 >= 1 + (n+1)a

    (1+a)^(n+1) >=1 + (n+1)a as required

    i'm pretty sure this is correct. Thanks for all your help lanedance
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof by induction
  1. Proof by induction (Replies: 2)

  2. Proof by induction (Replies: 9)

  3. Proof by induction (Replies: 32)

  4. Induction Proof (Replies: 14)

  5. Proof by Induction (Replies: 6)

Loading...