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Proof by induction for inequalities

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A sequence (Xn) is defined by X1=3 and Xn+1= (6Xn+1)/(2Xn+5) for all n[tex]\in[/tex] N.

    Prove by induction or otherwise that Xn-1 > 0 for all n [tex]\in[/tex] N.


    2. Relevant equations



    3. The attempt at a solution

    I'm not sure with what to do when dealing with inequalities in an induction proof. Initial i tried subing in the recursion formula when attempting the inductive step but i dont think it gets me anywhere. I'd really appreciate any guidance on where to start.

    Thanks
     
  2. jcsd
  3. Apr 18, 2009 #2

    Mark44

    Staff: Mentor

    The proposition is clearly true for n = 1 and n = 2, so suppose it's true for n = k. I.e., that xk > 1.

    For the induction step, you have to show that xk + 1 > 1.

    xk + 1 = (6xk + 1)/(2xk + 5)

    Carry out the division to get something that you can show is greater than 1. Is that enough of a start?
     
  4. Apr 18, 2009 #3
    It is easier if you write it as x_n > 1 instead of x_n - 1 >0.

    Suppose x_n > 1 [Show that x_(n+1) > 1]

    Mult both sides by 4.
    Add 2x to both sides.
    Add 1 to both sides.
    Divide both sides by the right side quantity (which is > 0, why?)

    And you should see x_(n+1) > 1.
     
  5. Apr 19, 2009 #4
    Thanks alot for the help. Its such a simple solution, no wonder i didnt get it
     
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