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Induction with inequalities

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove: 2n + 1 < 2n , with n >= 3


    2. Relevant equations



    3. The attempt at a solution

    2 (3) + 1 = 7 and 23 = 8.
    So 2 (3) + 1 < 23.
    Thus the inequality holds with n = 3:
    Suppose the inequality holds with n = k
    Then 2k+ 1 < 2k:
    So 2k + 1 + 2 < 2k + 2
    2k + 3 < 2k + 2k
    2k + 3 < 2(2k)
    2 (k + 1) + 1 < 2(k+1):
    So, the inequality holds with n = k + 1:






    Hi guys,

    some of the transition on the RHS, im blurred. Like the above there are 2 parts i dont understand,

    So 2k + 1 + 2 < 2k + 2
    2k + 3 < 2k + 2k

    on RHS, how to get from 2k+2 to 2k+2k. Arent when we do a change on LHS(e.g +2), is should be equal to RHS(e.g+2)? sry my understanding for induction is weak, can someone help elaborate the solution...

    Thanks very much.
     
    Last edited: Sep 6, 2009
  2. jcsd
  3. Sep 6, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    It's really simple.

    You already have that the LHS is less than the RHS

    2k + 3 < 2k + 2

    If you replace the 2 on the RHS with 2k, a number greater than 2 (k > 1 by assumption), the inequality still holds, therefore

    2k + 3 < 2k + 2k.
     
  4. Sep 6, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    2< 2k for any k> 1 and so, adding 2k to both sides, 2k+ 2< 2k+ 2k so 2k+ 2< 2(2k)= 2k+1.
     
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