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Math Help - Induction with Inequalities help

  1. #1
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    Induction with Inequalities help

    Hello, I am having some trouble with the following proofs by induction. I understand how to get the base case and induction hypothesis but its the actual induction that's bothering me. Sorry that it looks ugly but this is my first post and I haven't been able to figure out the formatting codes yet.

    1.) 2^n >= 1/8 n >= -3
    2.) (3/2)^n >= (1+ n/2) n>=1

    for 1 tried 2^(n+1) = 2*2^n >= 2*(1/8) > 1/8
    This however does not preserve the >=, just >. This makes sense to me on an instinctual level because n>= -3 means that n+1 will always be greater than 1/8

    for 2 I tried 1+(n+1)/2 = 1 + n/2 + 1/2 <= (3/2)^n + 1/2 < (3/2)^(n+1) but again this did not preserve <= but just <.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    1. Prove something stronger: 2^n > n (for some natural k n>k)
    or: check to n=-3,-2,-1,0,1 and then you will see that: 2^n is natural number.

    2. (3/2)^n >= (1+ n/2) n>=1

    or:

    (3/2)^n >={(n+2)/2} so...
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    I'm afraid I don't see where you are coming from in both cases. I do not see how to connect 2^n >= n to 2^n >= 1/8 for n >=-3. Do you mean to check n from -3 to 0 and prove 2^n > n for n >=1? Similarly I don't see what re-expressing 1 + n/2 as (n+2)/2 gets you
    Last edited by Qbit42; July 17th 2010 at 11:00 AM.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Qbit42 View Post
    I'm afraid I don't see where you are coming from in both cases. I do not see how to connect 2^n >= n to 2^n >= 1/8 for n >=-3. Do you mean to check n from -3 to 0 and prove 2^n > n for n >=1? Similarly I don't see what re-expressing 1 + n/2 as (n+2)/2 gets you
    Read again my post!
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  5. #5
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    Quote Originally Posted by Qbit42 View Post
    Hello, I am having some trouble with the following proofs by induction. I understand how to get the base case and induction hypothesis but its the actual induction that's bothering me. Sorry that it looks ugly but this is my first post and I haven't been able to figure out the formatting codes yet.

    1.) 2^n >= 1/8 n >= -3
    2.) (3/2)^n >= (1+ n/2) n>=1

    for 1 tried 2^(n+1) = 2*2^n >= 2*(1/8) > 1/8
    This however does not preserve the >=, just >. This makes sense to me on an instinctual level because n>= -3 means that n+1 will always be greater than 1/8

    for 2 I tried 1+(n+1)/2 = 1 + n/2 + 1/2 <= (3/2)^n + 1/2 < (3/2)^(n+1) but again this did not preserve <= but just <.
    That is not a concern.
    The "equal" for question 1 occurs only when n is -3.
    The "greater than" occurs for all n above -3.
    Hence n+1 is above -3 so you need not be concerned about losing the "equal" part.

    Same for the second question.

    You can do the proof as ">" for n>-3 and so on.
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