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Inequality Mathematical Induction

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove the Inequality for the indicated integer values of n.

    [tex]n!>2^n,n\geq4[/tex]

    2. Relevant equations

    3. The attempt at a solution

    For n=4 the formula is true because

    [tex]4!>2^4[/tex]

    Assume the n=k

    [tex]k!>2^k[/tex]

    Now I need to prove the equation for k+1

    I can multiply both sides by 2 and have

    [tex]2(k!)=2(k!)>(2)2^k=2^{k+1}[/tex]

    Is this what you would do next? I'm not quite sure what to do past this point.

    [tex]2(k!)>2^{k+1}>k+1[/tex]
     
  2. jcsd
  3. Jul 11, 2010 #2

    CompuChip

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    Why would you multiply by 2?
    It's not 2k! you want to make a statement about, but (k + 1)! = (k + 1) k!
     
  4. Jul 11, 2010 #3
    I dont know. When dealing with regular equations you would just set k=(k+1) but that's not the case here I think.

    Wouldn't you be trying to prove that [tex](k+1)!>2^{k+1}[/tex]? Why would I be making a statement about '(k + 1)! = (k + 1) k!'? Where is the 2 raised to the power of (k+1) and why is there 2 factorials when there isn't in the original problem?
     
  5. Jul 11, 2010 #4

    CompuChip

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    Sorry, you are approaching the problem from a different angle than I expected.

    Once you have 2 (k!) > 2k + 1, you only need to prove that (k + 1)! > 2 k! to be done with it.

    The hint I gave, that (k + 1)! is equal to (k + 1) times k!, is still useful.
     
  6. Jul 11, 2010 #5
    Ahh, so you could rewrite

    (k + 1)! > 2 k!

    as

    k!(k+1) > 2 k!

    Here you can see the when k > 1, k!(k+1) > 2 k! and since the original formula is only supposed to work for numbers greater than or equal to 4 its fine.

    Thanks.
     
  7. Jul 11, 2010 #6

    hunt_mat

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    We are required to show that [tex]n!>2^{n}[/tex], for the k+1 term, multiply by k+1 to obtain:
    [tex]
    (k+1)k!=(k+1)!>2^{k}(k+1)>2^{k+1}
    [/tex]
    Since k+1>2 for k>2 and we are done
     
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