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Mathematical Induction

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove the theorems using mathematical induction.

    [itex]\forall n[/itex] [itex]\in N[/itex], n[itex]\geq 4[/itex] [itex]\rightarrow[/itex]n2[itex]\leq n![/itex]

    Thanks in advance!
    2. Relevant equations



    3. The attempt at a solution
    First, check the base case which is n=4.
    [itex]\Rightarrow[/itex]n=4[itex]\geq[/itex]4-True

    [itex]\Rightarrow[/itex]42[itex]\leq[/itex]4*3*2*1

    [itex]\Rightarrow[/itex]16[itex]\leq[/itex]24-True

    Therefore, P(1) is true.

    Now, check for the case n=k+1, to prove P(k)[itex]\rightarrow[/itex]P(k+1).
    Assume P(k) is true.
    [itex]\Rightarrow[/itex]k[itex]\geq[/itex]4
    [itex]\Rightarrow[/itex]k2[itex]\leq[/itex]k!

    [itex]\Rightarrow[/itex]k+1[itex]\geq[/itex]4+1 -Add 1 on both side
    [itex]\Rightarrow[/itex]k+1[itex]\geq[/itex]5
    [itex]\Rightarrow[/itex](k+1)2=k2+2k+1
    [itex]\Rightarrow[/itex](k+1)!=(k+1)k!
    ...

    And I stop here can't get further to prove P(k+1) is true.
    Any help is appreciated.
     
  2. jcsd
  3. Apr 15, 2013 #2

    HallsofIvy

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    Rather than multiply the square note that what you want to prove is that
    [itex](k+1)^2= (k+1)(k+1)\le (k+1)k![/itex] which, because k+1 is positive, is
    the same as proving that [itex]k+1\le k![/itex]. You might want to do another induction to prove that is true first.

     
  4. Apr 15, 2013 #3
    Thanks HallsofIvy,
    So I should start at proving (n+1)(n+1)≤(n+1)n! [itex]\Rightarrow[/itex] n+1≤n!
    First, Check the base case n=4:
    [itex]\Rightarrow[/itex](4)+1≤4!
    [itex]\Rightarrow[/itex]5≤24
    So, the base case is true.

    Now, Assume P(k) is true, then prove prove P(k+1):
    [itex]\Rightarrow[/itex]k+1≤k! -is true (Premise)
    Also, let n=k+1:
    [itex]\Rightarrow[/itex]k+1+1≤(k+1)!
    [itex]\Rightarrow[/itex](k+1)+1≤(k+1)k!
    [itex]\Rightarrow[/itex]1+1/(k+1)≤k! [Since k+1 is positive, so the inequality sign won't change]
    Since 1+k≤k! is true and k>1/(k+1)
    Therefore, P(k+1) is true
    P(k)[itex]\rightarrow[/itex]P(k+1)
    And prove [itex]\forall n[/itex][itex]\in N[/itex]n≥4[itex]\rightarrow[/itex]n2≤n!

    Does this look alright??
     
  5. Apr 16, 2013 #4

    HallsofIvy

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    Staff Emeritus
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    No, that follows trivially by dividing both sides by n+1. What I said was that you should show that, for all [itex]n\ge 4[/itex], [itex]n+1\le n![/itex] (which is, in fact, what you are doing).

    You cannot assert this because you do not know if it is true yet. This is what you want to prove.

    Strictly speaking what you have done is prove "if P(k+1) is true then P(k) is true". Do it the other way around: if [itex]k+1\le k![/tex] then [itex](k+1)(k+1)\le k!(k+1)= (k+1)![/itex]

     
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