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Proof of (a^n)^m=a^(nm)

  1. Jun 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove (a[itex]^{n}[/itex])[itex]^{m}[/itex]=a[itex]^{nm}[/itex]

    2. Relevant equations

    Proof by induction
    a[itex]^{n}[/itex]*a=a[itex]^{n+1}[/itex]
    a[itex]^{n}[/itex]*a[itex]^{m}[/itex]=a[itex]^{nm}[/itex]

    3. The attempt at a solution

    Let a and n be fixed. I will induct on m.

    Suppose m=1. Then a[itex]^{(n)(m)}[/itex]=a[itex]^{n(1)}[/itex]=(a[itex]^{n}[/itex])[itex]^{1}[/itex]

    Now assume the hypothesis is true for any integer m in P. I will show this is true for m+1.

    a[itex]^{n(m+1)}[/itex]=(a[itex]^{n}[/itex])[itex]^{m+1}[/itex]

    Thus the hypothesis is true for m+1.

    Is this proof sufficient? I am once again struck by a problem that seems almost too simple to be proved.
     
  2. jcsd
  3. Jun 17, 2013 #2

    Office_Shredder

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    Your two equations in section 2 of your post seem fairly irrelevant to the problem at hand.

    Your inductive step seems to have zero content. Why is
    [tex] a^{n(m+1)} = \left( a^{n} \right)^{m+1}[/tex]
    true at all? The whole point is you're supposed to prove that using your inductive hypothesis. You need to use the fact that [tex] a^{nm} = \left(a^n\right)^m [/tex]
    at some point
     
  4. Jun 17, 2013 #3
    Your induction will still only prove that the relationship is true for integer m and n if you succeed. Clearly the relationship is more general than that.
     
  5. Jun 17, 2013 #4
    an(m+1)=an*(m+1)=(by the theorem) (an)(m+1)

    How else could I move from the left side of the equality to the right? If someone could give me a little direction I would love to solve it myself.
     
  6. Jun 17, 2013 #5
    Luckily I only need to prove it for integers right now. I don't think I could handle anything more atm.
     
  7. Jun 17, 2013 #6
    You need to show explicitly that the proposition is true for m+1 making use of the assumption that the proposition holds true for m, which you are clearly lacking here.
    Start with [itex](a^{n})^{(m+1)}[/itex], rearrange / rewrite it until you manage to obtain a term [itex](a^{n})^{m}[/itex] somewhere. Then you can replace that term with [itex]a^{nm}[/itex]
     
  8. Jun 17, 2013 #7

    Office_Shredder

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    Fightfish gives a good suggestion... alternatively from an*(m+1) do literally the only algebra you are allowed to do and re-write it as anm+n
     
  9. Jun 17, 2013 #8
    Here's the proof for m=1

    (an)m=(an)1=(By the definition)an

    Now for m+1
    Because xn+m=xn*xm

    Let x=(an)

    (x)(m+1)=(x)m*(x)1 and by the hypothesis and the case m=1 I have already proved and replacing x with (an)
    (x)m*(x)1=(an)m*(an)1=anm*an=anm+n=an(m+1)

    I'm pretty sure this proves it!? IM SO EXCITED THIS IS AWESOME HAHA!
     
  10. Jun 17, 2013 #9

    Office_Shredder

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    Yeah that looks good to me
     
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