Does This Proof Prove the Equation (a^{n})^{m}=a^{nm}?

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Homework Statement



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

Homework 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]

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.
 
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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
 
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.
 
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.
 
Fightfish said:
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.

Luckily I only need to prove it for integers right now. I don't think I could handle anything more atm.
 
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]
 
a_skier said:
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.

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
 
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!