# Suggestions on how to go about proving a^(m+n)=a^(m)a^(n)

1. Jul 14, 2013

### alianna

Let a be a nonzero number and m and n be integers. Prove the following equality. a^(m+n)=a^(m)a^(n)

Im not really sure what direction to go in. Im not sure if I need to show for n positive and negative separately or is there an easier way.

My attempt/ideas:
When n>0: a^(m)a^(n)= (a*a*...*a)(m times) *(a*a*a*a*...*a)(n times)
=a*a*...*a(m+n times)
=a^(m+n)
When n<0: a^(m)a^(n)=a^(m)=a....a(m times)/ a...a(n times)

2. Jul 14, 2013

### micromass

Staff Emeritus
Are you familiar with induction?

3. Jul 14, 2013

### alianna

Yes we just went over it but i wasn't sure how to go about induction with m and n being integers...

4. Jul 14, 2013

### micromass

Staff Emeritus
1) Take $m=0$. Prove that $a^{m + n} = a^m a^n$. This shouldn't be a problem.

2) Assume the result holds for $m$. Prove it holds for $m+1$. So you know that $a^{m+n} = a^m a^n$. You need to prove $a^{m + n + 1} = a^{m+1} a^n$.

5. Jul 14, 2013

### alianna

So would you be able to say: Since a^(m+n)=a^(m)a^(n).
Then this implies that: a^(m+n+1)= a^(m+1+n)= a^(m+1)a^(n)?

This doesn't seem right because we are assuming what we are trying to prove.
Why are you allowed to show for m+1 but dont have to for n+1 or do you just assume since m+1 works then n+1 works?
Im still confused how this deals with the negative values of m and n.

6. Jul 14, 2013

### Zondrina

Hint :

$a^{m+n+1} = aaaa.....a$

(m+n) + 1 times.

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