# I need someone to point the way... :-)

• Nov 29th 2007, 06:54 AM
Polyxendi
I need someone to point the way... :-)
Hey, guys!

I'm sure it's been done before, so if anyone could point me to the past thread or post it here, that'd be great!

My brother needs to prove by induction that

aˆm • aˆn = aˆ(n+m)

Thank you so much! Glad to be part of this forum :-)
• Nov 29th 2007, 07:27 AM
ThePerfectHacker
Quote:

Originally Posted by Polyxendi
Hey, guys!

I'm sure it's been done before, so if anyone could point me to the past thread or post it here, that'd be great!

My brother needs to prove by induction that

aˆm • aˆn = aˆ(n+m)

Thank you so much! Glad to be part of this forum :-)

Let \$\displaystyle m\$ be any positive integer. We shall prove, \$\displaystyle a^ma^n=a^{n+m}\$. If \$\displaystyle n=1\$ then we got \$\displaystyle a^ma=a^{m+1}\$ which is true. Now complete the inductive step. Can you do that?
• Nov 29th 2007, 09:08 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Let \$\displaystyle m\$ be any positive integer. We shall prove, \$\displaystyle a^ma^n=a^{n+m}\$. If \$\displaystyle n=1\$ then we got \$\displaystyle a^ma=a^{m+1}\$ which is true. Now complete the inductive step. Can you do that?

Actually you could start the base case as n = 0 (provided that \$\displaystyle a \neq 0\$, which is a condition you are likely using anyway.)

-Dan