1. ## proof by induction

show that (a^m)*(a^n)=a^(n+m) knowing that a^0=1 and a^n=a^(n-1)a. I know it's not a hard proof but i just find myself using the property i'm trying to prove to solve the proof which is deffinatly wrong, just need someone elses input if you know what i mean

2. Originally Posted by action259
show that (a^m)*(a^n)=a^(n+m) knowing that a^0=1 and a^n=a^(n-1)a. I know it's not a hard proof but i just find myself using the property i'm trying to prove to solve the proof which is deffinatly wrong, just need someone elses input if you know what i mean
We know that
1. a^n = a^(n-1) * a ==> a^(n + 1) = a^n * a
2. a^0 = 1
as givens.

So n = 0 case:

a^m * a^0 = a^m * 1, by 2
= a^m = a^(m + 0)

Now assume that a^m * a^n = a^(m + n) for some n. We need to show that a^m * a^(n + 1) = a^(m + n + 1).

So
a^m * a^(n + 1) = a^m * a^n * a, by 1
= a^(m + n) * a, by hypothesis
Now, m + n is just some integer x.
= a^x * a = a^(x + 1), by 1
= a^(m + n + 1)

Thus a^m * a^n = a^(m + n) for n = 0 and thus for all integer n >= 0.

-Dan

Frankly proof of this by induction is a bit of overkill as far as I'm concerned. We can prove it to be true just by applying the definition of a^n.

3. Originally Posted by action259
show that (a^m)*(a^n)=a^(n+m) knowing that a^0=1 and a^n=a^(n-1)a. I know it's not a hard proof but i just find myself using the property i'm trying to prove to solve the proof which is deffinatly wrong, just need someone elses input if you know what i mean
For any m we note that for any m,
(a^m)(a^0)=a^m=a^(m+0)

Thus, there is a k such as,
(a^m)(a^k)=a^(m+k)

Multiply both sides by "a":
(a^m)(a^k)(a)=a^(m+k)(a)
Thus,
a^m(a^{k+1})=a^{m+k+1}
Q.E.D.