Question

proof that (a^m ÷ a^n) = a^m-n​

Answer 1

Answer:

I have to prove (am)n=amn for all a∈G and m,n∈Z where G is a group. Is it enough to just expand (am)n=(am∗∗∗am)- n times. And then from here we can expand it a bit more to there there are mn amount of a′s? Or do I need to break it up into cases. I felt if I did I'll have atleast 3 cases and a few subcases. As of right now I have two cases one where m=n=0 and the other where a,b≠0.

Here is what I have so far.

New proof #2

Case 1: Let m>0 and n>0 We will proceed by induction. We fix m and induct on n. Base case: Let n=1. We see that am=am. Inductive case: Suppose that (am)k=amk We shall prove (am)k+1=am(k+1). It follows immediately from assumption that (am)k+1=am(k+1).

Case 2: m=n=0. It is immediately obvious that (am)n=amn

Case 3: m<0 and n<0. Let m=−t and n=−r where t,r>0. Then (am)n=(a−t)−r)=(a−1)t)r)−1=(a−1)rt)−1=(a−nt)−1=(ant)−1=an∗(−1)t=amn

Answer 2

Step-by-step explanation:

we can prove this formula with an help of an example

let take

$\frac{8}{4} = 2 \: so \: with \: this \: formulae \: we \: can \: write \: it \: as \\ \frac{ {2}^{3} }{ {2}^{2} } = {2}^{3 - 2} = {2}^{1} \\ so$

YOU CAN SEE THAT THE SOLUTION IS SAME IN BOTH SO HENCE PROVED

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