How to prove (a^c)^d = a^(cd) without knowing the values of c and d?

  • Thread starter Thread starter Llamas
  • Start date Start date
  • Tags Tags
    Proof
Llamas
Messages
4
Reaction score
0

Homework Statement



The problem is in the title: Prove (a^c)^d) = a^(cd)



Homework Equations



N is the set of natural numbers.

(ab)^n = (a^n)(b^n)
a^(p+n) = (a^p)*(a^n)
((a^p)(a^n)) * a = (a^p)(a^(n+1))

The Attempt at a Solution



c = p-q; d = j-k; p,q,j,kεN (by definition of integers)

(a^(p-q))^(j-k)

((a^(p-q))^j)/((a^(p-q))^k)

((a^(p+(-q)))^j)/((a^(p+ (-q)))^k)

((a^p)(a^(-q)))^j/((a^p)(a^(-q)))^k or (((a^p)/(a^q))^j)/(((a^p)/(a^q))^k)

I have no idea where to go from here. I've spent two hours on this and I've finished nearly all my other proofs on this assignment...This one is killing me though, and I have other work I need to get to at some point. Could anyone please give me a tip in the right direction? I feel like I have it all wrong.
 
Physics news on Phys.org
Welcome to PF!

Hi Llamas! Welcome to PF! :smile:

What are c and d, are they all natural (whole) numbers?

if so, just write ac = a times ac-1 :wink:
 


tiny-tim said:
Hi Llamas! Welcome to PF! :smile:

What are c and d, are they all natural (whole) numbers?

if so, just write ac = a times ac-1 :wink:

Thank you tiny-tim, the advice helped me finish the proof. Sorry I didn't respond until now, had other work as well.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top