# Proof of adding powers (real analysis)

1. Oct 1, 2011

### EV33

1. The problem statement, all variables and given/known data

Prove that br+s=brbs if r and s are rational.

2. Relevant equations

So far we know the basic field axioms and a a few other things related to powers.
1.) For every real x>0 and every integer n>0 there is one and only one positive real y such that yn=x
2.) if a and b are positive real numbers and n is a positive integer, then (ab)1/n=a1/nb1/n
3.)(ba)b=bab

3. The attempt at a solution

When I look at this problem I don't see any way to use the three facts above. The first thing that jumps at me is the field axiom of multiplicative associativity. So for me I see the proof as going as such.

Asssume r and s are rational. br+s=brbs due to multiplicative associativity. (QED)

Is the proof this simple or am I missing something?

Last edited: Oct 1, 2011
2. Oct 3, 2011

### EV33

Any help would be appreciated.

3. Oct 4, 2011

### Staff: Mentor

How can you represent a rational number r?

4. Oct 4, 2011

### EV33

I could say that r=m/n, s=u/v where m,n,u, and v are all integers and both n and v are not equal to zero. From here I still have the same problem though. I don't have any ideas for any inbetween steps. All I see is a simple regrouping ( associativity). Is there more to than that or is it really this simple?

5. Oct 4, 2011

### Staff: Mentor

OK, that's a start.

The right side of the equation you're trying to prove is brbs. Use the representations above of r and s, and #2 and #3 in your list of relevant equations.