Proof of adding powers (real analysis)

[EV33]

Homework Statement

Prove that b^r+s=b^rb^s if r and s are rational.

Homework 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=a^1/nb^1/n
3.)(b^a)^b=b^ab

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. b^r+s=b^rb^s due to multiplicative associativity. (QED)


Is the proof this simple or am I missing something?

 

[EV33]

Any help would be appreciated.

 

[Mark44]

How can you represent a rational number r?

 

[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?

 

[Mark44]

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.

OK, that's a start.

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

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?