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Proof of adding powers (real analysis)

  1. Oct 1, 2011 #1
    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. 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. jcsd
  3. Oct 3, 2011 #2
    Any help would be appreciated.
  4. Oct 4, 2011 #3


    Staff: Mentor

    How can you represent a rational number r?
  5. Oct 4, 2011 #4
    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?
  6. Oct 4, 2011 #5


    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.
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