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Spivak Calculus Proof: exponents distribute

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove:
    [itex]
    $\left(ab^{-1}\right)\left(cd^{-1}\right)^{-1}=\left(ad\right)\left(cb\right)^{-1}$.
    [/itex]

    I know how to distribute exponents to get both sides to look identical...

    2. Relevant equations

    ...but a step in the solution requires distributing exponents. But how do you prove that you can distribute exponents using only:
    1. association for addition and multiplication
    2. commutation for addition and multiplication
    3. identity for addition and multiplication
    4. additive and multiplicative inverses
    5. distribution
    6. trichotomy law
    7. closure for addition and multiplication

    3. The attempt at a solution
    This problem is presented before mathematical induction is introduced, so I would like to solve it without its use. But here is my attempt at induction:

    P(m) = (a^n)^m=a^(n*m)
    P(1) = (a^n)^1=a^n
    P(k+1)=(a^n)^(k+1)=a^(n*k+n*1)
     
  2. jcsd
  3. May 26, 2009 #2

    Mark44

    Staff: Mentor

    Math induction really doesn't seem called for here.

    Can you replace ab-1 by (a)(1/b)? Presumably if you have expressions with negative exponents, you can replace them by reciprocals with positive exponents.

    If not, you might use the fact that cd-1 and (cd-1)-1 are multiplicative inverses (i.e., multiply both sides by cd-1).
     
  4. May 26, 2009 #3
    Thanks Mark. I think I figured it out. It turns out I was creating my own problem all along.

    The actual statement Spivak wants me to prove is: (a/b)/(c/d)=(ad)/(bc)

    By defining this as: (ab^-1)(cd^-1)^-1=(ad)(cb)^-1, I created my own problem. All I have to do to "distribute exponents" is to understand the statement as:

    (ab^-1)(c^(-1)d)=(ad)(c^-1b^-1)

    after this, it's just commutation and association
     
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