# Spivak Calculus Proof: exponents distribute

• selfteaching9
In summary, the problem is to prove that (a/b)/(c/d)=(ad)/(bc) using only the listed properties of exponents: association, commutation, identity, additive and multiplicative inverses, distribution, trichotomy law, and closure. The solution involves understanding the statement as (ab^-1)(c^(-1)d)=(ad)(c^-1b^-1) and then using commutation and association. Mathematical induction is not needed.

## Homework Statement

Prove:
$\left(ab^{-1}\right)\left(cd^{-1}\right)^{-1}=\left(ad\right)\left(cb\right)^{-1}.$

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

## Homework 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
5. distribution
6. trichotomy law
7. closure for addition and multiplication

## 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)

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).

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: