# Multiplicative Inverse Proof

1. Feb 25, 2009

### jgens

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

Prove (ab)-1 = a-1b-1 for all a,b =! 0.

2. Relevant equations

Multiplicative inverse property: (a)(a-1) = 1
Commutivity: ab = ba
Associativity: (ab)c = a(bc)
Transitivity: If a = b and b = c then a = c

3. The attempt at a solution

Early today, my friend and I had a discussion regarding this statement and its proof using the principles stated above. The proof I provided is as follows:

(ab)(ab)-1 = 1 and 1 = (a)(a-1)(b)(b-1); therefore it follows by transitivity that (ab)(ab)-1 = (a)(a-1)(b)(b-1). By commutivity and associativity it follows that (ab)(ab)-1 = (ab)(a-1)(b-1), ultimately yielding (ab)-1 = (a-1)(b-1).

The proof my friend posted is as follows:

(ab)-1 = 1/(ab) = (1/a)(1/b) = (a-1)(b-1)

I argued that my friend's proof was not valid as it relies on the unproven (yet true) assumption that (ab)-1 = 1/(ab) which he would need to demonstrate was true before using it in a proof. I think my friend is right on this one and my criticisms aren't truly valid; however, I thought I would check here with people who are familiar with mathematics to ensure that any of the proofs, criticisms, etc. are valid or invalid.

Thanks!

2. Feb 25, 2009

### Tom Mattson

Staff Emeritus
Your proof is correct, and your friend's isn't. These proofs should only refer to addition, multiplication, and the axioms. Division is defined later in terms of multiplication. So things like $1/(ab)$ shouldn't even be showing up here.