How Do You Prove (ab)^-1 Equals a^-1b^-1?

[jgens]

Homework Statement

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

Homework 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

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!

 

[quantumdude]

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.