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!