1. The problem statement, all variables and given/known data Prove that if (a1, b1) ~ (a2, b2) and (c1, d1) ~ (c2, d2), then (a1, b1) + (c1, d1) ~ (a2, b2) + (c2, d2) and (a1, b1) [tex]\bullet[/tex] (c1, d1) ~ (a2, b2)[tex]\bullet[/tex] (c2, d2). Let [a, b] denote the equivalence class with respect to ~ of (a, b) in Z x (Z-{0}), and define Q to be the set of equivalence classes of Z x (Z-{0}). For all [a, b], [c, d] in Q define [a, b] + [c, d] = [(a, b) + (c, d)] and [a, b][tex]\bullet[/tex] [c, d] = [(a, b) (c, d)]; these definitions make sense, i.e., they do not depend on the choice of representatives. 2. Relevant equations (a, b) + (c, d) = (ad + bc, bd) and (a, b) [tex]\bullet[/tex] (c, d) = (ac, bd) (a, b) ~ (c, d) if and only if ad = bc 3. The attempt at a solution I tried using those definitions. I know you have to assume that (a1, b1) ~ (a2, b2) and (c1, d1) ~ (c2, d2). But I get stuck afterwards. Where do I go from there? Do I need something more?
Are you asking how to prove this? Write down what (a1, b1) ~ (a2, b2) means and what (c1, d1) ~ (c2, d2) means; these are given. Write down what (a1, b1) + (c1, d1) is, what (a2, b2) + (c2, d2) is, and then what (a1, b1) + (c1, d1) ~ (a2, b2) + (c2, d2) means. This is what you must prove. The proof practically writes itself once you write down what these statements mean. Repeat for *, which may require a tiny trick. Write out your work for us and ask questions if you get stuck.