(b) Proposition. For each integer m, 5 divides (m-m). Proof. Let m e Z. We will prove that 5 divides (m³ - m) by proving that (m - m) =0 (mod 5). We will use cases. For the first case, if m = 0 (mod 5), then m = 0 (mod 5) and, hence, (m3 – m) = 0 (mod 5). For the second case, if m = 1 (mod 5), then m = 1 (mod 5) and, hence, (ms -m) = (1 – 1) (mod 5), which means that (m3 - m) = 0 (mod 5). For the third case, if m = 2 (mod 5), then m = 32 (mod 5) and, hence, (m- m) = (32-2) (mod 5), which means that (m- m) = 0(mod 5).

I've been stuck on this one for a while. If the proposition is false, and the proof is incorrect, find the error in the proof and provide a counter example showing that it is false If the proposition is true, but the proof is wrong, fix the proof If everything is correct, nothing needs to be done