please explain clear show steps Throughout, a, b, c epsilon Z. Prove: If a | b and a is even, then b is even. The following statement is false; give a counterexample: If a | b and a is odd, then b is odd. Solution A) If a|b, then b = ak for some integer k. If a is even, then a = 2p for some integer p, so b = ak = (2p)k = 2(pk), which is even because b is equal to 2 times an integer pk. B) Again a|b so b = ak, but a is odd. If k is even, then even if a is odd b is still even; for example, if a = 3 and b = 6, a|b because b = 3(2) = ak where k = 2, but a = 3 is odd and b = 6 is even..