Proof the Lemma: Let R be an integral domain with a, b, c R. Then: • If a |b and b |c then a |c (transitivity of divisibility). • If a |b and a |c then a |xb + yc for any x, y Z (linearity of divisibility). Solution By definition, if a|b, there exists a k in Z such that b = ak. Similarly, if b|c, then there exists an l in Z such that c = bl. Substituting in b = ak to c = bl gives that c = akl. Since k and l are in Z, and it is closed under multiplication, kl is in Z. Therefore, by definition a|c. By definition, if a|b, there exists a k in Z such that b = ak. Similarly, if a|c, then there exists an l in Z such that c = al. Substituting these into xb + yc gives the expression xak + yal, or a(xk + yl). Since x, k, y, and l are all in Z, and Z is both closed under addition and multiplication, xk + yl is in Z. Therefore, by definition, a|a(xk+yl) = a|xb + yc..