[size=150]If k ∈ Z, then {n ∈ Z : n | k} ⊆ { n ∈ Z : n | k[sup]2[/sup] }. [br][br]Proof. Suppose k ∈ Z. We now need to show {n ∈ Z: n | k} ⊆ {n ∈ Z : n | k[sup]2[/sup] }.[br][br]Suppose a ∈ {n ∈ Z : n | k}. Then ii follows that a | k, so there is an integer c for which k = ac. [br][br]Then k[sup]2 [/sup]= a2 c[sup]2[/sup] . Therefore k[sup]2[/sup] = a(ac[sup]2[/sup] ),[br][br]and from this the definition of (a)__________________________ gives a | k[sup]2[/sup] . [br][br]But a | k means that a ∈ { n ∈ Z : n | k[sup]2[/sup] }. We have now shown {n ∈ Z : n | k} ⊆ {n∈ Z : n | k[sup]2[/sup] }. [/size]

[size=150]Use the methods introduced in this module to prove the following statements[br][br]Suppose A,B and C are sets. Prove that if A ⊆ B, then A −C ⊆ B −C[/size]