Given sets A and B, prove that the following two conditions are equivalent: (1) A / B = B / A, (2) A = B.

See proof below

Step-by-step explanation:

Remember that if we want to prove that two propositions P and Q are equivalent, we have to prove that P implies Q and Q implies P.

(1) → (2) Suppose that A / B = B / A, we will prove that A = B. Let x∈A, and for the sake of a contradiction suppose that x∉B. Then x∈A / B by definition of difference of sets. Since A / B = B / A and x∈A / B then x∈B / A, that is, x∈B and x∉A, which is a contradiction to the first assumption regarding x. Therefore x∈B, for all x∈A, which implies that A⊆B.

We can use a similar argument to prove that B⊆A, and both inclusions imply that A=B as we wanted to prove. Indeed, let x∈B, and suppose that x∉A Thus x∈B / A. But A / B = B / A, then x∈A / B, that is, x∈A and x∉B, which is absurd. Then x∈A, for all x∈B, that is, B⊆A.

(2) → (1). Suppose that A = B. We will prove that A / B = B / A. By definition, A / B is the set of elements of A that do not belong to B. However, A=B, so A / B is the set of elements of A that do not belon to A. There no exists such an element, since this condition is contradictory, thus A / B=∅, the empty set. Similarly, B / A = B / B = ∅. The empty set is unique, therefore A / B=∅ = B / A.

The previous parts show that (1) and (2) are logically equivalent.