Question

A and b are non-empty sets and a x b is a subset of b x

a. Prove that a =

b.

Answer 1

If A & B are two non- empty subsets and we have to prove AxB=BxA iff A=B

Proof:

We will prove this in two parts. In first part we will prove that if A=B then AxB=BxA and in the second part we will prove that if AxB=BxA then A=B.

i) For the first part let us assume that A=B

Then in AxB we can first replace first ‘A’ with ‘B’ (As by assumption A=B) so that it becomes BxB. Now we have AxB=BxB. In the next step we replace ‘B’ with ‘A’ so that BxB can be written as BxA.

Thus we have AxB=BxB=BxA

ii)For the second part we assume that AxB=BxA and then we will prove that A=B. We will prove this by double containment. We will prove that A is subset of B and then B is subset of A .

Let x∈ A and y∈B

Now (x, y) ∈ AxB

But by our assumption AxB and BxA are equal

So (x,y) ∈ BxA implying that x ∈B , sine x is an arbitrarily chosen element, so A is subset of B.

Now let x∈B and y∈A

So (x,y) ∈ BxA

But again since BxA=AxB; therefore x∈A. Thus implying as before that Bis subset of A

Thus from double containment viz. A being subset of B and B being subset of A; we get A=B.

Thus (i) and (ii) complete the proof.

If A & B are two non- empty subsets and we have to prove AxB=BxA iff A=B

Proof:

We will prove this in two parts. In first part we will prove that if A=B then AxB=BxA and in the second part we will prove that if AxB=BxA then A=B.

i) For the first part let us assume that A=B

Then in AxB we can first replace first ‘A’ with ‘B’ (As by assumption A=B) so that it becomes BxB. Now we have AxB=BxB. In the next step we replace ‘B’ with ‘A’ so that BxB can be written as BxA.

Thus we have AxB=BxB=BxA

ii)For the second part we assume that AxB=BxA and then we will prove that A=B. We will prove this by double containment. We will prove that A is subset of B and then B is subset of A .

Let x∈ A and y∈B

Now (x, y) ∈ AxB

But by our assumption AxB and BxA are equal

So (x,y) ∈ BxA implying that x ∈B , sine x is an arbitrarily chosen element, so A is subset of B.

Now let x∈B and y∈A

So (x,y) ∈ BxA

But again since BxA=AxB; therefore x∈A. Thus implying as before that Bis subset of A

Thus from double containment viz. A being subset of B and B being subset of A; we get A=B.

Thus (i) and (ii) complete the proof.

hope it helps you

Answer 2

Answer:

let X€A and Y€B

(x,y)€AxB

(x,y)€BxA. (AxB=BxA)

x€B and y€A

so, A=B