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| 1. |
If A and B are nonempty sets, prove that A×B=B×A A=B. |
| Answer» If A & B are two non- empty subsets and we have to prove AxB=BxA iff A=BProof: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=BThen 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=BxAii)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∈BNow (x, y) ∈ AxBBut by our assumption AxB and BxA are equalSo (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∈ASo (x,y) ∈ BxABut again since BxA=AxB; therefore x∈A. Thus implying as before that Bis subset of AThus 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=BProof: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=BThen 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=BxAii)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∈BNow (x, y) ∈ AxBBut by our assumption AxB and BxA are equalSo (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∈ASo (x,y) ∈ BxABut again since BxA=AxB; therefore x∈A. Thus implying as before that Bis subset of AThus 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. | |