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.
Test
Gaurav Seth 6 years, 1 month ago
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.
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