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Question

The cartesian product A×A has 9 elements among which are found (1,0) and (0,1). Find the set A and the remaining elements of A×A

Solution
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We know that if n(A)=p and n(B)=q, then n(A×B)=n(A)×n(B)=pq
n(A×A)=n(A)×n(A)
It is given that n(A×A)=9
n(A)×n(A)=9
n(A)=3
The ordered pairs (1,0) and (0,1) are two of the nine elements of A×A
Now, A×A={(a,a):aA}
Therefore 1,0 and 1 are elements of A
Since n(A)=3, so set A={1,0,1}
The remaining elements of set A×A are (1,1),(1,1),(0,1),(0,0),(1,1),(1,0) and (1,1)

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