The Cartesian product A times A has 9 element | vedclass

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(A) We know that if $n(A) = p,$ then $n(A 	imes A) = p^2.$Given $n(A 	imes A) = 9,$ so $p^2 = 9,$ which implies $n(A) = 3.$The elements of $A 	imes

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$(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)$

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(A) We know that if $n(A) = p,$ then $n(A 	imes A) = p^2.$Given $n(A 	imes A) = 9,$ so $p^2 = 9,$ which implies $n(A) = 3.$The elements of $A 	imes A$ are of the form $(a, b)$ where $a, b \in A.$Since $(-1, 0) \in A 	imes A,$ we have $-1 \in A$ and $0 \in A.$Since $(0, 1) \in A 	imes A,$ we have $0 \in A$ and $1 \in A.$Thus,the set $A = \{-1, 0, 1\}.$The set $A 	imes A$ consists of $3 	imes 3 = 9$ elements:$A 	imes A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}.$The given elements are $(-1, 0)$ and $(0, 1).$The remaining elements are $(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1).$

The Cartesian product $A \times A$ has $9$ elements,among which are found $(-1, 0)$ and $(0, 1)$. Find the set $A$ and the remaining elements of $A \times A$.

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