Question

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 $A \times A$.

Answer 1

Hint: Using the property of cartisian product of a set, let say P then $(P \times P)$ is defined as a relation, $(P \to P)$ where the elements of $(P \times P)$ will be in the form $({p_1},{p_2})$ where
${p_1},{p_2} \in P$
$P \times P = \{ ({p_1},{p_2}):{p_1},{p_2} \in P\} $
Using this we’ll get the elements of A and hence will get set A. from set A we can easily find $A \times A$.

Complete step by step solution: Given data: $n(A \times A) = 9$
$( - 1,0),(0,1) \in (A \times A)$
Now we know that if we a set let P then
$P \times P = \{ ({p_1},{p_2}):{p_1},{p_2} \in P\} $
Now since $( - 1,0),(0,1) \in (A \times A)$
We can say that $ - 1,0,1 \in A$
We know that if two sets let X and Y have m and n numbers of elements respectively then the number of elements in the sets $(X \times Y)$or $(Y \times X)$ will be the product of the number of elements in the respective sets i.e. mn
$n(A \times A) = n(A)n(A)$
Substituting the value of$n(A \times A) = 9$
\[ \Rightarrow {\left( {n(A)} \right)^2} = 9\]
\[\therefore n(A) = 3\]
And hence -1, 0, 1 are the only elements of A

$\therefore A = \left\{ { - 1,0,1} \right\}$
And \[A \times A = \left\{ {( - 1, - 1),( - 1,0),( - 1,1),(0, - 1),(0,0),(0,1),(1, - 1),(1,0),(1,1)} \right\}\]

Note: Most of the students think that in $( - 1, 0), (0, 1) \in (A \times A)$, element 0 is in two pair so the occurrence of 0 is will two times i.e. $A = \{ 0, 1, 0, - 1\} $ but it is wrong as we have found that $n\left( A \right) = 3$ and A contains only three elements and if we consider this set as set A then $n\left( A \right) = 4$ which is not true as we have proved that $n\left( A \right) = 3$