Question

If A={1,2} and B={x : x \[ \in \] N and \[{x^2} - 9 = 0\] }, then find \[A \times B\].

Answer 1

Hint:
Here we are given with two sets A and B. with elements and we are asked to find the product set of them. But first we have to find the elements of set B. Given the condition is there using it, find the elements first and then proceed!

Complete step by step solution:
Given that,
A={1,2}
 B={x : x \[ \in \] N and \[{x^2} - 9 = 0\] }
Here we have to find the values of set B elements.
If x=1, then \[{x^2} - 9 = 1 - 9 = - 8 \ne 0\]. So 1 is not the element.
If x=2, then \[{x^2} - 9 = 4 - 9 = - 5 \ne 0\]. So 2 is not the element.
If x=3, then \[{x^2} - 9 = 9 - 9 = 0\]. So this is the element.
Beyond this all elements will be greater than 0. So 3 is the only element of set B.
B={3}
Now \[A \times B\] = \[\left\{ {\left( {1,2} \right) \times \left( 3 \right)} \right\}\]
\[ \Rightarrow \left\{ {\left( {1,3} \right),\left( {2,3} \right)} \right\}\]
Further we can distribute the elements as,
\[ \Rightarrow \left\{ {\left( {1,2} \right),\left( {3,2} \right),\left( {1,3} \right),\left( {3,3} \right)} \right\}\]
So

\[A \times B \Rightarrow \left\{ {\left( {1,2} \right),\left( {3,2} \right),\left( {1,3} \right),\left( {3,3} \right)} \right\}\]

Note:
Here in set B students may take -3 as one of the elements because it also satisfies the condition but note that x belongs to natural numbers only. And -3 is not a natural number; it is an integer. So 3 is the only element that belongs to set B.