Question

Let \[\mathbb{Z}\] be the set of all integers and $A = \left\{ {\left( {a,b} \right):{a^2} + 3{b^2} = 28,a,b \in \mathbb{Z}} \right\}$ and $B = \left\{ {\left( {a,b} \right):a > b,a,b \in \mathbb{Z}} \right\}$. Then, the number of elements in $A \cap B$, is-
(1)1
(2) 4
(3) 6
(4) 5

Answer 1

Hint: First find the possible ordered pair of the set A and the set B and then try to understand the set the condition of the set that is asked in the problem, which contains all the elements that are in both the sets A and B. After the formation of the required set, we can count the element to get the required result.

Complete step-by-step answer:
It is given in the problem that let \[\mathbb{Z}\] be the set of all integers and $A = \left\{ {\left( {a,b} \right):{a^2} + 3{b^2} = 28,a,b \in \mathbb{Z}} \right\}$ and $B = \left\{ {\left( {a,b} \right):a > b,a,b \in \mathbb{Z}} \right\}$.
We have to find the number of elements in the set$A \cap B$.
First, we look into the set $A$.
$A = \left\{ {\left( {a,b} \right):{a^2} + 3{b^2} = 28,a,b \in \mathbb{Z}} \right\}$
It means that the set A contained the element of the ordered pair that follows the equation:
${a^2} + 3{b^2} = 28$
Put\[a = 1\]into the above equation,
${\left( 1 \right)^2} + 3{b^2} = 28$
$ \Rightarrow 1 + 3{b^2} = 28$
$ \Rightarrow 3{b^2} = 28 - 1$
$ \Rightarrow 3{b^2} = 27$
$ \Rightarrow {b^2} = 9$
$ \Rightarrow b = \pm 3$
So, the obtained points are $\left( {1,3} \right)$ and $\left( {1, - 3} \right)$.
Similarly, we find all the points that hold the given equations and the obtained given points are:
$A = \left\{ {\left( {5,1} \right),\left( { - 5,1} \right),\left( {5, - 1} \right),\left( { - 5, - 1} \right),\left( {4,2} \right),\left( {4, - 2} \right),\left( { - 4,2} \right),\left( { - 4, - 2} \right),\left( {1,3} \right),\left( { - 1,3} \right),\left( {1, - 3} \right),\left( { - 1, - 3} \right)} \right\}$
Now, take a look at the set B.
$B = \left\{ {\left( {a,b} \right):a > b,a,b \in \mathbb{Z}} \right\}$
The above set contains all the pairs of points $\left( {a,b} \right)$ such that $a > b$ and $a,b \in \mathbb{Z}$.
Now, take a look at the set that we have to find is:
$A \cap B$
This set contains all the elements that are in both the set $A$ and $B$.
All the elements of the set $A$ whose first element of the ordered pair is greater than the second element are the elements of the set $A \cap B$. So, the element of the set $A \cap B$ are:
$A \cap B = \left\{ {\left( {1,3} \right),\left( { - 1,3} \right),\left( { - 4, - 2} \right),\left( { - 4,2} \right),\left( { - 5, - 1} \right),\left( { - 5,1} \right)} \right\}$
So, there are 6 elements in the set \[A \cap B\].
Hence, the option (3) is correct.

Note: We can see the set B, it has the ordered pair of element such that the first element is greater than the second element, and both the element of the ordered pair is an integer number, thus it contains infinite number of elements, but the set A has the finite number of elements, so we can easily find the intersection of these sets.