Question

$$$$ Given $A = \\{ x: - 1 < x \leqslant - 5,x \in R\\}$ and $B = \\{ x: - 4 < x \leqslant 3,x \in R\\}$ Represent on different number lines:

i) $A \cap B$
ii) $A' \cap B$
iii) $A - B$

Answer 1

Hint- In order to solve this problem first we will make the set of $A$and $B$with the help of given data further by using these number of sets we will evaluate $A \cap B$ by taking the common elements of both the sets and $A' \cap B$ by taking the elements of $A'$ and $B$ and at last by $A - B$ removing the elements from set $A$which is also available in set $B$.

Complete step-by-step answer:
Given that $A = \\{ x: - 1 < x \leqslant - 5,x \in R\\}$ and $B = \\{ x: - 4 < x \leqslant 3,x \in R\\}$
By considering $A$ we have
A{\text{ }} = {\text{ }}\left\\{ {...{\text{ }}3,2,1,0{\text{ }}, - 5, - 6, - 7....} \right\\}
And by taking $B$ we get
B{\text{ }} = {\text{ }}\left\\{ {\;3,2,1,0, - 1, - 2, - 3} \right\\}
So we get two sets as
A{\text{ }} = {\text{ }}\left\\{ {...{\text{ }}3,2,1,0{\text{ }}, - 5, - 6, - 7....} \right\\}
and B{\text{ }} = {\text{ }}\left\\{ {\;3,2,1,0, - 1, - 2, - 3} \right\\}
Now step by step we will solve each part

i ) $A \cap B$
For finding $A \cap B$we will simply use the concept as $A \cap B$ is the set containing all elements of $A$ that also belong to $B$ (or equivalently, all elements of B that also belong to $A$).
As A{\text{ }} = {\text{ }}\left\\{ {...{\text{ }}3,2,1,0{\text{ }}, - 5, - 6, - 7....} \right\\}
and B{\text{ }} = {\text{ }}\left\\{ {\;3,2,1,0, - 1, - 2, - 3} \right\\}
So the common elements in both the sets are $3,2,1,0$
Hence $A \cap B = \\{ 3,2,1,0\\}$

ii) $A' \cap B$
For finding $A' \cap B$ first we have to find $A'$
Which can be given as $A'{\text{ }} = {\text{ }}R{\text{ }} - {\text{ }}A$
Or B{\text{ }} = {\text{ }}\left\\{ {\;3,2,1,0, - 1, - 2, - 3} \right\\}$$$$A'{\text{ }} = {\text{ }}R{\text{ }} - \;\left\\{ {...{\text{ }}3,2,1,0{\text{ }}, - 5, - 6, - 7....} \right\\}
A'{\text{ }} = {\text{ }}\left\\{ {{\text{ }} - 1, - 2, - 3, - 4,} \right\\}
and B{\text{ }} = {\text{ }}\left\\{ {\;3,2,1,0, - 1, - 2, - 3} \right\\}

Now again we will use same above property
So $A' \cap B = \\{ - 1, - 2, - 3\\}$

iii) $A - B$
As we know, the meaning of $A - B$is that all the elements which are in set $B$must not be available in set $A$. Or simply we can say we will remove the elements from set $A$ which is in set $B$
A{\text{ }} = {\text{ }}\left\\{ {...{\text{ }}3,2,1,0{\text{ }}, - 5, - 6, - 7....} \right\\}
B{\text{ }} = {\text{ }}\left\\{ {\;3,2,1,0, - 1, - 2, - 3} \right\\}
By using the above definition
A - B{\text{ }} = \;\left\\{ {{\text{ }}....5,4, - 5, - 6, - 7...} \right\\}

Note- Set theory is a branch of mathematical logic that sets studies, which are informally object collections. While any sort of object may be compiled into a set, the set theory is most commonly applied to objects related to mathematics.