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.