Question

If \[A=\left\{ 3,6,12,15,18,21 \right\}\], \[B=\left\{ 4,8,12,16,20 \right\}\], \[C=\left\{ 2,4,6,8,10,12,14,16 \right\}\] and \[D=\left\{ 5,10,15,20 \right\}\], then find:
(i) A-D
(ii) B-A

Answer 1

Hint: We know that the difference of any two sets say A and B i.e. (A-B) is a set which contains all the elements of A which are not present in set B. Using this concept, we can find the sets A-D and B-A.

Complete step-by-step answer:
We have been given sets as follows:
\[A=\left\{ 3,6,12,15,18,21 \right\}\]
\[B=\left\{ 4,8,12,16,20 \right\}\]
\[C=\left\{ 2,4,6,8,10,12,14,16 \right\}\]
\[D=\left\{ 5,10,15,20 \right\}\]
Now we know that the difference between any two sets A and B i.e. (A-B) is a set that contains all the elements of A which are not present in set B.
(i) A-D
\[\Rightarrow A-D=\left\{ 3,6,12,15,18,21 \right\}-\left\{ 5,10,15,20 \right\}\]
Since the element ’15’ of A is also present in set D. So we will exclude it to get (A-D).
\[\Rightarrow A-D=\left\{ 3,6,12,18,21 \right\}\]
(ii) B-A
\[\Rightarrow B-A=\left\{ 4,8,12,16,20 \right\}-\left\{ 3,6,12,15,18,21 \right\}\]
Since the element 12 of B is also present in the set A. So we will exclude it to get (B-A).
\[\Rightarrow B-A=\left\{ 4,8,16,20 \right\}\]
Therefore, we get,
(i) \[A-D=\left\{ 3,6,12,18,21 \right\}\]
(ii) \[B-A=\left\{ 4,8,16,20 \right\}\]

Note: Be careful while finding the difference of two sets and check that in (A-D) the sets only contain the elements of A which are not present in set D and similarly of (B-A) also. Also, remember that a set is a well-defined collection of distinct objects so make sure that our sets don’t contain any repetitive elements.