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) B-C
(ii) B-D

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 B-C and B-D.

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 have been given to find the following difference:
(i) B-C
\[\Rightarrow B-C=\left\{ 4,8,12,16,20 \right\}-\left\{ 2,4,6,8,10,12,14,16 \right\}\]
We know that (B-C) means a set of those elements of B which are not present in C.
Since, 4, 8, 12, 16 all elements of B are also present in set C.
\[\Rightarrow B-C=\left\{ 20 \right\}\]
(ii) B-D
\[\Rightarrow B-D=\left\{ 4,8,12,16,20 \right\}-\left\{ 5,10,15,20 \right\}\]
We know that (B-D) means a set of those elements of B which are not present in set D.
Since the element 20 from B is present in set D.
\[\Rightarrow B-D=\left\{ 4,8,12,16 \right\}\]
Therefore, the value of:
(i) \[B-C=\left\{ 20 \right\}\]
(ii) \[B-D=\left\{ 4,8,12,16 \right\}\]

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