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.