Question

f 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) C-A
(ii) D-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 C-A and D-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 have been given to find the following difference:
(i) C-A
\Rightarrow C-A=\left\\{ 2,4,6,8,10,12,14,16 \right\\}-\left\\{ 3,6,12,15,18,21 \right\\}
We know that (C-A) means a set of those elements of C which are not present in A.
Since the elements 6 and 12 are present in A also.
\Rightarrow C-A=\left\\{ 2,4,8,10,14,16 \right\\}
(ii) D-A
\Rightarrow D-A=\left\\{ 5,10,15,20 \right\\}-\left\\{ 3,6,12,15,18,25 \right\\}
We know that (D-A) means a set of those elements of D which are not present in A.
Since the elements 15 from D are also present in set A.
\Rightarrow D-A=\left\\{ 5,10,20 \right\\}
Therefore we get,
(i) C-A=\left\\{ 2,4,8,10,14,16 \right\\}
(ii) D-A=\left\\{ 5,10,20 \right\\}

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