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.