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-B
(ii) A-C

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-B and A-C.

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) A-B
\[\Rightarrow A-B=\left\{ 3,6,12,15,18,21 \right\}-\left\{ 4,8,12,16,20 \right\}\]
We know that (A-B) is equal to a set of those elements of A which are not present in B.
Since the element 12 from A is present in set B also. We can subtract that and write the remaining elements of A. Then we will get:
\[\Rightarrow A-B=\left\{ 3,6,15,18,21 \right\}\]
(ii) A-C
\[\Rightarrow A-C=\left\{ 3,6,12,15,18,21 \right\}-\left\{ 2,4,6,8,10,12,14,16 \right\}\]
We know that (A-C) means a set of those elements of A which are not present in set C.
Since the elements 6, 12 are present in set C also. We can subtract that and write the remaining elements of A. Then we will get:
\[\Rightarrow A-C=\left\{ 3,15,18,21 \right\}\]
Therefore, we get,
(i) \[A-B=\left\{ 3,6,15,18,21 \right\}\]
(ii) \[A-C=\left\{ 3,15,18,21 \right\}\]

Note: Be careful while finding the difference of two sets and check that in (A-B) the sets only contain the elements of A which are not present in set B and similarly of A-C. If we are not sure about the concepts and consider the other way round, we will get completely different sets.