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-D
(ii) B-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 A-D and B-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 know that the difference between any two sets A and B i.e. (A-B) is a set that contains all the elements of A which are not present in set B.
(i) A-D
\Rightarrow A-D=\left\\{ 3,6,12,15,18,21 \right\\}-\left\\{ 5,10,15,20 \right\\}
Since the element ’15’ of A is also present in set D. So we will exclude it to get (A-D).
\Rightarrow A-D=\left\\{ 3,6,12,18,21 \right\\}
(ii) B-A
\Rightarrow B-A=\left\\{ 4,8,12,16,20 \right\\}-\left\\{ 3,6,12,15,18,21 \right\\}
Since the element 12 of B is also present in the set A. So we will exclude it to get (B-A).
\Rightarrow B-A=\left\\{ 4,8,16,20 \right\\}
Therefore, we get,
(i) A-D=\left\\{ 3,6,12,18,21 \right\\}
(ii) B-A=\left\\{ 4,8,16,20 \right\\}

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