Question

If \[A=\left\{ 2,4,6,8,10,12 \right\}\] and \[B=\left\{ 3,4,5,6,7,8,10 \right\}\] then find:
(i) A-B
(ii) B-A
(iii) \[\left( A-B \right)\cup \left( B-A \right)\]

Answer 1

Hint: We know that the difference between two sets A and B i.e. (A-B) is a set of those elements of set A which are not the elements of set B. Also, the union \[\left( \cup \right)\] of two sets is a set which contains all elements that are present in the two sets which we have considered.

Complete step-by-step answer:
We have been given the sets \[A=\left\{ 2,4,6,8,10,12 \right\}\] and \[B=\left\{ 3,4,5,6,7,8,10 \right\}\].
Now, we know that (A-B) is a set of those elements of set A which are not the elements of set B.
\[\Rightarrow A-B=\left\{ 2,4,6,8,10,12 \right\}-\left\{ 3,4,5,6,7,8,10 \right\}\]
Since the elements 4, 6, 8 and 10 of A are common to the set B. So we will conclude these elements from set A in order to get (A-B).
\[\Rightarrow A-B=\left\{ 2,12 \right\}\]
Similarly, (B-A) represents the set which has elements of set B that are not present in set A. So we can write,
\[\Rightarrow B-A=\left\{ 3,4,5,6,7,8,10 \right\}-\left\{ 2,4,6,8,10,12 \right\}\]
We can see the elements 4, 6, 8 and 10 are common in both sets. So we will exclude them from set B to get (B-A) as,
\[\Rightarrow B-A=\left\{ 3,5,7 \right\}\]
Now we know that the union of two sets i.e. \[A\cup B\] means a set which contains all elements of A and B.
\[\Rightarrow \left( A-B \right)\cup \left( B-A \right)=\left\{ 2,12 \right\}\cup \left\{ 3,5,7 \right\}=\left\{ 2,12,3,5,7 \right\}\]
Therefore the value of \[\left( A-B \right)=\left\{ 2,12 \right\}\], \[\left( B-A \right)=\left\{ 3,5,7 \right\}\] and \[\left( A-B \right)\cup \left( B-A \right)=\left\{ 2,12,3,5,7 \right\}\].

Note: Be careful while finding the difference between the sets as there is a chance that we might include the elements of set B also in (A-B) which will be wrong. Also, remember that a set is a well-defined collection of distinct objects. Also, remember that union of any two sets must not contain the common element in repetition.