Question

If $A\Delta B = (A - B) \cup (B - A)$ and $A = \{ 1,2,3,4\} ,B = \{ 3,5,7\} $, then find $A\Delta B$.
A) $\{ 1,2,4,5,7\} $
B) $\{ 3\} $
C) $\{ 1,2,3,4,5,7\} $
D) None of these

Answer 1

Hint:We can solve this question if we have an idea about set operations. The union of two sets $A \cup B$ and difference $A - B$ are defined set theoretically, and using those, we can find the answer.If $A$ and $B$ are any two sets, then \[A \cup B\] contains all those elements which are either in $A$ or in $B$.$A - B$ contains all those elements contained in $A$ but does not contained in $B$.

Complete step-by-step answer:
Given $A = \{ 1,2,3,4\} $ and $B = \{ 3,5,7\} $
We need to find $A\Delta B$, which is defined as $A\vartriangle B = (A - B) \cup (B - A)$
$A - B$ contains all those elements contained in $A$ but does not contain in $B$.
$ \Rightarrow A - B = \{ 1,2,4\} $
$3$ does not belong to $A - B$ since $3 \in B$.
Similarly, $B - A = \{ 5,7\} $
Here also $3$ does not belong to $B = A$ since $3 \in A$.
$A \cup B$ contains all those elements which are either in $A$ or in $B$.
Therefore, $(A - B) \cup (B - A) = \{ 1,2,4\} \cup \{ 5,7\} = \{ 1,2,4,5,7\} $
$ \Rightarrow A\Delta B = \{ 1,2,4,5,7\} $

So, the correct answer is “Option A”.

Additional Information:There is another operation called Symmetric difference of two sets other than usual difference. It is defined as the set of all elements that is a member of exactly one of the sets (elements which are in one of the sets but not in both).
Let $A,B$ be any two sets.
If $\Delta $ represents the operation symmetric difference, then
$A\Delta B = (A \cup B) - (A \cap B)$
In other words, symmetric difference is the difference of Union and Intersection.

Note:We have to be careful while doing set operations. Union of two sets $A$ and $B$ are commutative, means $A \cup B = B \cup A$. But in the case of difference this is not true. That is, $A - B \ne B - A$. If $A$ is an empty set (or null set), then both union and difference is $A$ itself.