Question

If $A=\{a,b,c,d\}\text{and}B=\{c,d,e,f\},$ then select the "wrong" statement.

• A. $A\Delta B=B\Delta A$
• B. $A\Delta B=B-A$
• C. $A\Delta B=(A-B)\cap (B-A)$
• D. $A\Delta B=A-B$

Answer 1

Answer: (B), (C), (D)

$A\Delta B=A-B$

We know that $A\Delta B=(A-B)\cup (B-A)$
$=\left(\right\{a,b,c,d\}-\{c,d,e,f\left\}\right)\cup \left(\right\{c,d,e,f\}-\{a,b,c,d\left\}\right)$
$=\{a,b\}\cup \{e,f\}=\{a,b,e,f\}$

Now,
$B\Delta A=(B-A)\cup (A-B)$
$\left(\right\{c,d,e,f\}-\{a,b,c,d\left\}\right)\cup \left(\right\{a,b,c,d\}-\{c,d,e,f\left\}\right)$
$\{e,f\}\cup \{a,b\}=\{a,b,e,f\}$

$\therefore A\Delta B=B\Delta A$ is the correct option.
This result is also known as the commutative law for symmetric difference of $2$ sets.

Now, $B-A=\{c,d,e,f\}-\{a,b,c,d\}$
$\Rightarrow B-A=\{e,f\}$
And, $A\Delta B=\{a,b,e,f\}$
Clearly, $A\Delta B\ne B-A$
Hence, it's a wrong statement.

Now, $(A-B)\cap (B-A)=\{a,b\}\cap \{e,f\}=\varnothing$
And, $A\Delta B=\{a,b,e,f\}$
Clearly, $A\Delta B\ne (A-B)\cap (B-A)$
Hence, this statement is wrong too.

In the last $A-B=\{a,b\}\ne A\Delta B$
Hence, this statement is incorrect too.