Answer to Question #114730 in Algebra for ANJU JAYACHANDRAN

Question

For any two subsets A and B of a set U , we define their symmetric difference to
be A ∆ B = (A \ B) ∪ (B \ A)
i) Check whether ∆ distributes over ∩ .
ii) Show that A∆ fy =A
iii) Prove that A ∆ B =(A ∩ B compliment ) ∪(A compliment ∩ B).

Answer 1

1) Ans= $\Delta$ is not distribute over $\cap$ .

Counterexample : Let $A=\{ 1,2,3,4,5 \}$ ,$B=\{ 4,5 ,6,7,8\}$ ,

$C=\{ 7,8,9,10\}$

$B\cap C=\{ 7,8\}$ .

$A \Delta (B\cap C)=(A - (B\cap C )) \cup ((B\cap C)-A)$

$=\{1,2,3,4,5\} \cup \{ 7,8 \}$

$=\{ 1,2,3,4,5,7,8\}$ .

$A\Delta B=(A-B)\cup (B-A)=\{ 1,2,3\} \cup \{6,7,8\}$

$=\{ 1,2,3,6,7,8\}$

$A\Delta C=(A-C) \cup (C-A)=\{ 1,2,3,4,5\} \cup \{ 7,8,9,10\}$

$=\{ 1,2,3,4,5,7,8,9,10\}$

$(A\Delta B) \cap (A\Delta C)= \{ 1,2,3,7,8\}$

$\therefore \ A\Delta (B\cap C) \neq (A\Delta B) \cap (A\Delta C).$

2) ans : $A \Delta \phi=( A-\phi ) \cup (\phi - A)$ $=A\cup \phi=A$

3) ans:= Claim : $A\Delta B=(A\cap B') \cup (A'\cap B)$

Let $x\in A-B \iff x\in A \ and \ x\notin B \iff x\in A \ and \ x\in B'$

$\iff x\in A\cap B'$ .

Thus , $A-B=A\cap B'$

Similarly , $B-A= B\cap A'$ .

Hence , $A\Delta B=(A\cap B') \cup (B\cap A')$ .