Answer to Question #105438 in Algebra for Deepak

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 A∆ « = .
iii) Prove that A B (A B ) (A B)

Answer 1

$A\Delta B=(A\backslash B)\cup(B\backslash A)$

I) Check whether $\Delta$ distributes over $\cap$;

II) Show that $A\Delta \varnothing=A, A\Delta A=\varnothing;$

III) Prove that $A\Delta B=(A\cap B')\cup(A'\cap B).$

Solution:

I)

$A\Delta B$ does distribute over $A\cap B$ because symmetric difference is equivalent

to the union in both relative complements. The equality in this non-strict

inclusion has occurred because there are disjoint sets. The symmetric differenc

e is thus commutative and associative.

II)

An empty set is a neutral set and

$A\Delta \varnothing=(A\backslash \varnothing)\cup(\varnothing\backslash A)= A\cup\varnothing=A;$

$A\Delta A=(A\backslash A)\cup(A\backslash A)= \varnothing\cup\varnothing=\varnothing.$

III)

$A\Delta B=(A\backslash B)\cup(B\backslash A)=(A\cap B') \cup(B\cap A')=(A\cap B')\cup(A'\cap B)$

The symmetric difference in $A\Delta B$ means that all elements belong to $A\space or\space B$

but not in their intersection.