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Simplify $(A\cup B)\cap (A\cup B')\cap (A - B)$ for a given non empty sets $A$ and $B$, where $(A\cap B) = \varnothing .$

Given ,

$A\cap B=\phi$ which implies the two sets are disjoint.

Ven diagram of the problem given below:-

$S$ is the universal set .

$A\cup B$ is the set with all the element with $A$ and $B$.

$A\cup {B'}=B'$  as $B’=S-B$ which include $A$ as well so union will be $B’$

$A- {B}=A$  as their intersection is empty set.

So given expression is ,

$(A\cup B)\cap \left ( A\cup B' \right )\cap (A-B)$

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

=$A$

Method 2:-

Universal set $U=\left \{ 1,2,3,4,5,6,7,8,9,10 \right \}$

$A=\left \{ 1,2,3,4 \right \}$

$B=\left \{ 7,8,9,10 \right \}$

$B'=U-B=\left \{ 1,2,3,4,5,6 \right \}$

$A\cup B=\left \{ 1,2,3,4,7,8,9,10 \right \}$

$A\cup B'=\left \{ 1,2,3,4,5,6 \right \}$

$A- B=\left \{ 1,2,3,4 \right \}$

$(A\cup B)\cap \left ( A\cup B' \right )\cap (A-B)$

=$\left \{ 1,2,3,4,7,8,9,10 \right \}\cap \left \{ 1,2,3,4,5,6 \right \}\cap \left \{ 1,2,3,4 \right \}$

$= \left \{ 1,2,3,4 \right \}$

$=A$