Show that for any sets $\mathrm{A}$ and $\mathrm{B}$, $A=(A \cap B) \cup(A-B)$ and $A \cup(B-A)=(A \cup B).$
To show: $A=(A \cap B) \cup(A-B)$
Let $x \in A$
We have to show that $x \in(A \cap B) \cup(A-B)$
Case $I$
$x \in A \cap B$
Then, $x \in(A \cap B) \subset(A \cup B) \cup(A-B)$
Case $II$
$x \notin A \cap B$
$\Rightarrow x \notin A$ or $x \notin B$
$\therefore x \notin B[x \notin A]$
$\therefore x \notin A-B \subset(A \cup B) \cup(A-B)$
$\therefore A \subset(A \cap B) \cup(A-B)$ ...........$(1)$
It is clear that
$A \cap B \subset A$ and $(A-B) \subset A$
$\therefore(A \cap B) \cup(A-B) \subset A$ ..........$(2)$
From $(1)$ and $(2),$ we obtain
$A=(A \cap B) \cup(A-B)$
To prove: $A \cup(B-A) \subset A \cup B$
Let $x \in A \cup(B-A)$
$\Rightarrow x \in A$ or $(x \in B$ and $x \notin A)$
$ \Rightarrow (x \in A$ or $x \in B)$ and $(x \in A$ or $x \notin A)$
$\Rightarrow x \in(A \cup B)$
$\therefore A \cup(B-A) \subset(A \cup B) $ .........$(3)$
Next, we show that $(A \cup B) \subset A \cup(B-A)$
Let $y \in A \cup B$
$\Rightarrow y \in A$ or $y \in B$
$ \Rightarrow (y \in A$ or $y \in B)$ and $(y \in A{\rm{ }}$ or $y \notin A)$
$\Rightarrow y \in A$ or $(y \in B$ and $y \notin A)$
$\Rightarrow y \in A \cup(B-A)$
$\therefore A \cup B \subset A \cup(B-A)$ ...........$(4$)
Hence, from $(3)$ and $(4)$, we obtain $A \cup(B-A)=A \cup B$.
Consider the sets $A$ and $B$ of $A=\{2,4,6,8\}$ and $B=\{6,8,10,12\}$ Find $A \cap B .$
If $A=\{1,2,3,4\}, B=\{3,4,5,6\}, C=\{5,6,7,8\}$ and $D=\{7,8,9,10\} ;$ find
$A \cup B \cup C$
If $A=\{3,5,7,9,11\}, B=\{7,9,11,13\}, C=\{11,13,15\}$ and $D=\{15,17\} ;$ find
$A \cap D$
If $A=\{3,5,7,9,11\}, B=\{7,9,11,13\}, C=\{11,13,15\}$ and $D=\{15,17\} ;$ find
$A \cap B$
If $A$ and $B$ are disjoint, then $n(A \cup B)$ is equal to