Question

If $A\mathit{=}\mathit{\{}a\mathit{,}\mathit{}b\mathit{,}\mathit{}c\mathit{,}\mathit{}d\mathit{,}\mathit{}e\mathit{\}}$, $B\mathit{=}\mathit{\{}a\mathit{,}\mathit{}c\mathit{,}\mathit{}e\mathit{,}\mathit{}g\mathit{\}}$ and $C\mathit{=}\mathit{\{}b\mathit{,}\mathit{}e\mathit{,}\mathit{}f\mathit{,}\mathit{}g\mathit{\}}$, verify that:$A\cap \left(B-C\right)=\left(A\cap B\right)-\left(A\cap C\right)$

Answer 1

Prove the given condition

Given,

$$\begin{gathered} A\mathit{=}\mathit{\{}a\mathit{,}\mathit{}b\mathit{,}\mathit{}c\mathit{,}\mathit{}d\mathit{,}\mathit{}e\mathit{\}} \\ B\mathit{=}\mathit{\{}a\mathit{,}\mathit{}c\mathit{,}\mathit{}e\mathit{,}\mathit{}g\mathit{\}} \\ C\mathit{=}\mathit{\{}\mathit{b}\mathit{,}\mathit{}\mathit{e}\mathit{,}\mathit{}\mathit{f}\mathit{,}\mathit{}\mathit{g}\mathit{\}} \end{gathered}$$

We have to prove that,

$A\cap \left(B-C\right)=\left(A\cap B\right)-\left(A\cap C\right)$

For sets $P$ and $Q$, the notation $\left(P-Q\right)$ denotes the set which has all the elements of $P$ such that the elements are not elements of set $Q$. It is called their difference.

In general,

$\left(P-Q\right)=\left\{x:x\in P,x\notin Q\right\}$

For sets $P$ and $Q$, the notation $P\cap Q$ denotes the set which has all the elements of $P$ such that the elements are also elements of set $Q$. It is called their intersection.

In general,

$P\cap Q=\left\{x:x\in P,x\in Q\right\}$

So,

$\begin{array}{rcl}\mathrm{LHS}& =& A\cap \left(B-C\right)\\ & =& \left\{a,b,c,d,e\right\}\cap \left(\left\{a,c,e,g\right\}-\left\{b,e,f,g\right\}\right)\\ & =& \left\{a,b,c,d,e\right\}\cap \left\{a,c\right\}\\ & =& \left\{a,c\right\}\\ \mathrm{RHS}& =& \left(A\cap B\right)-\left(A\cap C\right)\\ & =& \left(\left\{a,b,c,d,e\right\}\cap \left\{a,c,e,g\right\}\right)-\left(\left\{a,b,c,d,e\right\}\cap \left\{b,e,f,g\right\}\right)\\ & =& \left\{a,c,e\right\}-\left\{b,e\right\}\\ & =& \left\{a,e\right\}\end{array}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence proved.