Question

If $A=\{x:x=6n,n\in W$ and $n<6\},$ $B=\{x:x=2n,n\in N$ and $2<n\le 9\}$ and $C=\{x:x=3n,n\in N$ and $4\le n<10\}$, then show that $A-(B\cap C)=(A-B)\cup (A-C)$

Answer 1

$A=\{x:x=6n,n\in W,n<6\}$
$x=6n$
$n=\{0,1,2,3,4,5\}$
$\Rightarrow x=6\times 0=0$
$x=6\times 1=6$
$x=6\times 2=12$
$x=6\times 3=18$
$x=6\times 4=24$
$x=6\times 5=30$
$\therefore A=\{0,6,12,18,24,30\}$
$B=\{x:x=2n,n\in N,2<n\le 9\}$
$n=\{3,4,5,6,7,8,9\}$
$x=2n$
$\Rightarrow x=2\times 3=6$
$2\times 4=8$
$2\times 5=10$
$2\times 6=12$
$2\times 7=14$
$2\times 8=16$
$2\times 9=18$
$\therefore B=\{6,8,10,12,14,16,18\}$
$C=\{x:x=3n,n\in N,4\le n<10\}$
$N=\{4,5,6,7,8,9\}$
$x=3\times 4=12$
$2x=3\times 5=15$
$x=3\times 6=18$
$x=3\times 7=21$
$x=3\times 8=24$
$x=3\times 9=27$
$x=2\times 9=18$
$\therefore C=\{12,15,18,21,24,27\}$
$A-(B\cap C)=(A-B)\cup (A-C)$
L.H.S $=$ R.H.S
$B\cap C=\{12,18\}$
$A-(B\cap C)=\{0,6,12,18,24,30\}-\{12,18\}$
$=\{0,6,24,30\}\dots \left(1\right)$
$\begin{array}{cc}& (A-B)=\{0,24,30\}\\ & (A-C)=\{0,6,30\}\\ & (A-B)\cup (A-C)=\{0,6,24,30\}\dots \left(2\right)\end{array}$
From (1) and (2), it is verified that
$A-(B\cap C)=(A-B)\cup (A-C)$