Question

If $A=\{y:y=\frac{a+1}{2},a\in W$ and $a\le 5\},$$B=\{y:y=\frac{2n-1}{2},n\in W$ and $n<5\}$ and $C=\{-1,-\frac{1}{2},1,\frac{3}{2},2\}$, then show that $A-(B\cup C)=(A-B)\cap (A-C)$

Answer 1

$A=\{y:y=\frac{a+1}{2},a\in W$ and $a\le 5\}$
$a=\{0,1,2,3,4,5\}$
$\Rightarrow y=\frac{a+1}{2}=\frac{1}{2}$
$y=\frac{1+1}{2}=\frac{2}{2}=1$
$y=\frac{2+1}{2}=\frac{3}{2}$
$y=\frac{3+1}{2}=\frac{4}{2}=2$
$y=\frac{4+1}{2}=\frac{5}{2}$
$y=\frac{5+1}{2}=\frac{6}{2}=3$
$\therefore A=\{\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},3\}$
$B=\{y:y=\frac{2n-1}{2},n\in W$ and $n<5\}$
$n=\{0,1,2,3,4\}$
$\Rightarrow y=\frac{2\times 0-1}{2}=\frac{-1}{2}$
$y=\frac{2\times 1-1}{2}=\frac{1}{2}$
$y=\frac{2\times 2-1}{2}=\frac{3}{2}$
$y=\frac{2\times 3-1}{2}=\frac{5}{2}$
$\therefore B=\{\frac{2\times 4-1}{2}=\frac{7}{2},\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}\}$
$$\begin{array}{cc}& C=\{-1,-\frac{1}{2},1,\frac{3}{2},2\}\\ & B\cup C=\{-1,-\frac{1}{2},\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},\frac{7}{2}\}\\ & A-(B\cup C)=\left\{3\right\}\dots \left(1\right)\\ & A-B=\{1,2,3\}\\ & A-C=\{\frac{1}{2},\frac{5}{2},3\}\\ & (A-B)\cap (A-C)=\left\{3\right\}\dots \left(2\right)\end{array}$$
From (1) and (2), it is verified that
$A-(B\cup C)=(A-B)\cap (A-C)$