Question

Let $A=\{1,2\},B=\{1,2,3,4\},C=\{5,6\}$ and $D=\{5,6,7,8\}.$ Verify that:
$\left(i\right)A\times (B\cap C)=(A\times B)\cap (A\times C)$
$\left(ii\right)A\times C$ is a subset of $B\times D$

Answer 1

$\left(i\right)$ Given : $A=\{1,2\},B=\{1,2,3,4\},C=\{5,6\}$ and $D=\{5,6,7,8\}.$

Finding intersection of two set $B$ and set $C$
$B\cap C=\{1,2,3,4\}\cap \{5,6\}$
$=$ no common element
$=\varphi$.

Finding cartesian product of $A$ and $(B\cap C)$
$A\times (B\cap C)=A\times \varphi$
$=\varphi \cdots \left(i\right)$

Finding cartesian product of set $A$ and $B$, and set $A$ and $C$
$A\times B=\{1,2\}\times \{1,2,3,4\}=\left\{\right(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4\left)\right\}$
And $A\times C=\{1,2\}\times \{5,6\}=\left\{\right(1,5),(1,6),(2,5),(2,6\left)\right\}$

Taking intersection
$(A\times B)\cap (A\times C)=$ no common element
$=\varphi \cdots \left(ii\right)$

Equation $\left(i\right)$ and equation $\left(ii\right)$ are equal i.e., $L.H.S=R.H.S.,$
So it`s proved that $A\times (B\cap C)=(A\times B)\cap (A\times C)$.


$\left(ii\right)$ Given : $A=\{1,2\},B=\{1,2,3,4\},C=\{5,6\}$ and $D=\{5,6,7,8\}.$

Cartesian product of set
$A\times C=\{1,2\}\times \{5,6\}=\left\{\right(1,5),(1,6),(2,5),(2,6\left)\right\}$

$B\times D=\{1,2,3,4\}\times \{5,6,7,8\}=\left\{\right(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8\left)\right\}$

All the elements of $A\times C$ are present in $B\times D$.
Hence, $A\times C$ is a subset of $B\times D$.