Question

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that :

(i) $A\times C\subset B\times D$

(ii) $A\times (B\cap C)=(A\times B)\cap (A\times C)$

Answer 1

(i) We have,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

$\therefore B\times D=\{1,2,3,4\}\times \{5,6,7,8\}$

= {(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)} ...(i)

and, $A\times C=(1,2)\times (5,6)$

= {(1, 5), (1, 6), (2, 5), (2, 6)}...(ii)

Clearly from equation (i) and equation (ii), we get

$A\times C\subset B\times D$

Hence verified.

(ii) We have,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

$\therefore B\cap C\{1,2,3,4\}\cap \{5,6\}=\varphi$

$A\times (B\cap C)=\{1,2\}\times \varphi =\varphi$ ...(i)

Now,

$A\times B=\{1,2\}\times \{1,2,3,4\}$

$=\left\{\right(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4\left)\right\}$

and, $A\times C=\{1,2\}\times \{5,6\}$

= {(1, 5), (1, 6), (2, 5), (2, 6)}

$\therefore (A\times B)\cap (A\times C)=\varphi$ ...(ii)

From equation (i) and equation (ii), we get

$A\times (B\cap C)=(A\times B)\cap (A\times C)$

Hence verified.