Question

If A = {1, 2, 3}, B = {4} and C = {5}, then verify that :

(i) $A\times (B\cup C)=(A\times B)\cup (A\times C)$

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

(iii) $A\times (B-C)=(A\times B)-(A\times C)$

Answer 1

(i) We have,

A = {1, 2, 3}, B = {4} and C = {5}

$\therefore B\cup C=\left\{4\right\}\cup \left\{5\right\}=\{4,5\}$

$\therefore A\times (B\cup C)=\{1,2,3\}=\{4,5\}$

$\Rightarrow A\times (B\cup C)$

$=\left\{\right(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)$ ....(i)

Now,

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

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

and, $A\times C=\{1,2,3\}\times \left\{5\right\}$

$=\left\{\right(1,5),(2,5),(3,5\left)\right\}$

$\therefore (A\times B)\cup (A\times C)=\left\{\right(1,4),(2,4),(3,4\left)\right\}\cup \left\{\right(1,5),(2,5),(3,5\left)\right\}$

$\Rightarrow (A\times B)\cup (A\times C)=\left\{\right(1,4),(1,5),(2,4),(2,5),(3,4),(3,5\left)\right\}$ ...(i)

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

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

Hence verified.

(ii) We have,

A = {1, 2, 3}, B = {4} and C = {5}

$\therefore B\cap C=\left\{4\right\}\cap \left\{5\right\}=\varphi$

$\therefore A\times (B\cap C)=\{1,2,3\}\times \varphi$

$\Rightarrow A\times (B\cap C)=\varphi$ ...(i)

Now,

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

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

and, $A\times C=\{1,2,3\}\times \left\{5\right\}$

$=\left\{\right(1,5),(2,5),(3,5\left)\right\}$

$\therefore (A\times B)\cap (A\times C)=\left\{\right(1,4),(2,4),(3,4\left)\right\}\cap \left\{\right(1,5),(2,5),(3,5\left)\right\}$

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

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

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

Hence verified.

(iii) We have,

A = {1, 2, 3}, B = {4} and C = {5}

$\therefore B-C=\left\{4\right\}$

$\therefore A\times (B-C)=\{1,2,3\}\times \left\{4\right\}$

$\Rightarrow A\times (B-C)=\left\{\right(1,4),(2,4),(3,4\left)\right\}$ ...(i)

Now,

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

$=(1,4),(2,4),(3,4)$

and $A\times C=\{1,2,3\}\times \left\{5\right\}$

$=\left\{\right(1,5),(2,5),(3,5\left)\right\}$

$\therefore (A\times B)-(A\times C)=\left\{\right(1,4),(2,4),(3,4\left)\right\}$ .....(ii)

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

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

Hence verified.