Question

Prove that : (i) $(A\cup B)\times C=(A\times C)\cup (B\times C)$

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

Answer 1

(i) Let (a, b) be an arbitary element of $(A\cup B)\times C.$ Then,

$(a,b)ϵ(A\cup B)\times C$

$\Rightarrow aϵA\cup B$ and $bϵC$ [By defination]

$\Rightarrow \left(aϵA\right)$ and $bϵC)$ or (a \epsilon A)\) and $bϵC)$

$\Rightarrow (a,b)ϵA\times C$ or (a, b) $ϵB\times C$

$\Rightarrow (a,b)ϵ(A\times C)\cup (B\times C)$

$\Rightarrow (a,b)ϵ(A\cup B)\times C$

$\Rightarrow (a,b)ϵ(A\times C)\cup (B\times C)$

$\Rightarrow (A\cup B)\times C\subseteq (A\times C)\cup (B\times C)$ ....(i)

Again, let (x, y) be an arbitary element of $(A\times C)\cup (B\times C)$. Then,

$(x,y)ϵ(A\times C)\cup (B\times C)$

$\Rightarrow (x,y)ϵA\times C$ or (x, y) \epsilon B \times C\)

$\Rightarrow xϵA$ and $yϵ$ C or $xϵB$ and $yϵC$

$\Rightarrow (xϵA\cup B)$ and $yϵC$

$\Rightarrow (x,y)ϵ(A\cup B)\times C$

$\Rightarrow (x,y)ϵ(A\times C)\cup (B\times C)$

$\Rightarrow (x,y)ϵ(A\cup B)\times C$

$\Rightarrow (A\times C)\cup (B\times C)\subseteq (A\cup B)\times C$ ....(i)

Using equation (i) and equation (ii), we get $(A\cup B)\times C=(A\times C)\cup (B\times C)$

Hence proved.

(ii) Let (a, b) be an arbitary element of $(A\cap B)\times C.$ Then,

$(a,b)ϵ(A\cap B)\times C$

$\Rightarrow aϵA\cap B$ and $bϵC$

$\Rightarrow (aϵA$ and $aϵB)$ and $bϵC$

[By defination]

$\Rightarrow (aϵA$ and $bϵC)$ and $(aϵB$ and $bϵC)$

$\Rightarrow (a,b)ϵA\times C$ and $(a,b)ϵB\times C$

$\Rightarrow (a,b)ϵ(A\times C)\cap (B\times C)$

$\Rightarrow (a,b)ϵ(A\cap C)\times C$

$\Rightarrow (a,b)ϵ(A\times C)\cap (B\times C)$

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

Let (x, y) be an arbitary element of $(A\times C)\cap (B\times C)$. Then,

$(x,y)ϵ(A\times C)\cap (B\times C)$

$\Rightarrow (x,y)ϵA\times C$ and $(x,y)ϵB\times C$

[By defination]

$\Rightarrow \left(xϵAandyϵC\right)$ and ($xϵB$ and $yϵC)$

$\Rightarrow (xϵA$ and $xϵB)$ and $yϵC$

$\Rightarrow xϵA\cap B$ and $yϵC$

$\Rightarrow (x,y)ϵ(A\cap B)\times C$

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

Using equation (i) and equation (ii), we get $(A\cap B)\times C=(A\times C)\cap (B\times C)$

Hence proved.