Question
Prove that:
$(\text{A}\cap\text{B})\times\text{C}=(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$

Answer

 Let (a, b) bean arbitrary element of $(\text{A}\cap\text{B})\times\text{C}.$ Then,

$(\text{a},\text{b})\in(\text{A}\cap\text{B})\times\text{C}$

$\Rightarrow\text{a}\in\text{A}\cap\text{B}$ and $\text{b}\in\text{C}$

$\Rightarrow(\text{a}\in\text{A}\text{ or a}\in\text{B})$ and $\text{b}\in\text{C}$ [By defination]

$\Rightarrow(\text{a}\in\text{A}\text{ and b}\in\text{C})\text{ or }(\text{a}\in\text{B}\text{ and b}\in\text{C})$

$\Rightarrow(\text{a},\text{b})\in\text{A}\times\text{C and (a, b)}\in\text{B}\times\text{C}$

$\Rightarrow(\text{a},\text{b})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$

$\Rightarrow(\text{a, b})\in(\text{A}\cap\text{B})\times\text{C}$

$\Rightarrow(\text{a, b})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$

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

Let (x, y) be an arbitrary element of $(\text{A}\times\text{C})\cap(\text{B}\times\text{C}).$ Then,

$(\text{x, y})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$

$\Rightarrow(\text{x, y)}\in\text{A}\times\text{C}\text{ and (x, y)}\in\text{B}\times\text{C}$ [By defination]

$\Rightarrow(\text{x}\in\text{A and y}\in\text{C) and (x}\in\text{B and y}\in\text{C})$

$\Rightarrow(\text{x}\in\text{A or x}\in\text{B})\text{ and y}\in\text{C}$

$\Rightarrow\text{x}\in\text{A}\cap\text{B and y}\in\text{C}$

$\Rightarrow(\text{x},\text{y})\in(\text{A}\cap\text{B})\times\text{C}$

$\Rightarrow(\text{x},\text{y})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$

$\Rightarrow(\text{x, y})\in(\text{A}\cap\text{B})\times\text{C}$

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

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

$(\text{A}\cap\text{B})\times\text{C}=(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$

Hence proved. 

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