Question
Prove that: $(\text{A}\cup\text{B})\times\text{C}=(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$
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Answer
Let (a, b) bean arbitrary element of $(\text{A}\cup\text{B})\times\text{C}.$ Then, $(\text{a},\text{b})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow\text{a}\in\text{A}\cup\text{B}$ and $\text{b}\in\text{C}$ [By defination] $\Rightarrow(\text{a}\in\text{A}\text{ or a}\in\text{B})$ and $\text{b}\in\text{C}$ [By defination] $\Rightarrow(\text{a}\in\text{A}\text{ and b}\in\text{C})\text{ or }(\text{a}\in\text{B}\text{ and b}\in\text{C})$ $\Rightarrow(\text{a},\text{b})\in\text{A}\times\text{C or (a, b)}\in\text{B}\times\text{C}$ $\Rightarrow(\text{a},\text{b})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{a, b})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow(\text{a, b})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{A}\cup\text{B})\times\text{C}\subseteq(\text{A}\times\text{C})\cup(\text{B}\times\text{C})\ ...(\text{i})$ Again, let (x, y) be an abitrary element of $(\text{A}\times\text{C})\cup(\text{B}\times\text{C}).$ Then, $(\text{x, y})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{x, y)}\in\text{A}\times\text{C}\text{ or (x, y)}\in\text{B}\times\text{C}$ $\Rightarrow\text{x}\in\text{A and y}\in\text{C or x}\in\text{B and y}\in\text{C}$ $\Rightarrow(\text{x}\in\text{A or x}\in\text{B})\text{ and y}\in\text{C}$ $\Rightarrow\text{x}\in\text{A}\cup\text{B and y}\in\text{C}$ $\Rightarrow(\text{x},\text{y})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow(\text{x},\text{y})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{x, y})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow(\text{A}\times\text{C})\cup(\text{B}\times\text{C})\subseteq(\text{A}\cup\text{B})\times\text{C}\ ...(\text{ii})$ Using equation (i) and equation (ii), we get $(\text{A}\cup\text{B})\times\text{C}=(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ Hence proved.
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