Question
For any two sets A and B, prove that
$\text{A}\cup(\text{B}-\text{A})=\text{A}\cup\text{B}$
$\text{A}\cup(\text{B}-\text{A})=\text{A}\cup\text{B}$
✓
Answer
Let $\text{x}\in \text{A} \cup \text{(B} - \text{A)}\Rightarrow\text{x}\in\text{A or x}\in\text{(B} - \text{A)}$
$\Rightarrow \text{x} \in \text{A or x} \in \text{B and x}\not\in \text{A}$
$\Rightarrow \text{x}\in \text{B}$
$\Rightarrow \text{x}\in \text{A}\cup\text{B}$ $[\because \text{B}\subset\text{(A}\cup\text{B})]$
This is true for all $\text{x}\in\text{A}\cup\text{(B} - \text{A)}$
$\therefore \text{A}\cup\text{(B} - \text{A)}\subset\text{(A}\cup \text{B})....\text{(i)}$
Conversely,
Let, $\text{x}\in \text{(A}\cup\text{B)}$
$\Rightarrow\text{x} \in \text{A or x} \in\text{B}$
$\Rightarrow \text{x}\in \text{A or x} \in \text{(B – A)}$ $[\because\text{B}\subset \text{(B – A)]}$
$\therefore\text{(A}\cup\text{B}) \subset \text{A}\cup \text{(B}- \text{A)}.....\text{(ii)}$
From (i) and (ii), we get
$\text{A}\cup\text{(B} - \text{A) = (A}\cup\text{B}).$
$\Rightarrow \text{x} \in \text{A or x} \in \text{B and x}\not\in \text{A}$
$\Rightarrow \text{x}\in \text{B}$
$\Rightarrow \text{x}\in \text{A}\cup\text{B}$ $[\because \text{B}\subset\text{(A}\cup\text{B})]$
This is true for all $\text{x}\in\text{A}\cup\text{(B} - \text{A)}$
$\therefore \text{A}\cup\text{(B} - \text{A)}\subset\text{(A}\cup \text{B})....\text{(i)}$
Conversely,
Let, $\text{x}\in \text{(A}\cup\text{B)}$
$\Rightarrow\text{x} \in \text{A or x} \in\text{B}$
$\Rightarrow \text{x}\in \text{A or x} \in \text{(B – A)}$ $[\because\text{B}\subset \text{(B – A)]}$
$\therefore\text{(A}\cup\text{B}) \subset \text{A}\cup \text{(B}- \text{A)}.....\text{(ii)}$
From (i) and (ii), we get
$\text{A}\cup\text{(B} - \text{A) = (A}\cup\text{B}).$
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