Question
Show that for any sets A and B,$\text{A}\cup(\text{B}-\text{A})=(\text{A}\cup\text{B})$

Answer

Let $\text{x}\in\text{A}\cup\text{(B - A)}$ $\Rightarrow\text{x}\in\text{A or x}\in(\text{B - A})$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B and x}\not\in\text{A}$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B}$ $\Rightarrow\text{x}\in\text{(A}\cup\text{B})$ $\therefore\text{A}\cup\text{(B - A)}\subset\text{(A}\cup\text{B}).....\text{(i)}$ Let and $\text{x}\in\text{(A}\cup\text{B})$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B}$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B and x}\not\in\text{A}$ $\Rightarrow\text{x}\in\text{A or x}\in\text{(B - A)}$ $\Rightarrow\text{x}\in\text{A}\cup\text{(B - A)}$ $\therefore(\text{A}\cup\text{B})\subset\text{A}\cup\text{(B - A)}.....\text{(ii)}$ From (i) and (ii), we get $\text{A}\cup\text{(B - A)}=\text{A}\cup\text{B.}$

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