Question

Show that for any sets A and B,
(i) $A=\left(A\cap B\right)\cup \left(A-B\right)$
(ii) $A\cup \left(B-A\right)=A\cup B$

Answer 1

(i)

$$\begin{gathered} \mathrm{RHS}=\left(A\cap B\right)\cup \left(A-B\right) \\ \Rightarrow \mathrm{RHS}=\left(A\cap B\right)\cup \left(A\cap B\text{'}\right) \\ \Rightarrow \mathrm{RHS}=\left[\left(A\cap B\right)\cup A\right]\cap \left(\left(A\cap B\right)\cup B\text{'}\right) \\ \Rightarrow \mathrm{RHS}=A\cap \left[\left(A\cup B\text{'}\right)\cap \left(B\cup B\text{'}\right)\right] \\ \Rightarrow \mathrm{RHS}=A\cap \left[\left(A\cup B\text{'}\right)\cap U\right] \\ \Rightarrow \mathrm{RHS}=A\cap \left(A\cup B\text{'}\right) \\ \Rightarrow \mathrm{RHS}=A=\mathrm{LHS} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{LHS}=A\cup \left(B-A\right) \\ \Rightarrow \mathrm{LHS}=A\cup \left(B\cap A\text{'}\right) \\ \Rightarrow \mathrm{LHS}=\left(A\cup B\right)\cap \left(A\cup A\text{'}\right) \\ \Rightarrow \mathrm{LHS}=\left(A\cup B\right)\cap \mathrm{U} \\ \Rightarrow \mathrm{LHS}=\mathrm{A}\cup \mathrm{B}=\mathrm{RHS} \end{gathered}$$