Question Number 3130 by Rasheed Soomro last updated on 05/Dec/15
$${If}\:{A},{B},{C}\:{and}\:{D}\:{are}\:{any}\:{four}\:{sets}\:{then} \\ $$$$\left({i}\right)\:\:\left({A}−{B}\right)\cup\left({C}−{D}\right)\overset{?} {=}\left({A}\cup{C}\right)−\left({B}\cup{D}\right) \\ $$$$\left({ii}\right)\:\left({A}−{B}\right)\cup\left({C}−{D}\right)\overset{?} {=}\left({A}\cup{C}\right)−\left({B}\cap{D}\right) \\ $$
Answered by prakash jain last updated on 05/Dec/15
$$\left({ii}\right) \\ $$$$\left({A}−{B}\right)\cup\left({C}−{D}\right)=\left({A}\cup{C}\right)−\left({B}\cap{D}\right) \\ $$$${A}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\} \\ $$$${B}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\} \\ $$$${C}=\left\{\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$${D}=\left\{\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$${A}−{B}=\emptyset \\ $$$${C}−{D}=\emptyset \\ $$$$\mathrm{LHS}=\left({A}−{B}\right)\cup\left({C}−{D}\right)=\emptyset \\ $$$${A}\cup{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$${B}\cap{D}=\emptyset \\ $$$$\mathrm{RHS}=\left({A}\cup{C}\right)−\left({B}\cap{D}\right)=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$$\mathrm{LHS}\neq\mathrm{RHS} \\ $$
Commented by Rasheed Soomro last updated on 06/Dec/15
$$\mathcal{T}^{\mathcal{HAN}} \mathcal{K}_{\mathcal{S}} \:! \\ $$
Answered by prakash jain last updated on 05/Dec/15
$$\left({i}\right) \\ $$$${A}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\} \\ $$$$\mathrm{B}=\left\{\mathrm{2},\mathrm{3},\mathrm{6},\mathrm{7}\right\} \\ $$$$\mathrm{C}=\left\{\mathrm{2},\mathrm{3},\mathrm{8},\mathrm{9}\right\} \\ $$$$\mathrm{D}=\left\{\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$$\mathrm{A}−\mathrm{B}=\left\{\mathrm{1},\mathrm{4},\mathrm{5}\right\} \\ $$$$\mathrm{C}−\mathrm{D}=\left\{\mathrm{2},\mathrm{3}\right\} \\ $$$$\mathrm{LHS}=\left({A}−{B}\right)\cup\left({C}−{D}\right)=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\} \\ $$$$\left(\mathrm{A}\cup\mathrm{C}\right)=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{8},\mathrm{9}\right\} \\ $$$$\left(\mathrm{B}\cup\mathrm{D}\right)=\left\{\mathrm{2},\mathrm{3},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$$\mathrm{RHS}=\left(\mathrm{A}\cup\mathrm{C}\right)−\left(\mathrm{B}\cup\mathrm{D}\right)=\left\{\mathrm{1},\mathrm{4},\mathrm{5}\right\} \\ $$$$\mathrm{LHS}\neq\mathrm{RHS} \\ $$
Commented by Rasheed Soomro last updated on 06/Dec/15
$$\mathcal{T}_{\mathcal{HAN}} \mathcal{K}^{\mathcal{S}\:!} \\ $$