Menu Close

If-A-B-C-and-D-are-any-four-sets-then-i-A-B-C-D-A-C-B-D-ii-A-B-C-D-A-C-B-D-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *