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Question Number 93567 by oustmuchiya@gmail.com last updated on 13/May/20

Given that A={0,1,3,5} B={1,2,4,7} and C={1,2,3,5,8} prove that  (i) (A∩B)∩C = A∩(B∩C)  (ii) (A∪B)∪C = A∪(B∪C)  (iii) (A∪B)∩C = (A∪C)∪(B∩C)  (iv) (A∩C)∪B = (A∪B)∩(C∪B)

$${Given}\:{that}\:{A}=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5}\right\}\:{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{7}\right\}\:{and}\:{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\}\:{prove}\:{that} \\ $$$$\left(\mathrm{i}\right)\:\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{C}\:=\:\mathrm{A}\cap\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{ii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C}\:=\:\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right) \\ $$$$\left(\mathrm{iii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\:=\:\left(\mathrm{A}\cup\mathrm{C}\right)\cup\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{iv}\right)\:\left(\mathrm{A}\cap\mathrm{C}\right)\cup\mathrm{B}\:=\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{C}\cup\mathrm{B}\right) \\ $$

Commented by Ritu Jain last updated on 13/May/20

( i) (A∩B)∩C = A∩(B∩C)         (A∩B)={0,1,3,5}∩{1,2,3,7}={1}  LHS= (A∩B)∩C=({1})∩{1,2,3,5,8}={1}            B∩C={1,2,4,7}∩{1,2,3,5,8}={1}   RHS=A∩(B∩C)={0,1,3,5}∩({1}={1}  LHS=RHS

$$\left(\:\mathrm{i}\right)\:\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{C}\:=\:\mathrm{A}\cap\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{A}\cap\mathrm{B}\right)=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5}\right\}\cap\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{7}\right\}=\left\{\mathrm{1}\right\} \\ $$$$\mathrm{LHS}=\:\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{C}=\left(\left\{\mathrm{1}\right\}\right)\cap\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\}=\left\{\mathrm{1}\right\} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{B}\cap\mathrm{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{7}\right\}\cap\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\}=\left\{\mathrm{1}\right\} \\ $$$$\:\mathrm{RHS}=\mathrm{A}\cap\left(\mathrm{B}\cap\mathrm{C}\right)=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5}\right\}\cap\left(\left\{\mathrm{1}\right\}=\left\{\mathrm{1}\right\}\right. \\ $$$$\mathrm{LHS}=\mathrm{RHS} \\ $$

Answered by otchereabdullai@gmail.com last updated on 14/May/20

ii.( AUB)= {0,1,2,3,4,5,7}       C={1,2,3,5,8}  ∴ {AUB}UC={0,1,2,3,4,5,7,8}  Also  A={0,1,3,5}  {BUC}={1,2,3,4,5,7,8}  ∴AU(BUC)={0,1,2,3,4,5,7,8}  Now since  (AUB)UC=AU(BUC) ={0,1,2,3,4,5,7,8}  we can conclude that  (AUB)UC=AU(BUC).       iii. (AUB)∩C=(AUC)U(B∩C)  (AUB)={0,1,2,3,4,5,7}  C={1,2,3,5,8}  ∴( AUB)∩C={0,1,2,3,4,5,7}∩{1,2,3,5,8}                     ={1,2,3,5}       also   ( AUC)= {0,1,2,3,5,8}    (B∩C)=  {1,2}  ∴ (AUC)U(B∩C)={0,1,2,3,5,8}  ⇒(AUB)∩C≠(AUC)U(B∩C)

$$\mathrm{ii}.\left(\:\mathrm{AUB}\right)=\:\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{7}\right\} \\ $$$$\:\:\:\:\:\mathrm{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\} \\ $$$$\therefore\:\left\{\mathrm{AUB}\right\}\mathrm{UC}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{7},\mathrm{8}\right\} \\ $$$$\mathrm{Also} \\ $$$$\mathrm{A}=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5}\right\} \\ $$$$\left\{\mathrm{BUC}\right\}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{7},\mathrm{8}\right\} \\ $$$$\therefore\mathrm{AU}\left(\mathrm{BUC}\right)=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{7},\mathrm{8}\right\} \\ $$$$\mathrm{Now}\:\mathrm{since} \\ $$$$\left(\mathrm{AUB}\right)\mathrm{UC}=\mathrm{AU}\left(\mathrm{BUC}\right)\:=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{7},\mathrm{8}\right\} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{conclude}\:\mathrm{that} \\ $$$$\left(\mathrm{AUB}\right)\mathrm{UC}=\mathrm{AU}\left(\mathrm{BUC}\right). \\ $$$$ \\ $$$$ \\ $$$$\:\mathrm{iii}.\:\left(\mathrm{AUB}\right)\cap\mathrm{C}=\left(\mathrm{AUC}\right)\mathrm{U}\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{AUB}\right)=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{7}\right\} \\ $$$$\mathrm{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\} \\ $$$$\therefore\left(\:\mathrm{AUB}\right)\cap\mathrm{C}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{7}\right\}\cap\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5}\right\} \\ $$$$\:\:\:\:\:\mathrm{also} \\ $$$$\:\left(\:\mathrm{AUC}\right)=\:\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\} \\ $$$$\:\:\left(\mathrm{B}\cap\mathrm{C}\right)=\:\:\left\{\mathrm{1},\mathrm{2}\right\} \\ $$$$\therefore\:\left(\mathrm{AUC}\right)\mathrm{U}\left(\mathrm{B}\cap\mathrm{C}\right)=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\} \\ $$$$\Rightarrow\left(\mathrm{AUB}\right)\cap\mathrm{C}\neq\left(\mathrm{AUC}\right)\mathrm{U}\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$

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