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Question-26889




Question Number 26889 by shiva123 last updated on 31/Dec/17
Answered by prakash jain last updated on 31/Dec/17
3. A∩(B−C)=(A∩B)−(A∩C)  x∈A∩(B−C)  ⇒x∈A ∧x∈(B−C)  ⇒x∈A∧x∈B∧x∉C  ⇒x∈(A∩B)∧x∉(A∩C)  ⇒x∈(A∩B)−(A∩C)  ⇒(A∩B)−(A∩C)⊇A∩(B−C)  (I)  x∈(A∩B)−(A∩C)  ⇒x∈(A∩B)∧x∉(A∩C)  ⇒x∈A∧x∈B∧x∉(A∩C)  ⇒x∈A∧x∈B∧x∉C  ⇒x∈A∧x∈(B−C)  ⇒x∈A∩(B−C)    ⇒(A∩B)−(A∩C)⊆A∩(B−C)  (II)  From (I) and (II)  (A∩B)−(A∩C)=A∩(B−C)  ■
$$\mathrm{3}.\:{A}\cap\left({B}−{C}\right)=\left({A}\cap{B}\right)−\left({A}\cap{C}\right) \\ $$$${x}\in{A}\cap\left({B}−{C}\right) \\ $$$$\Rightarrow{x}\in{A}\:\wedge{x}\in\left({B}−{C}\right) \\ $$$$\Rightarrow{x}\in{A}\wedge{x}\in{B}\wedge{x}\notin{C} \\ $$$$\Rightarrow{x}\in\left({A}\cap{B}\right)\wedge{x}\notin\left({A}\cap{C}\right) \\ $$$$\Rightarrow{x}\in\left({A}\cap{B}\right)−\left({A}\cap{C}\right) \\ $$$$\Rightarrow\left({A}\cap{B}\right)−\left({A}\cap{C}\right)\supseteq{A}\cap\left({B}−{C}\right)\:\:\left({I}\right) \\ $$$${x}\in\left({A}\cap{B}\right)−\left({A}\cap{C}\right) \\ $$$$\Rightarrow{x}\in\left({A}\cap{B}\right)\wedge{x}\notin\left({A}\cap{C}\right) \\ $$$$\Rightarrow{x}\in{A}\wedge{x}\in{B}\wedge{x}\notin\left({A}\cap{C}\right) \\ $$$$\Rightarrow{x}\in{A}\wedge{x}\in{B}\wedge{x}\notin{C} \\ $$$$\Rightarrow{x}\in{A}\wedge{x}\in\left({B}−{C}\right) \\ $$$$\Rightarrow{x}\in{A}\cap\left({B}−{C}\right)\:\: \\ $$$$\Rightarrow\left({A}\cap{B}\right)−\left({A}\cap{C}\right)\subseteq{A}\cap\left({B}−{C}\right)\:\:\left({II}\right) \\ $$$$\mathrm{From}\:\left({I}\right)\:\mathrm{and}\:\left({II}\right) \\ $$$$\left({A}\cap{B}\right)−\left({A}\cap{C}\right)={A}\cap\left({B}−{C}\right)\:\:\blacksquare \\ $$

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