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If-A-and-B-are-two-sets-and-U-is-a-universal-set-prove-that-A-B-B-A-A-B-




Question Number 1744 by Rasheed Ahmad last updated on 13/Sep/15
If A and B are two sets and U is  a universal set prove that  A ⊆ B  ⇒ B=A ∪ (A′ ∩ B)
$${If}\:\boldsymbol{\mathrm{A}}\:{and}\:\boldsymbol{\mathrm{B}}\:{are}\:{two}\:{sets}\:{and}\:\mathbb{U}\:{is} \\ $$$${a}\:{universal}\:{set}\:{prove}\:{that} \\ $$$$\boldsymbol{\mathrm{A}}\:\subseteq\:\boldsymbol{\mathrm{B}}\:\:\Rightarrow\:\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\:\cup\:\left(\boldsymbol{\mathrm{A}}'\:\cap\:\boldsymbol{\mathrm{B}}\right) \\ $$
Answered by Rasheed Ahmad last updated on 19/Sep/15
A ⊆ B  ⇒ B=A ∪ (A′ ∩ B)  ..........∗∗∗.......................  A ⊆ B  ⇒A∪B=B  Or   B=A∪B           =U∩(A∪B)     [∵ S=U∩S]           =(A∪A′)∩(A∪B)  [∵U=A∪A′]            =A∪(A′∩B)    [∵ (A∪B)∩(A∪C)                                                =A∪(B∩C)]
$$\boldsymbol{\mathrm{A}}\:\subseteq\:\boldsymbol{\mathrm{B}}\:\:\Rightarrow\:\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\:\cup\:\left(\boldsymbol{\mathrm{A}}'\:\cap\:\boldsymbol{\mathrm{B}}\right) \\ $$$$……….\ast\ast\ast………………….. \\ $$$$\boldsymbol{\mathrm{A}}\:\subseteq\:\boldsymbol{\mathrm{B}}\:\:\Rightarrow\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{B}} \\ $$$${Or}\:\:\:\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathbb{U}\cap\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right)\:\:\:\:\:\left[\because\:\boldsymbol{\mathrm{S}}=\mathbb{U}\cap\boldsymbol{\mathrm{S}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{A}}'\right)\cap\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right)\:\:\left[\because\mathbb{U}=\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{A}}'\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\boldsymbol{\mathrm{A}}\cup\left(\boldsymbol{\mathrm{A}}'\cap\boldsymbol{\mathrm{B}}\right)\:\:\:\:\left[\because\:\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right)\cap\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{C}}\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{\mathrm{A}}\cup\left(\boldsymbol{\mathrm{B}}\cap\boldsymbol{\mathrm{C}}\right)\right] \\ $$
Answered by arvind last updated on 18/Sep/15
$$ \\ $$
Answered by Rasheed Soomro last updated on 18/Sep/15
A ⊆ B  ⇒ B=A ∪ (A′ ∩ B)  RHS: A ∪ (A′ ∩ B)=(A∪A′)∩(A∪B)           [ Distributivity of   ∪   over   ∩  ]                            =U∩(A∪B)     [ A∪A′=U ]                            =A∪B          [ U ∩ S=S ]                            =B                [ ∵ A⊆B  ]                            =LHS
$$\boldsymbol{\mathrm{A}}\:\subseteq\:\boldsymbol{\mathrm{B}}\:\:\Rightarrow\:\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\:\cup\:\left(\boldsymbol{\mathrm{A}}'\:\cap\:\boldsymbol{\mathrm{B}}\right) \\ $$$${RHS}:\:\boldsymbol{\mathrm{A}}\:\cup\:\left(\boldsymbol{\mathrm{A}}'\:\cap\:\boldsymbol{\mathrm{B}}\right)=\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{A}}'\right)\cap\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\left[\:{Distributivity}\:{of}\:\:\:\cup\:\:\:{over}\:\:\:\cap\:\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathbb{U}\cap\left(\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\right)\:\:\:\:\:\left[\:\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{A}}'=\mathbb{U}\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}\:\:\:\:\:\:\:\:\:\:\left[\:\mathbb{U}\:\cap\:\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{S}}\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{\mathrm{B}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:\because\:\boldsymbol{\mathrm{A}}\subseteq\boldsymbol{\mathrm{B}}\:\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={LHS} \\ $$

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