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Question-115418




Question Number 115418 by mohammad17 last updated on 25/Sep/20
Commented by mohammad17 last updated on 25/Sep/20
help me sir i want (H.W)
$${help}\:{me}\:{sir}\:{i}\:{want}\:\left({H}.{W}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 25/Sep/20
For any Set  A∩∅=A  A∩A    =A  Suppose x∈A   So x∈A∩A  And A∩A=A  A∩C=C∩A  Suppose x∈A   and x∈C  x∈A∩C  x∈C∩A  So C∩A=A∩C
$$\mathrm{For}\:\mathrm{any}\:\mathrm{Set} \\ $$$$\mathrm{A}\cap\varnothing=\mathrm{A} \\ $$$$\mathrm{A}\cap\mathrm{A}\:\:\:\:=\mathrm{A} \\ $$$$\mathrm{Suppose}\:\mathrm{x}\in\mathrm{A}\: \\ $$$$\mathrm{So}\:\mathrm{x}\in\mathrm{A}\cap\mathrm{A} \\ $$$$\mathrm{And}\:\mathrm{A}\cap\mathrm{A}=\mathrm{A} \\ $$$$\mathrm{A}\cap\mathrm{C}=\mathrm{C}\cap\mathrm{A} \\ $$$$\mathrm{Suppose}\:\mathrm{x}\in\mathrm{A}\:\:\:\mathrm{and}\:\mathrm{x}\in\mathrm{C} \\ $$$$\mathrm{x}\in\mathrm{A}\cap\mathrm{C} \\ $$$$\mathrm{x}\in\mathrm{C}\cap\mathrm{A} \\ $$$$\mathrm{So}\:\mathrm{C}\cap\mathrm{A}=\mathrm{A}\cap\mathrm{C} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 25/Sep/20
x is an element of set A (x∈A)  And (x∈B)  And x∈( A∩B ) (By definition)  And (A∩B)⊆A   Also (A∩B)⊆B
$$\mathrm{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{element}\:\mathrm{of}\:\mathrm{set}\:\mathrm{A}\:\left(\mathrm{x}\in\mathrm{A}\right) \\ $$$$\mathrm{And}\:\left(\mathrm{x}\in\mathrm{B}\right) \\ $$$$\mathrm{And}\:\mathrm{x}\in\left(\:\mathrm{A}\cap\mathrm{B}\:\right)\:\left(\mathrm{By}\:\mathrm{definition}\right) \\ $$$$\mathrm{And}\:\left(\mathrm{A}\cap\mathrm{B}\right)\subseteq\mathrm{A}\: \\ $$$$\mathrm{Also}\:\left(\mathrm{A}\cap\mathrm{B}\right)\subseteq\mathrm{B} \\ $$$$ \\ $$
Commented by mohammad17 last updated on 25/Sep/20
yes sir i want other question in same this method
$${yes}\:{sir}\:{i}\:{want}\:{other}\:{question}\:{in}\:{same}\:{this}\:{method} \\ $$
Commented by Dwaipayan Shikari last updated on 25/Sep/20
It is better to use Venn diagram in Q7
$$\mathrm{It}\:\mathrm{is}\:\mathrm{better}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Venn}\:\mathrm{diagram}\:\mathrm{in}\:\mathrm{Q7} \\ $$
Answered by MWSuSon last updated on 25/Sep/20
by using the definition of set equality  and logical equivalence  X=Y iff  ∀x∈U, (x∈X⇔x∈Y) we can  show 6 and 7.  6)let x∈(A∩(B∪C))⇔x∈A∧x∈(B∪C)                            ⇔x∈A∧(x∈B∨x∈C)                      ⇔(x∈A∧x∈B)∨(x∈A∧x∈C)                        ⇔x∈(A∩B)∪(A∩C)  therefore A∩(B∪C)=(A∩B)∪(A∩C)  same way for 7
$$\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{set}\:\mathrm{equality} \\ $$$$\mathrm{and}\:\mathrm{logical}\:\mathrm{equivalence} \\ $$$$\mathrm{X}=\mathrm{Y}\:\mathrm{iff}\:\:\forall\mathrm{x}\in\mathrm{U},\:\left(\mathrm{x}\in\mathrm{X}\Leftrightarrow\mathrm{x}\in\mathrm{Y}\right)\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{show}\:\mathrm{6}\:\mathrm{and}\:\mathrm{7}. \\ $$$$\left.\mathrm{6}\right)\mathrm{let}\:\mathrm{x}\in\left(\mathrm{A}\cap\left(\mathrm{B}\cup\mathrm{C}\right)\right)\Leftrightarrow\mathrm{x}\in\mathrm{A}\wedge\mathrm{x}\in\left(\mathrm{B}\cup\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Leftrightarrow\mathrm{x}\in\mathrm{A}\wedge\left(\mathrm{x}\in\mathrm{B}\vee\mathrm{x}\in\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Leftrightarrow\left(\mathrm{x}\in\mathrm{A}\wedge\mathrm{x}\in\mathrm{B}\right)\vee\left(\mathrm{x}\in\mathrm{A}\wedge\mathrm{x}\in\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Leftrightarrow\mathrm{x}\in\left(\mathrm{A}\cap\mathrm{B}\right)\cup\left(\mathrm{A}\cap\mathrm{C}\right) \\ $$$$\mathrm{therefore}\:\mathrm{A}\cap\left(\mathrm{B}\cup\mathrm{C}\right)=\left(\mathrm{A}\cap\mathrm{B}\right)\cup\left(\mathrm{A}\cap\mathrm{C}\right) \\ $$$$\mathrm{same}\:\mathrm{way}\:\mathrm{for}\:\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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