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Set TheoryQuestion and Answers: Page 7

Question Number 1698 Answers: 2 Comments: 0

•Are A∪B=A∩B and A=B completely equivalent? •Simplify A∪B=A∩B to A=B using set operations and their properties.

$$\bullet{Are}\:\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\:\:{and}\:\:\boldsymbol{\mathrm{A}}=\boldsymbol{\mathrm{B}}\: \\ $$$${completely}\:{equivalent}? \\ $$$$\bullet{Simplify}\:\boldsymbol{\mathrm{A}}\cup\boldsymbol{\mathrm{B}}=\boldsymbol{\mathrm{A}}\cap\boldsymbol{\mathrm{B}}\:{to}\:\boldsymbol{\mathrm{A}}=\boldsymbol{\mathrm{B}} \\ $$$${using}\:{set}\:{operations}\:{and}\:{their} \\ $$$${properties}. \\ $$

Question Number 1694 Answers: 2 Comments: 0

proof that for two set A and B (or give a counter example) ∣A∪B∣≥∣A∩B∣

$$\mathrm{proof}\:\mathrm{that}\:\mathrm{for}\:\mathrm{two}\:\mathrm{set}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\left(\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\right) \\ $$$$\mid\mathrm{A}\cup\mathrm{B}\mid\geqslant\mid\mathrm{A}\cap\mathrm{B}\mid \\ $$

Question Number 1682 Answers: 0 Comments: 5

Let ∣ S ∣ denotes number of elements in a set S , N and R are sets of natural and real numbers respectively: ∣ N ∣=^(?) ∣ R ∣

$${Let}\:\mid\:\mathrm{S}\:\mid\:{denotes}\:{number}\:{of}\:{elements}\:{in}\:{a}\:{set}\:\mathrm{S}\:,\: \\ $$$$\mathbb{N}\:{and}\:\mathbb{R}\:{are}\:{sets}\:{of}\:{natural}\:{and}\:{real}\:{numbers}\: \\ $$$${respectively}: \\ $$$$\mid\:\mathbb{N}\:\mid\overset{?} {=}\mid\:\mathbb{R}\:\mid \\ $$

Question Number 1672 Answers: 2 Comments: 0

lets A and B be two finite sets, proof (or give a counter example) that ∣A∪B∣≤∣A∩B∣ ⇒ A=B

$$\mathrm{lets}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{be}\:\mathrm{two}\:\mathrm{finite}\:\mathrm{sets},\:\mathrm{proof}\:\left(\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\right)\:\mathrm{that} \\ $$$$\mid\mathrm{A}\cup\mathrm{B}\mid\leqslant\mid\mathrm{A}\cap\mathrm{B}\mid\:\Rightarrow\:\mathrm{A}=\mathrm{B} \\ $$

Question Number 1616 Answers: 0 Comments: 3

lets two sets A,B and take ∣X∣ the number of elements of the set X, them proof or give a counter example that if ∣A∪B∣=∞ and ∣A∩B∣=∞ then ∣A∣=∞ and ∣B∣=∞

$$\mathrm{lets}\:\mathrm{two}\:\mathrm{sets}\:\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{take}\:\mid\mathrm{X}\mid\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{the}\:\mathrm{set}\:\mathrm{X},\:\mathrm{them} \\ $$$$\mathrm{proof}\:\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\:\mathrm{that} \\ $$$$\mathrm{if}\:\mid\mathrm{A}\cup\mathrm{B}\mid=\infty\:\mathrm{and}\:\mid\mathrm{A}\cap\mathrm{B}\mid=\infty\:\mathrm{then}\:\mid\mathrm{A}\mid=\infty\:\mathrm{and}\:\mid\mathrm{B}\mid=\infty \\ $$

Question Number 1132 Answers: 0 Comments: 1

let S={1,2,3,4,5}, if A,B,C is such that A∩B∩C=∅ A∩B≠∅ A∩C≠∅ how many ways can be choose A,B and C

$$\mathrm{let}\:\mathrm{S}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\},\:\mathrm{if}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}=\varnothing \\ $$$$\mathrm{A}\cap\mathrm{B}\neq\varnothing \\ $$$$\mathrm{A}\cap\mathrm{C}\neq\varnothing \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{be}\:\mathrm{choose}\:\mathrm{A},\mathrm{B}\:\mathrm{and} \\ $$$$\mathrm{C} \\ $$

Question Number 245 Answers: 1 Comments: 0

A={0,2,4,6,8,10} B={0,1,3,4,6,7,9} C={1,2,4,5,7,8,10} ∣(A∪B)∩(B∪C)∣

$$\mathrm{A}=\left\{\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8},\mathrm{10}\right\} \\ $$$$\mathrm{B}=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{9}\right\} \\ $$$$\mathrm{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{7},\mathrm{8},\mathrm{10}\right\} \\ $$$$\mid\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{B}\cup\mathrm{C}\right)\mid \\ $$

Question Number 242 Answers: 1 Comments: 0

give the sets A={x∈N:0≤x≤9} B={0} C={1} D={2,3,5,7} E={4,6,8,9} compute ∣A\B∣+∣A\C∣+∣A\D∣+∣A\E∣ where ∣X∣=number of elements of X X\Y={x:x∈X e x∉Y} X−Y={x:x∈X e x∉Y}

$$\mathrm{give}\:\mathrm{the}\:\mathrm{sets} \\ $$$$\mathrm{A}=\left\{{x}\in\mathbb{N}:\mathrm{0}\leqslant{x}\leqslant\mathrm{9}\right\} \\ $$$$\mathrm{B}=\left\{\mathrm{0}\right\} \\ $$$$\mathrm{C}=\left\{\mathrm{1}\right\} \\ $$$$\mathrm{D}=\left\{\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7}\right\} \\ $$$$\mathrm{E}=\left\{\mathrm{4},\mathrm{6},\mathrm{8},\mathrm{9}\right\} \\ $$$$\mathrm{compute} \\ $$$$\mid\mathrm{A}\backslash\mathrm{B}\mid+\mid\mathrm{A}\backslash\mathrm{C}\mid+\mid\mathrm{A}\backslash\mathrm{D}\mid+\mid\mathrm{A}\backslash\mathrm{E}\mid \\ $$$$\mathrm{where} \\ $$$$\mid\mathrm{X}\mid=\mathrm{number}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{X} \\ $$$$\mathrm{X}\backslash\mathrm{Y}=\left\{{x}:{x}\in\mathrm{X}\:\boldsymbol{\mathrm{e}}\:{x}\notin\mathrm{Y}\right\} \\ $$$$\mathrm{X}−\mathrm{Y}=\left\{{x}:{x}\in\mathrm{X}\:\boldsymbol{\mathrm{e}}\:{x}\notin\mathrm{Y}\right\} \\ $$

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