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Set TheoryQuestion and Answers: Page 1

Question Number 212974    Answers: 0   Comments: 0

Question Number 212332    Answers: 0   Comments: 4

Please help 1.1.Let XUY=X for all sets X. Prove that Y=0(empty set). From Singler book "Excercises in set theory". I think this task is totaly wrong and cannot be proved. I would ask someone to provide me valid proof of that. I have sets X and Y such as Y is subset of X. For example. If Y={1} and X={1,2} then XUY=X is correct but that doesn't imply Y is empty. Another example when X=Y since X is any set. I can choose X=Y. Why not? Then YUY=Y is always true, but again, that doesnt imply Y is empty set Proof in book claim that is correct if we suppose Y is not empty and if we choose for instance X is empty set. Then 0UY=0 but this is wrong since 0UY=Y. Therefore, Y must be empty?

$$ \\ $$Please help 1.1.Let XUY=X for all sets X. Prove that Y=0(empty set). From Singler book "Excercises in set theory". I think this task is totaly wrong and cannot be proved. I would ask someone to provide me valid proof of that. I have sets X and Y such as Y is subset of X. For example. If Y={1} and X={1,2} then XUY=X is correct but that doesn't imply Y is empty. Another example when X=Y since X is any set. I can choose X=Y. Why not? Then YUY=Y is always true, but again, that doesnt imply Y is empty set Proof in book claim that is correct if we suppose Y is not empty and if we choose for instance X is empty set. Then 0UY=0 but this is wrong since 0UY=Y. Therefore, Y must be empty?

Question Number 211643    Answers: 1   Comments: 0

Let A={x ∈ R∣x^2 <4}and B={y ∈ Q∣y>−3}find A∩B

$$\mathrm{Let}\:\boldsymbol{{A}}=\left\{\boldsymbol{{x}}\:\in\:\mathbb{R}\mid\boldsymbol{{x}}^{\mathrm{2}} <\mathrm{4}\right\}\mathrm{and} \\ $$$$\boldsymbol{{B}}=\left\{\boldsymbol{{y}}\:\in\:\mathbb{Q}\mid\boldsymbol{{y}}>−\mathrm{3}\right\}\mathrm{find}\:\boldsymbol{{A}}\cap\boldsymbol{{B}} \\ $$

Question Number 209746    Answers: 1   Comments: 1

Q)Choose at least some members frome the set A={14,15,...,20,22,23,...,28} so that whith confidence includes three consecutive members?

$$\left.{Q}\right){Choose}\:{at}\:{least}\:{some}\:{members} \\ $$$${frome}\:{the}\:{set}\:{A}=\left\{\mathrm{14},\mathrm{15},...,\mathrm{20},\mathrm{22},\mathrm{23},...,\mathrm{28}\right\} \\ $$$${so}\:{that}\:{whith}\:{confidence}\:\:{includes}\:{three}\:{consecutive} \\ $$$${members}? \\ $$

Question Number 209246    Answers: 1   Comments: 0

Find f(x)=∫^( x) _( 0) (dt/(t+e^(f(t)) ))

$$\mathrm{Find}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{x}} \frac{\mathrm{dt}}{\mathrm{t}+\mathrm{e}^{\mathrm{f}\left(\mathrm{t}\right)} } \\ $$

Question Number 207864    Answers: 1   Comments: 0

$$\:\:\:\:\:\underbrace{\:} \\ $$

Question Number 207372    Answers: 0   Comments: 6

2 students are passing a test of n questions with the same chance to find each one Show the chance that they both don′t find a same question is ((3/4))^n

$$\mathrm{2}\:{students}\:{are}\:{passing}\: \\ $$$${a}\:{test}\:{of}\:\:{n}\:{questions}\:{with} \\ $$$${the}\:{same}\:{chance}\:{to}\:{find}\:{each}\:{one} \\ $$$${Show}\:\:{the}\:{chance}\:{that}\:{they}\:{both} \\ $$$$\:{don}'{t}\:{find}\:{a}\:{same}\:{question}\:{is}\:\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} \\ $$

Question Number 204141    Answers: 1   Comments: 0

Question Number 203737    Answers: 0   Comments: 0

Question Number 203634    Answers: 2   Comments: 0

$$\:\:\: \\ $$

Question Number 203490    Answers: 1   Comments: 0

1×3×5×7×9×...×2005 = ... (mod 1000)

$$\:\:\:\:\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{9}×...×\mathrm{2005}\:=\:...\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$

Question Number 199624    Answers: 0   Comments: 1

Question Number 198380    Answers: 0   Comments: 2

Question Number 196950    Answers: 1   Comments: 0

Prove that ∫^( (π/2)) _( 0) ((ln(1+αsint))/(sint))dt= (π^2 /8)−(1/2)(arccosα)^2

$$\mathrm{Prove}\:\mathrm{that}\:\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}+\alpha\mathrm{sin}{t}\right)}{\mathrm{sin}{t}}{dt}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arccos}\alpha\right)^{\mathrm{2}} \\ $$

Question Number 195393    Answers: 1   Comments: 0

prove that lim_(x→0) (((Σ_(k=1) ^n (1−(1/(2k)))^x )/n))^(1/( x )) = (1/4)(C_(2n) ^n )^(1/n)

$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\:\:\boldsymbol{{x}}\:\:}]{\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)^{{x}} }{{n}}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt[{\boldsymbol{{n}}}]{\mathrm{C}_{\mathrm{2}\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{n}}} } \\ $$

Question Number 200284    Answers: 1   Comments: 0

Prove that for any set A containing n elements, ∣P(A)∣=2^n .

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{set}\:{A}\:\mathrm{containing}\:{n} \\ $$$$\mathrm{elements},\:\mid\mathcal{P}\left({A}\right)\mid=\mathrm{2}^{{n}} . \\ $$

Question Number 195157    Answers: 1   Comments: 0

Prove that (x^3 /(2sin^2 ((1/2)arctan (x/y))))+(y^3 /(2cos^2 ((1/2)arctan (y/x))))=(x+y)(x^2 +y^2 )

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{3}} }{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\:\frac{{x}}{{y}}\right)}+\frac{{y}^{\mathrm{3}} }{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\:\frac{{y}}{{x}}\right)}=\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$

Question Number 194868    Answers: 1   Comments: 0

Prove that ∀n∈IN ∫^( 1) _( 0) t sin^(2n) (lnt)dt= (1/(1−e^(−2π) )) ∫^( π) _( 0) e^(−2t) sin^(2n) (t)dt

$$\mathrm{Prove}\:\mathrm{that}\:\forall{n}\in\mathrm{IN} \\ $$$$\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} {t}\:{sin}^{\mathrm{2}{n}} \left({lnt}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}\pi} }\:\underset{\:\mathrm{0}} {\int}^{\:\pi} {e}^{−\mathrm{2}{t}} {sin}^{\mathrm{2}{n}} \left({t}\right){dt} \\ $$

Question Number 194781    Answers: 0   Comments: 0

f_(n ) the general sentence is seqiencee fibonacci. prove that : f_(2n−1) =f_n ^2 +f_(n−1) ^2

$${f}_{{n}\:} \:\:{the}\:{general}\:{sentence}\:{is}\:{seqiencee} \\ $$$${fibonacci}.\: \\ $$$${prove}\:{that}\::\:\:{f}_{\mathrm{2}{n}−\mathrm{1}} ={f}_{{n}} ^{\mathrm{2}} +{f}_{{n}−\mathrm{1}} ^{\mathrm{2}} \\ $$$$ \\ $$

Question Number 194709    Answers: 1   Comments: 0

Show that in fibonacci sequence f_(3n) =f_n ^3 +f_(n+1) ^3 −f_(n−1) ^3

$${Show}\:{that}\:\:{in}\:{fibonacci}\:{sequence} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{f}_{\mathrm{3}{n}} ={f}_{{n}} ^{\mathrm{3}} +{f}_{{n}+\mathrm{1}} ^{\mathrm{3}} −{f}_{{n}−\mathrm{1}} ^{\mathrm{3}} \\ $$$$ \\ $$

Question Number 194693    Answers: 1   Comments: 0

if f_n =f_(n−1) +f_(n−2) ; f_1 =f_2 =1 then prove that 5∣f_(5n)

$${if}\:\:\:{f}_{{n}} ={f}_{{n}−\mathrm{1}} +{f}_{{n}−\mathrm{2}} \:\:;\:\:{f}_{\mathrm{1}} ={f}_{\mathrm{2}} =\mathrm{1} \\ $$$${then}\:\:\:{prove}\:{that}\:\:\:\mathrm{5}\mid{f}_{\mathrm{5}{n}} \:\: \\ $$

Question Number 194446    Answers: 0   Comments: 0

Question Number 191731    Answers: 1   Comments: 0

Use laws of algebra to prove the following (a)[(B−A)u(A−B)]=[(AuB)−(AnB)] (b)A▽(AnB)=A−B

$$\mathrm{Use}\:\mathrm{laws}\:\mathrm{of}\:\mathrm{algebra}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following} \\ $$$$\left(\mathrm{a}\right)\left[\left(\mathrm{B}−\mathrm{A}\right)\mathrm{u}\left(\mathrm{A}−\mathrm{B}\right)\right]=\left[\left(\mathrm{AuB}\right)−\left(\mathrm{AnB}\right)\right] \\ $$$$\left(\mathrm{b}\right)\mathrm{A}\bigtriangledown\left(\mathrm{AnB}\right)=\mathrm{A}−\mathrm{B} \\ $$

Question Number 191501    Answers: 1   Comments: 2

find a solution; e^x = ln(x)

$$ \\ $$$$\:\:\:\:{find}\:{a}\:{solution}; \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{e}}^{\boldsymbol{{x}}} \:=\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right) \\ $$$$ \\ $$

Question Number 190573    Answers: 1   Comments: 3

a^ 2+2ab+b^ 2

$$\hat {{a}}\mathrm{2}+\mathrm{2}{ab}+\hat {{b}}\mathrm{2} \\ $$

Question Number 189672    Answers: 0   Comments: 2

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