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Question Number 2393 by 123456 last updated on 19/Nov/15
f:[0,1]→R x=a_0 ,a_1 a_2 a_3 ... f(x)=Σ_(i=0) ^(+∞) a_i is f(x) continuous in all x∈[0,1] f(0,9999...)=?? f(1)=1
$${f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R} \\ $$$${x}={a}_{\mathrm{0}} ,{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} … \\ $$$${f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{+\infty} {\sum}}{a}_{{i}} \\ $$$$\mathrm{is}\:{f}\left({x}\right)\:\mathrm{continuous}\:\mathrm{in}\:\mathrm{all}\:{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}\left(\mathrm{0},\mathrm{9999}…\right)=?? \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$
Commented by prakash jain last updated on 19/Nov/15
f(x) is not continous is limit does not exists for irrational x and also for rational x= p/q where q≠2^n 5^m
$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{continous}\:\mathrm{is}\:\mathrm{limit}\:\mathrm{does}\:\mathrm{not} \\ $$$$\mathrm{exists}\:\mathrm{for}\:\mathrm{irrational}\:{x}\:\mathrm{and}\:\mathrm{also}\:\mathrm{for}\:\mathrm{rational} \\ $$$${x}=\:{p}/{q}\:\mathrm{where}\:{q}\neq\mathrm{2}^{{n}} \mathrm{5}^{{m}} \\ $$
Commented by prakash jain last updated on 19/Nov/15
I don′t think there is any real number like 0.99999.... (infinite times). It is same as 1.
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{any}\:\mathrm{real}\:\mathrm{number}\:\mathrm{like} \\ $$$$\mathrm{0}.\mathrm{99999}….\:\left(\mathrm{infinite}\:\mathrm{times}\right).\:\mathrm{It}\:\mathrm{is}\:\:\mathrm{same}\:\mathrm{as}\:\mathrm{1}. \\ $$
Commented by prakash jain last updated on 19/Nov/15
What i mean is 0.999.. is a recurring decimal and hence rational. The rational number representing this number is 1. This is different from other rationals like 1/3. which can be represented both in recurring decimal and fraction notation. So for example 0.49999999999...(infinite times)=1/2
$$\mathrm{What}\:\mathrm{i}\:\mathrm{mean}\:\mathrm{is}\:\mathrm{0}.\mathrm{999}..\:\mathrm{is}\:\mathrm{a}\:\mathrm{recurring}\:\mathrm{decimal} \\ $$$$\mathrm{and}\:\mathrm{hence}\:\mathrm{rational}.\:\mathrm{The}\:\mathrm{rational}\:\mathrm{number} \\ $$$$\mathrm{representing}\:\mathrm{this}\:\mathrm{number}\:\mathrm{is}\:\mathrm{1}.\: \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{different}\:\mathrm{from}\:\mathrm{other}\:\mathrm{rationals} \\ $$$$\mathrm{like}\:\mathrm{1}/\mathrm{3}.\:\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{represented}\:\mathrm{both} \\ $$$$\mathrm{in}\:\mathrm{recurring}\:\mathrm{decimal}\:\mathrm{and}\:\mathrm{fraction}\:\mathrm{notation}. \\ $$$$\mathrm{So}\:\mathrm{for}\:\mathrm{example} \\ $$$$\mathrm{0}.\mathrm{49999999999}…\left(\mathrm{infinite}\:\mathrm{times}\right)=\mathrm{1}/\mathrm{2} \\ $$
Answered by Filup last updated on 19/Nov/15
0.9^− is exactly equal to 1. Because of the infinite number of 9′s being appended, this makes it entirely equal. Think about it like this: If 1≠0.9^− , then: 0.9^− =0.999...9990000^− ... which is contradictory against the definition of recurring decimals. ALL recurring decimals are rational, thus: f(0.9^− )=f(1) iff 0.9^− is appended infinitly!
$$\mathrm{0}.\overset{−} {\mathrm{9}}\:\mathrm{is}\:\boldsymbol{{exactly}}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1}. \\ $$$$\mathrm{Because}\:\mathrm{of}\:\mathrm{the}\:\mathrm{infinite}\:\mathrm{number}\:\mathrm{of}\:\mathrm{9}'\mathrm{s} \\ $$$$\mathrm{being}\:\mathrm{appended},\:\mathrm{this}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{entirely} \\ $$$$\mathrm{equal}. \\ $$$$ \\ $$$$\mathrm{Think}\:\mathrm{about}\:\mathrm{it}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\mathrm{If}\:\mathrm{1}\neq\mathrm{0}.\overset{−} {\mathrm{9}},\:{then}: \\ $$$$\mathrm{0}.\overset{−} {\mathrm{9}}=\mathrm{0}.\mathrm{999}…\mathrm{999000}\overset{−} {\mathrm{0}}…\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{contradictory}\:\mathrm{against}\:\mathrm{the}\:\mathrm{definition} \\ $$$$\mathrm{of}\:\mathrm{recurring}\:\mathrm{decimals}.\:\boldsymbol{{ALL}}\:\mathrm{recurring} \\ $$$$\mathrm{decimals}\:\mathrm{are}\:\mathrm{rational},\:{thus}: \\ $$$${f}\left(\mathrm{0}.\overset{−} {\mathrm{9}}\right)={f}\left(\mathrm{1}\right)\:\:\:{iff}\:\:\:\mathrm{0}.\overset{−} {\mathrm{9}}\:\mathrm{is}\:\mathrm{appended}\:\mathrm{infinitly}! \\ $$
Commented by prakash jain last updated on 19/Nov/15
Correct. 0.49^− =1/2 0.9^− =1. 1/2 =.5
$$\mathrm{Correct}. \\ $$$$\mathrm{0}.\mathrm{4}\overset{−} {\mathrm{9}}=\mathrm{1}/\mathrm{2} \\ $$$$\mathrm{0}.\overset{−} {\mathrm{9}}=\mathrm{1}. \\ $$$$\mathrm{1}/\mathrm{2}\:\:=.\mathrm{5} \\ $$
Commented by 123456 last updated on 19/Nov/15
0,49^� =(1/2) 0,4+x=(1/2) (2/5)+x=(1/2) x=(1/2)−(2/5)=((5−4)/(10))=(1/(10))=0,1
$$\mathrm{0},\mathrm{4}\bar {\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{0},\mathrm{4}+{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}+{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{5}−\mathrm{4}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{10}}=\mathrm{0},\mathrm{1} \\ $$
Commented by prakash jain last updated on 20/Nov/15
0.1111×9=.9999 (1/9)=0.1111..... (1/9)×9=0.99999..... But 0.9^− and 1 are same numbers. f(x)=Σ_(i=0) ^∞ a_i if x=a_0 .a_1 a_2 .... f(0.9^− )=?
$$\mathrm{0}.\mathrm{1111}×\mathrm{9}=.\mathrm{9999} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{0}.\mathrm{1111}….. \\ $$$$\frac{\mathrm{1}}{\mathrm{9}}×\mathrm{9}=\mathrm{0}.\mathrm{99999}….. \\ $$$$\mathrm{But}\:\mathrm{0}.\overset{−} {\mathrm{9}}\:\mathrm{and}\:\mathrm{1}\:\mathrm{are}\:\mathrm{same}\:\mathrm{numbers}. \\ $$$${f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} \:\:\mathrm{if}\:{x}={a}_{\mathrm{0}} .{a}_{\mathrm{1}} {a}_{\mathrm{2}} …. \\ $$$${f}\left(\mathrm{0}.\overset{−} {\mathrm{9}}\right)=? \\ $$
Commented by RasheedAhmad last updated on 19/Nov/15
One difference between 0.9_(−) ^(−) and 0.3_(−) ^(−) 0.3^(−) can be achieved by actual division process. 1÷3=0.3^(−) But 0.9^(−) cannot be achieved by actual process of division of any two integers. 0.9^(−) =1=(1/1) 1÷1 doesn′t yield 0.9999... it is simply 1 0.49^(−) is also like 0.9^(−) It cannot be achieved by actual process of division of any two integers. 1÷2 doesn′t yield 0.49999... it is simply 0.5
$$\underset{−} {\mathcal{O}{ne}\:{difference}\:{between}\:\mathrm{0}.\mathrm{9}} \\ $$$$\underset{−} {{and}\:\mathrm{0}.\mathrm{3}} \\ $$$$\mathrm{0}.\overline {\mathrm{3}}\:{can}\:{be}\:{achieved}\:{by}\:{actual} \\ $$$${division}\:{process}. \\ $$$$\mathrm{1}\boldsymbol{\div}\mathrm{3}=\mathrm{0}.\overline {\mathrm{3}} \\ $$$${But}\:\mathrm{0}.\overline {\mathrm{9}}\:{cannot}\:{be}\:{achieved}\:{by} \\ $$$${actual}\:{process}\:{of}\:{division} \\ $$$${of}\:{any}\:{two}\:{integers}. \\ $$$$\mathrm{0}.\overline {\mathrm{9}}=\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}} \\ $$$$\mathrm{1}\boldsymbol{\div}\mathrm{1}\:{doesn}'{t}\:{yield}\:\mathrm{0}.\mathrm{9999}… \\ $$$${it}\:{is}\:{simply}\:\mathrm{1} \\ $$$$\mathrm{0}.\mathrm{4}\overline {\mathrm{9}}\:{is}\:{also}\:{like}\:\mathrm{0}.\overline {\mathrm{9}} \\ $$$${It}\:{cannot}\:{be}\:{achieved}\:{by}\:{actual} \\ $$$${process}\:{of}\:{division}\:{of}\:{any}\:{two} \\ $$$${integers}. \\ $$$$\mathrm{1}\boldsymbol{\div}\mathrm{2}\:{doesn}'{t}\:{yield}\:\mathrm{0}.\mathrm{49999}… \\ $$$${it}\:{is}\:{simply}\:\mathrm{0}.\mathrm{5} \\ $$
Commented by 123456 last updated on 19/Nov/15
(a/9)=0,a^� a∈{0,1,2,3,4,5,6,7,8,9?} (9/9)=0,9999... 9∣9 0 1 9∣9 90⌊0,99... 81 90 81 ⋮ 4∣4 40 ⌊0,9... 36 40 ⋮
$$\frac{{a}}{\mathrm{9}}=\mathrm{0},\bar {{a}}\:\:\:{a}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}?\right\} \\ $$$$\frac{\mathrm{9}}{\mathrm{9}}=\mathrm{0},\mathrm{9999}… \\ $$$$\mathrm{9}\mid\mathrm{9} \\ $$$$\mathrm{0}\:\:\mathrm{1} \\ $$$$\mathrm{9}\mid\mathrm{9} \\ $$$$\mathrm{90}\lfloor\mathrm{0},\mathrm{99}… \\ $$$$\mathrm{81} \\ $$$$\:\:\mathrm{90} \\ $$$$\:\:\mathrm{81} \\ $$$$\:\:\:\:\vdots \\ $$$$\mathrm{4}\mid\mathrm{4} \\ $$$$\mathrm{40}\:\lfloor\mathrm{0},\mathrm{9}… \\ $$$$\mathrm{36} \\ $$$$\:\:\:\mathrm{40} \\ $$$$\:\:\:\:\vdots \\ $$
Commented by Rasheed Soomro last updated on 20/Nov/15
′V Nice!′ for your deep approach! You have successfully showed/proved in various ways that 0.9^(−) =1 Actually I don′t deny the above. I only pointed out the difference between 0.3^(−) and 0.9^(−) 0.3^(−) is the result of ordinary division of two integers 1÷3=0.3333... whereas 0.9^(−) is not. You have also tried to show that 0.9^(−) can be achieved by ordinary division (9÷9) but in this connection you haven′t followed some rules of ordinary division. So I think 0.9^(−) ,0.49^(−) etc aren′t achievable by ordinary_(−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−) division of two integers._(−−−−−−−−−−−−−−−−−−−)
$$'{V}\:{Nice}!'\:\:{for}\:{your}\:{deep}\:{approach}! \\ $$$${You}\:{have}\:{successfully}\:{showed}/{proved}\:{in}\:{various}\:{ways}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{0}.\overline {\mathrm{9}}=\mathrm{1} \\ $$$${Actually}\:{I}\:{don}'{t}\:{deny}\:{the}\:{above}.\:{I}\:{only}\:{pointed}\:{out} \\ $$$${the}\:{difference}\:{between}\:\mathrm{0}.\overline {\mathrm{3}}\:\:{and}\:\:\mathrm{0}.\overline {\mathrm{9}} \\ $$$$\mathrm{0}.\overline {\mathrm{3}}\:{is}\:{the}\:{result}\:{of}\:{ordinary}\:\:{division}\:{of}\:{two}\:{integers} \\ $$$$\mathrm{1}\boldsymbol{\div}\mathrm{3}=\mathrm{0}.\mathrm{3333}…\:\:{whereas}\:\:\mathrm{0}.\overline {\mathrm{9}}\:\:{is}\:{not}. \\ $$$${You}\:{have}\:{also}\:{tried}\:{to}\:{show}\:{that}\:\mathrm{0}.\overline {\mathrm{9}}\:{can}\:{be}\:{achieved} \\ $$$${by}\:{ordinary}\:\:{division}\:\left(\mathrm{9}\boldsymbol{\div}\mathrm{9}\right)\:{but}\:{in}\:{this}\:{connection}\:{you} \\ $$$${haven}'{t}\:{followed}\:{some}\:{rules}\:{of}\:{ordinary}\:{division}. \\ $$$$\underset{−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−} {{So}\:{I}\:{think}\:\mathrm{0}.\mathrm{9},\mathrm{0}.\mathrm{49}\:\:{etc}\:{aren}'{t}\:{achievable}\:{by}\:{ordinary}} \\ $$$$\underset{−−−−−−−−−−−−−−−−−−−} {{division}\:{of}\:{two}\:{integers}.} \\ $$$$ \\ $$
Commented by 123456 last updated on 19/Nov/15
0,9999... =9/10+9/10^2 +... =((9/10)/(1−1/10))=((9/10)/(9/10))=1 (infinite pg) or s=0,999.... 10s=9,999.... 9s=9 s=1 ps: 1=0,11111111...._2 since Σ_(i=1) ^(+∞) (1/2^i )=((1/2)/(1−1/2))=1 if f doenst is continuous them (dont sure) lim_(n→+∞) f(x_n )≠f(lim_(n→+∞) x_n ) in general them this have something about f(0,99...) and f(1) or ⌊0,999...⌋ and ⌊1⌋ or i think it have :v
$$\mathrm{0},\mathrm{9999}… \\ $$$$=\mathrm{9}/\mathrm{10}+\mathrm{9}/\mathrm{10}^{\mathrm{2}} +… \\ $$$$=\frac{\mathrm{9}/\mathrm{10}}{\mathrm{1}−\mathrm{1}/\mathrm{10}}=\frac{\mathrm{9}/\mathrm{10}}{\mathrm{9}/\mathrm{10}}=\mathrm{1}\:\left(\mathrm{infinite}\:\mathrm{pg}\right) \\ $$$$\mathrm{or} \\ $$$$\mathrm{s}=\mathrm{0},\mathrm{999}…. \\ $$$$\mathrm{10s}=\mathrm{9},\mathrm{999}…. \\ $$$$\mathrm{9s}=\mathrm{9} \\ $$$$\mathrm{s}=\mathrm{1} \\ $$$$\mathrm{ps}: \\ $$$$\mathrm{1}=\mathrm{0},\mathrm{11111111}…._{\mathrm{2}} \\ $$$$\mathrm{since} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} }=\frac{\mathrm{1}/\mathrm{2}}{\mathrm{1}−\mathrm{1}/\mathrm{2}}=\mathrm{1} \\ $$$$\mathrm{if}\:{f}\:\mathrm{doenst}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{them}\:\left(\mathrm{dont}\:\mathrm{sure}\right) \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:{f}\left({x}_{{n}} \right)\neq{f}\left(\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{x}_{{n}} \right)\:\mathrm{in}\:\mathrm{general} \\ $$$$\mathrm{them}\:\mathrm{this}\:\mathrm{have}\:\mathrm{something}\:\mathrm{about} \\ $$$${f}\left(\mathrm{0},\mathrm{99}…\right)\:\mathrm{and}\:{f}\left(\mathrm{1}\right)\:{or} \\ $$$$\lfloor\mathrm{0},\mathrm{999}…\rfloor\:\mathrm{and}\:\lfloor\mathrm{1}\rfloor \\ $$$$\mathrm{or}\:\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{have}\::\mathrm{v} \\ $$
Commented by Rasheed Soomro last updated on 20/Nov/15
0.9^(−) =1⇒f(0.9^(−) )=f(1)=1 On the other hand f(0.9^(−) )=9.∞=∞ That means to avoid contadiction we should deny the separte existence of 0.9^(−) and think ′ it is nothing but 1′ ? Similarly f(0.49^(−) )=f(0.5)=5?
$$\mathrm{0}.\overline {\mathrm{9}}=\mathrm{1}\Rightarrow{f}\left(\mathrm{0}.\overline {\mathrm{9}}\right)={f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${On}\:{the}\:{other}\:{hand}\:{f}\left(\mathrm{0}.\overline {\mathrm{9}}\right)=\mathrm{9}.\infty=\infty \\ $$$${That}\:{means}\:{to}\:{avoid}\:{contadiction}\:{we}\:{should} \\ $$$${deny}\:{the}\:{separte}\:{existence}\:{of}\:\mathrm{0}.\overline {\mathrm{9}}\:{and}\:{think}\:'\:{it}\:{is} \\ $$$${nothing}\:{but}\:\mathrm{1}'\:? \\ $$$${Similarly}\:{f}\left(\mathrm{0}.\mathrm{4}\overline {\mathrm{9}}\right)={f}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{5}? \\ $$
Commented by 123456 last updated on 20/Nov/15
one day in some random math i found about ⌊0,99...⌋=0 and ⌊1⌋=1 but 0,9..=1 the answer them give are about continuity of ⌊x⌋, i will search the thing later.
$$\mathrm{one}\:\mathrm{day}\:\mathrm{in}\:\mathrm{some}\:\mathrm{random}\:\mathrm{math}\:\mathrm{i}\:\mathrm{found} \\ $$$$\mathrm{about}\:\lfloor\mathrm{0},\mathrm{99}…\rfloor=\mathrm{0}\:\mathrm{and}\:\lfloor\mathrm{1}\rfloor=\mathrm{1}\:\mathrm{but}\:\mathrm{0},\mathrm{9}..=\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{them}\:\mathrm{give}\:\mathrm{are}\:\mathrm{about}\:\mathrm{continuity} \\ $$$$\mathrm{of}\:\lfloor{x}\rfloor,\:\mathrm{i}\:\mathrm{will}\:\mathrm{search}\:\mathrm{the}\:\mathrm{thing}\:\mathrm{later}. \\ $$
Commented by Rasheed Soomro last updated on 21/Nov/15
I ′ll wait for update!
$$\mathcal{I}\:'{ll}\:{wait}\:{for}\:{update}! \\ $$

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