About 0.999… = 1












23












$begingroup$


I've just happened to read this question on MO (that of course has been closed) and some of the answers to a similar question on MSE.



I know almost nothing of nonstandard analysis and was asking myself if something like the sentence « $1- 0.999 dots$ is a nonzero positive infinitesimal» could be easily expressed and proved in nonstandard analysis.



First of all, what is 0.999... ? If we take the usual definition as a series or as a limit of a sequence of rationals, then it will still be a real number and equal to $1$ (I guess by "transfer principle", but please correct me if I'm wrong).



Instead, let's define



$$0.9_N:=sum_{i=1}^N 9cdot 10^{-i} $$



where $Nin{}^*mathbb{N}setminusmathbb{N}$ is an infinite nonstandard natural number. This $0.9_N$ is a legitimate element of ${}^*mathbb{R}$, expressed as $0.$ followed by an infinite number of "$9$" digits.



What can be said about $epsilon_N:=1-0.9_N$ ? Is there an elementary proof that $epsilon_N$ is a positive infinitesimal of ${}^*mathbb{R}$ ? (by "elementary" I mean just order and field axioms and the intuitive facts about infinitesimals, like that for $x$ infinite $1/x$ is infinitesimal etc.; no nonprincipal ultrafilters & C).






nonstandard-analysis decimal-expansion







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    +1 for not making the mistake $0.overline{9} < 1$, and recognizing that you need a terminating decimal to get something less than 1.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 16:42


















23












$begingroup$


I've just happened to read this question on MO (that of course has been closed) and some of the answers to a similar question on MSE.



I know almost nothing of nonstandard analysis and was asking myself if something like the sentence « $1- 0.999 dots$ is a nonzero positive infinitesimal» could be easily expressed and proved in nonstandard analysis.



First of all, what is 0.999... ? If we take the usual definition as a series or as a limit of a sequence of rationals, then it will still be a real number and equal to $1$ (I guess by "transfer principle", but please correct me if I'm wrong).



Instead, let's define



$$0.9_N:=sum_{i=1}^N 9cdot 10^{-i} $$



where $Nin{}^*mathbb{N}setminusmathbb{N}$ is an infinite nonstandard natural number. This $0.9_N$ is a legitimate element of ${}^*mathbb{R}$, expressed as $0.$ followed by an infinite number of "$9$" digits.



What can be said about $epsilon_N:=1-0.9_N$ ? Is there an elementary proof that $epsilon_N$ is a positive infinitesimal of ${}^*mathbb{R}$ ? (by "elementary" I mean just order and field axioms and the intuitive facts about infinitesimals, like that for $x$ infinite $1/x$ is infinitesimal etc.; no nonprincipal ultrafilters & C).






nonstandard-analysis decimal-expansion







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    +1 for not making the mistake $0.overline{9} < 1$, and recognizing that you need a terminating decimal to get something less than 1.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 16:42
















23












23








23


4



$begingroup$


I've just happened to read this question on MO (that of course has been closed) and some of the answers to a similar question on MSE.



I know almost nothing of nonstandard analysis and was asking myself if something like the sentence « $1- 0.999 dots$ is a nonzero positive infinitesimal» could be easily expressed and proved in nonstandard analysis.



First of all, what is 0.999... ? If we take the usual definition as a series or as a limit of a sequence of rationals, then it will still be a real number and equal to $1$ (I guess by "transfer principle", but please correct me if I'm wrong).



Instead, let's define



$$0.9_N:=sum_{i=1}^N 9cdot 10^{-i} $$



where $Nin{}^*mathbb{N}setminusmathbb{N}$ is an infinite nonstandard natural number. This $0.9_N$ is a legitimate element of ${}^*mathbb{R}$, expressed as $0.$ followed by an infinite number of "$9$" digits.



What can be said about $epsilon_N:=1-0.9_N$ ? Is there an elementary proof that $epsilon_N$ is a positive infinitesimal of ${}^*mathbb{R}$ ? (by "elementary" I mean just order and field axioms and the intuitive facts about infinitesimals, like that for $x$ infinite $1/x$ is infinitesimal etc.; no nonprincipal ultrafilters & C).






nonstandard-analysis decimal-expansion







share|cite|improve this question











$endgroup$




I've just happened to read this question on MO (that of course has been closed) and some of the answers to a similar question on MSE.



I know almost nothing of nonstandard analysis and was asking myself if something like the sentence « $1- 0.999 dots$ is a nonzero positive infinitesimal» could be easily expressed and proved in nonstandard analysis.



First of all, what is 0.999... ? If we take the usual definition as a series or as a limit of a sequence of rationals, then it will still be a real number and equal to $1$ (I guess by "transfer principle", but please correct me if I'm wrong).



Instead, let's define



$$0.9_N:=sum_{i=1}^N 9cdot 10^{-i} $$



where $Nin{}^*mathbb{N}setminusmathbb{N}$ is an infinite nonstandard natural number. This $0.9_N$ is a legitimate element of ${}^*mathbb{R}$, expressed as $0.$ followed by an infinite number of "$9$" digits.



What can be said about $epsilon_N:=1-0.9_N$ ? Is there an elementary proof that $epsilon_N$ is a positive infinitesimal of ${}^*mathbb{R}$ ? (by "elementary" I mean just order and field axioms and the intuitive facts about infinitesimals, like that for $x$ infinite $1/x$ is infinitesimal etc.; no nonprincipal ultrafilters & C).






nonstandard-analysis decimal-expansion




nonstandard-analysis decimal-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:58









Community♦

1




1










asked Jan 18 '13 at 16:40









QfwfqQfwfq

324112




324112








  • 5




    $begingroup$
    +1 for not making the mistake $0.overline{9} < 1$, and recognizing that you need a terminating decimal to get something less than 1.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 16:42
















  • 5




    $begingroup$
    +1 for not making the mistake $0.overline{9} < 1$, and recognizing that you need a terminating decimal to get something less than 1.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 16:42










5




5




$begingroup$
+1 for not making the mistake $0.overline{9} < 1$, and recognizing that you need a terminating decimal to get something less than 1.
$endgroup$
– Hurkyl
Jan 18 '13 at 16:42






$begingroup$
+1 for not making the mistake $0.overline{9} < 1$, and recognizing that you need a terminating decimal to get something less than 1.
$endgroup$
– Hurkyl
Jan 18 '13 at 16:42












3 Answers
3






active

oldest

votes


















9












$begingroup$

We can use the geometric series formula:



$$0.9_N = sum_{i=1}^N 9 cdot 10^{-i} = 9 cdot 10^{-1} cdot frac{1 - 10^{-N}}{1 - 10^{-1}} = (1 - 10^{-N})$$



Since $N$ is infinite, $epsilon_N = 10^{-N} = 1 / 10^N$ is infinitesimal.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N approx infty$, but I think a lot of people would dislike that notation.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 16:47












  • $begingroup$
    You type faster than I do. :-) Infinite integer is the usual term, so it doesn’t bother me.
    $endgroup$
    – Brian M. Scott
    Jan 18 '13 at 16:51










  • $begingroup$
    But why the formula $sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
    $endgroup$
    – Qfwfq
    Jan 18 '13 at 16:51






  • 1




    $begingroup$
    @Qfwfq Hint: Transfer principle.
    $endgroup$
    – Ian Mateus
    Jan 18 '13 at 16:52






  • 4




    $begingroup$
    @Qfwfq: One key thing to note is that, internally, this is a finite sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 17:02



















7












$begingroup$

$$1-sum_{k=1}^N9cdot10^{-k}=sum_{kge N+1}9cdot 10^{-k}=9sum_{kge N+1}10^{-k}=frac{9cdot 10^{-(N+1)}}{1-10^{-1}}=10^{-N}=frac1{10^N};,$$



which is surely intuitively a positive infinitesimal.



Added: There is a nice elementary axiomatization of the hyperreals in Jerry Keisler’s Elementary Calculus: An infinitesimal approach, which is freely available here; it is intended for students in a first calculus course and neatly avoids the transfer principle as such. His Foundations of infinitesimal calculus contains a slightly more sophisticated version, since it’s intended for instructors using the undergraduate text. It’s freely available here, and its version of the axiomatic development can also be found in Section $15$ of this PDF. What he calls the Function Axiom (Axiom $C$ in the PDF) justifies the standard calculation:




For each real function $f$ of $n$ variables there is a corresponding hyperreal function ${}^*f$ of $n$ variables, called the natural extension of $f$.




The function in question here is the function that takes $n$ to $sum_{k=1}^n9cdot10^{-k}$.



A slightly different version of this approach is found in Keith Stroyan’s notes here, especially Section $2.3$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (Thank you for your valid answer. I accepted the other one because of the comment thread following it)
    $endgroup$
    – Qfwfq
    Jan 18 '13 at 17:08










  • $begingroup$
    @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful.
    $endgroup$
    – Brian M. Scott
    Jan 18 '13 at 17:36










  • $begingroup$
    Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^ast Bbb N$ but restricted to $Bbb N$ wouldn't be identical to the decimal expansion ranged over $Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction.
    $endgroup$
    – Metta World Peace
    May 19 '13 at 21:19










  • $begingroup$
    It's not that it "wouldn't be identical" to the decimal expansion ranged over $mathbb{N}$. Rather, a "decimal expansion ranged over $mathbb{N}$" does not exist since $mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets.
    $endgroup$
    – Mikhail Katz
    May 20 '13 at 8:44






  • 1




    $begingroup$
    @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*Bbb R$, you can’t talk about a sum over $Bbb N$, because you can’t even talk about $Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model.
    $endgroup$
    – Brian M. Scott
    May 20 '13 at 16:54



















3












$begingroup$

In addition to the fine answers given earlier, I would like to mention also the somewhat unsung hero, A. H. Lightstone, and his extended decimal notation in which your infinitesimal $epsilon_N$ can be written as $0.000ldots;ldots 0001$, where the first nonzero digit occurs precisely at infinite rank $N$. The notation is explained in his article in the American Mathematical Monthly (see especially page 246).






share|cite|improve this answer











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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    We can use the geometric series formula:



    $$0.9_N = sum_{i=1}^N 9 cdot 10^{-i} = 9 cdot 10^{-1} cdot frac{1 - 10^{-N}}{1 - 10^{-1}} = (1 - 10^{-N})$$



    Since $N$ is infinite, $epsilon_N = 10^{-N} = 1 / 10^N$ is infinitesimal.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N approx infty$, but I think a lot of people would dislike that notation.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 16:47












    • $begingroup$
      You type faster than I do. :-) Infinite integer is the usual term, so it doesn’t bother me.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 16:51










    • $begingroup$
      But why the formula $sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 16:51






    • 1




      $begingroup$
      @Qfwfq Hint: Transfer principle.
      $endgroup$
      – Ian Mateus
      Jan 18 '13 at 16:52






    • 4




      $begingroup$
      @Qfwfq: One key thing to note is that, internally, this is a finite sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 17:02
















    9












    $begingroup$

    We can use the geometric series formula:



    $$0.9_N = sum_{i=1}^N 9 cdot 10^{-i} = 9 cdot 10^{-1} cdot frac{1 - 10^{-N}}{1 - 10^{-1}} = (1 - 10^{-N})$$



    Since $N$ is infinite, $epsilon_N = 10^{-N} = 1 / 10^N$ is infinitesimal.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N approx infty$, but I think a lot of people would dislike that notation.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 16:47












    • $begingroup$
      You type faster than I do. :-) Infinite integer is the usual term, so it doesn’t bother me.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 16:51










    • $begingroup$
      But why the formula $sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 16:51






    • 1




      $begingroup$
      @Qfwfq Hint: Transfer principle.
      $endgroup$
      – Ian Mateus
      Jan 18 '13 at 16:52






    • 4




      $begingroup$
      @Qfwfq: One key thing to note is that, internally, this is a finite sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 17:02














    9












    9








    9





    $begingroup$

    We can use the geometric series formula:



    $$0.9_N = sum_{i=1}^N 9 cdot 10^{-i} = 9 cdot 10^{-1} cdot frac{1 - 10^{-N}}{1 - 10^{-1}} = (1 - 10^{-N})$$



    Since $N$ is infinite, $epsilon_N = 10^{-N} = 1 / 10^N$ is infinitesimal.






    share|cite|improve this answer









    $endgroup$



    We can use the geometric series formula:



    $$0.9_N = sum_{i=1}^N 9 cdot 10^{-i} = 9 cdot 10^{-1} cdot frac{1 - 10^{-N}}{1 - 10^{-1}} = (1 - 10^{-N})$$



    Since $N$ is infinite, $epsilon_N = 10^{-N} = 1 / 10^N$ is infinitesimal.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 18 '13 at 16:46









    HurkylHurkyl

    112k9120264




    112k9120264








    • 2




      $begingroup$
      P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N approx infty$, but I think a lot of people would dislike that notation.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 16:47












    • $begingroup$
      You type faster than I do. :-) Infinite integer is the usual term, so it doesn’t bother me.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 16:51










    • $begingroup$
      But why the formula $sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 16:51






    • 1




      $begingroup$
      @Qfwfq Hint: Transfer principle.
      $endgroup$
      – Ian Mateus
      Jan 18 '13 at 16:52






    • 4




      $begingroup$
      @Qfwfq: One key thing to note is that, internally, this is a finite sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 17:02














    • 2




      $begingroup$
      P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N approx infty$, but I think a lot of people would dislike that notation.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 16:47












    • $begingroup$
      You type faster than I do. :-) Infinite integer is the usual term, so it doesn’t bother me.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 16:51










    • $begingroup$
      But why the formula $sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 16:51






    • 1




      $begingroup$
      @Qfwfq Hint: Transfer principle.
      $endgroup$
      – Ian Mateus
      Jan 18 '13 at 16:52






    • 4




      $begingroup$
      @Qfwfq: One key thing to note is that, internally, this is a finite sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies.
      $endgroup$
      – Hurkyl
      Jan 18 '13 at 17:02








    2




    2




    $begingroup$
    P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N approx infty$, but I think a lot of people would dislike that notation.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 16:47






    $begingroup$
    P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N approx infty$, but I think a lot of people would dislike that notation.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 16:47














    $begingroup$
    You type faster than I do. :-) Infinite integer is the usual term, so it doesn’t bother me.
    $endgroup$
    – Brian M. Scott
    Jan 18 '13 at 16:51




    $begingroup$
    You type faster than I do. :-) Infinite integer is the usual term, so it doesn’t bother me.
    $endgroup$
    – Brian M. Scott
    Jan 18 '13 at 16:51












    $begingroup$
    But why the formula $sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
    $endgroup$
    – Qfwfq
    Jan 18 '13 at 16:51




    $begingroup$
    But why the formula $sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
    $endgroup$
    – Qfwfq
    Jan 18 '13 at 16:51




    1




    1




    $begingroup$
    @Qfwfq Hint: Transfer principle.
    $endgroup$
    – Ian Mateus
    Jan 18 '13 at 16:52




    $begingroup$
    @Qfwfq Hint: Transfer principle.
    $endgroup$
    – Ian Mateus
    Jan 18 '13 at 16:52




    4




    4




    $begingroup$
    @Qfwfq: One key thing to note is that, internally, this is a finite sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 17:02




    $begingroup$
    @Qfwfq: One key thing to note is that, internally, this is a finite sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies.
    $endgroup$
    – Hurkyl
    Jan 18 '13 at 17:02











    7












    $begingroup$

    $$1-sum_{k=1}^N9cdot10^{-k}=sum_{kge N+1}9cdot 10^{-k}=9sum_{kge N+1}10^{-k}=frac{9cdot 10^{-(N+1)}}{1-10^{-1}}=10^{-N}=frac1{10^N};,$$



    which is surely intuitively a positive infinitesimal.



    Added: There is a nice elementary axiomatization of the hyperreals in Jerry Keisler’s Elementary Calculus: An infinitesimal approach, which is freely available here; it is intended for students in a first calculus course and neatly avoids the transfer principle as such. His Foundations of infinitesimal calculus contains a slightly more sophisticated version, since it’s intended for instructors using the undergraduate text. It’s freely available here, and its version of the axiomatic development can also be found in Section $15$ of this PDF. What he calls the Function Axiom (Axiom $C$ in the PDF) justifies the standard calculation:




    For each real function $f$ of $n$ variables there is a corresponding hyperreal function ${}^*f$ of $n$ variables, called the natural extension of $f$.




    The function in question here is the function that takes $n$ to $sum_{k=1}^n9cdot10^{-k}$.



    A slightly different version of this approach is found in Keith Stroyan’s notes here, especially Section $2.3$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (Thank you for your valid answer. I accepted the other one because of the comment thread following it)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 17:08










    • $begingroup$
      @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 17:36










    • $begingroup$
      Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^ast Bbb N$ but restricted to $Bbb N$ wouldn't be identical to the decimal expansion ranged over $Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction.
      $endgroup$
      – Metta World Peace
      May 19 '13 at 21:19










    • $begingroup$
      It's not that it "wouldn't be identical" to the decimal expansion ranged over $mathbb{N}$. Rather, a "decimal expansion ranged over $mathbb{N}$" does not exist since $mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets.
      $endgroup$
      – Mikhail Katz
      May 20 '13 at 8:44






    • 1




      $begingroup$
      @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*Bbb R$, you can’t talk about a sum over $Bbb N$, because you can’t even talk about $Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model.
      $endgroup$
      – Brian M. Scott
      May 20 '13 at 16:54
















    7












    $begingroup$

    $$1-sum_{k=1}^N9cdot10^{-k}=sum_{kge N+1}9cdot 10^{-k}=9sum_{kge N+1}10^{-k}=frac{9cdot 10^{-(N+1)}}{1-10^{-1}}=10^{-N}=frac1{10^N};,$$



    which is surely intuitively a positive infinitesimal.



    Added: There is a nice elementary axiomatization of the hyperreals in Jerry Keisler’s Elementary Calculus: An infinitesimal approach, which is freely available here; it is intended for students in a first calculus course and neatly avoids the transfer principle as such. His Foundations of infinitesimal calculus contains a slightly more sophisticated version, since it’s intended for instructors using the undergraduate text. It’s freely available here, and its version of the axiomatic development can also be found in Section $15$ of this PDF. What he calls the Function Axiom (Axiom $C$ in the PDF) justifies the standard calculation:




    For each real function $f$ of $n$ variables there is a corresponding hyperreal function ${}^*f$ of $n$ variables, called the natural extension of $f$.




    The function in question here is the function that takes $n$ to $sum_{k=1}^n9cdot10^{-k}$.



    A slightly different version of this approach is found in Keith Stroyan’s notes here, especially Section $2.3$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (Thank you for your valid answer. I accepted the other one because of the comment thread following it)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 17:08










    • $begingroup$
      @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 17:36










    • $begingroup$
      Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^ast Bbb N$ but restricted to $Bbb N$ wouldn't be identical to the decimal expansion ranged over $Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction.
      $endgroup$
      – Metta World Peace
      May 19 '13 at 21:19










    • $begingroup$
      It's not that it "wouldn't be identical" to the decimal expansion ranged over $mathbb{N}$. Rather, a "decimal expansion ranged over $mathbb{N}$" does not exist since $mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets.
      $endgroup$
      – Mikhail Katz
      May 20 '13 at 8:44






    • 1




      $begingroup$
      @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*Bbb R$, you can’t talk about a sum over $Bbb N$, because you can’t even talk about $Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model.
      $endgroup$
      – Brian M. Scott
      May 20 '13 at 16:54














    7












    7








    7





    $begingroup$

    $$1-sum_{k=1}^N9cdot10^{-k}=sum_{kge N+1}9cdot 10^{-k}=9sum_{kge N+1}10^{-k}=frac{9cdot 10^{-(N+1)}}{1-10^{-1}}=10^{-N}=frac1{10^N};,$$



    which is surely intuitively a positive infinitesimal.



    Added: There is a nice elementary axiomatization of the hyperreals in Jerry Keisler’s Elementary Calculus: An infinitesimal approach, which is freely available here; it is intended for students in a first calculus course and neatly avoids the transfer principle as such. His Foundations of infinitesimal calculus contains a slightly more sophisticated version, since it’s intended for instructors using the undergraduate text. It’s freely available here, and its version of the axiomatic development can also be found in Section $15$ of this PDF. What he calls the Function Axiom (Axiom $C$ in the PDF) justifies the standard calculation:




    For each real function $f$ of $n$ variables there is a corresponding hyperreal function ${}^*f$ of $n$ variables, called the natural extension of $f$.




    The function in question here is the function that takes $n$ to $sum_{k=1}^n9cdot10^{-k}$.



    A slightly different version of this approach is found in Keith Stroyan’s notes here, especially Section $2.3$.






    share|cite|improve this answer











    $endgroup$



    $$1-sum_{k=1}^N9cdot10^{-k}=sum_{kge N+1}9cdot 10^{-k}=9sum_{kge N+1}10^{-k}=frac{9cdot 10^{-(N+1)}}{1-10^{-1}}=10^{-N}=frac1{10^N};,$$



    which is surely intuitively a positive infinitesimal.



    Added: There is a nice elementary axiomatization of the hyperreals in Jerry Keisler’s Elementary Calculus: An infinitesimal approach, which is freely available here; it is intended for students in a first calculus course and neatly avoids the transfer principle as such. His Foundations of infinitesimal calculus contains a slightly more sophisticated version, since it’s intended for instructors using the undergraduate text. It’s freely available here, and its version of the axiomatic development can also be found in Section $15$ of this PDF. What he calls the Function Axiom (Axiom $C$ in the PDF) justifies the standard calculation:




    For each real function $f$ of $n$ variables there is a corresponding hyperreal function ${}^*f$ of $n$ variables, called the natural extension of $f$.




    The function in question here is the function that takes $n$ to $sum_{k=1}^n9cdot10^{-k}$.



    A slightly different version of this approach is found in Keith Stroyan’s notes here, especially Section $2.3$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 18 '13 at 17:36

























    answered Jan 18 '13 at 16:50









    Brian M. ScottBrian M. Scott

    460k40516917




    460k40516917












    • $begingroup$
      (Thank you for your valid answer. I accepted the other one because of the comment thread following it)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 17:08










    • $begingroup$
      @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 17:36










    • $begingroup$
      Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^ast Bbb N$ but restricted to $Bbb N$ wouldn't be identical to the decimal expansion ranged over $Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction.
      $endgroup$
      – Metta World Peace
      May 19 '13 at 21:19










    • $begingroup$
      It's not that it "wouldn't be identical" to the decimal expansion ranged over $mathbb{N}$. Rather, a "decimal expansion ranged over $mathbb{N}$" does not exist since $mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets.
      $endgroup$
      – Mikhail Katz
      May 20 '13 at 8:44






    • 1




      $begingroup$
      @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*Bbb R$, you can’t talk about a sum over $Bbb N$, because you can’t even talk about $Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model.
      $endgroup$
      – Brian M. Scott
      May 20 '13 at 16:54


















    • $begingroup$
      (Thank you for your valid answer. I accepted the other one because of the comment thread following it)
      $endgroup$
      – Qfwfq
      Jan 18 '13 at 17:08










    • $begingroup$
      @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful.
      $endgroup$
      – Brian M. Scott
      Jan 18 '13 at 17:36










    • $begingroup$
      Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^ast Bbb N$ but restricted to $Bbb N$ wouldn't be identical to the decimal expansion ranged over $Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction.
      $endgroup$
      – Metta World Peace
      May 19 '13 at 21:19










    • $begingroup$
      It's not that it "wouldn't be identical" to the decimal expansion ranged over $mathbb{N}$. Rather, a "decimal expansion ranged over $mathbb{N}$" does not exist since $mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets.
      $endgroup$
      – Mikhail Katz
      May 20 '13 at 8:44






    • 1




      $begingroup$
      @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*Bbb R$, you can’t talk about a sum over $Bbb N$, because you can’t even talk about $Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model.
      $endgroup$
      – Brian M. Scott
      May 20 '13 at 16:54
















    $begingroup$
    (Thank you for your valid answer. I accepted the other one because of the comment thread following it)
    $endgroup$
    – Qfwfq
    Jan 18 '13 at 17:08




    $begingroup$
    (Thank you for your valid answer. I accepted the other one because of the comment thread following it)
    $endgroup$
    – Qfwfq
    Jan 18 '13 at 17:08












    $begingroup$
    @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful.
    $endgroup$
    – Brian M. Scott
    Jan 18 '13 at 17:36




    $begingroup$
    @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful.
    $endgroup$
    – Brian M. Scott
    Jan 18 '13 at 17:36












    $begingroup$
    Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^ast Bbb N$ but restricted to $Bbb N$ wouldn't be identical to the decimal expansion ranged over $Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction.
    $endgroup$
    – Metta World Peace
    May 19 '13 at 21:19




    $begingroup$
    Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^ast Bbb N$ but restricted to $Bbb N$ wouldn't be identical to the decimal expansion ranged over $Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction.
    $endgroup$
    – Metta World Peace
    May 19 '13 at 21:19












    $begingroup$
    It's not that it "wouldn't be identical" to the decimal expansion ranged over $mathbb{N}$. Rather, a "decimal expansion ranged over $mathbb{N}$" does not exist since $mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets.
    $endgroup$
    – Mikhail Katz
    May 20 '13 at 8:44




    $begingroup$
    It's not that it "wouldn't be identical" to the decimal expansion ranged over $mathbb{N}$. Rather, a "decimal expansion ranged over $mathbb{N}$" does not exist since $mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets.
    $endgroup$
    – Mikhail Katz
    May 20 '13 at 8:44




    1




    1




    $begingroup$
    @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*Bbb R$, you can’t talk about a sum over $Bbb N$, because you can’t even talk about $Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model.
    $endgroup$
    – Brian M. Scott
    May 20 '13 at 16:54




    $begingroup$
    @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*Bbb R$, you can’t talk about a sum over $Bbb N$, because you can’t even talk about $Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model.
    $endgroup$
    – Brian M. Scott
    May 20 '13 at 16:54











    3












    $begingroup$

    In addition to the fine answers given earlier, I would like to mention also the somewhat unsung hero, A. H. Lightstone, and his extended decimal notation in which your infinitesimal $epsilon_N$ can be written as $0.000ldots;ldots 0001$, where the first nonzero digit occurs precisely at infinite rank $N$. The notation is explained in his article in the American Mathematical Monthly (see especially page 246).






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      In addition to the fine answers given earlier, I would like to mention also the somewhat unsung hero, A. H. Lightstone, and his extended decimal notation in which your infinitesimal $epsilon_N$ can be written as $0.000ldots;ldots 0001$, where the first nonzero digit occurs precisely at infinite rank $N$. The notation is explained in his article in the American Mathematical Monthly (see especially page 246).






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        In addition to the fine answers given earlier, I would like to mention also the somewhat unsung hero, A. H. Lightstone, and his extended decimal notation in which your infinitesimal $epsilon_N$ can be written as $0.000ldots;ldots 0001$, where the first nonzero digit occurs precisely at infinite rank $N$. The notation is explained in his article in the American Mathematical Monthly (see especially page 246).






        share|cite|improve this answer











        $endgroup$



        In addition to the fine answers given earlier, I would like to mention also the somewhat unsung hero, A. H. Lightstone, and his extended decimal notation in which your infinitesimal $epsilon_N$ can be written as $0.000ldots;ldots 0001$, where the first nonzero digit occurs precisely at infinite rank $N$. The notation is explained in his article in the American Mathematical Monthly (see especially page 246).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '17 at 10:18

























        answered Apr 16 '13 at 12:11









        Mikhail KatzMikhail Katz

        30.7k14399




        30.7k14399






























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