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I was wondering why epsilon, the smallest positive number, isn't a rational number. I was watching a video a few days ago about surreal numbers, and I've learned that, in the field of surreal numbers, o.(9) is not equal to 1, in contrast to the field of the real numbers, where they represent the same number. In the field of surreal numbers, you would get epsilon by subtracting 0,(9) from 1. If you were to do this in the rationals, you would just get 0. But I think there is a method do get epsilon even in the rationals, you would just take the following limit:

$$\lim\limits_{n \to\infty} \sum_{i=0}^n {1\over 10^i}$$ Am I making a wrong mathematical assumption or...? Is there a reason for which epsilon couldn't be a rational number?

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    $\begingroup$ The value of your limit is $0$. $\endgroup$
    – user856
    Dec 29, 2015 at 18:33
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    $\begingroup$ Although in your case it's $0$, the limit of a sequence of rational numbers is not necessarily rational! Consider $3, 3.1, 3.14, 3.141, \ldots$, with a limit of $\pi$. $\endgroup$
    – Théophile
    Dec 29, 2015 at 18:37
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    $\begingroup$ @Aphotesis What does 'rational' mean in the context of surreal numbers? $\endgroup$
    – Marc
    Dec 29, 2015 at 18:39
  • $\begingroup$ Does that limit even exist in the surreal numbers (at least in the order topology)? If we take $\varepsilon=\{0|1,\frac{1}2,\frac{1}4,\ldots\}$ then if $\varepsilon'=\{\varepsilon|1,\frac{1}2,\frac{1}4,\ldots\}$, the interval $(0,\varepsilon')$ contains $\varepsilon$, but no term of the sequence. $\endgroup$
    – Milo Brandt
    Dec 29, 2015 at 19:26
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    $\begingroup$ There is no clear way to give sense to $0.(9)$ in the surreals so the video you watched probably just stated $1-\varepsilon$ was this number because $1 - \varepsilon = \{0;0.9;0.99;...\} \ | \ \{1\}$ but this is just a convention. $\endgroup$
    – nombre
    Dec 31, 2015 at 10:27

3 Answers 3

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See Surreal number :

Consider the smallest positive number in $S_ω$:

$\varepsilon =\{S_{-}\cup S_{0}|S_{+}\}=\{0|1,{\tfrac {1}{2}},{\tfrac {1}{4}},{\tfrac {1}{8}},...\}=\{0|y\in S_{*}:y>0\}$.

This number is larger than zero but less than all positive dyadic fractions. It is therefore an infinitesimal number, often labeled $ε$.

Thus epsilon is, "by definition" less than (and so different from) all rational in the $(0,1)$ interval.

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    $\begingroup$ And it surely isn't the smallest positive number. $\endgroup$
    – egreg
    Dec 29, 2015 at 18:53
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    $\begingroup$ @AndresMejia: Yes, it's not in $\Bbb R$, it is in the field of surreal numbers, as the OP clearly mentions. $\endgroup$
    – Alex M.
    Dec 29, 2015 at 18:55
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    $\begingroup$ @AndresMejia The surreal numbers are not archimedean. $\endgroup$
    – egreg
    Dec 29, 2015 at 19:02
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    $\begingroup$ @ASKASK You are right, of course!! And now I read a bit on the linked Wikipedia page, and $S_\omega$ does not designate all surreal numbers, and $S_\omega$ is just a "step" in the process of constructing all the surreal numbers. In particular $S_\omega$ is not a field. So as egreg commented, this $\epsilon$ is not "the smallest positive number" (which cannot exist). It is the smallest positive number in $S_\omega$, I believe. Also, your comment shows that no ordered field can have a least positive element. $\endgroup$
    – Jeppe Stig Nielsen
    Dec 30, 2015 at 8:16
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    $\begingroup$ @egreg: While the surreal numbers are not archimedean, they are "ordinal-archimedean": For every surreal number, there exists an ordinal number that is larger. And thus for any positive surreal $r$, there exists an ordinal $\alpha$ such that $1/\alpha < r$. $\endgroup$
    – celtschk
    Dec 13, 2017 at 21:42
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It seems you are assuming that "multiplying infinitely many rational numbers yields a rational number," and this is not necessarily true. Doing things "infinitely many times" (really, taking a limit of a process) is a pretty good way to leave the rationals, even when we stay within the realm of real numbers.

Every (nonnegative) decimal number you can write down is a sum of rational numbers of the form $\frac{d}{10^n}$, where $d$ is an integer between $0$ and $9$, and $n$ is some integer. So, for example,

$$\pi = \frac{3}{10^0} + \frac{1}{10^1} + \frac{4}{10^2} + \frac{1}{10^3} + \ldots,$$

but $\pi$ certainly isn't rational, despite being the sum of infinitely many rational numbers.

For products, we can use Euler's product formula for the Riemann zeta function to write

$$\frac{\pi^2}{6} = \left( \frac{1}{1 - 2^{-2}} \right)\left( \frac{1}{1 - 3^{-2}} \right) \left( \frac{1}{1 - 5^{-2}} \right) \left( \frac{1}{1 - 7^{-2}} \right) \cdot \ldots$$

a clearly irrational number as a product of infinitely many rational numbers, although the value of $\pi^2/6$ is highly nontrivial.

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    $\begingroup$ The OP works with surreal, not with real numbers, and here things are a little bit different. (Yours is the fourth answer under which I post this comment.) $\endgroup$
    – Alex M.
    Dec 29, 2015 at 18:52
  • $\begingroup$ I know it doesn't yield a rational number necessarily, but when I'm looking at that limit, for me it resembles epsilon, an infinity of zeroes and, somehow, a 1 at the end. $\endgroup$
    – Asix
    Dec 29, 2015 at 18:54
  • $\begingroup$ I do understand, @AlexM. , I was merely pointing out one mathematical assumption I thought I identified. I know essentially nothing about surreal numbers, so perhaps my comment is not worth anything here. If I accrue enough downvotes, I'll take that to be the case :) $\endgroup$
    – pjs36
    Dec 29, 2015 at 18:55
  • $\begingroup$ "The OP works with surreal, not with real numbers" Which is not relevant. The reals can be considered a subset of the surreals and the rationals of subset of the reals. Neither the reals nor the rationals have a smallest positive number. So if the surreals do, it can not be rational. $\endgroup$
    – fleablood
    Dec 29, 2015 at 21:35
  • $\begingroup$ "The OP works with surreal, not with real numbers" Wrong. The OP is asking about rational nuumbers which are real. His question is "why isn't epsilon rational". Well, it's not rational because it isn't real. And to explain why it isn't real we pretty much have to talk about real numbers, don't we? $\endgroup$
    – fleablood
    Dec 29, 2015 at 22:04
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Okay. I do not know much about the surreal numbers. So I looked on wikipedia and these were the VERY FIRST two sentences.

"In mathematics, the surreal number system is an arithmetic continuum containing the real numbers as well as infinite and infinitesimal numbers, respectively larger or smaller in absolute value than any positive real number. The surreals share many properties with the reals, including a total order ≤ and the usual arithmetic operations (addition, subtraction, multiplication, and division); as such, they form an ordered field."

This answers why epsilon, the smallest positive (surreal) number, can not be rational. The rational numbers and the real numbers have the archimedian property that between any two rationals and between any two reals there exist a third. So it is impossible to have a smallest positive real or rational number. So if a smallest positive number is possible (as apparently it is) it can not be real or rational otherwise it would be the smallest real or rational number which is impossible.

So, I interpret the OP question to be why doesn't his construction of $\lim_{n \rightarrow \infty} \prod_{i=1}^n\frac 1{10^i} $ produce a least rational number?

Because the lim = 0. (It also needn't be rational {although it is}. But it does need to be real-- and there is no least positive real either.)

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Okay, I wasn't aware that epsilon exists in the field of surreal numbers and does have a definition as the smallest positive surreal number (I guess).

But here, fundamentally, the surreals are not the reals. The rationals are the rationals and as $0 < 1/(n+1) < 1/n$ there can be no smallest positive rationals. Thus any system that allows a smallest positive number (which the reals do not), such a number can not be rational.

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1) epsilon is not the smallest positive number. epsilon is not a set number at all. It is a variable to represent possible numbers. It is no more or less a number than, x, n, or i are numbers for exactly the same reasons and in exactly the same way.

2) There is no smallest positive |real| number. This is fundamental to mathematics real analysis. [ed: and so any smallest positive number can not be real.]

3) The limit of a sequence (or sum) of rational numbers need not be rational at all. Indeed it is a fundamental principal of real [ed: emphasis added later] numbers that all real [ed: but not surreal] numbers (rational and irrationals) are limits of sequences of rational numbers.

4) The limit of the infinite product you gave is 0.

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    $\begingroup$ The OP works with surreal, not with real numbers, and here things are a little bit different. (Yours is the third answer under which I post this comment - the other two got deleted by their authors.) $\endgroup$
    – Alex M.
    Dec 29, 2015 at 18:46
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    $\begingroup$ The OP is asking about rational numbers. Which are real. The smallest positive surreal number is possible but it most certainly is neither rational nor real. $\endgroup$
    – fleablood
    Dec 29, 2015 at 22:15
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    $\begingroup$ The OP asks about a surreal number being rational or not. $\endgroup$
    – Alex M.
    Dec 29, 2015 at 22:22
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    $\begingroup$ And this particular surreal number is not rational. It's not rational because it is the smallest positive number. Such a number can not possibly be rational nor even real. To explain why it can not be rational I DO have to talk about real numbers $\endgroup$
    – fleablood
    Dec 29, 2015 at 22:31
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    $\begingroup$ All of which has nothing to do with the OP question "why can't the smallest positive be rational?" $\endgroup$
    – fleablood
    Dec 30, 2015 at 0:23

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