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For instance, we can certainly represent π in this fashion.

$$ \frac{\pi}{4} \;=\; \sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1} .\! $$

$\ln(2)$ is also irrational. And even that can be represented as an infinite sum of a sequence of rational numbers:

$$ \ln (1+x) \;=\; \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n. $$ with $x=1$.

And also, $\sqrt2$:

$$ \sqrt2 \;=\; \sum_{k=0}^\infty\frac{(2k-1)!!}{4^kk!}\tag{2} $$

I'm curious if this applies to all irrational numbers? Is so, how do you go about proving it?

Peter Woolfitt
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shreedhar
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    See also: http://math.stackexchange.com/questions/338328/irrational-and-rational-sequence-proof – Martin Sleziak Oct 06 '15 at 07:28
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    As you can see from the answers, how you go about proving it will depend on exactly how you define the real numbers. I was taught a converging series definition first, which trivially answers your question, and only later learned about Dedekind cuts, which lead to a somewhat less trivial answer. – David K Oct 06 '15 at 13:15
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    This is not at all a duplicate of the linked question; the linked question asks for an infinite sum that is some way "finitely expressible" (just take a look at its accepted answer, which would certainly not be a very definitive answer to this question!). If anything, that question should be closed as a duplicate of this one, which has more extensive answers (and Lucian's answer which does address the main thrust of the other question). – Eric Wofsey Oct 08 '15 at 12:42

10 Answers10

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Since that series $$ \sum_{n=0}^\infty\frac{(-1)^n}{2n+1} $$ converges conditionally, the Riemann Rearrangement Theorem says that we can get every real number, rational or irrational, by rearranging the terms of that series

So, yes, every irrational number can be written as the limit of the sum of rational numbers.

robjohn
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    That answer should have more upvotes :) – Taladris Oct 06 '15 at 11:16
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    The odd denominators come from the question, I know, but apart from that one can just take $\sum \frac{(-1)^n}{n}$ of course. – Jeppe Stig Nielsen Oct 06 '15 at 12:26
  • @JeppeStigNielsen: yes, one can use any conditionally convergent series. – robjohn Oct 06 '15 at 12:57
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    this is a very nice answer. – Konstantinos Gaitanas Oct 06 '15 at 20:43
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    That's all very fine, but isn't it overkill? After all, every real has a decimal expansion (which is really the same statement) is equally accurate and far more well known and acceptable to mathematicians with all levels of comprehension. Still, it is very true. – fleablood Oct 07 '15 at 00:47
  • This seems slightly overkill. – Akiva Weinberger Oct 07 '15 at 01:13
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    @fleablood I don't actually see this as overkill compared to some other answers here... This answer at least leads to an interesting theorem! – Brevan Ellefsen Oct 07 '15 at 01:14
  • Well, not compared to other answers... :) And, yes, it's an interesting theorem. All of these answer are overkill really. And they all are sort of circular (mine included) in that whatever reason they give relies upon real numbers always having convergent sequences which is really what we are claiming. On the other hand, as Lucien pointed out, maybe the OP meant a sum series in which the terms are determined by a finite expressed definition. Which is another issue altogether. – fleablood Oct 07 '15 at 02:49
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    If anyone's curious, the original exposition of this theorem is in section 3 of Riemann's habilitation thesis (wherein is also described something that is not the Riemann integral, but apparently gives Riemann the credit for inventing it), Über die Darstellbarkeit einer Function durch eine trigonometrische Reihe. – Chappers Oct 07 '15 at 17:55
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Yes, an easy way to see this is to look at decimal expansions, as we actually use this fact daily when we say that a number is equal to its expansion. For example, $$\pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+\cdots$$ $$e = 2 + 0.7 + 0.01 + 0.008 + 0.0002 +0.00008 +0.000001 +0.0000008+\cdots$$ Each partial sum is a rational number, and you can break apart any other irrational number the same way.

JTP - Apologise to Monica
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Brevan Ellefsen
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  • This is exactly what I was thinking, but I spent too much time trying to formalize it into a 'closed form' with some sort of digit-selecting function. Very nice response. – Sean Henderson Oct 06 '15 at 05:03
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    This is not a proof. You still need to prove that every real number has a decimal expansion which is in fact a stronger statement than the one in the question. – Alexander Belopolsky Oct 06 '15 at 05:08
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    @AlexanderBelopolsky You have to start somewhere... – Peter Woolfitt Oct 06 '15 at 05:10
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    The only reason we know decimals work as because the reals are defined so that every real number is the limit of a sequence of rationals. That's the definition. There is nothing to prove. – fleablood Oct 06 '15 at 05:15
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    @AlexanderBelopolsky It is true that I assume this, but I figured that it is a pretty well known fact that numbers have decimal expansions. As Peter said, you have to start somewhere. I could have made it a much deeper proof, but this seemed much more intuitive. – Brevan Ellefsen Oct 06 '15 at 05:16
  • What "numbers"? If I define a number as a limit of a series of rationals then the statement asked about is a tautology. This is a mathematics site. Try to be rigorous. – Alexander Belopolsky Oct 06 '15 at 05:21
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    @AlexanderBelopolsky I feel like you are forgetting scope here. I understand completely what you are saying, but this is not MathOverflow, this is Math.SE. The purpose of this site is to be understandable for novice mathematicians who might not understand things such as Dedekind cuts; I proposed an answer that someone like this could understand. We don't teach children set theory in order to find $2+2$.... we sometimes just accept things as being true in math. In fact, fundamentally we are accepting something as true at the base of math. Reals can be defined this way, so why not do so here? – Brevan Ellefsen Oct 06 '15 at 05:31
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    The definition of a real number as a decimal expansion is probably the most difficult definition to work with. Try to prove the distributive property starting from that definition. Note that it is not clear what OP is asking. A possible interpretation of the question may include only analytical series where each term is a rational expression. In this case, I believe the answer is negative. – Alexander Belopolsky Oct 06 '15 at 05:40
  • @BrevanEllefsen, is 3+0.1+0.04+0.001+0.0005+0.00009+0.000002+⋯ a representation of $\pi$ ? – Anonymous Coward Oct 06 '15 at 19:41
  • @JoseAntonioDuraOlmos Yes, it is. Without getting too deep, just think about how our counting system works. We have that $2454 = 210^3 + 410^2 + 510^1 + 410^0$. We expand this definition to get the decimal part of a number, with negative powers of $10$. Therefore, $\pi = 310^0 + 110^{-1} + 410^{-2} + 110^{-3} + 5*10^{-4} +...$. This is often how we define real numbers (there are more complex ways to prove this, but these are expressed in other, less elementary answers) – Brevan Ellefsen Oct 06 '15 at 21:24
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    That other expression truly represents 2454. I can compute the 7th decimal for 2454 from that expression; it is 0. But I can't compute the 7th decimal for $\pi$ from (3+0.1+0.04+0.001+0.0005+0.00009+0.000002+⋯), so it does not represent $\pi$ . My point is that, yes, all real numbers are the sum of an infinite series of rationals. But, no, we can't represent all of those series. Because our representations are limited to finite statements composed from a finite alphabet, and those are countable. – Anonymous Coward Oct 06 '15 at 22:19
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    @BrevanEllefsen I believe Jose is talking about the fact that although there are uncountably many reals and uncountably many countable sequences of rationals, there are only countably many computable real numbers or definable sequences of rationals. – Mario Carneiro Oct 07 '15 at 00:32
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    Actually, I think we can still demonstrate the with a fuzzy definition of "the real numbers are the rationals with the holes plugged in". All reals are infinitely close to some rationals. (Because between any two numbers, irrational or not, we can find a rational number between them [well that might not to be shown]). Thus even by that fuzzy definition there must a sequence of rationals the converge to every real number. – fleablood Oct 07 '15 at 00:41
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    OP was looking for a formulaic summation. His summation resulting in pi, for example, is all I need to spend some time to get N digits. Your answer doesn't offer that at all. – JTP - Apologise to Monica Oct 07 '15 at 04:04
  • $e = 2 + 0.1 + 0.08 + 0.002 +0.0008 +0.00004 +0.000005+\cdots$? – robjohn Oct 07 '15 at 11:08
  • @Brevan Ellefsen. For an explanation of indefinable/unrepresentable reals see : http://math.stackexchange.com/a/617751/251953 . (at)Mario Carneiro. Yes, that is my point. – Anonymous Coward Oct 07 '15 at 11:51
  • @robjohn - good catch. I was trying to ignore that. Focusing on the (in my opinion) non-answer. – JTP - Apologise to Monica Oct 07 '15 at 12:14
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    I couldn't look at it any longer, edited e. – JTP - Apologise to Monica Oct 07 '15 at 12:37
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Every real number can be represented as an infinite sum of rationals.

Proof: Let $a\in\mathbb{R}$ and $a_1,a_2,\dots$ be a sequence of rationals converging to $a$.

Then

$$a=a_1+\sum\limits_{n=1}^\infty(a_{n+1}-a_n)$$

Peter Woolfitt
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  • You still have to prove that for every real number there exists a sequence of rational numbers that converges to it. – Alexander Belopolsky Oct 06 '15 at 05:03
  • I would call this a well-known fact, @Brevan, but for the existence of such sequences, you can always use say the one in your own answer :P – Peter Woolfitt Oct 06 '15 at 05:05
  • @Alexander See above comment – Peter Woolfitt Oct 06 '15 at 05:06
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    "You still have to prove that for every real number there exists a sequence of rational numbers that converges to it." That's the definition of a real number. – fleablood Oct 06 '15 at 05:11
  • Neither of you have presented a proof. – Alexander Belopolsky Oct 06 '15 at 05:11
  • @AlexanderBelopolsky what is your definition of a real number? – Peter Woolfitt Oct 06 '15 at 05:12
  • "how does one show that there exists a sequence of rationals that converge to $a$? Isn't that just an assumption unless you have already seen it proven, in which case you get circular logic?" The fundament theorem of analysis defines the real numbers to be the extension of the rationals where this is true. It's the definition of a real number. – fleablood Oct 06 '15 at 05:13
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    @Peter: https://en.m.wikipedia.org/wiki/Dedekind_cut – Alexander Belopolsky Oct 06 '15 at 05:14
  • @AlexanderBelopolsky After thinking about is fleablood is correct here, and I realized I used the same intuition implicitly in my answer... my mistake in the comment above :P It is well known that reals are defined by limiting values, and I thought that the easiest way to get these limiting values is using decimal expansions (as my answer uses) – Brevan Ellefsen Oct 06 '15 at 05:19
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    @AlexanderBelopolsky The elements on the lower side of the cut, in increasing order and starting from anywhere, form a sequence of rationals converging to the real number corresponding to the cut. – user253751 Oct 06 '15 at 05:20
  • @immibis - that's closer to a proof. – Alexander Belopolsky Oct 06 '15 at 05:25
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    If we accept that every real number has an infinite decimal expansion (which is more or less how it is defined in high school) than we can always use the decimal expansion to define the sum. – fleablood Oct 06 '15 at 05:31
  • "The elements on the lower side of the cut, in increasing order and starting from anywhere, form a sequence of rationals converging to the real number corresponding to the cut." Now that packs a lot of punch into a short comment! Good job. That's basically the heart of the fund. th. and def of real numbers. – fleablood Oct 06 '15 at 05:32
  • Okay, if between any two irrational numbers is a rational and between any two rationals is an irrational, it follows that for every irrational numbers there are rationals that are infinitely close to the irrational numbers. So we can take a sequence of these rational numbers and the converge to the irrational number. (But now have to prove that between any to irrationals is a rational number.) – fleablood Oct 06 '15 at 06:13
  • @PeterWoolfitt: Not only 'every irrational number' but every real number can be represented as an infinite sum of rationals. For example $1=\sum_{n=1}^\infty 2^{-n}$ – CiaPan Oct 06 '15 at 07:31
  • @CiaPan $1=1+0+0+0+0+0+\cdots$ also works – user253751 Oct 06 '15 at 08:06
  • @CiaPan Certainly true. I just said irrational because that was what the question asked for. Indeed the proof is for every real. – Peter Woolfitt Oct 06 '15 at 08:07
  • Isn't this related to the density of the rationals in the reals? Or.. the density of the irrationals? Or both? – Todd Wilcox Oct 06 '15 at 11:33
  • @fleablood: If you start with decimal expansions as your definition of the reals, you have to deal with artifacts thereof before you can even begin to reason about them. – Kevin Oct 06 '15 at 15:29
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    "The elements on the lower side of the cut, in increasing order and starting from anywhere, form a sequence of rationals converging to the real number corresponding to the cut." The elements on the lower side of the cut cannot be arranged in increasing order in general, or at least cannot be enumerated in increasing order. (What's the very first rational number larger than 1, for instance?). On the other hand, an increasing subsequence of them CAN be chosen. But it might converge to something less than the cut-value. – John Hughes Oct 06 '15 at 20:31
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    Consider the Dedikind cut $C$ associated to the irrational $s$. For each positive integer $n$, pick an element $q_n$ of $C$ with $s - \frac{1}{n} < q_n < s - \frac{1}{n+1}$. Then the $q_n$ form an increasing sequence of rationals whose limit is $s$. Now let $r_1 = 0$, and $r_n = q_{n+1} - q_n$ for $n > 1$. The the sums $\sum_1^n r_i = q_{n+1} - q_1= q_{n=1}$ have $s$ as their limit. – John Hughes Oct 06 '15 at 20:31
  • @JohnHughes I interpret immibis's comment as saying that we take at each stage the first rational number (under some sequential ordering of $\Bbb Q$) that is greater than the one at this stage and in the lower cut. No regularizing bounds like your $1/n$ sequence are needed, because the lower cut has no maximum and we eventually test every element of the lower cut (so every element in the lower cut is less than some element in the sequence). – Mario Carneiro Oct 07 '15 at 00:42
  • That's an extremely generous reading, @MarioCarneiro, and I don't really agree that this is what was said, but your rewriting of it is far more convincing to me. :) – John Hughes Oct 07 '15 at 01:24
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    "

    @fleablood: If you start with decimal expansions as your definition of the reals" Okay. I never said we can use the decimal expansion as a definition. We really really shouldn't and to do so kind of sickens me. I said most of our impressions of the reals are based on such a concept that many think of it as a definition. As a definition it is a crappy one.

     – fleablood Oct 07 '15 at 02:59
  • "Isn't this related to the density of the rationals in the reals?" I think of it as being more or less equivalent. If you define the reals as having the least upper bound property then you can prove density. Otherwise, I think, you have to assume density from which... well, you can prove every real has a rational sequence converging to it. I'm not 100% certain you can prove the reals have the least upper bound property but maybe you can. – fleablood Oct 07 '15 at 03:21
  • @fleablood No, density isn't enough. After all, $\Bbb Q$ is dense in itself and it satisfies all the other ordered field axioms, but it's not complete. – Mario Carneiro Oct 07 '15 at 04:23
  • Mario Carneiro. D'oh. Of course you are right. (However every rational does have a rational sequence that converges to it. It's just not that all bounded rational sequence converge to a rational. So I think density is enough to show all reals have a rational sequence that converges to it, but density isn't enough to show all bounded rational sequences converge to a real.) – fleablood Oct 07 '15 at 05:44
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If I understand you correctly, then the answer is no. Notice that all the sequences in question have general terms of a “regular” form. However, since the number of such “regular” expressions is countable, whereas the number of irrationals is not, the logical conclusion would be that it is simply impossible. Rob John mentioned rearranging the “regular” terms of a conditionally convergent expression to obtain every real number imaginable. True, but in this case the rearrangement itself would be “irregular”, thus disrupting the “regularity” of the expressions you gave as examples.

Lucian
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    +1 for pointing out that some real numbers cannot be represented by any finite-length expression. – Keen Oct 06 '15 at 16:27
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    This is the true answer. Since finite expressions are countable and real numbers are uncountable we have to conclude that not all real numbers can be represented. And us humans can't create expressions of infinite length. – Anonymous Coward Oct 06 '15 at 19:46
  • "However, since the number of such “regular” expressions is countable" This is not true. "Since finite expressions are countable". We are not talking about finite expressions. We are talking about infinite expressions. And those are uncountable. ($2^\infty$ is uncountable) Any finite expression would obviously be rational. But all reals are an infinite sum of rationals. – fleablood Oct 06 '15 at 21:00
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    "+1 for pointing out that some real numbers cannot be represented by any finite-length expression" Well, duh! Any finite expression would be rational. But we are talking about infinite expressions. – fleablood Oct 06 '15 at 21:01
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    "However, since the number of such “regular” expressions is countable" This is simply not true. And it's not true because of Cantor arguments. A decimal number is a regular expression. And all the decimal points with a finite number of terms is countable. But all the decimals with an infinite number of terms are not. By Cantor's diagonal. A more basic result is that [0,1]^i = [0,1]X[0,1]X...X[0,1] has are countable. But [0, 1]^$\infty$ is not. – fleablood Oct 06 '15 at 21:14
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    @fleablood: I am afraid that the expressions for the general term used by the OP are slightly more precise than the vague $\displaystyle\sum_{k>0}\frac{a_k}{10^k}$ , where each $a_k$ is some random decimal digit. The latter are indeed uncountable, the former, however, are not. – Lucian Oct 06 '15 at 22:54
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    @fleablood: A decimal number is a regular expression. - Your use of the word “regular” differs from mine. $($I just wanted to point that out$)$. – Lucian Oct 06 '15 at 23:04
  • "I am afraid that the expressions for the general term used by the OP are slightly more precise than the vague". How so? Okay, what does "regular" mean to you? And why do you think the OP question require that concept of "regular". The OP asked simply of limits of sums of rationals. I suppose the the phrase "represented" by is ambiguous. It could be taken to mean that terms will be described in a way that they are all predictably through finite information. If so (but that is NOT what the OP asked) than maybe not... – fleablood Oct 07 '15 at 00:29
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    @fleablood: Obviously, I cannot read someone else's mind. If my intuition is off, no biggie: there are all the other highly-upvoted answers out there, which tackle precisely that other possible interpretation. But if by any chance my reading is correct, and the OP is indeed asking if such “nice formulas” exist for all reals, then someone should leave an answer explaining why that isn't the case. – Lucian Oct 07 '15 at 01:02
  • I will admit it never occurred to me that the OP was talking about "nice" formulas. "nice" formulas would require a definition of what we mean by "nice" (I've never heard the term "regular" for this). In general you are right of course. There are only countably many ways to "describe" numbers finitely. But we have to define what "describe" means. A decimal expansion "describes" although we have no way to predict an expansion but we can circularly "describe" a specific real by defining the sum terms in term of the real. – fleablood Oct 07 '15 at 02:56
  • @fleablood Perhaps saying "finite-complexity" rather than "finite-length" would have made the statement easier to interpret. – Keen Oct 07 '15 at 14:26
  • @keen. Yes, that would be better. Although analyzing complexity is ... er.... complex. – fleablood Oct 07 '15 at 16:26
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    @fleablood. "Well, duh! Any finite expression would be rational. But we are talking about infinite expressions." We don't disagree but rather are entangled in terminology. All expresions in OP are what I call finite expressions in that they are represented with a finite number of symbols. They express a sum of inifinite rationals; which can result in an irrational. Since OP used only those kind of expressions I assume he is only asking about those kinds. And we can't get all irrationals with those expressions expressions. As explained in the answer of the recently marked duplicate – Anonymous Coward Oct 08 '15 at 17:53
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    Of course, if we are talking about expressions of infinite length, like the decimal expansion of an irrational, then yes, all irrationals can be obtained that way. But there is no human way to completely represent such infinite length expression. – Anonymous Coward Oct 08 '15 at 17:55
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====new edit====

It comes to my attention through Lucien's answer that " represented as an infinite sum of a sequence of rational numbers" can be interpreted two ways. It can be simply $x = \sum q_n $ where each $q_n$ is a rational number. This is the way I interpreted it and it's this interpretation that the rest of this answer is based on.

Or it could be interpreted as $x = \sum $(some nice rule that gives a rational number based on n). The examples of the OP are of this type and have a predictive quality. We can use them to calculate the value of the real number. My interpretation has no predictive quality as to what the $q_n$ terms will be; just that there are a series of rational terms that will converge to the real irrational x.

By my interpretation, all irrationals can be so represented (answer below). By Lucien's interpretation, they can not. His/Her reason is there only countably many rules. I'm not sure of that, but I believe irrationals being uncountable make them "arbitrary" and unpredictable. But I'd have a very difficult time formalizing that.

========== end of new edit ===========

Short answer: That is the definition of a real number.

Long answer:

The fundamental thereom of analysis is that there is an ordered field that extends the rationals such that the field has the least lower bound property. We define the real numbers to be that field.

This means, by definition, every real number is the limit of a convergent sequence of rationals. By definition.

Infinite sums are the limit of finite sums. Hence every real can be written as an infinite sum of rationals. This equivalent to the definition of real number.

The proof of the fundamental thereom is kind of tedious and long. It's not hard but the point is you do the proof before the reals are defined and the definition comes out during the proof.

Longer answer:

Outline of Fund Theorem:

Step 1: Define a "cut" to be a set of rationals with properties:

i) a cut is not empty. ii) if p is in a cut then every rational number less than p is in the cut iii) for any p in the cut you can find a larger rational that is in the cut

So a cut could be all the rationals less than but not equal to 3. Or all rational numbers whose squares are less than 2. (The first is going to eventually be equivalent to 3, and the latter is going to eventually be equivalent to $\sqrt 2$

Step 2: Define a < b to mean the cut a is a subset of the cut b.

Step 3: Show that the set of all cuts, let's call it R~ has the least upper bound property.

Sheesh. This is where it gets abstract. The least upper bound property means every bounded set in a Universal Set (such as what the Reals will be once we define them) has a distinct limit that is in the universal set. Example: Q does not have the least upper bound property.

So we can have a set of cuts called A. It can be bounded above meaning the is a cut, b, such that all the cuts in A are subsets of b. (Remember "smaller" means "is a subset of"). The union of all the cuts in A is bigger or equal to all cuts in A. the union is a cut itself. The union it the smallest cut that is bigger than all the cuts in A. So the union is a least upper bound and R~ has the least upper bound property.

Step 4: Define cut a "+" cut b to be the cut that contains the sums of elements from a plus elements from b. Define 0~ to be the cut that contains all the negative numbers. This satisfies addition properties.

Step 5: More about field and additive and order properties than you'd care to think about.

Step 6-8: Show that R~ a field.

Step 9: Show the Q~ = all the cuts that are defined to be all points less than a rational number is equivalent to Q. so Q has an extension that is equivalent to R~. We call the R, the real numbers.

So..... So each real number is simply the limit of all the rational numbers in some cut. The cut provides sequences of rational numbers that converge to that real number.

So every real number is the limit of a convergent sequence of rational numbers.

fleablood
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  • What is "the fundamental theorem of analysis"? – Alexander Belopolsky Oct 06 '15 at 05:17
  • This is absolutely true, and is much more rigid than my answer or Peter's. Nevertheless, this is probably way above the OP's knowledge (I've barely touched this myself, generally just accepting this as a truth rather than going through the proof). Good answer though! – Brevan Ellefsen Oct 06 '15 at 05:21
  • "There exists an ordered field R which has the least-upper-bound property. Moreover, R contains Q, the rational numbers, as a subfield." – fleablood Oct 06 '15 at 05:22
  • Brevan, it may be above the OP's knowledge, but you can't prove something that is given in a definition. Basically we have to ask what is a real number. Maybe our concept is a fuzzy "it's the rationals with all the holes filled in". That's fine but... we don't need to know how to proof the fundamental theorem, but we should realize that what the OP discovered is actually the definition of real numbers. – fleablood Oct 06 '15 at 05:27
  • @fleablood wonderful response. I see what you mean here. I'll agree that the simplest answer is that it is the definition of real numbers, which is what I imply in my answer (using decimal expansions, a high-school definition for constructing these real numbers). You bring up a good point about how we can't prove a definition... While I think this is above the OP's scope, I'm interested in the proof and I see your point entirely. +1 – Brevan Ellefsen Oct 06 '15 at 05:35
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    You misspell "theorem" as "thereom" so consistently that I suspect this is a novel term meaning a crossbread between a theorem and axiom and a definition. :-) – Alexander Belopolsky Oct 06 '15 at 05:47
  • @fleablood A beautiful proof outline. Thanks for writing that out here. – Brevan Ellefsen Oct 06 '15 at 06:07
  • I suck at spelling. Okay, I guess I got carried away, but I think the real question and answer, is to incorporate the OP's property as the definition of real numbers. It was a good question and a good observation on the OP's part. – fleablood Oct 06 '15 at 06:08
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    I like Alexander's new word "crossbread" -- sounds like an angry pumpernickel. – John Hughes Oct 06 '15 at 20:36
  • I like the answer but why not just refer to the Dedekind cut process as a definition? It's been some years but I think that is the way I learned the formal way to construct real numbers; and consequently taken as a definition. – rrogers Oct 07 '15 at 13:10
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Here's a simple answer.

Pick an irrational x.

Find a rational in (x - 1, x). Call this $a_0$. Find another rational in (x - 1/2, x). Call this $a_1$. Keep finding rational $a_n$ in $(x - 1/2^n , x)$. These $a_n$ converge to x.

$x=a_0+\sum\limits_{n=1}^\infty(a_{n}-a_{n-1})$

fleablood
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    @PeterWoolfitt: Should it? I tried ... Oh, darn it. Yes, because a_n is supposed to tend to ... argh. I hate indexing errors .. I'll fix it. – fleablood Oct 07 '15 at 21:57
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Let $\alpha$ be an irrational number. Let $n_0:=\lfloor\alpha\rfloor$ and define inductively $$n_{k+1}:=\left\lfloor10^{k+1}\left(\alpha-\sum_{i=0}^k\dfrac{n_i}{10^i}\right)\right\rfloor.$$ Then $$\sum_{k=0}^\infty\dfrac{n_k}{10^k}=\alpha.$$

Note that $\alpha$ need not be irrational.

CIJ
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Yes, that depends on the fact that $\mathbb Q$ is dense in $\mathbb R$ and closed under subtraction. For a dense set $D$ in $X$ that is also closed under subtraction then any $x\in X$ can be expressed as a sum of elements in $D$.

Proof: Since $D$ is dense in $X$ we have for each element $x\in X$ that $x = \lim_{n\to\infty}d_n$. Since $D$ is closed under subtraction $\delta_n = d_{n}-d_{n-1} \in D$ and $d_n = d_0+\sum_1^n\delta_n$, so $x = d_0+\sum_1^\infty\delta_n$.

skyking
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First, every rational $p/q$ can be represented as the limit of a series of rational numbers. The simplest is $a_0 = \frac{p}{q}$, $a_k = O, \forall k \neq 0$. You can do that in an infine number of ways: $a_0 = \frac{p-1}{q}$, $a_1= \frac{1}{q}$, and others $a_k$ are $0$, and you can build your own easily.

Continued fractions are often considered as more "mathematically natural" representations of any real number than other representations such as decimal representations. You can produce a series of "best" rational approximations to any read number $\alpha$, in the shape of: $$\alpha \sim b_0 + \frac{1}{b_1+\frac{1}{b_2+\frac{1}{b_3+\ldots}}}\,. $$ Those representations have great properties. And of course as they become rational when you stop at $b_k$, the above property for rationals holds.

So yes, any real is the limit of an infinite quantity of sums of rational series.

Laurent Duval
  • 6,452
-3

The rational numbers are dense in the real numbers.

Therefore every irrational number x greater than 0, there exists a rational number y such that

$$0 < y < x $$

and

$$y + \frac{x-y}{2} + \frac{\frac{x-y}{2}}{2} + \frac{\frac{\frac{x-y}{2}}{2}}{2} ...$$

converges to x.

(This can easily be generalized for x < 0.)

QED

Brevan Ellefsen
  • 14,793
batsplatsterson
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