# Is it possible to represent every irrational number as a (limit of) an infinite sum of rational numbers? [duplicate]

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?

• Oct 6, 2015 at 7:28
• 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. Oct 6, 2015 at 13:15
• 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). Oct 8, 2015 at 12:42

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.

• That answer should have more upvotes :) Oct 6, 2015 at 11:16
• The odd denominators come from the question, I know, but apart from that one can just take $\sum \frac{(-1)^n}{n}$ of course. Oct 6, 2015 at 12:26
• this is a very nice answer. Oct 6, 2015 at 20:43
• 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. Oct 7, 2015 at 0:47
• @fleablood I don't actually see this as overkill compared to some other answers here... This answer at least leads to an interesting theorem! Oct 7, 2015 at 1:14

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.

• 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. Oct 6, 2015 at 5:08
• @AlexanderBelopolsky You have to start somewhere... Oct 6, 2015 at 5:10
• 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. Oct 6, 2015 at 5:15
• @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? Oct 6, 2015 at 5:31
• 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. Oct 6, 2015 at 5:40

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

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

• 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? Oct 6, 2015 at 5:02
• "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. Oct 6, 2015 at 5:11
• Oct 6, 2015 at 5:14
• @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. Oct 6, 2015 at 5:20
• 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. Oct 6, 2015 at 5:31

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.

• +1 for pointing out that some real numbers cannot be represented by any finite-length expression.
– Keen
Oct 6, 2015 at 16:27
• 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. Oct 6, 2015 at 19:46
• "+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. Oct 6, 2015 at 21:01
• "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. Oct 6, 2015 at 21:14
• @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. Oct 7, 2015 at 1:02

====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.

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.

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.

• What is "the fundamental theorem of analysis"? Oct 6, 2015 at 5:17
• 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! Oct 6, 2015 at 5:21
• "There exists an ordered field R which has the least-upper-bound property. Moreover, R contains Q, the rational numbers, as a subfield." Oct 6, 2015 at 5:22
• 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. :-) Oct 6, 2015 at 5:47
• I like Alexander's new word "crossbread" -- sounds like an angry pumpernickel. Oct 6, 2015 at 20:36

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.

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

• @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. Oct 7, 2015 at 21:57

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.

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 $x = d_0+\sum_1^\infty\delta_n$.

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.

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

• After the first term, every term in your series is irrational. Oct 6, 2015 at 22:04
• You're right of course epimophic. Oct 6, 2015 at 22:49
• You can modify it by not picking the midpoint but picking a q_i in the segment. Oct 7, 2015 at 3:24