I tried to search in the internet about this but did not get any exciting answers. So my question is: How is construction of transcendental numbers like $\pi$ and $e$ explained via Dedekind cuts?

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    $\begingroup$ What do you think there is to explain? $\endgroup$ – Brian M. Scott Sep 9 '13 at 20:57
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    $\begingroup$ @BrianM.Scott I think he means to ask for a predicate $P$ such that $\pi = \{x\in \Bbb Q\colon P(x)\}$ and such that this set is a dedekind cut. $\endgroup$ – Git Gud Sep 9 '13 at 20:58
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    $\begingroup$ For instance, to obtain $e$ take $\{x\in \mathbb{Q} \mid \forall n\in \mathbb{N},\ \sum_{k=0}^n \frac{1}{k!} < x\}$ as one of the sets. $\endgroup$ – njguliyev Sep 9 '13 at 21:01
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    $\begingroup$ Because the premiss behind Dedekind Cuts is that we have not yet defined real numbers, so it does not make sense to say the series converges to something, since that something isn't defined yet :) $\endgroup$ – Daron Sep 9 '13 at 21:03
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    $\begingroup$ @Daron How can you even try to define $\pi$ without defining the real numbers first? Even if you do, that's unusual to say the least. $\endgroup$ – Git Gud Sep 9 '13 at 21:04

Here is a simple description for $e$. The left set consists of all rationals $r$ such that $$r\lt 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}$$ for some $n$. This description is close in spirit to one of the many definitions of $e$.

One can give a similar description for $\pi$, though there is nothing as natural. We could use the following variant of the "Leibniz" series, using for the left set all rationals $r$ such that $$r\lt 4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+ \cdots +\frac{4}{4n+1}-\frac{4}{4n+3}$$ for some $n$. Note that we stop with a "$-$" because we want to make sure we are below $\pi$.

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    $\begingroup$ Nicolas, could you please elaborate? How does your description of $\pi$ and $e$ satisfy the definition of cut? In particular, how do you argue that term in the RHS is a rational number and still satisfy the property that it is $>$ r? $\endgroup$ – math Sep 9 '13 at 21:28
  • $\begingroup$ Take the $e$ case. Tiny details depend on how we define cut. Let's define $A|B$ by saying a rational is in $A$ if it is $\lt 1+1+\frac{1}{2!}+\cdots +\frac{1}{n!}$ for some $n$. Let $B$ be the rest of the rationals. It is easy to verify $A|B$ is a cut. For example, immediately from the definition, if $r\in A$ and $s\lt r$ then $s\in A$. The set $A$ is non-empty, and not equal to the set of rationals, since we can prove by induction that no number $1+1+\cdots +\frac{1}{n!}$ is greater than $3$. $\endgroup$ – André Nicolas Sep 9 '13 at 21:55
  • $\begingroup$ Okay, I understand this. But are you implicitly using the idea of convergence of series by taking $n$ to be infinity to define $e$? If yes, we have never defined what series, convergence, limits mean, and on top of that, the infinite sum won't be rational number! If not, I still cannot figure out why with finite $n$, $A|B$ is going to define $e$ (as we did not take rational numbers less and more than the finite sum)! My knowledge of Analysis is very limited. Perhaps, this is hindering my understanding. $\endgroup$ – math Sep 10 '13 at 5:30
  • $\begingroup$ I am using what I know to decide what to use. But the definition, and verification of properties, does not require any convergence argument. For $A$ we took the rationals $r$ such that there is a finite sum $s_n$ with $r\lt s_n$. Purely finite definition, no limits. $\endgroup$ – André Nicolas Sep 10 '13 at 5:39
  • $\begingroup$ Is this a nice way to write $\pi$? $$\{ x \in \mathbb{Q}: \exists n \in \mathbb{N}, \forall m \in \mathbb{N}, m \ge n \implies x < 4 \sum_{i=1}^m{(-1)^{i+1}/(2i-1)}\}$$ $\endgroup$ – Alex Oct 22 '16 at 23:54

Based on what other people said. Let $q_n = 1 + 1/1! + 1/2! + ... + 1/n!$. This is a rational number, thus the associated cut $q_n^* = \{ x \in \mathbb{Q} ~ | ~ x < q_n \}$. The number $e$ is the supremum of all of these. Thus, we can say, $$ e = \bigcup_{n\geq 1} q_n^* $$


If one feels uneasy to use series to define the Dedekind cut for $e$, we can instead take the set $Q_e$ of every rational approximations of $e$ and define the cut as the set $R_e$ of every rational $q$ such as $q < p$ for some $p \in Q_e$.

The series provides a method to compute some members of the cut, but you don't need the series to define the cut itself. The same goes for $\pi$ or any real number.


We invoke the power of abstraction. If we construct the real numbers as Dedekind cuts of the rationals, then we use this method to show that the methods of calculus and real analysis work properly.

Then, we use our considerable experience in calculus to construct $e$ and $\pi$.

If we were so inclined, we could take the entire construction in terms of calculus, and rewrite every individual part in terms of Dedekind cuts. However, nobody ever does this: there is no benefit in carrying out such a tedious exercise.

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    $\begingroup$ It seems to be that you're only saying it is possible to do it and in no way saying how it can be done. $\endgroup$ – Git Gud Sep 9 '13 at 21:09

André Nicolas posted a very natural Dedekind cut for e, but lamented that the equivalent for $\pi$ is less natural. So this answer covers a "natural" Dedekind cut for $\pi$.

The left set consists of all rational numbers which are less than the perimeter of some polygon inscribed in a circle of unit diameter. The right set is the complement of the left set with respect to $\mathbb{Q}$. Equivalently, but more usefully, the right set is the set of all rational numbers which are greater than or equal to the perimeter of some polygon circumscribed about a circle of unit diameter.

It is not hard to prove that this satisfies the requirements of the Dedekind cut axiom:

  1. The left set is nonempty, because it contains all the negative rationals.
  2. The right set is nonempty, because you can construct an arbitrary circumscribed polygon, which must be of finite perimeter and thus shorter than at least one rational number.
  3. The left set is closed downwards, because if it contains $x \in \mathbb{Q}$, then $x$ is less than the perimeter of some polygon $P$, which is also true of every rational number less than $x$.
  4. The left set lacks a greatest element, because, returning to our example of $x$ and $P$, we require that $x$ is strictly less than the perimeter of $P$, and so there must be another rational number which is closer to the perimeter than $x$ is. $x$ cannot be the greatest element, and so there can be no greatest element at all.
  5. Proving that our definition of the right set does indeed produce the complement of the left set is a bit trickier, but as discussed below, the general idea has been understood since Archimedes (albeit not in the specific context of set theory).

Furthermore, we even have an algorithm for classifying rationals (which is not required, as the Dedekind cut axiom does not say that the sets must be recursively enumerable, but merely that they must exist). To determine whether a given rational belongs to the left set or the right set, iteratively construct regular polygons both inscribed in and circumscribed about the circle, with an increasing number of sides each iteration. Eventually, the rational number will either exceed the perimeter of a circumscribed polygon, or be exceeded by the perimeter of an inscribed polygon, and then you have your answer.

Unfortunately, the perimeter of a polygon is not necessarily rational. So on its face, this definition is in terms of other irrational numbers, which may feel less pure. On the other hand, you don't strictly need to work in the reals, so long as you can determine whether a given rational number is larger or smaller than a given polygon's perimeter, which means you just need an arbitrarily-accurate numerical approximation of a large-enough set of polygons (e.g. just the regular ones) to classify all rationals. The exact formulas, and methods for their rational approximation, were described by Archimedes, and this was apparently considered a reasonable method of approximating $\pi$ as recently as 1630.

Overall, this definition has the advantage of being extremely well-studied, and "obviously correct." However, we should acknowledge that modern numerical methods converge faster and are easier to calculate to a higher degree of precision. This definition is great if you want an easy proof that $\pi \in \mathbb{R}$, or if you want an intuitive feel of "what $\pi$'s Dedekind cut actually looks like," but this is not necessarily the best way of calculating $\pi$'s digits.


The real numbers are really troublesome due to the infinities that enter the whole game. It is possible to show that $\sqrt2$ is a cut, because you can easily go to a rational by squaring. With transcendental numbers this is much more difficult, and actually impossible. You simply have to believe that transcendentals are good numbers. It is a choice so to say.

What is done above actually, is that $e$ and $\pi$ are represented by some finite rational interval that can be chosen arbitrary small. But rational intervals do not obey the field rules, since multiplication is not distributive over addition in an unambiguous way no matter how small you choose them. Also the interval that contains zero causes serious problems in elaborate calculations. Calculations based on intervals always can result in an interval that contains zero. Personally I would rather consider real values instead of real numbers, since there is no clarity whether real values are good numbers in general. It is still debated among mathematicians whether the reals are good numbers.

  • $\begingroup$ Welcome to MSE, here is a mathjax tutorial math.meta.stackexchange.com/questions/5020/… $\endgroup$ – joshuaheckroodt Nov 25 '17 at 12:22
  • $\begingroup$ it is true that generally we cannot construct cuts for every real number. But in the case of something as easily definable as $e$ or $\pi$? That's fairly easy. No idea what "real values" means here, though. And who still "debates" that thing you say at the end? Do you have any references? $\endgroup$ – Asaf Karagila Nov 25 '17 at 12:40
  • $\begingroup$ (And fringe ultrafinitists like Zeilberger or WIldberger do not count as any serious debate in this context.) $\endgroup$ – Asaf Karagila Nov 25 '17 at 12:41
  • $\begingroup$ Maybe mathematicians like De Brouwer are valid enough for you? $\endgroup$ – Jan Nov 25 '17 at 15:34
  • $\begingroup$ But then again Wildberger also mentions De Brouwer, so maybe this famous mathemtician is now also fringe becouse of assiociation? $\endgroup$ – Jan Nov 25 '17 at 15:35

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