When defining irrational numbers like $e$ and $\pi$ via Dedekind cuts (like for example here),

  1. how can one prove that these numbers are in fact irrational? All irrationality proofs of $e$ that I know of involve infinite series and all irrationality proofs of $\pi$ that I know of involve a definition of $\pi$ via trigonometric functions.
  2. how can one construct the decimal expansion of these numbers? I am aware of the procedure of how to get the decimal expansion from a Dedekind cut, but to get this you need to determine biggest digits such that the resulting rational number lies in the cut. This is easy to do for, say, $\sqrt{2}$, but I do not see how to do that in the cases of $e$ and $\pi$.

The background is that I want to have a treatment of irrational numbers that avoids the notion of convergence.


For example, we can define $e$ to be the Dedekind cut with lower set $A = \left\lbrace x \in \mathbb{Q} : \exists n : x < \sum_{i=0}^n \frac{1}{i!} \right\rbrace$. It is easy to show that $2 \in A$ since $2 < \sum_{i=0}^2 \frac{1}{i!}$ but how does one show $3 \notin A$ without using infinite sums (if this is even possible)?

  • $\begingroup$ Dedekind cuts are more used as a convenient way to describe how to complete $\mathbb{Q}$ to make $\mathbb{R}$, than to define specific real numbers. But anyway, any Cauchy sequence of rational numbers can be transformed into a Dedekind cut of rational numbers, and vice versa. So I don't get what kind of answer you expect... Please could you clarify ? $\endgroup$ Commented Apr 3, 2023 at 20:20
  • $\begingroup$ I edited the question to (hopefully) clarify what I am lokking for. $\endgroup$
    – Martin
    Commented Apr 4, 2023 at 17:48
  • $\begingroup$ To prove that $3 \notin A$, you use another characterization than the series: we need decreasing upper bounds. This can be done with the continued fraction: $2+\frac{1}{1+\dots}$ is already sufficient to prove that $3\notin A$. One can also use the decreasing sequence $(1+\frac{1}{n})^{n+1}$. Of course it requires non obvious algebraic tricks to prove that all these Dedekind cuts are the same. But that's exactly the same problem with a series. Nobody ever computes infinite sums: we know $e$ is $2.7\dots$ because someone has found an algebraic trick to frame it between $2.7$ and $2.8$. $\endgroup$ Commented Apr 4, 2023 at 21:19
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    $\begingroup$ $$1+1+{1\over2}+{1\over6}+{1\over24}+{1\over120}+\dots+{1\over n!}<1+1+{1\over2}+{1\over4}+{1\over8}+{1\over16}+\dots+{1\over2^{n-1}}<3.$$ $\endgroup$ Commented Apr 5, 2023 at 12:58
  • 1
    $\begingroup$ @Martin The series for $\pi$ can also be bounded, with telescoping: $\frac {4}{1} -\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+ \cdots +\frac{4}{4n+1}-\frac{4}{4n+3}$ $< \frac {4}{1} -\frac{4}{3} + (\frac{4}{3} - \frac{4}{5}) + \frac{4}{5}-\frac{4}{7} + \cdots +(\frac{4}{4n-1}-\frac{4}{4n+1}) +\frac{4}{4n+1}-\frac{4}{4n+3}$ $=4 - \frac {4}{4n+3} < 4$. And, as you already noted, to get decimal digits one just has to use as is some beginning part of the series, and bound the rest. $\endgroup$ Commented Apr 7, 2023 at 16:12

1 Answer 1


(Following the discussion in comments)

For a Dedekind cut, we see that a definition of $A$ (the lower part of the cut) with an increasing sequence makes for an easy proof of $x \in A$, but not of $x \notin A$. To get around this fact we can have a separate definition for the set $B$ which is complementary to $A$ in $\mathbb{Q}$ (i.e. $\mathbb{Q}$ is the disjoint union of $A$ and $B$, and all elements of $A$ are below all elements of $B$), using a decreasing sequence.

For example we can use two sequences $a_n$ and $b_n$, which both converge to $e$ (as we were talking about $e$), one increasing below $e$ and one decreasing above $e$, and define
$A = \left\lbrace x \in \mathbb{Q} : \exists N, \forall n \ge N, x < a_n \right\rbrace = \left\lbrace x \in \mathbb{Q} : \exists n, x < a_n \right\rbrace$ (because $(a_n)_{n \in \mathbb{N}}$ is increasing)
$B = \left\lbrace x \in \mathbb{Q} : \exists N, \forall n \ge N, x > b_n \right\rbrace = \left\lbrace x \in \mathbb{Q} : \exists n, x > b_n \right\rbrace$ (because $(b_n)_{n \in \mathbb{N}}$ is decreasing)

(Of course one needs to prove that sequences $a_n$ and $b_n$ have the correct properties. This can consist in proving that, for example: $a_n$ is increasing, $b_n$ is decreasing, $a_0 < b_0$, and $\forall \varepsilon>0, \exists n, b_n < a_n+\varepsilon$.)

With the sequence I quoted in comments, we can use $a_n = (1+\frac{1}{n})^n$ and $b_n = (1+\frac{1}{n})^{n+1}$.
With continued fraction, we can use the partial fractions $f_n$: even ones are below $e$, odd ones are above $e$, so let's have $a_n=f_{2n}, b_n=f_{2n+1}$.
In both cases, given a rational number, it is easy to find which set, $A$ or $B$, it belongs to.

But anyway, the main point is: there is no real difference in practice between the Cauchy sequence approach and the Dedekind cut approach. At least to construct $\mathbb{R}$ from $\mathbb{Q}$, because for the Dedekind cut approach we need a total order in the entry set ($\mathbb{Q}$), that is compatible with the topology we want: it would not work to complete $\mathbb{Q}^n$ into $\mathbb{R}^n$.

There is no real difference because, in order to know for example that the two first digits in $e$ are $2.7$, we need to have a proof that $2.7 \le e < 2.8$. This kind of proof is as difficult in theory with a Cauchy sequence as with a Dedekind cut. The only difference is that, in practice, usual Cauchy sequences are given with a mathematical expression which makes it tractable to prove inequalities such as $2.7 \le e < 2.8$, while usual Dedekind cuts are reserved to abstract proofs in the general case, such as completing $\mathbb{Q}$ into $\mathbb{R}$, so we get the impression that Dedekind cuts are untractable.

If we have a Cauchy sequence of rational numbers $(a_n)_{n \in \mathbb{N}}$, we can define an equivalent Dedekind cut with $A = \left\lbrace x \in \mathbb{Q} : \exists N, \forall n \ge N, x \le a_n \right\rbrace$.

Reciprocally, if we have a Dedekind cut defined by its lower set $A$, we can derive an equivalent Cauchy sequence by the following process:

  • $a_0$ is any element in $A$.
  • $a_{2n+1}$ is the lowest of $\left\lbrace a_{2n}+\frac k {2^{2n}}, k \in \mathbb{N}, k \ge 0 \right\rbrace$ not in $A$.
  • $a_{2n+2}$ is the highest of $\left\lbrace a_{2n+1}-\frac k {2^{2n+1}}, k \in \mathbb{N}, k \ge 0 \right\rbrace$ in $A$.
    This defines a Cauchy sequence alternatively above and below the limit.

EDIT: in order to find it more easily, I copy here the last comment I made to the main question.

The series for $\pi$ can be bounded with telescoping: $\frac {4}{1} -\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+ \cdots +\frac{4}{4n+1}-\frac{4}{4n+3}$ $< \frac {4}{1} -\frac{4}{3} + (\frac{4}{3} - \frac{4}{5}) + \frac{4}{5}-\frac{4}{7} + \cdots +(\frac{4}{4n-1}-\frac{4}{4n+1}) +\frac{4}{4n+1}-\frac{4}{4n+3}$ $=4 - \frac {4}{4n+3} < 4$.
This enables to compute decimal digits: compute a partial sum, and bound the remaining terms by telescoping such as above. By computing longer partial sums and bounding the rest, this gives a decreasing sequence above the limit; which can be used (as explained in the beginning of this answer) to give a tractable Dedekind cut for $\pi$.

In comments to the main question, Intelligenti pauca has shown how the series for $e$ can be bounded too.

  • $\begingroup$ thanks for your answer. I think the critical point here will be to show that $b_n < a_n + \varepsilon$. For the mentioned sequences one would have to show that $\frac{(\frac{n+1}{n})^n}{n} < \varepsilon$ $\endgroup$
    – Martin
    Commented Apr 7, 2023 at 10:26
  • $\begingroup$ @Martin To show that $\exists n, (1+\frac{1}{n})^{n+1}-(1+\frac{1}{n})^n < \varepsilon$, we can write $\forall n, (1+\frac{1}{n})^n < (1+\frac{1}{n})^{n+1} < 4$, because $(1+\frac{1}{n})^{n+1}$ is decreasing (derivative is $(\log(1+\frac 1 n) - \frac 1 n)(1+\frac 1 n)^{n+1}$ which is negative). So $(1+\frac{1}{n})^{n+1}-(1+\frac{1}{n})^n = \frac{1}{n} (1+\frac{1}{n})^n < \frac 4 n$, so we can take $n > \frac 4 {\varepsilon}$. $\endgroup$ Commented Apr 7, 2023 at 12:22
  • $\begingroup$ @Martin May I ask why you did not give the bounty (or some part of it) to my answer? What is missing? $\endgroup$ Commented Apr 9, 2023 at 22:37
  • $\begingroup$ You're answer (before the edit that I just saw) still relied on convergence to determine the decimal expansion $\endgroup$
    – Martin
    Commented Apr 10, 2023 at 17:52
  • $\begingroup$ @Martin You can still validate it and vote for it. $\endgroup$ Commented Apr 10, 2023 at 20:49

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