# Dedekind cuts: How to prove irrationality and how to compute decimal expansion in practice

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.

EDIT:

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)?

• 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 ? Commented Apr 3, 2023 at 20:20
• I edited the question to (hopefully) clarify what I am lokking for. Commented Apr 4, 2023 at 17:48
• 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 $2.8$. Commented Apr 4, 2023 at 21:19
• $$1+1+{1\over2}+{1\over6}+{1\over24}+{1\over120}+\dots+{1\over n!}<1+1+{1\over2}+{1\over4}+{1\over8}+{1\over16}+\dots+{1\over2^{n-1}}<3.$$ Commented Apr 5, 2023 at 12:58
• @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. Commented Apr 7, 2023 at 16:12

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.

• 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 $\frac{(\frac{n+1}{n})^n}{n} < \varepsilon$ Commented Apr 7, 2023 at 10:26
• @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 $n > \frac 4 {\varepsilon}$. Commented Apr 7, 2023 at 12:22
• @Martin May I ask why you did not give the bounty (or some part of it) to my answer? What is missing? Commented Apr 9, 2023 at 22:37
• You're answer (before the edit that I just saw) still relied on convergence to determine the decimal expansion Commented Apr 10, 2023 at 17:52
• @Martin You can still validate it and vote for it. Commented Apr 10, 2023 at 20:49