I took the liberty to attempt to construct Dedekind cuts for $e$ and $\pi.$ That is, come up with a set $\alpha$ of rational numbers (that would correspond to the reals $e$ and $\pi$) such that,

If $x \in \alpha$ and $y \in \mathbb Q : y < x$, then $y \in \alpha$

$\alpha \neq \varnothing$

$\alpha \neq \mathbb Q$

$\alpha$ has no greatest element

I came up with the following (hopefully valid) rational Dedekind cuts,

$e = \left\{a\in\mathbb Q \, | \, a <0 \lor \left( \exists n \in \mathbb N : a < \left(1 + \frac{1}{n}\right)^n \right) \right\}$

$\pi = \left\{a\in\mathbb Q \, | \, a <0 \lor \left( \exists n \in \mathbb N : a^2 < \displaystyle\sum_{i=1}^n \frac{6}{i^2} \right) \right\}$

The justification for these seemly arbitrary cuts is the simple fact that

$$e := \displaystyle\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$$


$$\pi^2 = \displaystyle\lim_{n \to \infty} \displaystyle\sum_{i=1}^n \frac{6}{i^2}$$

Are these cuts valid, and how would one attempt to show that they indeed satisfy the requirements for $\alpha$?

  • $\begingroup$ You can't take a sum from $i=1$ to $i=x$ for non-integer $x$.... Replace $x$ by $\lfloor x \rfloor$ to make it a valid sum. $\endgroup$ – Alex G. Nov 2 '16 at 22:48
  • $\begingroup$ @AlexG. Although, my condition is only that some $x$ should exist such that the inequality holds. The sum doesn't need to be defined for every $x$ for that. $\endgroup$ – Markus Klyver Nov 2 '16 at 22:50
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    $\begingroup$ Yes, you're correct. So yes, that is a correct Dedekind cut. I would still suggest changing $x$ to $n$ and $\Bbb Q$ to $\Bbb N$ to clarify the idea. $\endgroup$ – Alex G. Nov 2 '16 at 23:25
  • $\begingroup$ @AlexG. Great. So it only has to be proven. The hardest one would to show that the cut truly has no greatest element. $\endgroup$ – Markus Klyver Nov 3 '16 at 5:28

The first one is not a valid cut because you can't in general take a rational power of a negative number, so it should be changed to $e = \left\{a\in\mathbb Q \, | \, a <0 \lor \left( \exists n \in \mathbb N : a < \left(1 + \frac{1}{n}\right)^n \right) \right\}$.

The second expression is meaningless since to extract square roots you need irrational numbers, which you presumably don't have yet if you are trying to build the reals out of the rationals. But you can set

$\pi = \left\{a\in\mathbb Q \, | \, a <0 \lor \left( \exists n \in \mathbb N \colon a^2 <{\displaystyle\sum_{i=1}^n \frac{6}{i^2}} \right) \right\}$

Showing that these cuts are not rational is of course not a simple matter. Proof of irrationality of $e$ is a bit easier than the proof of irrationality of $\pi$.

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  • $\begingroup$ Why not just use $a^2\lt\sum_{i=1}^n{6\over i^2}$? $\endgroup$ – Barry Cipra Nov 3 '16 at 14:02
  • $\begingroup$ Thank you for your input. However, regarding the cut for $e$, I only state that there should exist some x such that the inequality holds. Regarding the second one, will it not always exist some rational $x$ such that the square root is well-defined? $\endgroup$ – Markus Klyver Nov 3 '16 at 14:11
  • $\begingroup$ @BarryCipra, good point, thanks. $\endgroup$ – Mikhail Katz Nov 3 '16 at 14:22
  • $\begingroup$ @MarkusKlyver, I don't see why there should exist such a number. $\endgroup$ – Mikhail Katz Nov 3 '16 at 14:23
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    $\begingroup$ @MarkusKlyver, the real problem with using $x\in\mathbb{Q}$ instead of $n\in\mathbb{N}$ in the Dedekind cut for $e$ is that it allows for values of $x$ that are close to $-1$, which means that all values of $a$ are in the cut. For example, $a=50\lt100^{100/99}=(1+{1\over x})^x$ with $x=-100/99$. $\endgroup$ – Barry Cipra Nov 3 '16 at 14:36

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