# Can $\pi$ be defined using Dedekind cuts?

I have read that Dedekind cuts allow you to define the real numbers from the rationals. For example, $$\sqrt{2}$$ can be defined in the following way:

• Partition the rational numbers into two sets $$A$$ and $$B$$, such that all of the elements of $$A$$ are less than the elements of $$B$$
• $$A=\{a \in \mathbb{Q}:x<0 \text{ or } x^2 < 2$$}
• $$B=\{b \in \mathbb{Q}: x > 0 \text{ and } x^2 \geq 2$$}
• Because $$B$$ has no lower bound, there is a 'gap' in the number line
• We define $$\sqrt{2}$$ to fill this gap: $$\sqrt{2}$$ is the unique number such that $$x^2=2, x>0$$. $$\sqrt{2}$$ sits on the partition line that we previously drew (hence why we say $$\sqrt{2}$$ is that cut)

Hopefully, I understand Dedekind cuts well enough to ask this question. It seems to me that because $$\pi$$ is transcendental, it cannot fill a gap in the same way that $$\sqrt{2}$$ does. There is no polynomial equation that $$\pi$$ helps solve. By contrast, $$\sqrt{2}$$ solves the equation $$x^2=2, x \geq 0$$. Nevertheless, $$\pi$$ is a real number, and Dedekind cuts define the real numbers. So, can $$\pi$$ be defined using Dedekind cuts, or is more work needed? Moreover, does the usual geometric definiton of $$\pi$$ have a well-defined mathematical meaning if you have not constructed the real numbers from the rationals?

We know that $$\pi=4\int_0^1\frac{dx}{1+x^2}.$$ A lower Riemann sum of this integral is $$L_n=\frac4n\sum_{k=1}^{n}\frac1{1+k^2/n^2}.$$ An upper Riemann sum of this integral is $$U_n=\frac4n\sum_{k=0}^{n-1}\frac1{1+k^2/n^2}.$$ We can define $$A=\{a\in\Bbb Q:a and $$B=\{b\in\Bbb Q:b>U_n\text{ for some }n\in\Bbb N\}$$.

Any sequence $$c_n$$ that converges to $$\pi$$ can be used to define such a cut, with

$$A=\{a\in\mathbb Q\mid a\lt c_n\text{ infinitely often}\}\hphantom{\;.}$$

and

$$B=\{b\in\mathbb Q\mid b\gt c_n\text{ infinitely often}\}\;.$$

• Thank you for responding. What does 'infinitely often' mean in this context?
– Joe
May 4, 2020 at 8:28
• @Joe: That there are infinitely many $n\in\mathbb N$ for which $a\lt c_n$ (or $b\gt c_n$). May 4, 2020 at 8:29

There are many ways to do this, but my preference is to use Euler's solution to the Basel problem (https://en.wikipedia.org/wiki/Basel_problem). Here's how I would define the lower set of the Dedekind cut:

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

The upper set can be defined this way:

$$\left\{ a \in \Bbb Q | (a > 0) \land \forall n \in \Bbb N : a^2 > \sum_{i=1}^n\frac{6}{i^2} \right\}$$