Title: Notes on Niven's proof that π is irrational
Date: 2024-05-03 13:00

<script defer src="./files/katex.min.js" type="text/javascript"></script>
<link rel="stylesheet" href="./files/katex.min.css">
<script defer src="./files/auto-render.min.js" type="text/javascript"></script>
<script defer src="./files/katex_config.js" type="text/javascript"></script>

<noscript>
This article uses the (amazing) [KaTeX]( https://github.com/Khan/KaTeX )
javascript framework. It seems that you've disabled javascript,
so you won't have a sexy display of the equations, but since they're written
in LaTeX, I'm quite sure that you can figure how to render them locally
in a proper way ;)
</noscript>

π is a fascinating number, with a lot of cool properties. This is why I'm still salty 
from when my [university math teacher](http://utbmbz.free.fr/)
declared, after showing us the classical "[$$\sqrt{2}$$ is
irrational](https://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality)",
that the same goes for $$\pi$$ but that since the proof was more involved, we won't
waste time on it. It's indeed a tad more involved, but nothing insurmountable.

Something like 10 years later, I finally spent the time to read a proof of
it, namely [Ivan Niven](https://en.wikipedia.org/wiki/Ivan_M._Niven)'s [A simple proof that pi is
irrational](https://www.ams.org/journals/bull/1947-53-06/S0002-9904-1947-08821-2/S0002-9904-1947-08821-2.pdf)
article from November 1946. This article is a cleaned up version of my personal notes,
making it a bit more hand-holding than the original article.

[Irrational numbers](https://en.wikipedia.org/wiki/Irrational_number) are
number that can't be written as a ratio of two integers. The way we're going to
prove it here is by doing a [proof by
contradiction](https://en.wikipedia.org/wiki/Proof_by_contradiction): we''ll assume that $$\pi=\frac{a}{b}$$
is true, do some calculations based on this assumption, and end up with
something impossible, meaning that our assumption was wrong.

We start by defining a function: $$f(x) = \frac{x^n(a - bx)^n}{n!}$$,
and warm up by proving that $$f(x) = f(\pi -x)$$:

%%
\begin{equation}
\begin{split}
f(\pi - x) &= f(\frac{a}{b} - x) \\\\
           &= \frac{(\frac{a}{b} - x)^n(a - b(\frac{a}{b} - x))^n}{n!} \\\\
           &= \frac{(\frac{a}{b} - x)^n(a - (a - bx))^n}{n!} \\\\
           &= \frac{(\frac{a}{b} - x)^n(bx)^n}{n!} \\\\
           &= \frac{(\frac{1}{b})^n(a-bx)^n(bx)^n}{n!} \\\\
           &= \frac{x^n(a - bx)^n}{n!} \\\\
           &= f(x)
\end{split}
\end{equation}
%%

With this out of the way, let's use the [binomial theorem](https://en.wikipedia.org/wiki/Binomial_theorem)
to transform $$(a - bc)^n$$ into $$\sum_{k=0}^{n}\binom{n}{k}a^{n-k}(-bx)^k$$
and move it back in $$f(x)$$ to see if we can simplify things a bit:

%%
\begin{split}
f(x) &= \frac{x^n(a - bx)^n}{n!} \\\\
     &= \frac{x^n(\sum_{k=0}^{n}\binom{n}{k}a^{n-k}(-bx)^k)}{n!} \\\\
     &= \frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}a^{n-k}(-bx)^kx^n \\\\
     &= \frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}a^{n-k}(-b)^kx^{n+k} \\\\
\end{split}
%%

Since we're mostly interested about $$x$$, let's simplify this further by rewriting $$j=k+n$$:

%%
\begin{split}
f(x) &= \frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}a^{n-k}(-b)^{k}x^{n+k} \\\\
     &= \frac{1}{n!}\sum_{j=n}^{2n}\binom{n}{j-n}a^{2n-j}(-b)^{j-n}x^{j} \\\\
\end{split}
%%

Way better, we can now clearly see that the coefficients of $$x^0$$, $$x^1$$, …, $$x^{n−1}$$ are all zero,
and that the degree of $$f$$ is at most $$2n$$, meaning that, with $$f^{(j)}$$
being the $$j$$<sup>th</sup> [derivative](https://en.wikipedia.org/wiki/Derivative) of $$f$$:

%%
\begin{equation}
f^{(j)}(0) = \begin{cases}
   0 &\text{if } j < n \\\\
   0 &\text{if } j > 2n
\end{cases}
\end{equation}
%%

But what if $$n ≤ j ≤ 2n$$?

%%
\begin{equation\*}
\begin{split}
f^{(j)}(0) &= (\frac{1}{n!}\sum_{j=n}^{2n}\binom{n}{j-n}a^{2n-j}(-b)^{j-n}x^{j})^{(j)} \\\\
           &= \frac{j!}{n!}{\binom{n}{j-n}}a^{2n-j}(-b)^{j-n} \\\\
\end{split}
\end{equation*}
%%

Since:

%%
\begin{rcases}
n≤j \implies \frac{j!}{n!} \in \N^\*\\\\
\binom{n}{j-n} \in \N^\*\\\\
a, b \in \|, j≤2n \implies a^{2n-j} \in \N^\*\\\\
a, b \in \Z, n≤j \implies -b^{j-n} \in \Z\\\\
\end{rcases} \implies \forall (n≤j≤2n), f^{(j)}(0), f^{(j)}(\pi) \in \Z
%%

Thus from $$\eqref{1}$$ and $$\eqref{2}$$:

%%
\boxed{f^{(j)}(0), f^{(j)}(\pi) \in \Z}
\label{3}\tag{3}
%%

Let's build another function based on $$f(x)$$: $$F(x)$$, the alternating sum of $$f$$ and its first $$n$$ even derivatives.

%%
F(x) = \sum_{j=0}^{n}(-1)^jf^{(2j)}(x) = f(x) - f^{(2)}(x) + f^{(4)}(x) - ... + (-1)^jf^{(2j)}(x) + ... + (-1)^nf^{(2n)}(x)
%%


From $$\eqref{3}$$, we can deduce that $$F(0), F(\pi) \in \Z$$, and thus

%%
\boxed{F(0) + F(\pi) \in \Z}
\label{4}\tag{4}
%%

We can also remark that:

%%
\begin{equation\*}
\begin{split}
F''(x) &= \sum_{j=0}^{n}(-1)^jf^{(2j+2)}(x) \\\\
       &= f^{(2)}(x) - f^{(4)}(x) + f^{(6)}(x) - ... + (-1)^jf^{(2j+2)}(x) + ... + (-1)^nf^{(2n+2)}(x) \\\\
       &= f(x) - (f(x) - f^{(2)}(x) + f^{(4)}(x) - ... + (-1)^jf^{(2j)}(x) + ... + (-1)^nf^{(2n)}(x)) \\\\
       &= f(x) - F(x) \\\\
\end{split}
\end{equation\*}
%%

Time to introduce our last equation: $$\frac{d}{dx}\\{F'(x)sin(x) -
F(x)cos(x)\\}$$, and try to express it in terms of $$f(x)$$, using the [Leibniz rule](https://en.wikipedia.org/wiki/Product_rule):

%%
\begin{equation\*}
\begin{split}
\frac{d}{dx}\\{F'(x)sin(x) - F(x)cos(x)\\} &= F''(x)sin(x) + F'(x)cos(x) - (F'(x)cos(x) - F(x)sin(x)) \\\\
                                         &= F''(x)sin(x) + F(x)sin(x) \\\\
                                         & = (F''(x) - F(x)) sin(x) \\\\
                                         & = f(x)sin(x) \\\\
\end{split}
\end{equation\*}
%%

By integrating this result, we get an interesting equality:

%%
\begin{equation\*}
\begin{split}
\int^{\pi}_{0}f(x)sin(x)dx &= [F'(x)sin(x) - F(x)cos(x)]_0^\pi \\\\
                           &= F'(\pi)sin(\pi) - F(\pi)cos(\pi) - (F'(0)sin(0) - F(0)cos(0)) \\\\
                           &= F'(\pi)sin(\pi) - F(\pi)cos(\pi) - F'(0)sin(0) + F(0)cos(0) \\\\
                           &= F(\pi) + F(0) \\\\
\end{split}
\end{equation\*}
%%

Meaning that from $$\eqref{4}$$:

%%
\boxed{\int^{\pi}_{0}f(x)sin(x)dx \in \Z}
\label{5}\tag{5}
%%


With $$0 < x < \pi$$, both $$f(x)$$ and $$sin(x)$$ are positive, giving:

%%
\boxed{0 < x < \pi, 0 < f(x)sin(x)}
\label{6}\tag{6}
%%

from $$\eqref{5}$$ and $$\eqref{6}$$, we can deduce:

%%
\boxed{0 < x < \pi, 1 ≤ \int^{\pi}_{0}f(x)sin(x)dx}
\label{7}\tag{7}
%%


Still with $$0 < x < \pi$$, we can rewrite $$\eqref{6}$$ as:

%%
\begin{equation\*}
\begin{split}
f(x)sin(x) &= \frac{x^n(a - bx)^n}{n!}sin(x) \\\\
           &≤ \frac{x^n(a - bx)^n}{n!} \\\\
           &≤ \frac{x^na^n}{n!} \\\\
\end{split}
\end{equation\*}
%%

therefore, from the [inequality properties of integrals](https://en.wikipedia.org/wiki/Integral#Inequalities):

%%
\begin{equation\*}
\begin{split}
\int^{\pi}_{0}f(x)sin(x)dx &≤ \int^{\pi}\_{0} \frac{x^{n}a^{n}}{n!} \\\\
                              &≤ \frac{\pi^{n+1}a^n}{n!}
\end{split}
\end{equation\*}
%%

we can summarize this and $$\eqref{7}$$ as:

%%
\boxed{
    0 < x < \pi,
    1 ≤ \int^{\pi}_{0} f(x) sin(x) dx ≤ \frac{\pi^{n+1}a^n}{n!}
}
\label{8}\tag{8}
%%

But now we have a problem: because we can make $$n$$ as big as we want, 
making $$\frac{\pi^{n+1}a^n}{n!} < 1$$, which is a direct contradiction
with $$\eqref{8}$$. Thus our initial supposition, $$\pi = \frac{a}{b}$$ is wrong,
meaning that $$\pi$$ can't be a rational ∎.