Lecture 31: Irrationality of pi

Rational and irrational numbers

To start this talk about $\pi$, it would be best to extend the integers to the rational and irrational numbers since this is a number theory class.

We denote the set of all rational numbers by $\mathbb{Q} = \{\frac{a}{b} : a,b \in \mathbb{Z}, b \neq 0, (a, b) = 1\}$

• $2, \frac{2}{3}, \frac{2}{5} \in \mathbb{Q}$

Note that a rational number's decimal part either terminates or is infinitely repeating.

We define the set of irrational numbers as $\{x : x \notin \mathbb{Q}\}$

• $\sqrt{2}, \sqrt{3}, e \notin \mathbb{Q}$

An irrational number has a nonterminating decimal expansion with no repeating pattern.
An irrational number is the limit of rational numbers from above and below:

• $1, \frac{14}{10}, \frac{141}{100}, \frac{1414}{1000}, \cdots \rightarrow \sqrt{2}$ from below
• $2, \frac{15}{10}, \frac{142}{100}, \frac{1415}{1000}, \cdots \rightarrow \sqrt{2}$ from above

The set of irrational numbers can be denoted $\mathbb{R}\backslash\mathbb{Q}$ since a number cannot be rational and irrational.

An intro to $\pi$

$\pi := \frac{c}{d}$ where $c$ is the circumfrence of a circle and $d$ is the diameter of the same circle.
Since all circles are similiar, $\pi$ is a constant.

Approximations

Egypt (1650 BCE): It was noticed that the area of a square with side length 8 and the area of a circle with diameter 9 were approximately equal, so by setting them equal and solving for $\pi$, they found

(1)

\begin{align} \pi \approx \left( \frac{16}{9} \right)^{2} = \left( \frac{4}{3} \right)^{4} \approx 3.16 \end{align}

Babylonia (1600 BCE): $\pi$ was approximated by equating a circle to a hexagon and relating the ratio of the perimeter to the area.

(2)

\begin{align} \pi \approx 3\;\frac{1}{8} = 3.125 \end{align}

Archimedes (250 BCE): Circumscribed and inscribed a circle with a regular polygon of n sides. $\pi$ is bounded between the ratio of the perimeters to the areas

(3)

\begin{align} 3.1408 \approx \frac{223}{71} < \pi < \frac{22}{7} \approx 3.142 \end{align}

China: Improved Archimedes approximation to 3.1416 with a 96-gon (260 CE) and later $3.1415926 < \pi < 3.1415927$ using a 24576-gon (480 CE)
China (480 CE): $\pi \approx \frac{355}{113}$ which is suprisingly close to pi given only three digits in the numerator and denomenator
Leibniz Series (1600's):

(4)

\begin{align} \pi = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots = 4\sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1} \end{align}

Proving pi is irrational

Niven's Proof

This will be a proof by contradiction, so suppose $\pi \in \mathbb{Q} \Longleftrightarrow \pi = \frac{a}{b}$ where $a,b \in \mathbb{N}, b \neq 0$ and no we are left to show something goes wrong by choosing this assumption.

Now we define 2 functions as

(5)

\begin{align} f(x) := \frac{x^{n}(a-bx)^{n}}{n!} \end{align}

(6)

\begin{align} F(x) := f(x) + \cdots + (-1)^{j}f^{(2j)}(x) + \cdots + (-1)^{n}f^{(2n)}(x) \end{align}

Notice that $F(x)$ is just the alternating sum of $f(x)$ and its first n even derivatives.

Claim 1: $F(0) = F(\pi)$

Proof: by pulling out a $b^{n}$, we get

(7)

\begin{align} f(x) = \frac{b^{n}}{n!}x^{n}(\pi-x)^{n} = f(\pi - x) \end{align}

Using the chain rule and induction yields $f^{(j)}(x) = (-1)^{j}f^{(j)}(\pi - x)$.
So $f^{(2j)}(0) = (-1)^{2j}f^{(2j)}(\pi) = f^{(2j)}(\pi)$
and thus from the definition of $F(x)$, Claim 1 must be true.

Claim 2: $F(0) \in \mathbb{Z}$

Proof: Using the binomial expansion theorem on $(a-bx)^{n}$, we get

(8)

\begin{align} (a-bx)^{n} = \sum_{k=0}^n \, {n \choose k}a^{n-k}(-bx)^k \end{align}

Making the substitution $j = k + n$ and plugging this value into $f(x)$ we get

(9)

\begin{align} f(x)=\frac{1}{n!}\sum_{j=n}^{2n}{n \choose j-n}a^{2n-j}(-b)^{j-n}x^{j} \end{align}

Using properties of derivatives, it can be shown quite easily that for $n \leq m \leq 2n$

(10)

\begin{align} f^{(m)}(0) = \frac{m!}{n!}{n \choose m-n}a^{2n-m}(-b)^{m-n} \end{align}

and $f^{(m)}(0) = 0$ otherwise.
Now notice that since $m \geq n; \frac{m!}{n!}, {n \choose m-n} \in \mathbb{Z}$
This implies $f^{(j)}(0) \in \mathhbb{Z}$ and $F(0) \in \mathbb{Z}$ follows immediately.

Claim 3: $\frac{1}{2} \int_0^\pi \, f(x)\sin(x) dx = F(0)$

Proof: Since $f(x)$ is a degree $2n$ polynomial, $f^{(2n + 2)}(x) \equiv 0$
Because of this, it follows that $F'' + F = f$
Sparing the gory details, using this and integrating by parts $2n$ times we obtain

(11)

\begin{align} (F'(x)\sin(x) - F(x)\cos(x))' = f(x)\sin(x) \end{align}

Therefore

(12)

\begin{align} \frac{1}{2} \int_0^\pi \, f(x)\sin(x) dx = \frac{1}{2}(F'(x)\sin(x) - F(x)\cos(x))|_0^\pi \end{align}

$\frac{1}{2} \int_0^\pi \, f(x)\sin(x) dx = \frac{F(0) + F(\pi)}{2} = \frac{2F(0)}{2} = F(0)$

Conclusion
Since $f(x) > 0$ and $\sin(x) > 0$ on he interval $(0, \pi), \; \frac{1}{2} \int_0^\pi \, f(x)\sin(x) dx > 0$
This result, Claim 2, and Claim 3 imply that

(13)

\begin{align} F(0) \geq 1 \end{align}

Note that $x(\pi-x) = \left(\frac{\pi}{2}\right)^{2} - \left(x-\frac{\pi}{2}\right)^{2} \leq \left(\frac{\pi}{2}\right)^{2}$ which yields

(14)

\begin{align} f(x) \leq \frac{b^{n}}{n!}\left(\frac{\pi}{2}\right)^{2n} \end{align}

Since $\sin(x) \leq 1$

(15)

\begin{align} f(x)\sin(x) \leq \frac{b^{n}}{n!}\left(\frac{\pi}{2}\right)^{2n} \end{align}

This means that

(16)

\begin{align} F(0) = \frac{1}{2} \int_0^\pi \, f(x)\sin(x) dx \leq \frac{1}{2} \int_0^\pi \, \frac{b^{n}}{n!}\left(\frac{\pi}{2}\right)^{2n} dx = \frac{b^{n}}{n!}\left(\frac{\pi}{2}\right)^{2n + 1} \end{align}

So we now know that $F(0) \leq \frac{b^{n}}{n!}\left(\frac{\pi}{2}\right)^{2n + 1}$
But because $n!$ dominates $b^{n}\left(\frac{\pi}{2}\right)^{2n + 1}$,
$\exists \;n \;\;s.t. \;\;\frac{b^{n}}{n!}\left(\frac{\pi}{2}\right)^{2n + 1} < 1$
From this we get

(17)

\begin{align} \exists \;n \;\;s.t.\;\; F(0) < 1 \end{align}

But we know

(18)

\begin{align} \forall \;n, F(0) \geq 1 \end{align}

So the law of Trichotomy says there must be a contradiction which means one of our hypothesis is wrong, and the only hypothesis is that $\pi \in \mathbb{Q}$ which must be wrong
Therefore

(19)

\begin{align} \pi \notin \mathbb{Q}. \end{align}

The "Non-Proof"

Requisite background knowledge:

$\forall \; x : x \in \mathbb{R}$, we can write $x$ in terms of a specific base, $b : b \in \mathbb{N}$ in the form:

(20)

\begin{align} x \equiv a_n*b^n+a_{n-1}*b^{n-1}+ \cdots + a_1*b+a_0+a_{-1}*b^{-1}+a_{-2}*b^{-2}+ \cdots \end{align}

Note that $\forall \; i, \; a_i \in \mathbb{Z}_b$. That is, the $a_i$ take on values in the set $\{ 0, \; 1, \; \ldots, \; b-1 \}$. Also note that we place the "decimal" point between $a_0$ and $a_{-1}$. (Please note that "decimal", in the more general sense, refers to the powers of $b^k : b^k < 1$, and not specifically to base 10.)
For example, in base 10, $178.3 = 1*10^2+7*10+8+3*10^{-1}$.

We can simplify our expansion in base $b$ if we only write the $a_i$ portion of each term (and, if it is not clear, be sure to write b_$k$, where $k$ is the given base).
For example, in base 5, $167 \frac{1}{3} = 1132.131313 \ldots \;$ b₅.

A number, $y \in \mathbb{R}$ is said to be normal in base $b$ if and only if the expansion of $y$ is non-terminating and as $i \rightarrow - \infty$, the frequencies of the occurence of each $a_i$ converge.
For example, $.01234567890123456789 \ldots$ is normal in base 10.

The "Non-Proof" Itself

It is known that $\forall \; q \in \mathbb{Q}, \; q$ is not normal in every base (in particular, if $q = \frac{a}{b}$, then in base $b, \; q$ terminates).
For example, $\frac{1}{7} = .1$ b₇; even in base 10, $\frac{1}{7} = 0.142857 \ldots$ lacks each of 0, 3, 6, and 9.

Now consider $x : x \in \mathbb{R}$.
We know that $x : x \in \mathbb{Q} \subset \mathbb{R}, \Leftrightarrow x$ is not normal in every base.
However, some $x : x \in \mathbb{R} \backslash \mathbb{Q} \subset \mathbb{R}$ have a chance of being normal in every base. If we could show that a number, $x$, was normal in every base, then it would be obvious that $x$ is irrational.

It is conjectured that $\pi$ is normal in every base.
$\pi = 3.141592653589793238462643383279 \ldots$ b₁₀ (decimal). The first 0 (the only possible coefficient of $10^k : k \in \mathbb{Z}, \; k < 0$ not yet named) occurs within the first 40 digits.
$\pi = 3.243F6A8885A308D1319 \ldots$ b₁₆ (hexadecimal). (I promise that if you expand far enough you get the "B", "C", and "E" digits.)
$\pi = 11.001001000011111110110 \ldots$ b₂ (binary). If we group pairs of 0's and 1's in the decimal, .00100100001111110110 we see that we get 5 unpaired 0's and 5 unpaired 1's (all underlined), which pair up nicely…

This demonstration indicates, for bases $b : 2 \leq b \leq 16$, counting $n : 10 \leq n \leq 10000$ digits, that as $n \rightarrow$ a large value, $\pi$ approaches normality. However, mathematicians have discovered very little regarding how to prove that a number is normal in a very large number of bases, much less in every base.

Thus, as we cannot prove that $\pi$ is normal in every base (although it appears to be so), we cannot use this method to prove that $\pi$ is irrational.

Odds of choosing 2 relatively prime numbers

$\pi$ can turn up in unusual places. For example, we will show how it simply "falls out" of the Reimann zeta function, $\zeta(s)$.
Let $p$ be a prime number.

We can show that the probability that a specifc $p$ divides a randomly selected $x : x \in \mathbb{N}$ is $\frac{1}{p}$.

First, we assume that we have a collection of the first $n : n >> p$ elements of $\mathbb{Z}$, which is the set $A := \{ 1, \; 2, \; 3, \; \ldots , \; n-1, \; n \}$. Let $x : x \in A$ be randomly selected.
If $p | n$, then we have $r*p = n$. Thus, we can partition the elements of $A$ into $r$ ordered groups, call them the $A_i : 1 \leq i \leq r$ of $p$ elements each:
$A_1 = \{ 1, \; \ldots , \; p \}$
$A_2 = \{ p+1, \; \ldots , \; 2p \}$
$\vdots$
$A_i = \{ i*p+1, \; \ldots , \; (i+1)*p \}$
$\vdots$
$A_r = \{ r*(p-1)+1, \; \ldots , \; r*p \}$
Thus, $x$ must fall into one of these $A_i$. Within each $A_i$, only 1 element out of $p$ is divisble by $p$, namely $i*p$. Furthermore, $x$ randomly selected $\Rightarrow$ the odds that $x = i*p$ for some $i$ is the same as the odds that $x = p$ for $i = 1$.
Thus, there is a $\frac{1}{p}$ chance that $p|x$.

If $n = p*r + q$, with $0 < q < p$, then we do the same thing for the first $r-1 \; A_i$. We add the extra $q$ elements to $A_r$.
From before, we see that the odds that $p | x$ is $\frac{(r-1)+1}{((r-1)*p)+(p+q)} = \frac{r}{r*p+q}$. As $n \rightarrow \infty, \; r \rightarrow \infty \Rightarrow \frac{r}{r*p+q} \rightarrow \frac{1}{p}$ (as $p, \; q$ are held constant).

With this result in hand, we see that as $n \rightarrow \infty$, the odds that $p | x \rightarrow \frac{1}{p}$.

Corollary: The odd that $p$ divides two randomly selected integers is $\frac{1}{p^2}$

Proof: The odds of $p$ dividing one of these numbers is $\frac{1}{p}$, and since these are independent events, we can multiply these odds, so we get the odds to be $\left(\frac{1}{p}\right)\left(\frac{1}{p}\right)$ = $\frac{1}{p^2}$ as desired.

It can then be seen easily that $p$ fails to divide both numbers is $1 - \frac{1}{p^2}$.

Because each prime is distinct, the odds that no prime will divide both of these randomly chosen integers is

(21)

\begin{align} \prod_{p \; prime} (1-p^{-2}) \end{align}

Using the Euler product formula we learned in class, we get that

(22)

\begin{align} \prod_{p \; prime} (1-p^{-2}) = \frac{1}{\zeta(2)} \end{align}

So the odds that two randomly chosen integers share no common prime factors is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$

Now note that sharing no common prime factors is the samre thing as being relatively prime and we are done.

This last calculation relates number theory to $\pi$ which may seem a little strange.