## 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

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

**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

**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):

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

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

(10)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

Therefore

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

**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

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)Since $\sin(x) \leq 1$

(15)This means that

(16)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

But we know

(18)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

## 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)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, $167 \frac{1}{3} = 1132.131313 \ldots \;$ b_{5}.

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, $\frac{1}{7} = .1$ b_{7}; 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.

$\pi = 3.141592653589793238462643383279 \ldots$ b_{10} (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.

$\pi = 3.243F6A8885A308D1319 \ldots$ b_{16} (hexadecimal). (I promise that if you expand far enough you get the "B", "C", and "E" digits.)

$\pi = 11.001001000011111110110 \ldots$ b_{2} (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)Using the Euler product formula we learned in class, we get that

(22)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.