Irrationally Happy it’s π Day!

Achievements Unlocked

Today is a big day for akickinthediscovery.com! Not only is today $\pi$ day, basically the holiday that celebrates math(s), but this $\pi$ day article will be the $50^{th}$ article written on akickinthediscovery.com. Also, just as significant, it’s Einstein‘s birthday! Last $\pi$ day feels like yesterday, when we went through one of Euler’s solutions to the Basel problem. Today, we will keep building on our knowledge of $\pi$ and prove that $\pi$ is an irrational number. We will follow a brilliant proof devised by Ivan Niven (also, there is a typo in his paper, so we fixed it!), but before we show that, we will show $\sqrt{2}$ and $e$ (Euler’s number) are irrational as well. This is so that there is one short resource that has proof of all three!

Setting the Stage

Before we move on, just to make sure that we are all on the same page, let’s recall what a rational number is.

(Incomplete) Definition (Rational Numbers $\mathbb{Q}$ ): We call the set of integer fractions, denoted $\mathbb{Q},$ the set of rational numbers. That is,

$\mathbb{Q} = \left\{\frac{a}{b}\;:\;a,b\in \Z \;\;\And \;\;b\neq 0\right\}.$

Remark: The reason why the above definition is incomplete has to do with the fact that we are not giving structure to $\mathbb{Q}$ that it ought to have. For example, we know that $\frac{1}{2} = \frac{3}{6} = \frac{-8\;\;}{16} = \frac{43}{86} = \cdots,$ however, this is not deducible from our “definition” of $\mathbb{Q}.$ But, for our purposes, we will think of $\mathbb{Q}$ as the set of integer fractions and assume that all the properties and rules that we know about manipulating fractions hold without issue.

Now that we know what rational numbers are, we can define irrational numbers in a similarly incomplete way.

(Incomplete) Definition (Irrational Numbers $\R/\mathbb{Q}$ ): We call real numbers that are not rational numbers irrational numbers. That is,

$\R/\mathbb{Q} = \left\{r\in \R \;:\; r\notin \mathbb{Q}\right\}.$

Remark: This definition is similarly incomplete. For instance, we are assuming that we know that the real numbers are. But this discussion will take us too much into the weeds. So, for those interested, you can search for the construction of the real numbers on Google. The two classic ways are to use Dedekind cuts or to use equivalence classes of Cauchy sequences of rational numbers.

We are now set up to show three things: (i) $\sqrt{2}$ is irrational; (ii) $e$ is irrational; and the coup de grâce (iii) $\pi$ is irrational.

$\sqrt{2}$ Using Fermat’s Infinite Descent

Theorem: The square root of 2 is not a rational number,

$\sqrt{2}\notin \mathbb{Q}.$

Proof: (Click in the Discovery)

Key Idea: We will see a trend in all the proofs presented here; in particular, every proof is by contradiction. It’s usually very difficult to prove that something is impossible, and in our case, it’s very difficult to show it’s impossible to find integers whose ratio equals $\sqrt{2}$ , i.e., that it’s impossible to find $a,b\in \Z,$ such that $\frac{a}{b} =\sqrt{2}.$ It’s far easier to assume there are two integers $a,b\in \Z$ such that $\frac{a}{b} =\sqrt{2},$ and then deduce a contradiction. This will be our strategy.

Beginning of proof: Assume for the hope of a contradiction that $\sqrt{2}$ is a rational number. It follows that we can find two integers $a,b\in \Z$ such that $\frac{a}{b} =\sqrt{2}.$ Note that we can safely assume that $a$ and $b$ are both positive. Since working with square roots and fractions can be annoying, let’s square and rearrange everything to deduce,

a^2 = 2b^2.

Intuitively, we know that if $a^2$ is even, then $a$ must also be even. Rigorously, this is a consequence of Euclid’s lemma (or Euler’s Lemma). (Or, you can do a mini proof by contrapositive: if $a$ is not even (odd), then $a^2$ is not even (odd).) In any case, we conclude that $a$ is even, and therefore $a = 2a_0,$ for some $a_0\in \Z.$ Let’s plug this into our equation relating $a$ and $b,$

(2a_0)^2 = 4a_0^2 = 2b^2

Canceling a factor of 2 gives,

2a_0^2 = b^2.

Similarly, we see that $b^2$ is even, and hence $b$ is even. Therefore, $b = 2b_0$ for some $b_0\in \Z.$ It follows,

2a_0^2 = (2b_0)^2 = 4b_0^2.

Again, canceling a factor of 2 gives,

a_0^2 = 2b_0^2.

Observe that the above equation implies that $\sqrt{2}=\frac{a_0}{b_0} .$ This means that we have found a new way to express the square root of 2 as a fraction of positive integers:

\sqrt{2}=\frac{a}{b} =\frac{a_0}{b_0}, \qquad\qquad \mathrm{where} \;\; 0<a_0<a\;\; \mathrm{and} \;\; 0<b_0<b.

Note that $0<a_0<a$ and $0<b_0<b$ follow from $a = 2a_0$ and $b = 2b_0.$ Interesting…

Taking a look back at $a_0^2 = 2b_0^2,$ you might notice that we are in the exact same situation that we started, except we are using the letters $a_0$ and $b_0$ instead of $a$ and $b.$ Thus, by going through the same argument, we can find two more positive integers $a_1$ and $b_1$ such that: (i) $a_1^2 = 2b_1^2;$ (ii) $\sqrt{2}=\frac{a_1}{b_1};$ and (iii) $a_0 = 2a_1$ and $b_0 = 2b_1.$ In particular, (ii) and (iii) imply that we have found yet another way to express $\sqrt{2}$ as a fraction of smaller positive integers: $0<a_1<a_0<a$ and $0<b_1<b_0<b.$ And by starting with (i), we can go through the same reasoning yet again… and then again, and again indefinitely, always finding smaller and smaller numbers $a_n,b_n\in \N$ to express $\sqrt{2}$ as a fraction of integers.

\sqrt{2}=\frac{a}{b} =\frac{a_0}{b_0} =\frac{a_1}{b_1} \cdots =\frac{a_n}{b_n}\cdots, \qquad\qquad \begin{matrix}\mathrm{where} \;\; 0<\cdots< a_n<\cdots<a_1<a_0<a\;\;\\ \;\,\mathrm{and} \;\;\; 0<\cdots< b_n<\cdots<b_1<b_0<b. \end{matrix}

But, this process must stop at some point since there are only a finite number of positive integers that $a_n$ and $b_n$ could be. In particular, the numerator $a_n$ can only be one of the numbers $1,\.2,\.3,\cdots, \,a-1$ and the denominator $b_n$ can only be one of the numbers $1,\,2,\,3,\,\cdots, \.b-1.$ This is our contradiction and hence $\sqrt{2}\notin \mathbb{Q}.$

Remark: Fermat’s infinite descent is the particular type of contradiction that we reached. That is, assuming $\sqrt{2}$ is rational, let us find an infinite sequence of fractions to represent $\sqrt{2},$ where each new representation is constructed from the previous one and uses smaller positive integers. However, there is not an infinite number of positive integers less than $a$ or $b.$ Thus, a contradiction.

$\square$

Euler’s Number $e$ is not rational

Theorem: Define Euler’s number $e,$ to be equal to,

$e = \sum_{n=0}^{\infty} \frac{1}{n!}.$

Then, Euler’s number is an irrational number.

This proof is a classic!

One quick fact that we will use is that $2<e< 3.$ This follows from the following two observations: $e>\frac{1}{0!}+ \frac{1}{1!} =2$ and,

\;\;\;\;\;\;\;\;\;e = \sum_{n=0}^{\infty} \frac{1}{n!} = \frac{1}{0!}+\sum_{n=1}^{\infty} \frac{1}{n!}

\lt 1 +\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}

= 1 + \sum_{n=0}^{\infty} \frac{1}{2^n} \;\;\;\;

= 1+ \frac{1}{1-1/2} \;\;\;

= 3 .\qquad\qquad\;\;\;\;

Where we used that $2^{1-1} =1!,$ and $2^{2-1} =2!,$ and $2^{n-1}\lt n!$ for all $n\geq 3$ in going from the first line to the second. (I challenge you to prove this inequality, have fun!)

Proof: (Click in the Discovery)

Key Idea: Factorials grow super-duper fast. So fast that the series that defines the number $e$ converges quickly. So quick that, in some sense, $e$ cannot be a rational number. Let’s make this idea more precise.

Beginning of proof: Assume for the hope of a contradiction that $e$ were rational. It follows that we can find two integers $a,b\in \Z$ such that $\frac{a}{b} =e.$ First, note that we can assume that $a$ and $b$ are both positive. Second, note that $b\geq2$ by observing that $2<e< 3$ and there is no way to express a number greater than 2 and less than 3 as a fraction of positive integers with the denominator equal to 1. So, we have $\frac{a}{b} =e$ for $a,b\in \N$ and $b\geq2 .$

Now, consider $e \cdot (b!).$

e\cdot b! = \sum_{n=0}^{\infty} \frac{b!}{n!} = \frac{b!}{0!}+\frac{b!}{1!}+ \frac{b!}{2!} + \cdots + \frac{b!}{b!} + \frac{b!}{(b+1)!} + \cdots.

Observe that the denominator is less than the numerator for the first $b+1.$ Consequently, the first $b+1$ terms are integers, and hence the sum of the first $b+1$ terms is an integer, which we denote by $N.$ To highlight this observation, let’s group the integer terms,

e\cdot b! = \underbrace{\Bigg( \frac{b!}{0!}+\frac{b!}{1!}+ \frac{b!}{2!} + \cdots + \frac{b!}{b!}\Bigg)}_{=\;N \in \Z} + \frac{b!}{(b+1)!} + \frac{b!}{(b+2)!} + \cdots.

After some rearrangement, we have

e\cdot b! – N= \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \cdots.

Moreover, $e\cdot b!$ is an integer as well. Indeed, since $e = \frac{a}{b}$ it follows $e\cdot b! = a\cdot(b-1)!\in \Z.$ Thus, the left-hand side is an integer, and we conclude

\mathrm{Integer}\;=\; \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \cdots.

However, we claim that the right-hand side is not an integer, and once we show this, we will have found our contradiction. Observe that,

0<\frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \cdots \lt \frac{1}{b} + \frac{1}{b^2} + \cdots = \sum_{n=1}^{\infty} \frac{1}{b^n} = \frac{1/b}{1-1/b} = \frac{1}{b-1}\leq 1 \qquad \mathrm{for} \; b\geq 2.

So that the tail of the sum is greater than 0 but less than 1. We have found our contradiction! We conclude that $e$ is irrational.

$\square$

The Coup de Grâce $\pi$ is not rational.

Theorem: $\pi$ is irrational.

Again, we follow a brilliant proof by Ivan Niven.

Proof: (Click in the Discovery)

Key Idea: The idea behind this proof is to find a quantity that we can show to be an integer under the assumption that π is rational, which must be greater than 0 but less than 1. The difficulty and brilliance come from actually finding that quantity.

Beginning of proof: Assume for the hope of a contradiction that $\pi$ is a rational number. It follows that we can find two integers $a,b\in \Z$ such that $\frac{a}{b} =\pi.$ We can safely assume that $a$ and $b$ are both positive. Now consider the polynomials,

f(x) = \frac{x^n(a-bx)^n}{n!} ,

and

F(x) = \sum_{k=0}^{n}(-1)^{k}f^{(2k)}(x) .

Our goal is to find some $n\in \N$ that leads to a contradiction. With this goal in mind, we make some observations regarding $f(x)$ and $F(x).$

8 Key Observations regarding $f(x)$ and $F(x)$ :

$\bigstar$ Observation (1): $f(x) \cdot n!$ is a polynomial with integer coefficients, i.e., $f(x) \cdot n! \in \Z[x].$

$\bigstar$ Observation (2): $f(x) \gt 0$ for all $x\in (0,\pi).$ Indeed, since $x<\pi=\frac{a}{b}$ implies $a>bx.$

$\bigstar$ Observation (3): $f(x) \lt \frac{(\pi a)^n}{n!}$ for all $x\in (0,\pi).$ This is seen by noting that $x \lt \pi$ for $x\in (0,\pi)$ and $a-bx \lt a$ for $x\in (0,\pi).$

$\bigstar$ Observation (4): the lowest power of $x$ that shows up in $f(x)$ is $x^n.$

$\bigstar$ Observation (5): $f(x) =b^n f(\pi -x).$ Indeed, since $\frac{a}{b} =\pi$ we have,

b^nf(\pi -x) =b^nf\left(\frac{a}{b} -x\right) = b^n\frac{\left(\frac{a}{b}-x \right)^n\left(a-b\left(\frac{a}{b}-x\right)\right)^n}{n!}

\qquad= \frac{ \left(a-bx \right)^n \left(a-a+x\right)^n}{n!} = \frac{x^n(a-bx)^n}{n!}

= f(x) .\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;

A corollary of this result is the following: $f^{(k)}(x) =b^n f^{(k)}(\pi -x).$ Which we will make use of.

$\bigstar$ Observation (6): $f^{(k)}(0) =f^{(k)}(\pi) =0$ for all $0\leq k\lt n.$ This follows from observations (4) and (5). Since the lowest power of $x$ in $f(x)$ is $x^n,$ the $k^{th}$ derivative for $0\leq k\lt n$ will not yield a constant term, that is, it leaves an $x$ in every term in $f^{(k)}(x) .$ Hence, $f^{(k)}(0) =0.$ Now, using (5) we conclude that $f^{(k)}(0) =b^nf^{(k)}(\pi) =0.$

$\bigstar$ Observation (7): $f^{(k)}(0) =b^nf^{(k)}(\pi) \in \Z$ for all $n\leq k\leq 2n.$ This observation requires more justification.

Note that when we expand the product in $f(x) ,$ we get

f(x) = \sum_{\ell=0}^n \frac{c_{\ell}}{n!} \cdot x^{n+\ell},

where $c_{\ell}\in \Z.$ We are interested in determining $f^{(k)}(0) .$ This means we can focus on the constant term in $f^{(k)}(0)$ because all other terms will be 0. Observe that the constant term in $f^{(k)}(0)$ will be the term with $x^{k}$ in $f(x) .$ Then, by taking the $k^{th}$ derivative, we bring down a factor of $k!$ and we deduce that the constant term in $f^{(k)}(0)$ is equal to $\frac{k!\cdot c_{k-n}}{n!}.$ And, since $n\leq k\leq 2n$ we see that $\frac{k!\cdot c_{k-n}}{n!}\in \Z.$

Now, using observation (5) we deduce $f^{(k)}(0) =b^nf^{(k)}(\pi) \in \Z.$

$\bigstar$ Observation (8): $F(0)$ and $F(\pi)$ are integers. This follows from the previous items (6) and (7). In particular, the sum $F(0) + F(\pi)$ is an integer.

Now, let’s set up a differential equation in $F(x) ,$

\frac{d}{dx} \Bigg(F'(x) \;\sin(x) – F(x) \;\cos(x) \Bigg) = 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) \;\sin(x).\qquad\qquad\qquad\;\;\,

Therefore,

\int_0^{\pi}f(x) \,\sin(x)\;dx =\Big(F'(x) \;\sin(x) – F(x) \;\cos(x)\Big){\Huge |}_0^{\pi} = F(\pi) + F(0)\in \Z.

We will use the fact that the integral above is an integer to find a contradiction. Our strategy is to use our observations about $f(x)$ to bound the integrand, then use this bound to find an $n\in \N$ such that the integral cannot be an integer.

Observe that the integrand, $f(x) \sin(x),$ is strictly bounded by,

0\lt\;f(x) \sin(x)\lt\; \frac{(\pi a)^n}{n!}

for $0\lt x\lt \pi.$ These follow from Observation (2) and Observation (3). It follows that the integral, $\int_0^{\pi}f(x) \,\sin(x)\;dx ,$ is positive and bounded by

0<\int_0^{\pi}f(x) \,\sin(x)\;dx \leq \sup_{x\in [0,\pi]} {(f(x) \,\sin(x))} \cdot \int_0^{\pi} \;dx \leq\pi\frac{(\pi a)^n}{n!} .

for any choice of $n.$ However, we can see from the upper bound above that the integral can be made to be strictly less than by choosing a sufficiently large $n.$

0<\int_0^{\pi}f(x) \,\sin(x)\;dx <1 \qquad \mathrm{for}\;\mathrm{sufficiently }\;\mathrm{large} \;n.

This is our contradiction! The integral cannot both be an integer and be in the interval $(0,1).$

Thus, $\pi$ is irrational!

$\square$

Wow…

The number π is incredibly fascinating to people who study mathematics. There are an infinity of reasons why. It might be because π is defined to be equal to the ratio of the circumference of a circle to its diameter, and yet somehow π shows up mysteriously all over the place! For instance, π shows up in the sums,

\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6},

\sum_{n=1}^{\infty} \frac{(-1)^{\alpha(n)}}{n} = \pi\qquad\And \qquad \sum_{n=1}^{\infty} \frac{(-1)^{\beta(n)}}{n} = \frac{\pi}{2},

where $\alpha(n) = \#\{\mathrm{prime}\;\mathrm{factors}\;\mathrm{of}\;n\;\mathrm{congruent}\;\mathrm{to}\;1 \;\;(\mathrm{mod}\;{4})\}$ , and $\beta(n) = \#\{\mathrm{prime}\;\mathrm{factors}\;\mathrm{of}\;n\;\mathrm{congruent}\;\mathrm{to}\; 3\;\;(\mathrm{mod}\;{4})\}.$ Or, π shows up in,

\Bigg(\prod_{p\equiv 1 \;(\mathrm{mod}\;4)}\frac{p}{p-1} \Bigg)\cdot \Bigg(\prod_{p\equiv 3 \;(\mathrm{mod}\;4)}\frac{p}{p+1} \Bigg) = \frac{\pi}{4},

And those are formulas only discovered by Euler! We also have the Wallis product

\prod_{n=1}^{\infty} \frac{(2n)^2}{(2n-1)(2n+1)} = \Big(\frac{2}{1}\cdot\frac{2}{3}\Big) \Big(\frac{4}{3}\cdot\frac{4}{5}\Big) \Big(\frac{6}{5}\cdot\frac{6}{7}\Big)\cdots = \frac{\pi}{2}

If sums and products aren’t enough,

\int_{-\infty}^{\infty}e^{-x^2}\;dx = \sqrt{\pi}.

And last but not least,

e^{i\pi}+1 =0.

And, other than that last identity (Euler’s Identity), it’s not at all obvious why π, a number defined in terms of circles, would show up! Furthermore, π also has some mysteries left for us to solve. For instance, we don’t know if π is a normal number. Or, whether $\pi^{{\pi}^{\pi^{\pi}}}$ is an integer! How amazing! Maybe you will solve these problems!

Be Kind. Be Curious. Be Compassionate. Be Creative.

And Have Fun!

A Kick in the Discovery