A Proof of Irrationality of π Based on Nested Radicals with Roots of 2

1. Introduction

In 1714, the English mathematician Roger Cotes discovered a remarkable identity [1,2]

$$ix=ln\left(cos\left(x\right)+isin\left(x\right)\right).$$

A few decades later, Swiss mathematician Leonardo Euler found a reformulated form of this identity as

$$e^{ix}=cos\left(x\right)+isin\left(x\right)$$

from which it follows that

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

This equation, also known as Euler’s identity, is commonly considered as the most beautiful formula in mathematics as it relates the ubiquitous constants $\pi$ and e to each other [2]. Sometimes these constants $\pi$ and e are also regarded as Archimedes’ constant and Euler’s number, respectively.

A proof of irrationality of the constant e may not be difficult (see for example [3,4]). However, it was not easy to find a proof of irrationality of $\pi$; a long time passed since discovery of $\pi$ by ancient Babylonians and Egyptians [5,6,7] to prove its irrationality.

A first proof that $\pi$ is irrational was given by Swiss mathematician Johann Heinrich Lambert in 1761 [6,8] (see also [9]). In his work Lambert showed that if $x\ne 0$ in the following infinite continuous fraction

$$tan\left(x\right)={\displaystyle \frac{x}{1-\frac{x^2}{3-\frac{x^2}{5-\frac{x^2}{7-\frac{x^2}{9-\dots }}}}}},$$

then value of x cannot be rational when its expansion on the right side is rational. Therefore, in the equation

$$tan\left({\displaystyle \frac{\pi }{4}}\right)=1$$

the constant $\pi$ must be irrational.

A first proof of irrationality of $\pi$ by contradiction was found in 1873 by French mathematician Charles Hermite [10]. There are several other proofs of irrationality of $\pi$ [11,12,13,14,16,17]. One of them, published by Niven in 1947, is particularly interesting and attracts much attention. In his work [12], Niven proved the irrationality of $\pi$ also by contradiction. In particular, with the help of the series expansion

$$F\left(x\right)=\sum _{m=0}^n{(-1)}^m{\displaystyle \frac{d^{2m}}{d{x}^{2m}}}f\left(x\right),$$

where

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

he showed that it is impossible to represent $\pi$ as a ratio of two positive integers a and b. Despite a long history, research on the irrationality of $\pi$ still remains interesting [8,15,16,17].

In this work, we present a proof of irrationality of $\pi$ based on nested radicals of kind $c_k=\sqrt{2+c_{k-1}}$, where $c_0=0$. These nested radicals have been used in our earlier publications [18,19] to generate the Machin-like formulas for $\pi$. To the best of our knowledge, this approach is new and has never been reported.

2. Preliminaries

The identity (1) below has been used in our previous publications [18,19] as a starting point to generate the Machin-like formulas for $\pi$. The following theorem shows how this identity can be derived.

Theorem 1. 

The following equation [20]

$${\displaystyle \frac{\pi }{4}}=2^{k-1}arctan\left({\displaystyle \frac{\sqrt{2-c_{k-1}}}{c_k}}\right),k\ge 1,$$

(1)

where k is an integer, holds.

Proof. 

Using the double angle identity

$$cos\left(2x\right)=2{cos}^2\left(x\right)-1,$$

by induction it follows that

$$cos\left({\displaystyle \frac{\pi }{2^2}}\right)={\displaystyle \frac{1}{2}}\sqrt{2}={\displaystyle \frac{1}{2}}{c}_1,$$

$$cos\left({\displaystyle \frac{\pi }{2^3}}\right)={\displaystyle \frac{1}{2}}\sqrt{2+\sqrt{2}}={\displaystyle \frac{1}{2}}{c}_2,$$

$$cos\left({\displaystyle \frac{\pi }{2^{k+1}}}\right)={\displaystyle \frac{1}{2}}\underset{n\mathrm{square}\mathrm{roots}}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\dots +\sqrt{2}}}}}}={\displaystyle \frac{1}{2}}{c}_k.$$

(2)

Therefore, we get

$$sin\left({\displaystyle \frac{\pi }{2^{k+1}}}\right)=\sqrt{1-{cos}^2\left({\displaystyle \frac{\pi }{2^{k+1}}}\right)}.$$

(3)

Thus, using equations (2) and (3) we obtain

$$\begin{array}{cc}\hfill tan\left({\displaystyle \frac{\pi }{2^{k+1}}}\right)& =tan\left({\displaystyle \frac{\sqrt{1-{cos}^2\left(\frac{\pi }{2^{k+1}}\right)}}{cos\left(\frac{\pi }{2^{k+1}}\right)}}\right)\hfill \\ & =tan\left({\displaystyle \frac{\sqrt{2-cos\left(\frac{\pi }{2^k}\right)}}{cos\left(\frac{\pi }{2^{k+1}}\right)}}\right)=tan\left({\displaystyle \frac{\sqrt{2-c_{k-1}}}{c_k}}\right)\hfill \end{array}$$

or

$${\displaystyle \frac{\pi }{2^{k+1}}}=arctan\left({\displaystyle \frac{\sqrt{2-c_{k-1}}}{c_k}}\right)$$

and this completes the proof. □

Since the integer k can be arbitrarily large, we can also write

$${\displaystyle \frac{\pi }{4}}=\underset{k\to \infty }{lim}{2}^{k-1}arctan\left({\displaystyle \frac{\sqrt{2-c_{k-1}}}{c_k}}\right).$$

(4)

Using the limit (4) we can derive a well-known formula for $\pi$ [21]

$$\pi =\underset{k\to \infty }{lim}{2}^k\sqrt{2-{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\dots +\sqrt{2}}}}}}_{k-1\mathrm{square}\mathrm{roots}}}=\underset{k\to \infty }{lim}{2}^k\sqrt{2-c_{k-1}}.$$

Another formula for $\pi$ that can also be derived from the limit (4) is given by (see [22] and literature therein)

$$\pi =\underset{k\to \infty }{lim}{2}^k\sum _{n\ge k}{\displaystyle \frac{\sqrt{2-c_{n-1}}}{c_n}}.$$

It should be noted that this limit can be further simplified as

$$\pi =\underset{k\to \infty }{lim}{2}^{k-1}\sum _{n\ge k}\sqrt{2-c_{n-1}}$$

or

$$\pi =\underset{k\to \infty }{lim}{2}^k\sum _{n\ge k}\sqrt{2-c_n}$$

since

$$\underset{n\to \infty }{lim}{c}_n=\underset{n\to \infty }{lim}\underset{n\mathrm{square}\mathrm{roots}}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\dots +\sqrt{2}}}}}}=2.$$

3. Irrationality of $\pi$

Consider three theorems below.

Theorem 2. 

The following limit

$$\pi =\underset{k\to \infty }{lim}{\displaystyle \frac{2^{k+1}}{{\alpha }_k}},$$

(5)

where

$${\alpha }_k=⌊{\displaystyle \frac{2^{k+1}}{\pi }}⌋$$

(6)

holds.

Proof. 

According to equations (4), (5) and (6) the constant ${\alpha }_k$ represents the integer part of the arctangent function as follows

$${\alpha }_k=⌊{\displaystyle \frac{1}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}⌋.$$

Therefore, we can express the reciprocal of the arctangent function as

$${\displaystyle \frac{1}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}={\alpha }_k+{\beta }_k,$$

where ${\beta }_k$ is the fractional part given by

$${\beta }_k=\left\{{\displaystyle \frac{1}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}\right\}=\left\{{\displaystyle \frac{2^{k+1}}{\pi }}\right\}.$$

Thus, equation (1) can be represented in form

$$\pi ={\displaystyle \frac{2^{k+1}}{{\alpha }_k+{\beta }_k}},k\ge 1.$$

(7)

Since the fractional part ${\beta }_k$ cannot be smaller than zero and greater than one while the integer part ${\alpha }_k$ tends to infinity with increasing k, it follows that

$$\underset{k\to \infty }{lim}{\displaystyle \frac{{\alpha }_k}{\frac{1}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}}=\underset{k\to \infty }{lim}{\displaystyle \frac{{\alpha }_k}{{\alpha }_k+{\beta }_k}}=1.$$

Therefore, from this limit and equation (7) we have

$$\underset{k\to \infty }{lim}{\displaystyle \frac{2^{k+1}}{{\alpha }_k+{\beta }_k}}=\underset{k\to \infty }{lim}{\displaystyle \frac{2^{k+1}}{{\alpha }_k}}$$

and this completes the proof. □

Theorem 3. 

The following inequality

$$\pi <{\displaystyle \frac{2^{k+1}}{{\alpha }_k}},1\le k<\infty$$

(8)

holds.

Proof. 

We can show that the fractional part ${\beta }_k$ cannot be equal to zero and, therefore, is given by the following inequality

$$0<{\beta }_k<1.$$

(9)

The constant ${\beta }_k$ is always greater than zero because the reciprocal of the arctangent function in equation

$${\displaystyle \frac{1}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}={\displaystyle \frac{2^{k+1}}{\pi }}$$

cannot be an integer. In particular, since $\pi$ is not an even integer, the ratio on the right side of the equation above is not an integer. For example, it can be shown that $\pi$ is a number located between $3.1408$ and $3.1429$ due to inequality [23,24]

$${\displaystyle \frac{223}{71}}<\pi <{\displaystyle \frac{22}{7}}.$$

More explicitly, as it follows from equation (7)

$${\beta }_k={\displaystyle \frac{2^{k+1}}{\pi }}-{\alpha }_k,k\ge 1$$

the coefficient ${\beta }_k$ cannot be equal to zero as $\pi$ is not an even integer. Thus, according to equation (7) and inequality (9) the theorem is proved. □

Theorem 4. 

The following equation

$${\alpha }_{k+1}=\left\{\begin{array}{ccc}& 2{\alpha }_k,\hfill & \hfill 0<{\gamma }_k<1/2\\ & 2{\alpha }_k+1,\hfill & \hfill 1/2\le {\gamma }_k<1\end{array}\right.$$

holds.

Proof. 

Since the ratio

$${\displaystyle \frac{2^{k+2}arctan\left(\frac{\sqrt{2-c_k}}{c_{k+1}}\right)}{2^{k+1}arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}={\displaystyle \frac{\pi }{\pi }}=1,$$

we can write

$${\displaystyle \frac{2arctan\left(\frac{\sqrt{2-c_k}}{c_{k+1}}\right)}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}={\displaystyle \frac{\frac{1}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}{\frac{1}{2}\frac{1}{arctan\left(\frac{\sqrt{2-c_k}}{c_{k+1}}\right)}}}=1.$$

This equation leads to

$${\displaystyle \frac{\frac{1}{arctan\left(\frac{\sqrt{2-c_k}}{c_{k+1}}\right)}}{\frac{1}{arctan\left(\frac{\sqrt{2-c_{k-1}}}{c_k}\right)}}}=2$$

or

$${\displaystyle \frac{{\alpha }_{k+1}+{\beta }_{k+1}}{{\alpha }_k+{\beta }_k}}=2.$$

(10)

From equation (10) it follows that

$${\alpha }_{k+1}+{\beta }_{k+1}=2\left({\alpha }_k+{\beta }_k\right).$$

Taking the floor function from the both sides leads to

$$⌊{\alpha }_{k+1}+{\beta }_{k+1}⌋=⌊2\left({\alpha }_k+{\beta }_k\right)⌋$$

and since ${\alpha }_{k+1}$ is an integer while $0<{\beta }_{k+1}<1$, we get

$$⌊{\alpha }_{k+1}⌋=⌊2\left({\alpha }_k+{\beta }_k\right)⌋$$

or

$${\alpha }_{k+1}=⌊2\left({\alpha }_k+{\beta }_k\right)⌋=2{\alpha }_k+⌊2{\beta }_k⌋.$$

As we can see from this equation, ${\alpha }_{k+1}$ is equal to $2{\alpha }_k$ when $0<{\beta }_k<1/2$ and equal to $2{\alpha }_k+1$ when $1/2\le {\beta }_k<1$. This completes the proof. □

Finally, the lemma below shows how the limit (5) and inequality (8) lead to the irrationality of $\pi$

Lemma 1. 

The constant π is irrational.

Proof. 

Define the following integers

$${\gamma }_k=\left\{\begin{array}{cccc}& {\gamma }_{k-1}+1,\hfill & & \mathrm{if}{\alpha }_k\mathrm{is}\mathrm{odd},\hfill \\ & {\gamma }_{k-1},\hfill & & \mathrm{if}{\alpha }_k\mathrm{is}\mathrm{even},\hfill \end{array}\right.$$

where ${\gamma }_1=1$. Consequently, we can construct the sequences for positive integers k, ${\gamma }_k$, ${\alpha }_k$ and ${\alpha }_{{\gamma }_k}$ as follows

$$\begin{array}{cccc}& {\left\{k\right\}}_{k=1}^{\infty }\hfill & & =\{1,2,3,4,5,6,7,8,9,10,\dots \},\hfill \\ & {\left\{{\alpha }_k\right\}}_{k=1}^{\infty }\hfill & & =\{1,2,5,10,20,40,81,162,325,651,\dots \},\hfill \\ & {\left\{{\gamma }_k\right\}}_{k=1}^{\infty }\hfill & & =\{1,1,3,3,3,3,7,7,9,10,\dots \}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\left\{{\alpha }_{{\gamma }_k}\right\}}_{k=1}^{\infty }& =\left\{{\alpha }_1,{\alpha }_1,{\alpha }_3,{\alpha }_3,{\alpha }_3,{\alpha }_3,{\alpha }_7,{\alpha }_7,{\alpha }_9,{\alpha }_{10},\dots \right\}\hfill \\ \hfill & =\{1,1,5,5,5,5,81,81,325,651,\dots \}.\hfill \end{array}$$

(11)

The numbers ${\alpha }_k$ from the sequence ${\left\{{\alpha }_k\right\}}_{k=1}^{\infty }$ can be found in [25]. While the integers in the sequence ${\left\{{\alpha }_k\right\}}_{k=1}^{\infty }$ can be even and odd, the integers in the sequence ${\left\{{\alpha }_{{\gamma }_k}\right\}}_{k=1}^{\infty }$ are always odd. It means that if an integer ${\alpha }_k$ is an even number, then it has a common factor with integer $2^{k+1}$ as both of them are divisible by 2.

Thus, due to divisibility by 2 when ${\alpha }_k$ is an even number, we can rearrange the inequality (8) and limit (5) as

$$\pi <{\displaystyle \frac{2^{{\gamma }_{{\gamma }_k}+1}}{{\alpha }_{{\gamma }_k}}},{\alpha }_{{\gamma }_k}\in 2\mathbb{N}+1$$

(12)

and

$$\pi =\underset{k\to \infty }{lim}{\displaystyle \frac{2^{{\gamma }_k+1}}{{\alpha }_{{\gamma }_k}}},{\alpha }_{{\gamma }_k}\in 2\mathbb{N}+1,$$

(13)

respectively.

Assume that $\pi$ can be represented as a ratio of two positive integers p and q. Then, according to inequality (12) and equation (13), we immediately get a contradiction with our assumption that $\pi$ can be represented as a ratio of two integers.

If we assume that starting from some integer $k_0$ the equation

$${\alpha }_{k_0+n+1}\stackrel{?}{=}2{\alpha }_{k_0+n},\forall n\ge 0$$

always holds, then the limit (13) converges in the form

$$\underset{k\to \infty }{lim}{\displaystyle \frac{2^{{\gamma }_k+1}}{{\alpha }_{{\gamma }_k}}}\stackrel{?}{=}{\displaystyle \frac{2^{{\gamma }_{max}+1}}{{\alpha }_{{\gamma }_{max}}}}$$

(14)

where ${\gamma }_{max}$ is presumably the largest integer in the sequence ${\left\{{\gamma }_k\right\}}_{k=1}^{\infty }$ and ${\alpha }_{{\gamma }_{max}}\stackrel{?}{=}{\alpha }_{k_0}$ is presumably the largest odd integer in the sequence ${\left\{{\alpha }_{{\gamma }_k}\right\}}_{k=1}^{\infty }$. However, the equation (14) contradicts the inequality (12) as its right side must be greater than $\pi$. Therefore, such integers ${\gamma }_{max}$ and ${\alpha }_{{\gamma }_{max}}$ do not exist and equation (14) is incorrect.

On the other hand, if we assume that despite absence of the numbers ${\gamma }_{max}$ and ${\alpha }_{{\gamma }_{max}}$ the limit (13) still can converge as a ratio of two integers p and q such that

$$\underset{k\to \infty }{lim}{\displaystyle \frac{2^{{\gamma }_k+1}}{{\alpha }_{{\gamma }_k}}}\stackrel{?}{=}{\displaystyle \frac{p}{q}},$$

(15)

then it contradicts the fact that even numerator $2^{{\gamma }_k+1}$ and odd denominator ${\alpha }_{{\gamma }_k}$ are always relative primes at any value of k and, therefore, these two numbers do not have a common divisor except 1 at any value of k. As a result, it cannot converge as a ratio of two integers p and q. Thus, we can conclude that the limit (15) is also incorrect. This completes the proof that the constant $\pi$ is irrational. □

4. Rational Approximation of $\pi$

The limit (5) shows that we can approximate $\pi$ in form of the rational approximation as given by

$$\pi \approx {\displaystyle \frac{2^{k+1}}{{\alpha }_k}},k\gg 1.$$

(16)

Consider the following examples (a link for the extended table showing values of ${\alpha }_k$ can be found in [25])

$$\begin{array}{cccccc}& {\alpha }_{70}\hfill & & ={\alpha }_{{\gamma }_{70}}\hfill & & =751,587,968,840,192,313,983\hfill \\ & {\alpha }_{71}\hfill & & =2{\alpha }_{{\gamma }_{70}}\hfill & & =1,503,175,937,680,384,627,966\hfill \\ & {\alpha }_{72}\hfill & & =4{\alpha }_{{\gamma }_{70}}\hfill & & =3,006,351,875,360,769,255,932\hfill \\ & {\alpha }_{73}\hfill & & =8{\alpha }_{{\gamma }_{70}}\hfill & & =6,012,703,750,721,538,511,864\hfill \\ & {\alpha }_{74}\hfill & & =16{\alpha }_{{\gamma }_{70}}\hfill & & =12,025,407,501,443,077,023,728\hfill \end{array}$$

Although the values of the coefficient from ${\alpha }_{70}$ to ${\alpha }_{74}$ increases by a factor of 2, the corresponding ratios

$$\begin{array}{cc}\hfill {\displaystyle \frac{2^{75}}{{\alpha }_{74}}}& ={\displaystyle \frac{2^{74}}{{\alpha }_{73}}}={\displaystyle \frac{2^{73}}{{\alpha }_{72}}}={\displaystyle \frac{2^{72}}{{\alpha }_{71}}}={\displaystyle \frac{2^{71}}{{\alpha }_{70}}}={\displaystyle \frac{2^{{\gamma }_{71}}}{{\alpha }_{{\gamma }_{70}}}}\hfill \\ & =\underset{22\mathrm{correct}\mathrm{digits}\mathrm{of}\pi }{\underbrace{3.141592653589793238462}}80398052\dots \hfill \end{array}$$

remain unchanged. This occurs because the ratio of two adjacent values is

$${\alpha }_{k+1}=2{\alpha }_k,70\le k\le 74.$$

However, at $k=75$ we get

$${\alpha }_{75}=(2+1){\alpha }_{74}.$$

As the values

$$\begin{array}{cc}& {\alpha }_{75}={\alpha }_{{\gamma }_{75}}=24,050,815,002,886,154,047,457\hfill \\ & {\alpha }_{76}=2{\alpha }_{75}=48,101,630,005,772,308,094,914\hfill \\ & {\alpha }_{77}=4{\alpha }_{75}=96,203,260,011,544,616,189,828\hfill \end{array}$$

we get

$${\displaystyle \frac{2^{78}}{{\alpha }_{77}}}={\displaystyle \frac{2^{77}}{{\alpha }_{76}}}={\displaystyle \frac{2^{76}}{{\alpha }_{75}}}={\displaystyle \frac{2^{{\gamma }_{76}}}{{\alpha }_{{\gamma }_{75}}}}=\underset{23\mathrm{correct}\mathrm{digits}\mathrm{of}\pi }{\underbrace{3.1415926535897932384626}}7335739\dots .$$

These examples showing the relations between the positive integers k, ${\alpha }_k$, ${\gamma }_k$ and ${\alpha }_{{\gamma }_k}$ help us to understand how the rational approximation (16) tend to $\pi$ with increasing the integer k.

5. Conclusion

Four theorems that can be used to prove the irrationality of $\pi$ are considered. These theorems are related to nested radicals consisting of square roots of 2 of kind $c_k=\sqrt{2+c_{k-1}}$ and $c_0=0$. Examples of the rational approximation tending to $\pi$ with increasing the integer k are provided.