Proof that π is irrational

mathematical constant $π$
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In the 18th century, Johann Heinrich Lambert proved that the number $π$ (pi) is irrational: that is, it cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer. In the 19th century, Charles Hermite found a proof that requires no prerequisite knowledge beyond basic calculus. Three simplifications of Hermite's proof are due to Mary Cartwright, Ivan Niven and Bourbaki. Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich.

In 1882, Ferdinand von Lindemann proved that $π$ is not just irrational, but transcendental as well.^[1]

Lambert's proof

Scan of formula on page 288 of Lambert's "Mémoires sur quelques propriétés remarquables des quantités transcendantes, circulaires et logarithmiques", Mémoires de l'Académie royale des sciences de Berlin (1768), 265–322.

In 1761, Lambert proved that $π$ is irrational by first showing that this continued fraction expansion holds:

\tan(x) = \cfrac{x}{1 - \cfrac{x^2}{3 - \cfrac{x^2}{5 - \cfrac{x^2}{7 - {}\ddots}}}}.

Then Lambert proved that if x is non-zero and rational then this expression must be irrational. Since tan( $π$ /4) = 1, it follows that $π$ /4 is irrational and therefore that $π$ is irrational.^[2] A simplification of Lambert's proof is given below.

Hermite's proof

This proof uses the characterization of $π$ as the smallest positive number whose half is a zero of the cosine function and it actually proves that $π$ ² is irrational.^[3]^[4] As in many proofs of irrationality, the argument proceeds by reductio ad absurdum.

Consider the sequences (A_n)_n ≥ 0 and (U_n)_n ≥ 0 of functions from R into R thus defined:

$A_0(x)=\sin(x);\,$
$(\forall n\in\mathbb{Z}_+):A_{n+1}(x)=\int_0^xyA_n(y)\,dy;$
$U_0(x)=\frac{\sin(x)}x;$
$(\forall n\in\mathbb{Z}_+):U_{n+1}(x)=-\frac{U_n'(x)}x.$

It can be proved by induction that

(\forall n\in\mathbb{Z}_+):A_n(x)=\frac{x^{2n+1}}{(2n+1)!!}-\frac{x^{2n+3}}{2\times(2n+3)!!}+\frac{x^{2n+5}}{2\times4\times(2n+5)!!}\mp\cdots

and that

(\forall n\in\mathbb{Z}_+):U_n(x)=\frac1{(2n+1)!!}-\frac{x^2}{2\times(2n+3)!!}+\frac{x^4}{2\times4\times(2n+5)!!}\mp\cdots

and therefore that

U_n(x)=\frac{A_n(x)}{x^{2n+1}}.

\frac{A_{n+1}(x)}{x^{2n+3}}=U_{n+1}(x)=-\frac{U_n'(x)}x=-\frac1x\frac d{dx}\left(\frac{A_n(x)}{x^{2n+1}}\right),

which is equivalent to

A_{n+1}(x)=(2n+1)A_n(x)-xA_n'(x)=(2n+1)A_n(x)-x^2A_{n-1}(x).\,

It follows by induction from this, together with the fact that A₀(x) = sin(x) and that A₁(x) = −x cos(x) + sin(x), that A_n(x) can be written as P_n(x²)sin(x) + xQ_n(x²)cos(x), where P_n and Q_n are polynomial functions with integer coefficients and where the degree of P_n is smaller than or equal to ⌊n/2⌋. In particular, A_n( $π$ /2) = P_n( $π$ ²/4).

Hermite also gave a closed expression for the function A_n, namely

A_n(x)=\frac{x^{2n+1}}{2^nn!}\int_0^1(1-z^2)^n\cos(xz)\,dz.

He did not justify this assertion, but it can be proved easily. First of all, this assertion is equivalent to

\frac{1}{2^nn!}\int_0^1(1-z^2)^n\cos(xz)\,dz=\frac{A_n(x)}{x^{2n+1}}=U_n(x).

Proceeding by induction, take n = 0.

\int_0^1\cos(xz)\,dz=\frac{\sin(x)}x=U_0(x)

and, for the inductive step, consider any n ∈ Z₊. If

\frac{1}{2^nn!}\int_0^1(1-z^2)^n\cos(xz)\,dz=U_n(x),

then, using integration by parts and Leibniz's rule, one gets

{\begin{aligned}&{}\quad {\frac {1}{2^{{n+1}}(n+1)!}}\int _{0}^{1}(1-z^{2})^{{n+1}}\cos(xz)\,dz\\&={\frac {1}{2^{{n+1}}(n+1)!}}{\Biggl (}\overbrace {\left.(1-z^{2})^{{n+1}}{\frac {\sin(xz)}x}\right|_{{z=0}}^{{z=1}}}^{{=\,0}}+\int _{0}^{1}2(n+1)(1-z^{2})^{n}z{\frac {\sin(xz)}x}\,dz{\Biggr )}\\[8pt]&={\frac 1x}\cdot {\frac 1{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}z\sin(xz)\,dz\\[8pt]&=-{\frac 1x}\cdot {\frac d{dx}}\left({\frac 1{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,dz\right)\\[8pt]&=-{\frac {U_{n}'(x)}x}=U_{{n+1}}(x).\end{aligned}}

If $π$ ²/4 = p/q, with p and q in N, then, since the coefficients of P_n are integers and its degree is smaller than or equal to ⌊n/2⌋, q^⌊n/2⌋P_n( $π$ ²/4) is some integer N. In other words,

\begin{align}N&=q^{\left\lfloor\frac n2\right\rfloor}A_n\left(\frac\pi2\right)\\&=q^{\left\lfloor\frac n2\right\rfloor}\frac{\left(\frac pq\right)^{n+\frac 12}}{2^nn!}\int_0^1(1-z^2)\cos\left(\frac\pi2z\right)\,dz.\end{align}

But this number is clearly greater than 0; therefore, N ∈ N. On the other hand,

\lim_{n\in\mathbb{N}}q^{\left\lfloor\frac n2\right\rfloor}\frac{\left(\frac pq\right)^{n+\frac 12}}{2^nn!}=0.

and so, if n is large enough, N < 1. Thereby, a contradiction is reached.

Hermite did not present his proof as an end in itself but as an afterthought within his search for a proof of the transcendence of $π$ . He discussed the recurrence relations to motivate and to obtain a convenient integral representation. Once this integral representation is obtained, there are various ways to present a succinct and self-contained proof starting from the integral (as in Cartwright's, Bourbaki's or Niven's presentations), which Hermite could easily see (as he did in his proof of the transcendence of e^[5]).

Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, A_n(x) is the "residue" (or "remainder") of Lambert's continued fraction for tan(x).^[6]

Cartwright's proof

Harold Jeffreys wrote that this proof was set as an example in an exam at Cambridge University in 1945 by Mary Cartwright, but that she had not traced its origin.^[7]

Consider the integrals

I_{n}(x)=\int _{{-1}}^{1}(1-z^{2})^{n}\cos(xz)\,dz,

where n is a non-negative integer.

Two integrations by parts give the recurrence relation

x^{2}I_{n}(x)=2n(2n-1)I_{n-1}(x)-4n(n-1)I_{n-2}(x).\qquad (n\geq 2)

J_n(x)=x^{2n+1}I_n(x),\,

then this becomes

J_{n}(x)=2n(2n-1)J_{{n-1}}(x)-4n(n-1)x^{2}J_{{n-2}}(x).

Furthermore, J₀(x) = 2sin(x) and J₁(x) = −4x cos(x) + 4sin(x). Hence for all n ∈ Z₊,

J_n(x)=x^{2n+1}I_n(x)=n!\bigl(P_n(x)\sin(x)+Q_n(x)\cos(x)\bigr),\,

where P_n(x) and Q_n(x) are polynomials of degree ≤ 2n, and with integer coefficients (depending on n).

Take x = $π$ /2, and suppose if possible that $π$ /2 = a/b, where a and b are natural numbers (i.e., assume that $π$ is rational). Then

\frac{a^{2n+1}}{n!}I_n\left(\frac\pi2\right) = P_n\left(\frac\pi2\right)b^{2n+1}.

The right side is an integer. But 0 < I_n( $π$ /2) < 2 since the interval [−1, 1] has length 2 and the function that is being integrated takes only values between 0 and 1. On the other hand,

\frac{a^{2n+1}}{n!} \to 0\text{ as }n \to \infty.

Hence, for sufficiently large n

0 < \frac{a^{2n+1}I_n\left(\frac\pi2\right)}{n!} < 1,

that is, we could find an integer between 0 and 1. That is the contradiction that follows from the assumption that $π$ is rational.

This proof is similar to Hermite's proof. Indeed,

\begin{align}J_n(x)&=x^{2n+1}\int_{-1}^1 (1 - z^2)^n \cos(xz)\,dz\\&=2x^{2n+1}\int_0^1 (1 - z^2)^n \cos(xz)\,dz\\&=2^{n+1}n!A_n(x).\end{align}

However, it is clearly simpler. This is achieved by passing the inductive definition of the functions A_n and taking as a starting point their expression as an integral.

Niven's proof

This proof uses the characterization of $π$ as the smallest positive zero of the sine function.^[8]

Suppose that $π$ is rational, i.e. $π$ = a /b for some integers a and b ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function f from R into R defined by

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

and, for each x ∈ R denote by

F(x)=f(x)-f''(x)+f^{{(4)}}(x)+\cdots +(-1)^{j}f^{{(2j)}}(x)+\cdots +(-1)^{n}f^{{(2n)}}(x)

the alternating sum of f and its first n derivatives of even order.

Claim 1: F(0) + F( $π$ ) is an integer.

Proof: Expanding f as a sum of monomials, the coefficient of x^k is a number of the form c_k /n! where c_k is an integer, which is 0 if k < n. Therefore, f^(k)(0) is 0 when k < n and it is equal to (k! /n!) c_k if n ≤ k ≤ 2n; in each case, f^(k)(0) is an integer and therefore F(0) is an integer.

On the other hand, f( $π$ – x) = f(x) and so (–1)^kf^(k)( $π$ – x) = f^(k)(x) for each non-negative integer k. In particular, (–1)^kf^(k)( $π$ ) = f^(k)(0). Therefore, f^(k)( $π$ ) is also an integer and so F( $π$ ) is an integer (in fact, it is easy to see that F( $π$ ) = F(0), but that is not relevant to the proof). Since F(0) and F( $π$ ) are integers, so is their sum.

Claim 2:

\int_0^\pi f(x)\sin(x)\,dx=F(0)+F(\pi)

Proof: Since f^(2n + 2) is the zero polynomial, we have

F'' + F = f.\,

The derivatives of the sine and cosine function are given by sin' = cos and cos' = −sin. Hence the product rule implies

(F'\cdot\sin - F\cdot\cos)' = f\cdot\sin\!

By the fundamental theorem of calculus

\int_0^\pi f(x)\sin(x)\,dx= \bigl(F'(x)\sin x - F(x)\cos x\bigr)\Big|_{0}^{\pi}.\!

Since sin 0 = sin $π$ = 0 and cos 0 = – cos $π$ = 1 (here we use the above-mentioned characterization of $π$ as a zero of the sine function), Claim 2 follows.

Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < $π$ (because $π$ is the smallest positive zero of the sine function), Claims 1 and 2 show that F(0) + F( $π$ ) is a positive integer. Since 0 ≤ x(a – bx) ≤ $π$ a and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ $π$ , we have, by the original definition of f,

\int_0^\pi f(x)\sin(x)\,dx\le\pi\frac{(\pi a)^n}{n!}

which is smaller than 1 for large n, hence F(0) + F( $π$ ) < 1 for these n, by Claim 2. This is impossible for the positive integer F(0) + F( $π$ ).

The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula

\begin{align} \int_0^\pi f(x)\sin(x)\,dx &=\sum_{j=0}^n (-1)^j \bigl(f^{(2j)}(\pi)+f^{(2j)}(0)\bigr)\\ &\qquad+(-1)^{n+1}\int_0^\pi f^{(2n+2)}(x)\sin(x)\,dx, \end{align}

which is obtained by 2n + 2 integrations by parts. Claim 2 essentially establishes this formula, where the use of F hides the iterated integration by parts. The last integral vanishes because f^(2n + 2) is the zero polynomial. Claim 1 shows that the remaining sum is an integer.

Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight.^[6] In fact,

\begin{align}J_n(x)&=x^{2n+1}\int_{-1}^1(1-z^2)^n\cos(xz)\,dz\\&=\int_{-1}^1\bigl(x^2-(xz)^2\bigr)^nx\cos(xz)\,dz.\end{align}

Therefore, the substitution xz = y turns this integral into

\int_{-x}^x(x^2-y^2)^n\cos(y)\,dy.

In particular,

\begin{align}J_n\left(\frac\pi2\right)&=\int_{-\pi/2}^{\pi/2}\left(\frac{\pi^2}4-y^2\right)^n\cos(y)\,dy\\ &=\int_0^\pi\left(\frac{\pi^2}4-\left(y-\frac\pi2\right)^2\right)^n\cos\left(y-\frac\pi2\right)\,dy\\ &=\int_0^\pi y^n(\pi-y)^n\sin(y)\,dy\\&=\frac{n!}{b^n}\int_0^\pi f(x)\sin(x)\,dx.\end{align}

Another connection between the proofs lies in the fact that Hermite already mentions^[3] that if f is a polynomial function and

F=f-f^{(2)}+f^{(4)}\mp\cdots,

then

\int f(x)\sin(x)\,dx=F'(x)\sin(x)-F(x)\cos(x),

from which it follows that

\int_0^\pi f(x)\sin(x)\,dx=F(\pi)+F(0).

Bourbaki's proof

Bourbaki's proof is outlined as an exercise in his Calculus treatise.^[9] For each natural number b and each non-negative integer n, define

A_{n}(b)=b^{n}\int _{0}^{\pi }{\frac {x^{n}(\pi -x)^{n}}{n!}}\sin(x)\,dx.

Since A_n(b) is the integral of a function which defined on [0, $π$ ] that takes the value 0 on 0 and on $π$ and which is greater than 0 otherwise, A_n(b) > 0. Besides, for each natural number b, A_n(b) <1 if n is large enough, because

(\forall x\in [0,\pi ]):x(\pi -x)\leq \left({\frac \pi 2}\right)^{2}

and therefore

A_{n}(b)\leq \pi b^{n}{\frac 1{n!}}\left({\frac \pi 2}\right)^{{2n}}=\pi {\frac {(b\pi ^{2}/4)^{n}}{n!}}.

On the other hand, recursive integration by parts allows us to deduce that, if a and b are natural number such that $π$ = a/b and f is the polynomial function from [0, $π$ ] into R defined by

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

then:

{\begin{aligned}A_{n}(b)&=\int _{0}^{\pi }f(x)\sin(x)\,dx\\&=\left[-f(x)\cos(x)\right]_{{x=0}}^{{x=\pi }}-\left[-f'(x)\sin(x)\right]_{{x=0}}^{{x=\pi }}+\cdots \pm \left[f^{{(2n)}}(x)\cos(x)\right]_{{x=0}}^{{x=\pi }}\pm \int _{0}^{\pi }f^{{(2n+1)}}(x)\cos(x)\,dx.\end{aligned}}

This last integral is 0, since f^(2n + 1) is the null function (because f is a polynomial function whose degree is 2n). Since each function f^(k) (with 0 ≤ k ≤ 2n) takes integer values on 0 and on $π$ (see Claim 1 from Niven's proof) and since the same thing happens with the sine and the cosine functions, this proves that A_n(b) is an integer. Since it is also greater than 0, it must be a natural number. But it was also proved that A_n(b) <1 if n is large enough, thereby reaching a contradiction.

This proof is quite close to Niven's proof, the main difference between them being the way of proving that the numbers A_n(b) are integers.

Laczkovich's proof

Miklós Laczkovich's proof is a simplification of Lambert's original proof.^[10] He considers the functions

\begin{align} f_k(x) & = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \\ & {} \quad (k\notin\{0,-1,-2,\ldots\}). \end{align}

These functions are clearly defined for all x ∈ R. Besides

f_{1/2}(x)=\cos(2x)\text{ and }f_{3/2}(x)=\frac{\sin(2x)}{2x}.

Claim 1: The following recurrence relation holds:

(\forall x\in\mathbb{R}):\frac{x^2}{k(k+1)}f_{k+2}(x)=f_{k+1}(x)-f_k(x).

Proof: This can be proved by comparing the coefficients of the powers of x.

Claim 2: For each x ∈ R, $\lim_{k\to+\infty}f_k(x)=1.$

Proof: In fact, the sequence x²ⁿ/n! is bounded (since it converges to 0) and if C is an upper bound and if k > 1, then

\bigl|f_k(x)-1\bigr|\leqslant\sum_{n=1}^\infty\frac C{k^n}=C\frac{1/k}{1-1/k}=\frac C{k-1}.

Claim 3: If x ≠ 0 and if x² is rational, then

(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)\neq0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\notin\mathbb{Q}.

Proof: Otherwise, there would be a number y ≠ 0 and integers a and b such that f_k(x) = ay and f_k + 1(x) = by. In order to see why, take y = f_k + 1(x), a = 0 and b = 1 if f_k(x) = 0; otherwise, choose integers a and b such that f_k + 1(x)/f_k(x) = b/a and define y = f_k(x)/a = f_k + 1(x)/b. In each case, y cannot be 0, because otherwise it would follow from claim 1 that each f_k + n(x) (n ∈ N) would be 0, which would contradict claim 2. Now, take a natural number c such that all three numbers bc/k, ck/x² and c/x² are integers and consider the sequence

g_n=\begin{cases}f_k(x)&\text{ if }n=0\\ \frac{c^n}{k(k+1)\cdots(k+n-1)}f_{k+n}(x)&\text{ otherwise.}\end{cases}

Then

g_0=f_k(x)=ay\in\mathbb{Z}y\text{ and }g_1=\frac ckf_{k+1}(x)=\frac{bc}ky\in\mathbb{Z}y.

On the other hand, it follows from claim 1 that

\begin{align}g_{n+2}&=\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}\cdot\frac{x^2}{(k+n)(k+n+1)}f_{k+n+2}(x)\\ &=\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}f_{k+n+1}(x)-\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}f_{k+n}(x)\\ &=\frac{c(k+n)}{x^2}g_{n+1}-\frac{c^2}{x^2}g_n\\ &=\left(\frac{ck}{x^2}+\frac c{x^2}n\right)g_{n+1}-\frac{c^2}{x^2}g_n,\end{align}

which is a linear combination of g_n + 1 and g_n with integer coefficients. Therefore, each g_n is an integer multiple of y. Besides, it follows from claim 2 that each g_n is greater than 0 (and therefore that g_n ≥ |y|) if n is large enough and that the sequence of all g_n's converges to 0. But a sequence of numbers greater than or equal to |y| cannot converge to 0.

Since f_1/2( $π$ /4) = cos( $π$ /2) = 0, it follows from claim 3 that $π$ ²/16 is irrational and therefore that $π$ is irrational.

On the other hand, since

\tan x=\frac{\sin x}{\cos x}=x\frac{f_{3/2}(x/2)}{f_{1/2}(x/2)},

another consequence of claim 3 is that, if x ∈ Q\{0}, then tan x is irrational.

Laczkovich's proof is really about the hypergeometric function. In fact, f_k(x) = ₀F₁(k; −x²) and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation.^[11] This allowed Laczkovich to find a new and simpler proof of the fact that the tangent function has the continued fraction expansion that Lambert had discovered.

Laczkovich's result can also be expressed in Bessel functions of the first kind J_ν(x). In fact, Γ(k)J_k − 1(2x) = x^k − 1f_k(x). So Laczkovich's result is equivalent to: If x ≠ 0 and if x² is rational, then

(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):\frac{x J_k(x)}{J_{k-1}(x)}\notin\mathbb{Q}.

References

↑ Lindemann, Ferdinand von (2004) [1882], "Ueber die Zahl $π$ ", in Berggren, Lennart; Borwein, Jonathan M.; Borwein, Peter B., Pi, a source book (3rd ed.), New York: Springer-Verlag, pp. 194–225, ISBN 0-387-20571-3
↑ Lambert, Johann Heinrich (2004) [1768], "Mémoire sur quelques propriétés remarquables des quantités transcendantes circulaires et logarithmiques", in Berggren, Lennart; Borwein, Jonathan M.; Borwein, Peter B., Pi, a source book (3rd ed.), New York: Springer-Verlag, pp. 129–140, ISBN 0-387-20571-3
1 2 Hermite, Charles (1873), "Extrait d'une lettre de Monsieur Ch. Hermite à Monsieur Paul Gordan", Journal für die reine und angewandte Mathematik (in French), 76, pp. 303–311
↑ Hermite, Charles (1873), "Extrait d'une lettre de Mr. Ch. Hermite à Mr. Carl Borchardt", Journal für die reine und angewandte Mathematik (in French), 76, pp. 342–344
↑ Hermite, Charles (1912) [1873], "Sur la fonction exponentielle", in Picard, Émile, Œuvres de Charles Hermite (in French), III, Gauthier-Villars, pp. 150–181
1 2 Zhou, Li (2011), "Irrationality proofs à la Hermite", Math. Gazette (November), arXiv:0911.1929
↑ Jeffreys, Harold (1973), Scientific Inference (3rd ed.), Cambridge University Press, p. 268, ISBN 0-521-08446-6
↑ Niven, Ivan (1947), "A simple proof that $π$ is irrational" (PDF), Bulletin of the American Mathematical Society, 53 (6), p. 509, doi:10.1090/s0002-9904-1947-08821-2
↑ Bourbaki, Nicolas (1949), Fonctions d'une variable réelle, chap. I–II–III, Actualités Scientifiques et Industrielles (in French), 1074, Hermann, pp. 137–138
↑ Laczkovich, Miklós (1997), "On Lambert's proof of the irrationality of $π$ ", American Mathematical Monthly, 104 (5), pp. 439–443, doi:10.2307/2974737, JSTOR 2974737
↑ Gauss, Carl Friedrich (1811–1813), "Disquisitiones generales circa seriem infinitam", Commentationes Societatis Regiae Scientiarum Gottingensis recentiores (in Latin), 2

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