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Proof that π is irrational

In the 1760s, Johann Heinrich Lambert was the first to prove that the number πisirrational, meaning it cannot be expressed as a fraction ${\displaystyle a/b}$, where ${\displaystyle a}$ and ${\displaystyle b}$ are both integers. 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 Nicolas Bourbaki. Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich. Many of these are proofs by contradiction.

In 1882, Ferdinand von Lindemann proved that ${\displaystyle \pi }$ is not just irrational, but transcendental as well.^[1]

Lambert's proof[edit]

In 1761, Johann Heinrich Lambert proved that ${\displaystyle \pi }$ is irrational by first showing that this continued fraction expansion holds:

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

Then Lambert proved that if ${\displaystyle x}$ is non-zero and rational, then this expression must be irrational. Since ${\displaystyle \tan {\tfrac {\pi }{4}}=1}$, it follows that ${\displaystyle {\tfrac {\pi }{4}}}$ is irrational, and thus ${\displaystyle \pi }$ is also irrational.^[2] A simplification of Lambert's proof is given below.

Hermite's proof[edit]

Written in 1873, this proof uses the characterization of ${\displaystyle \pi }$ as the smallest positive number whose half is a zero of the cosine function and it actually proves that ${\displaystyle \pi ^{2}}$ is irrational.^[3][4] As in many proofs of irrationality, it is a proof by contradiction.

Consider the sequences of real functions ${\displaystyle A_{n}}$ and ${\displaystyle U_{n}}$ for ${\displaystyle n\in \mathbb {N} _{0}}$ defined by:

${\displaystyle {\begin{aligned}A_{0}($x)&=\sin(x),&&A_{n+1}(x)=\int _{0}^{x}yA_{n}(y)\,dy\\[4pt]U_{0}(x)&={\frac {\sin(x)}{x}},&&U_{n+1}(x)=-{\frac {U_{n}'(x)}{x}}\end{aligned}}}

Using induction we can prove that

${\displaystyle {\begin{aligned}A_{n}($x)&={\frac {x^{2n+1}}{(2n+1)!!}}-{\frac {x^{2n+3}}{2\times (2n+3)!!}}+{\frac {x^{2n+5}}{2\times 4\times (2n+5)!!}}\mp \cdots \\[4pt]U_{n}(x)&={\frac {1}{(2n+1)!!}}-{\frac {x^{2}}{2\times (2n+3)!!}}+{\frac {x^{4}}{2\times 4\times (2n+5)!!}}\mp \cdots \end{aligned}}}

and therefore we have:

${\displaystyle U_{n}($x)={\frac {A_{n}(x)}{x^{2n+1}}}.\,}

So

${\displaystyle {\begin{aligned}{\frac {A_{n+1}($x)}{x^{2n+3}}}&=U_{n+1}(x)=-{\frac {U_{n}'(x)}{x}}=-{\frac {1}{x}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {A_{n}(x)}{x^{2n+1}}}\right)\\[6pt]&=-{\frac {1}{x}}\left({\frac {A_{n}'(x)\cdot x^{2n+1}-(2n+1)x^{2n}A_{n}(x)}{x^{2(2n+1)}}}\right)\\[6pt]&={\frac {(2n+1)A_{n}(x)-xA_{n}'(x)}{x^{2n+3}}}\end{aligned}}}

which is equivalent to

${\displaystyle A_{n+1}($x)=(2n+1)A_{n}(x)-x^{2}A_{n-1}(x).\,}

Using the definition of the sequence and employing induction we can show that

${\displaystyle A_{n}($x)=P_{n}(x^{2})\sin(x)+xQ_{n}(x^{2})\cos(x),\,}

where ${\displaystyle P_{n}}$ and ${\displaystyle Q_{n}}$ are polynomial functions with integer coefficients and the degree of ${\displaystyle P_{n}}$ is smaller than or equal to ${\displaystyle {\bigl \lfloor }{\tfrac {1}{2}}n{\bigr \rfloor }.}$ In particular, ${\displaystyle A_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=P_{n}{\bigl (}{\tfrac {1}{4}}\pi ^{2}{\bigr )}.}$

Hermite also gave a closed expression for the function ${\displaystyle A_{n},}$ namely

${\displaystyle A_{n}($x)={\frac {x^{2n+1}}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z.\,}

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

${\displaystyle {\frac {1}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos($xz)\,\mathrm {d} z={\frac {A_{n}(x)}{x^{2n+1}}}=U_{n}(x).}

Proceeding by induction, take ${\displaystyle n=0.}$

${\displaystyle \int _{0}^{1}\cos($xz)\,\mathrm {d} z={\frac {\sin(x)}{x}}=U_{0}(x)}

and, for the inductive step, consider any natural number ${\displaystyle n.}$If

${\displaystyle {\frac {1}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos($xz)\,\mathrm {d} z=U_{n}(x),}

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

${\displaystyle {\begin{aligned}&{\frac {1}{2^{n+1}(n+1)!}}\int _{0}^{1}\left(1-z^{2}\right)^{n+1}\cos($xz)\,\mathrm {d} z\\&\qquad ={\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)\left(1-z^{2}\right)^{n}z{\frac {\sin(xz)}{x}}\,\mathrm {d} z{\Biggr )}\\[8pt]&\qquad ={\frac {1}{x}}\cdot {\frac {1}{2^{n}n!}}\int _{0}^{1}\left(1-z^{2}\right)^{n}z\sin(xz)\,\mathrm {d} z\\[8pt]&\qquad =-{\frac {1}{x}}\cdot {\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {1}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z\right)\\[8pt]&\qquad =-{\frac {U_{n}'(x)}{x}}\\[4pt]&\qquad =U_{n+1}(x).\end{aligned}}}

If${\displaystyle {\tfrac {1}{4}}\pi ^{2}=p/q,}$ with ${\displaystyle p}$ and ${\displaystyle q}$in${\displaystyle \mathbb {N} }$, then, since the coefficients of ${\displaystyle P_{n}}$ are integers and its degree is smaller than or equal to ${\displaystyle {\bigl \lfloor }{\tfrac {1}{2}}n{\bigr \rfloor },}$ ${\displaystyle q^{\lfloor n/2\rfloor }P_{n}{\bigl (}{\tfrac {1}{4}}\pi ^{2}{\bigr )}}$ is some integer ${\displaystyle N.}$ In other words,

${\displaystyle N=q^{\lfloor n/2\rfloor }{A_{n}}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=q^{\lfloor n/2\rfloor }{\frac {1}{2^{n}n!}}\left({\dfrac {p}{q}}\right)^{n+{\frac {1}{2}}}\int _{0}^{1}(1-z^{2})^{n}\cos \left({\tfrac {1}{2}}\pi z\right)\,\mathrm {d} z.}$

But this number is clearly greater than ${\displaystyle 0.}$ On the other hand, the limit of this quantity as ${\displaystyle n}$ goes to infinity is zero, and so, if ${\displaystyle n}$ is large enough, ${\displaystyle N<1.}$ 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 ${\displaystyle \pi .}$ 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 ${\displaystyle e}$^[5]).

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

Cartwright's proof[edit]

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] It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University.^[8]

Consider the integrals

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

where ${\displaystyle n}$ is a non-negative integer.

Two integrations by parts give the recurrence relation

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

If

${\displaystyle J_{n}($x)=x^{2n+1}I_{n}(x),}

then this becomes

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

Furthermore, ${\displaystyle J_{0}($x)=2\sin x} and ${\displaystyle J_{1}($x)=-4x\cos x+4\sin x.} Hence for all ${\displaystyle n\in \mathbb {Z} _{+},}$

${\displaystyle J_{n}($x)=x^{2n+1}I_{n}(x)=n!{\bigl (}P_{n}(x)\sin(x)+Q_{n}(x)\cos(x){\bigr )},}

where ${\displaystyle P_{n}($x)} and ${\displaystyle Q_{n}($x)} are polynomials of degree ${\displaystyle \leq n,}$ and with integer coefficients (depending on ${\displaystyle n}$).

Take ${\displaystyle x={\tfrac {1}{2}}\pi ,}$ and suppose if possible that ${\displaystyle {\tfrac {1}{2}}\pi =a/b}$ where ${\displaystyle a}$ and ${\displaystyle b}$ are natural numbers (i.e., assume that ${\displaystyle \pi }$ is rational). Then

${\displaystyle {\frac {a^{2n+1}}{n!}}I_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=P_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}b^{2n+1}.}$

The right side is an integer. But ${\displaystyle 0<I_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}<2}$ since the interval ${\displaystyle [-1,1]}$ has length ${\displaystyle 2}$ and the function being integrated takes only values between ${\displaystyle 0}$ and ${\displaystyle 1.}$ On the other hand,

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

Hence, for sufficiently large ${\displaystyle n}$

${\displaystyle 0<{\frac {a^{2n+1}I_{n}\left({\frac {\pi }{2}}\right)}{n!}}<1,}$

that is, we could find an integer between ${\displaystyle 0}$ and ${\displaystyle 1.}$ That is the contradiction that follows from the assumption that ${\displaystyle \pi }$ is rational.

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

${\displaystyle {\begin{aligned}J_{n}($x)&=x^{2n+1}\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\[5pt]&=2x^{2n+1}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\[5pt]&=2^{n+1}n!A_{n}(x).\end{aligned}}}

However, it is clearly simpler. This is achieved by omitting the inductive definition of the functions ${\displaystyle A_{n}}$ and taking as a starting point their expression as an integral.

Niven's proof[edit]

This proof uses the characterization of ${\displaystyle \pi }$ as the smallest positive zero of the sine function.^[9]

Suppose that ${\displaystyle \pi }$ is rational, i.e. ${\displaystyle \pi =a/b}$ for some integers ${\displaystyle a}$ and ${\displaystyle b}$ which may be taken without loss of generality to both be positive. Given any positive integer ${\displaystyle n,}$ we define the polynomial function:

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

and, for each ${\displaystyle x\in \mathbb {R} }$ let

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

Claim 1: ${\displaystyle F(0)+F(\pi )}$ is an integer.

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

On the other hand, ${\displaystyle f(\pi -x)=f($x)} and so ${\displaystyle (-1)^{k}f^{($k)}(\pi -x)=f^{(k)}(x)} for each non-negative integer ${\displaystyle k.}$ In particular, ${\displaystyle (-1)^{k}f^{($k)}(\pi )=f^{(k)}(0).} Therefore, ${\displaystyle f^{($k)}(\pi )} is also an integer and so ${\displaystyle F(\pi )}$ is an integer (in fact, it is easy to see that ${\displaystyle F(\pi )=F(0)}$). Since ${\displaystyle F(0)}$ and ${\displaystyle F(\pi )}$ are integers, so is their sum.

Claim 2:

${\displaystyle \int _{0}^{\pi }f($x)\sin(x)\,dx=F(0)+F(\pi )}

Proof: Since ${\displaystyle f^{(2n+2)}}$ is the zero polynomial, we have

${\displaystyle F''+F=f.}$

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

${\displaystyle (F'\cdot \sin {}-F\cdot \cos {})'=f\cdot \sin }$

By the fundamental theorem of calculus

${\displaystyle \left.\int _{0}^{\pi }f($x)\sin(x)\,dx={\bigl (}F'(x)\sin x-F(x)\cos x{\bigr )}\right|_{0}^{\pi }.}

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

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

${\displaystyle \int _{0}^{\pi }f($x)\sin(x)\,dx\leq \pi {\frac {(\pi a)^{n}}{n!}}}

which is smaller than ${\displaystyle 1}$ for large ${\displaystyle n,}$ hence ${\displaystyle F(0)+F(\pi )<1}$ for these ${\displaystyle n,}$ by Claim 2. This is impossible for the positive integer ${\displaystyle F(0)+F(\pi ).}$ This shows that the original assumption that ${\displaystyle \pi }$ is rational leads to a contradiction, which concludes the proof.

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

${\displaystyle \int _{0}^{\pi }f($x)\sin(x)\,dx=\sum _{j=0}^{n}(-1)^{j}\left(f^{(2j)}(\pi )+f^{(2j)}(0)\right)+(-1)^{n+1}\int _{0}^{\pi }f^{(2n+2)}(x)\sin(x)\,dx,}

which is obtained by ${\displaystyle 2n+2}$ integrations by parts. Claim 2 essentially establishes this formula, where the use of ${\displaystyle F}$ hides the iterated integration by parts. The last integral vanishes because ${\displaystyle f^{(2n+2)}}$ 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,

${\displaystyle {\begin{aligned}J_{n}($x)&=x^{2n+1}\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz\\&=\int _{-1}^{1}\left(x^{2}-(xz)^{2}\right)^{n}x\cos(xz)\,dz.\end{aligned}}}

Therefore, the substitution ${\displaystyle xz=y}$ turns this integral into

${\displaystyle \int _{-x}^{x}(x^{2}-y^{2})^{n}\cos($y)\,dy.}

In particular,

${\displaystyle {\begin{aligned}J_{n}\left({\frac {\pi }{2}}\right)&=\int _{-\pi /2}^{\pi /2}\left({\frac {\pi ^{2}}{4}}-y^{2}\right)^{n}\cos($y)\,dy\\[5pt]&=\int _{0}^{\pi }\left({\frac {\pi ^{2}}{4}}-\left(y-{\frac {\pi }{2}}\right)^{2}\right)^{n}\cos \left(y-{\frac {\pi }{2}}\right)\,dy\\[5pt]&=\int _{0}^{\pi }y^{n}(\pi -y)^{n}\sin(y)\,dy\\[5pt]&={\frac {n!}{b^{n}}}\int _{0}^{\pi }f(x)\sin(x)\,dx.\end{aligned}}}

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

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

then

${\displaystyle \int f($x)\sin(x)\,dx=F'(x)\sin(x)-F(x)\cos(x)+C,}

from which it follows that

${\displaystyle \int _{0}^{\pi }f($x)\sin(x)\,dx=F(\pi )+F(0).}

Bourbaki's proof[edit]

Bourbaki's proof is outlined as an exercise in his calculus treatise.^[10] For each natural number b and each non-negative integer ${\displaystyle n,}$ define

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

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

${\displaystyle x(\pi -x)\leq \left({\frac {\pi }{2}}\right)^{2}}$

and therefore

${\displaystyle A_{n}($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, repeated integration by parts allows us to deduce that, if ${\displaystyle a}$ and ${\displaystyle b}$ are natural numbers such that ${\displaystyle \pi =a/b}$ and ${\displaystyle f}$ is the polynomial function from ${\displaystyle [0,\pi ]}$ into ${\displaystyle \mathbb {R} }$ defined by

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

then:

${\displaystyle {\begin{aligned}A_{n}($b)&=\int _{0}^{\pi }f(x)\sin(x)\,dx\\[5pt]&={\Big [}{-f(x)\cos(x)}{\Big ]}_{x=0}^{x=\pi }\,-{\Big [}{-f'(x)\sin(x)}{\Big ]}_{x=0}^{x=\pi }+\cdots \\[5pt]&\ \qquad \pm {\Big [}f^{(2n)}(x)\cos(x){\Big ]}_{x=0}^{x=\pi }\,\pm \int _{0}^{\pi }f^{(2n+1)}(x)\cos(x)\,dx.\end{aligned}}}