In the 1760s, Johann Heinrich Lambert was the first to prove that the number π is irrational, meaning it cannot be expressed as a fraction {\displaystyle a/b}

, where {\displaystyle a}

and {\displaystyle b}

(资料图片仅供参考)

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 π is not just irrational, but transcendental as well.

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:

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 thus π is also irrational. A simplification of Lambert's proof is given below.

Hermite's proof

Written in 1873, 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 π2 is irrational. As in many proofs of irrationality, it is a proof by contradiction.

Consider the sequences of functions An and Un from {\displaystyle \mathbb {R} }

into {\displaystyle \mathbb {R} }

for {\displaystyle n\in \mathbb {N} _{0}}

defined by:

Using induction we can prove that

and therefore we have:

So

which is equivalent to

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

where Pn and Qn are polynomial functions with integer coefficients and the degree of Pn is smaller than or equal to ⌊n/2⌋. In particular, An(π/2) = Pn(π2/4).

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

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

If π2/4 = p/q, with p and q in {\displaystyle \mathbb {N} }

, then, since the coefficients of Pn are integers and its degree is smaller than or equal to ⌊n/2⌋, q⌊n/2⌋Pn(π2/4) is some integer N. In other words,

But this number is clearly greater than 0. On the other hand, the limit of this quantity as n goes to infinity is zero, 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).

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

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. It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University. Consider the integrals

where n is a non-negative integer.

Two integrations by parts give the recurrence relation

If

then this becomes

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

where Pn(x) and Qn(x) are polynomials of degree ≤ n, 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

The right side is an integer. But 0 < In(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,

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

Hence, for sufficiently large n

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,

{\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 An 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.

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:

and, for each x ∈ ℝ let

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

Claim 1:F(0) + F(π) is an integer.

Proof:Expanding f as a sum of monomials, the coefficient of xk is a number of the form ck /n! where ck 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!) ck 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:

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

Proof:Since 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 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) ≤ πa and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have, by the original definition of f,

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). This shows that the original assumption that π 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 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. 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 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 that if f is a polynomial function and

then

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

from which it follows that

Bourbaki's proof

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

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

and therefore

On the other hand, repeated integration by parts allows us to deduce that, if a and b are natural numbers such that π = a/b and f is the polynomial function from [0,π] into Rdefined 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 \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}}}

This last integral is 0, since f (2n + 1) is the null function (because f is a polynomial function of degree 2n). Since each function f (k) (with 0 ≤ k ≤ 2n) takes integer values on 0 and on π and since the same thing happens with the sine and the cosine functions, this proves that An(b) is an integer. Since it is also greater than 0, it must be a natural number. But it was also proved that An(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 An(b) are integers.

Laczkovich's proof

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

{\displaystyle 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 \}).}

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

{\displaystyle f_{\frac {1}{2}}(x)=\cos(2x),}

{\displaystyle f_{\frac {3}{2}}(x)={\frac {\sin(2x)}{2x}}.}

Claim 1:The following recurrence relation holds:

{\displaystyle \forall x\in \mathbb {R} :\qquad {\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, {\displaystyle \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

{\displaystyle \left|f_{k}(x)-1\right|\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

{\displaystyle \forall k\in \mathbb {Q} \smallsetminus \{0,-1,-2,\ldots \}:\qquad f_{k}(x)\neq 0\quad {\text{ and }}\quad {\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 fk(x) = ay and fk + 1(x) = by. In order to see why, take y = fk + 1(x), a = 0 and b = 1 if fk(x) = 0; otherwise, choose integers a and b such that fk + 1(x)/fk(x) = b/a and define y = fk(x)/a = fk + 1(x)/b. In each case, y cannot be 0, because otherwise it would follow from claim 1 that each fk + 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

{\displaystyle g_{n}={\begin{cases}f_{k}(x)&n=0\\{\dfrac {c^{n}}{k(k+1)\cdots (k+n-1)}}f_{k+n}(x)&n\neq 0\end{cases}}}

Then

{\displaystyle g_{0}=f_{k}(x)=ay\in \mathbb {Z} y\quad {\text{ and }}\quad g_{1}={\frac {c}{k}}f_{k+1}(x)={\frac {bc}{k}}y\in \mathbb {Z} y.}

On the other hand, it follows from claim 1 that

{\displaystyle {\begin{aligned}g_{n+2}&={\frac {c^{n+2}}{x^{2}k(k+1)\cdots (k+n-1)}}\cdot {\frac {x^{2}}{(k+n)(k+n+1)}}f_{k+n+2}(x)\\[5pt]&={\frac {c^{n+2}}{x^{2}k(k+1)\cdots (k+n-1)}}f_{k+n+1}(x)-{\frac {c^{n+2}}{x^{2}k(k+1)\cdots (k+n-1)}}f_{k+n}(x)\\[5pt]&={\frac {c(k+n)}{x^{2}}}g_{n+1}-{\frac {c^{2}}{x^{2}}}g_{n}\\[5pt]&=\left({\frac {ck}{x^{2}}}+{\frac {c}{x^{2}}}n\right)g_{n+1}-{\frac {c^{2}}{x^{2}}}g_{n},\end{aligned}}}

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

Since f1/2(π/4) = cos(π/2) = 0, it follows from claim 3 that π2/16 is irrational and therefore that π is irrational.

On the other hand, since

{\displaystyle \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, fk(x) = 0F1(k; −x2) and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation. 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 Jν(x). In fact, Γ(k)Jk − 1(2x) = xk − 1fk(x). So Laczkovich's result is equivalent to: If x ≠ 0 and if x2 is rational, then

{\displaystyle \forall k\in \mathbb {Q} \smallsetminus \{0,-1,-2,\ldots \}:\qquad {\frac {xJ_{k}(x)}{J_{k-1}(x)}}\notin \mathbb {Q} .}

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