Proof that pi is irrational

During the 1760s, Johann Heinrich Lambert pioneered the demonstration that the number π is irrational, signifying its inability to be represented as a fraction a / b , {\displaystyle a/b,} where both a {\displaystyle a} and b {\displaystyle b} are integers. Subsequently, in the 19th century, Charles Hermite developed a proof that necessitates only a foundational understanding of calculus. Mary Cartwright, Ivan Niven, and "Nicolas Bourbaki" each contributed simplifications of Hermite's original proof. Miklós Laczkovich also provided a distinct proof, which serves as a simplification of Lambert's method. A significant number of these proofs employ the technique of contradiction.

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 ${\displaystyle b}$ 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 established that ${\displaystyle \pi }$ is not merely an irrational number, but also a transcendental one.

Lambert's Proof

In 1761, Johann Heinrich Lambert demonstrated the irrationality of ${\displaystyle \pi }$ by initially presenting its continued fraction expansion as valid:

Lambert subsequently demonstrated that if ${\displaystyle x}$ is a non-zero rational number, the corresponding expression is necessarily irrational. Given that tan ⁡<!-- ⁡ --> π<!-- π --> §3334§ = §4041§ {\displaystyle \tan {\tfrac {\pi }{4}}=1} , it logically follows that ${\displaystyle {\tfrac {\pi }{4}}}$ must be irrational, which in turn implies that ${\displaystyle \pi }$ is also irrational. A simplified version of Lambert's proof is presented subsequently.

Hermite's Proof

Developed in 1873, this proof leverages the definition of π<!-- π --> {\displaystyle \pi } as the minimal positive value for which its half constitutes a zero of the cosine function. It further establishes the irrationality of ${\displaystyle \pi ^{2}}$. Consistent with numerous irrationality proofs, this demonstration employs a proof by contradiction.

Let us consider the sequences of real functions, ${\displaystyle A_{n}}$ and ${\displaystyle U_{n}}$, for n ∈<!-- ∈ --> N §6162§ {\displaystyle n\in \mathbb {N} _{0}} , defined as follows:

15§ ( x ) = sin ⁡ ( x ) , A n + §5556§ ( x ) = ∫ §7273§ x y A n ( y ) d y U §110111§ ( x ) = sin ⁡ ( x ) x , U n + §159160§ ( x ) = − U n ′ ( x ) x {\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}}}

The subsequent relationships can be formally established through mathematical induction:

43§ ( §4956§ ) ! ! − x §75 §86 + x §118 §129 ∓ ⋯ U n ( x ) = §190191§ ( §195202§ ) ! ! − x §221 §226 + x §258259§ §263 ∓ ⋯ {\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}}}

Consequently, the following relationship is established:

U n ( x ) = A n ( x ) x §4445§ n + §5051§ . {\displaystyle U_{n}(x)={\frac {A_{n}(x)}{x^{2n+1}}}.\,}

Consequently,

22§ ( x ) x §36 = U n + §6263§ ( x ) = − U n ′ ( x ) x = − §108109§ x d d x ( A n ( x ) x §155162§ ) = − §185186§ x ( A n ′ ( x ) ⋅ x §220227§ − ( §235242§ ) x §249 A n ( x ) x §274285§ ) ) = ( §310317§ ) A n ( x ) − x A n ′ ( x ) x §360 {\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}}}

This relationship is equivalent to the following expression:

A n + §1415§ ( x ) = ( §2829§ n + §3435§ ) A n ( x ) − x §5960§ A n − §7273§ ( x ) . {\displaystyle A_{n+1}(x)=(2n+1)A_{n}(x)-x^{2}A_{n-1}(x).\,}

By applying the sequence's definition and utilizing mathematical induction, it can be demonstrated that:

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

These are polynomial functions, ${\displaystyle P_{n}}$ and ${\displaystyle Q_{n}}$, which possess integer coefficients. The degree of ${\displaystyle P_{n}}$ is less than or equal to ⌊§8081§§8283§n⌋.{\displaystyle {\bigl \lfloor }{\tfrac {1}{2}}n{\bigr \rfloor }.} Specifically, An(§128129§§130131§π)=Pn(§163164§§165166§π§176177§).{\displaystyle A_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=P_{n}{\bigl (}{\tfrac {1}{4}}\pi ^{2}{\bigr )}.}

Hermite additionally provided a closed-form expression for the function ${\displaystyle A_{n},}$, which is defined as:

An(x)=x§2829§n+§3435§§4041§nn!∫§5859§§6263§(§6869§−z§7778§)ncos⁡(xz)dz.{\displaystyle A_{n}(x)={\frac {x^{2n+1}}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z.\,}

Although Hermite did not provide a justification for this assertion, its proof is straightforward. Fundamentally, this assertion is equivalent to

The function Un(x) is formally defined by the following integral expression: §89§ §1213§ n n ! ∫ §3132§ §3536§ ( §4142§ − z §5051§ ) n cos ⁡ ( x z ) d z = A n ( x ) x §107108§ n + §113114§ = U n ( x ) . {\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).}

To initiate the inductive proof, consider the base case where n=§1011§. ${\displaystyle n=0.}$

For this specific instance, the integral evaluates to: ∫<!-- ∫ --> §1112§ §1516§ cos ⁡ ( x z ) d z = sin ⁡ ( x ) x = U §6768§ ( x ) {\displaystyle \int _{0}^{1}\cos(xz)\,\mathrm {d} z={\frac {\sin(x)}{x}}=U_{0}(x)}

Subsequently, for the inductive step, we consider an arbitrary natural number n. ${\displaystyle n.}$ If

the following relationship is assumed to hold for Un(x): §89§ §1213§ n n ! ∫ §3132§ §3536§ ( §4142§ − z §5051§ ) n cos ⁡ ( x z ) d z = U n ( x ) , {\displaystyle {\frac {1}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z=U_{n}(x),}

Subsequently, through the application of integration by parts and Leibniz's rule, one obtains

Assuming that

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

However, this quantity is unequivocally greater than ${\displaystyle 0.}$ Conversely, the limit of this quantity as ${\displaystyle n}$${\displaystyle n}$${\displaystyle N<1.}$ This leads to a logical contradiction.

Hermite's proof was not an isolated objective but rather a secondary outcome of his quest to demonstrate the transcendence of ${\displaystyle \pi .}$ He utilized recurrence relations to develop and derive a suitable integral representation. Upon establishing this integral representation, several approaches exist for constructing a concise and self-contained proof, as exemplified by the presentations of Cartwright, Bourbaki, and Niven. Hermite could readily discern these methods, mirroring his approach in proving the transcendence of ${\displaystyle e}$.

Furthermore, Hermite's proof bears a closer resemblance to Lambert's proof than initially apparent. Specifically, ${\displaystyle A_{n}(x)}$ represents the "residue" or "remainder" of Lambert's continued fraction for ${\displaystyle \tan x.}$

Cartwright's Proof

Harold Jeffreys documented that Mary Cartwright introduced this proof as an examination problem at Cambridge University in 1945, though she had not identified its original source. This proof continues to be featured on the fourth problem sheet for the Analysis IA course at Cambridge University.

The following integrals are considered:

In(x)=∫<!-- ∫ -->−<!-- − -->§3031§§3435§(§4041§−z§4950§)ncos⁡(xz)dz,{\displaystyle I_{n}(x)=\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz,}

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

Applying integration by parts twice yields the following recurrence relation:

x§1011§In(x)=§3031§n(§3637§n−<!-- − -->§4344§)In−§5657§(x)−§6970§n(n−§8081§)In−§9394§(x).(n≥§114115§){\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

The relationship is expressed as Jn(x)=x§2627§n+§3233§In(x),{\displaystyle J_{n}(x)=x^{2n+1}I_{n}(x),}.

Subsequently, this expression transforms into:

Jn(x)=§2223§n(§2829§n−<!-- − -->§3536§)Jn−§4849§(x)−§6162§n(n−§7273§)x§8081§Jn−§9394§(x).{\displaystyle J_{n}(x)=2n(2n-1)J_{n-1}(x)-4n(n-1)x^{2}J_{n-2}(x).}

Furthermore, the initial conditions are defined as J§1011§(x)=§2223§sin⁡<!-- ⁡ -->x{\displaystyle J_{0}(x)=2\sin x} and J§4950§(x)=−§6465§xcos⁡x+§7778§sin⁡x.{\displaystyle J_{1}(x)=-4x\cos x+4\sin x.} These conditions are valid for all ${\displaystyle n\in \mathbb {Z} _{+},}$.

Jn(x)=x§2627§n+§3233§In(x)=n!(Pn(x)sin⁡(x)+Qn(x)cos⁡(x)),{\displaystyle J_{n}(x)=x^{2n+1}I_{n}(x)=n!{\bigl (}P_{n}(x)\sin(x)+Q_{n}(x)\cos(x){\bigr )},}

These polynomials, specifically ${\displaystyle P_{n}(x)}$ and ${\displaystyle Q_{n}(x)}$, possess a degree of ${\displaystyle \leq n,}$ and feature integer coefficients that are contingent upon ${\displaystyle n}$.

Let x=§1314§§1516§π,{\displaystyle x={\tfrac {1}{2}}\pi ,} and assume, for the sake of contradiction, that §4243§§4445§π=a/b{\displaystyle {\tfrac {1}{2}}\pi =a/b}, where ${\displaystyle a}$ and ${\displaystyle b}$ are natural numbers (implying that ${\displaystyle \pi }$ is rational). Then

a§1213§n+§1819§n!In(§4748§§4950§π)=Pn(§8283§§8485§π)b§102103§n+§108109§.{\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-hand side of the equation is an integer. However, the inequality §27§<In(§2728§§29π)<§44${\displaystyle 0 holds true because the interval${\displaystyle [-1,1]} has a length of {\displaystyle 2}, and the integrand's values are exclusively confined between{\displaystyle 0} and{\displaystyle 1.} Conversely,

a§1213§n+§1819§n!→§3334§ as n→∞.{\displaystyle {\frac {a^{2n+1}}{n!}}\to 0\quad {\text{ as }}n\to \infty .}

Hence, for sufficiently large {\displaystyle n},

The following inequality is established: 7§ < a §1724§ I n ( π §43 ) n ! < §6162§ , {\displaystyle 0<{\frac {a^{2n+1}I_{n}\left({\frac {\pi }{2}}\right)}{n!}}<1,}

This implies the existence of an integer between §67§ {\displaystyle 0} and ${\displaystyle 1.}$ Such a finding constitutes a contradiction, which arises from the initial assumption that ${\displaystyle \pi }$ is a rational number.

This proof is analogous to Hermite's demonstration; indeed,

41§ ∫ − §5253§ §5657§ ( §6263§ − z §71 ) n cos ⁡ ( x z ) d z = §112125§ ∫ §133134§ §137138§ ( §143144§ − z §152 ) n cos ⁡ ( x z ) d z = §194202§ n ! A n ( x ) . {\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}}}

Nevertheless, this approach offers a distinct simplification, which is realized by omitting the inductive definition of the functions ${\displaystyle A_{n}}$ and instead commencing with their integral expression.

Niven's Proof

This proof utilizes the characterization of ${\displaystyle \pi }$ as the minimal positive root of the sine function.

Let us assume that ${\displaystyle \pi }$ is a rational number, implying it can be expressed as ${\displaystyle \pi =a/b}$, where ${\displaystyle a}$ and ${\displaystyle b}$ are positive integers, without loss of generality. For any positive integer ${\displaystyle n,}$ the polynomial function is defined as:

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

Additionally, for every ${\displaystyle x\in \mathbb {R} }$, we define:

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

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

Proof:

Conversely, given ${\displaystyle f(\pi -x)=f(x)}$, it follows that ( −<!-- − --> §4950§ ) k f ( k ) ( π − x ) = f ( k ) ( x ) {\displaystyle (-1)^{k}f^{(k)}(\pi -x)=f^{(k)}(x)} for every non-negative integer ${\displaystyle k.}$ Specifically, ( −<!-- − --> §140141§ ) k f ( k ) ( π ) = f ( k ) ( §185186§ ) . {\displaystyle (-1)^{k}f^{(k)}(\pi )=f^{(k)}(0).} Consequently, ${\displaystyle f^{(k)}(\pi )}$ is likewise an integer, which implies that ${\displaystyle F(\pi )}$ is an integer (indeed, it can be readily demonstrated that F ( π<!-- π --> ) = F ( §277278§ ) {\displaystyle F(\pi )=F(0)} ). As both F ( §300301§ ) {\displaystyle F(0)} and ${\displaystyle F(\pi )}$ are integers, their sum is also an integer.

Claim 2:

∫<!-- ∫ --> §1011§ π f ( x ) sin ⁡ ( x ) d x = F ( §5152§ ) + F ( π ) {\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx=F(0)+F(\pi )}

Proof: Given that f(§1415§n+§2021§){\displaystyle f^{(2n+2)}} represents the zero polynomial, it follows that:

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

The derivatives of the sine and cosine functions are defined as sin' = cos and cos' = −sin, respectively. Consequently, applying the product rule yields:

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

According to the fundamental theorem of calculus,

∫<!-- ∫ -->§1617§πf(x)sin⁡(x)dx=(F′(x)sin⁡x−F(x)cos⁡x)|§106107§π.{\displaystyle \left.\int _{0}^{\pi }f(x)\sin(x)\,dx={\bigl (}F'(x)\sin x-F(x)\cos x{\bigr )}\right|_{0}^{\pi }.}

Given that sin⁡<!-- ⁡ -->§1112§=sin⁡π=§2526§{\displaystyle \sin 0=\sin \pi =0} and cos⁡<!-- ⁡ -->§4647§=−cos⁡π=§6364§{\displaystyle \cos 0=-\cos \pi =1} (utilizing the previously established characterization of ${\displaystyle \pi }$ as a root of the sine function), Claim 2 is consequently substantiated.

Conclusion: Given that 19§ {\displaystyle f(x)>0} and 44§ {\displaystyle \sin x>0} for the interval 60§ < x < π {\displaystyle 0 (as ${\displaystyle \pi }$ represents the smallest positive root of the sine function), Claims 1 and 2 collectively demonstrate that F ( §107108§ ) + F ( π ) {\displaystyle F(0)+F(\pi )} constitutes a positive integer. Furthermore, considering that §138139§ ≤ x ( a − b x ) ≤ π a {\displaystyle 0\leq x(a-bx)\leq \pi a} and §180181§ ≤ sin ⁡ x ≤ §195196§ {\displaystyle 0\leq \sin x\leq 1} for the range §211212§ ≤ x ≤ π , {\displaystyle 0\leq x\leq \pi ,} it follows from the original definition of ${\displaystyle f,}$

∫<!-- ∫ --> §1112§ π f ( x ) sin ⁡ ( x ) d x ≤ π ( π a ) n n ! {\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx\leq \pi {\frac {(\pi a)^{n}}{n!}}}

This value is less than §67§{\displaystyle 1} for sufficiently large ${\displaystyle n,}$. Consequently, 45§)+F(π)<§6162§{\displaystyle F(0)+F(\pi )<1} holds for these specific values of ${\displaystyle n,}$ according to Claim 2. However, this outcome is impossible for the positive integer F(§99100§)+F(π).{\displaystyle F(0)+F(\pi ).} This demonstrates that the initial premise, asserting that ${\displaystyle \pi }$ is rational, results in a contradiction, thereby concluding the proof.

The preceding proof represents a refined version, intentionally simplified regarding its prerequisite knowledge, derived from an analysis of the following formula:

∫<!-- ∫ -->§1112§πf(x)sin⁡(x)dx=∑j=§5657§n(−§6970§)j(f(§8990§j)(π)+f(§112113§j)(§122123§))+(−§137138§)n+§147148§∫§155156§πf(§170(x)sin⁡(x)dx,{\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,}

This result is derived through ${\displaystyle 2n+2}$ successive integrations by parts. Claim 2 substantiates this formula, where the variable ${\displaystyle F}$ implicitly accounts for the iterative integration process. The final integral term becomes zero because ${\displaystyle f^{(2n+2)}}$ constitutes the zero polynomial. Furthermore, Claim 1 demonstrates that the remaining summation yields an integer value.

Niven's proof exhibits a closer resemblance to Cartwright's (and consequently Hermite's) methodology than initially perceived. Specifically,

41§ ∫ − §5253§ §5657§ ( §6263§ − z §71 ) n cos ⁡ ( x z ) d z = ∫ − §120121§ §124125§ ( x §137 − ( x z ) §154 ) n x cos ⁡ ( x z ) d z . {\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}}}

Consequently, the application of the substitution ${\displaystyle xz=y}$ converts this integral into the following form:

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

Specifically,

142§ π ( π §162 §166167§ − ( y − π §188 ) §197 ) n cos ⁡ ( y − π §229 ) d y = ∫ §258259§ π y n ( π − y ) n sin ⁡ ( y ) d y = n ! b n ∫ §343344§ π f ( x ) sin ⁡ ( x ) d x . {\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}}}

A further connection between these proofs is evident in Hermite's observation that if ${\displaystyle f}$ represents a polynomial function, and

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

then the following integral identity holds:

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

Consequently, it can be deduced that

∫<!-- ∫ --> §1112§ π f ( x ) sin ⁡ ( x ) d x = F ( π ) + F ( §6263§ ) . {\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx=F(\pi )+F(0).}

Bourbaki's Proof

The proof attributed to Bourbaki is presented as an exercise within his comprehensive calculus treatise. For every natural number b and each non-negative integer ${\displaystyle n,}$ the following is defined:

A n ( b ) = b n ∫<!-- ∫ --> §3435§ π x n ( π − x ) n n ! sin ⁡ ( x ) d x . {\displaystyle A_{n}(b)=b^{n}\int _{0}^{\pi }{\frac {x^{n}(\pi -x)^{n}}{n!}}\sin(x)\,dx.}

Given that ${\displaystyle A_{n}(b)}$ represents the integral of a function defined over the interval [ §3637§ , π ] {\displaystyle [0,\pi ]} , which evaluates to §5960§ {\displaystyle 0} at both §7576§ {\displaystyle 0} and ${\displaystyle \pi }$ but is strictly greater than §108109§ {\displaystyle 0} elsewhere, it follows that ${\displaystyle A_{n}(b)>0.}$. Furthermore, for any natural number ${\displaystyle b,}$ the condition 191§ {\displaystyle A_{n}(b)<1} holds when ${\displaystyle n}$ is sufficiently large, because

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

Consequently,

A n ( b ) ≤<!-- ≤ --> π<!-- π --> b n §3536§ n ! ( π §5455§ ) §5960§n = π ( b π §8081§ <!-- π -->/ §8586§ ) n n ! .{\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!}}.}

Conversely, through the iterative application of integration by parts, it can be established that if ${\displaystyle a}$ and ${\displaystyle b}$ are natural numbers such that ${\displaystyle \pi =a/b}$, and if ${\displaystyle f}$ denotes the polynomial function mapping from the interval [ §8384§ , π ] {\displaystyle [0,\pi ]} to ${\displaystyle \mathbb {R} }$, defined as follows:

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

then the following holds:

36§ π f ( x ) sin ⁡ ( x ) d x = [ − f ( x ) cos ⁡ ( x ) ] x = §121122§ x = π − [ − f ′ ( x ) sin ⁡ ( x ) ] x = §185186§ x = π + ⋯   ± [ f ( §228 ( x ) cos ⁡ ( x ) ] x = §265266§ x = π ± ∫ §288289§ π f ( §303310§ ) ( x ) cos ⁡ ( x ) d x . {\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}}}

The final integral evaluates to §67§ , {\displaystyle 0,} because f ( §3031§ n + §3637§ ) {\displaystyle f^{(2n+1)}} represents the null function. This is due to ${\displaystyle f}$ being a polynomial of degree ${\displaystyle 2n}$. Given that each function ${\displaystyle f^{(k)}}$ (for §119120§ ≤ k ≤ §129130§ n {\displaystyle 0\leq k\leq 2n} ) yields integer values at §148149§ {\displaystyle 0} and ${\displaystyle \pi }$, and considering the analogous behavior of sine and cosine functions, it is established that ${\displaystyle A_{n}(b)}$ is an integer. Furthermore, since this value is also greater than §209210§ , {\displaystyle 0,} it must be a natural number. However, it has also been demonstrated that 244§ {\displaystyle A_{n}(b)<1} for sufficiently large ${\displaystyle n}$, which presents a contradiction.

This particular proof bears a strong resemblance to Niven's proof; their primary distinction lies in the methodology employed to demonstrate that the values ${\displaystyle A_{n}(b)}$ are integers.

Laczkovich's Proof

Miklós Laczkovich developed a proof that simplifies Lambert's original methodology. His approach involves considering specific functions.

f k ( x ) = §2223§ − x §3334§ k + x §4950§ §5455§ ! k ( k + §6667§ ) − x §8283§ §8788§ ! k ( k + §99100§ ) ( k + §109110§ ) + ⋯ ( k ∉ { §132133§ , − §139140§ , − §146147§ , …<!-- … --> } ) . {\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 explicitly defined for any real number ${\displaystyle x.}$ Furthermore,

f §1011§ / §1617§ ( x ) = cos ⁡ ( §3536§ x ) , {\displaystyle f_{1/2}(x)=\cos(2x),}

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

Claim 1: The subsequent recurrence relation is valid for any real number ${\displaystyle x}$:

{\displaystyle {\frac {x^2}{k(k+1)}}f_{k+2}(x)=f_{k+1}(x)-f_k(x).}

The proof for this assertion involves a comparative analysis of the coefficients associated with the powers of {\displaystyle x.}

Claim 2 posits that for every real number {\displaystyle x,} the following holds:

{\displaystyle \lim _{k\to +\infty }f_k(x)=1.}

Proof: The sequence {\displaystyle x^{2n}/n!} is demonstrably bounded, as it converges to{\displaystyle 0}. Consequently, if{\displaystyle C} represents an upper bound and

The following inequality holds: {\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}}.} This demonstrates that the absolute difference between fk(x) and 1 is bounded by an infinite series that converges to C divided by (k minus 1).

Claim 3 asserts that if x is not equal to §1314§ (i.e., {\displaystyle x\neq 0,}), and if x squared (i.e., {\displaystyle x^{2}}) is a rational number, and furthermore, if k belongs to the set of rational numbers excluding 0 and negative integers (i.e., {\displaystyle k\in \mathbb {Q} \smallsetminus \{0,-1,-2,\ldots \}}), then the following conditions apply:

Specifically, fk(x) will not equal §2324§, and the ratio of fk+§4445§(x) to fk(x) will not be a rational number, as formally stated: {\displaystyle f_{k}(x)\neq 0\quad {\text{ and }}\quad {\frac {f_{k+1}(x)}{f_{k}(x)}}\notin \mathbb {Q} .}

Proof: Otherwise, assume there exists a non-zero number y and integersa andb satisfying the conditionsf_k(x)=ay andf_{k+1}(x)=by.

The function 44§cnk(k+§6768§)⋯(k+n−§7980§)fk+n(x)n≠§103104§{\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}}} is defined as follows:

It then follows that

The initial terms are given by g§1011§=fk(x)=ay∈Zy and g§5657§=ckfk+§7778§(x)=bcky∈Zy.{\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.}.

Conversely, it is established by Claim 1 that

60§ ) ⋯ ( k + n − §7778§ ) ⋅ x §93 ( k + n ) ( k + n + §118119§ ) f k + n + §137 ( x ) = c n + §167 x §176 k ( k + §188189§ ) ⋯ ( k + n − §206207§ ) f k + n + §225226§ ( x ) − c n + §248 x §257 k ( k + §269270§ ) ⋯ ( k + n − §287288§ ) f k + n ( x ) = c ( k + n ) x §342 g n + §356357§ − c §369 x §377 g n = ( c k x §417 + c x §433 n ) g n + §453454§ − c §466 x §474 g n , {\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}}}

This expression represents a linear combination of g n + §1415§ {\displaystyle g_{n+1}} and ${\displaystyle g_{n}}$, featuring integer coefficients. Consequently, each term ${\displaystyle g_{n}}$ constitutes an integer multiple of ${\displaystyle y.}$ Furthermore, based on Claim 2, it is established that every ${\displaystyle g_{n}}$ exceeds §116117§ {\displaystyle 0} (implying that ${\displaystyle g_{n}\geq |y|}$) for sufficiently large ${\displaystyle n}$, and that the sequence comprising all ${\displaystyle g_{n}}$ converges towards ${\displaystyle 0.}$ However, a sequence of numbers that are consistently greater than or equal to ${\displaystyle |y|}$ cannot converge to ${\displaystyle 0.}$

Given that f §1011§ / §1617§ ( §2526§ §2728§ π ) = cos ⁡ §4748§ §4950§ π = §5960§ , {\displaystyle f_{1/2}({\tfrac {1}{4}}\pi )=\cos {\tfrac {1}{2}}\pi =0,} , Claim 3 logically demonstrates that §8081§ §8283§ π §9293§ {\displaystyle {\tfrac {1}{16}}\pi ^{2}} is irrational, which in turn proves that ${\displaystyle \pi }$ is irrational.

Conversely, given that

tan ⁡<!-- ⁡ --> x = sin ⁡<!-- ⁡ --> x cos ⁡<!-- ⁡ --> x = x f §4849§ / §5455§ ( x / §6667§ ) f §7677§ / §8283§ ( x / §9495§ ) , {\displaystyle \tan x={\frac {\sin x}{\cos x}}=x{\frac {f_{3/2}(x/2)}{f_{1/2}(x/2)}},}

A further implication of Claim 3 is that if x ∈<!-- ∈ --> Q ∖<!-- ∖ --> { §2021§ } , {\displaystyle x\in \mathbb {Q} \smallsetminus \{0\},} then ${\displaystyle \tan x}$ is an irrational number.

Laczkovich's proof centers on the hypergeometric function. Specifically, f k ( x ) = §2627§ F §3435§ ( k − x §4950§ ) {\displaystyle f_{k}(x)={}_{0}F_{1}(k-x^{2})} . Gauss had previously established a continued fraction expansion for the hypergeometric function by utilizing its functional equation. This foundational work enabled Laczkovich to develop a novel and more straightforward proof for the continued fraction expansion of the tangent function, a discovery originally attributed to Lambert.

Laczkovich's findings can alternatively be articulated using Bessel functions of the first kind, specifically ${\displaystyle J_{\nu }(x)}$. Specifically, the relationship is defined by Γ<!-- Γ --> ( k ) J k −<!-- − --> §5354§ ( §5960§ x ) = x k −<!-- − --> §7677§ f k ( x ) {\displaystyle \Gamma (k)J_{k-1}(2x)=x^{k-1}f_{k}(x)} , where ${\displaystyle \Gamma }$ denotes the gamma function. Consequently, Laczkovich's finding is equivalent to the assertion that if x ≠<!-- ≠ --> §130131§ , {\displaystyle x\neq 0,} and ${\displaystyle x^{2}}$ is rational, with k ∈<!-- ∈ --> Q ∖<!-- ∖ --> { §184185§ , − §191192§ , − §198199§ , …<!-- … --> } {\displaystyle k\in \mathbb {Q} \smallsetminus \{0,-1,-2,\ldots \}} , then

x J k ( x ) J k −<!-- − --> §3637§ ( x ) ∉ Q . {\displaystyle {\frac {xJ_{k}(x)}{J_{k-1}(x)}}\notin \mathbb {Q} .}

Demonstration of the irrationality of e

• Proof that e is irrational
• Demonstration of the transcendence of π

References