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Proof that π is irrational
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<h2 id="lamberts-proof">Lambert's proof</h2> <p>In 1761, <a href="/facts/Johann_Heinrich_Lambert/CG3IHVB9">Johann Heinrich Lambert</a> proved that π {\displaystyle \pi } is irrational by first showing that this <a href="/facts/Continued_fraction/Wg1cMPxN">continued fraction</a> expansion holds: </p> tan ( x ) = x 1 − x 2 3 − x 2 5 − x 2 7 − ⋱ . {\displaystyle \tan(x)={\cfrac {x}{1-{\cfrac {x^{2}}{3-{\cfrac {x^{2}}{5-{\cfrac {x^{2}}{7-{}\ddots }}}}}}}}.} <p>Then Lambert proved that if x {\displaystyle x} is non-zero and rational, then this expression must be irrational. Since tan π 4 = 1 {\displaystyle \tan {\tfrac {\pi }{4}}=1} , it follows that π 4 {\displaystyle {\tfrac {\pi }{4}}} is irrational, and thus π {\displaystyle \pi } is also irrational.<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> A simplification of Lambert's proof is given below. </p> <h2 id="hermites-proof">Hermite's proof</h2> <p>Written in 1873, this proof uses the characterization of π {\displaystyle \pi } as the smallest positive number whose half is a <a href="/facts/Zero_of_a_function/mlK3dhoT">zero</a> of the <a href="/facts/Cosine/czej7ur0">cosine</a> function and it actually proves that π 2 {\displaystyle \pi ^{2}} is irrational.<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a><a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a> As in many proofs of irrationality, it is a <a href="/facts/Proof_by_contradiction/vHtFzFQ5">proof by contradiction</a>. </p><p>Consider the sequences of <a href="/facts/Real_function/bbNG3DnN">real functions</a> A n {\displaystyle A_{n}} and U n {\displaystyle U_{n}} for n ∈ N 0 {\displaystyle n\in \mathbb {N} _{0}} defined by: </p> A 0 ( x ) = sin ( x ) , A n + 1 ( x ) = ∫ 0 x y A n ( y ) d y U 0 ( x ) = sin ( x ) x , U n + 1 ( 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}}} <p>Using <a href="/facts/Mathematical_induction/KxAPqNrH">induction</a> we can prove that </p> A n ( x ) = x 2 n + 1 ( 2 n + 1 ) ! ! − x 2 n + 3 2 × ( 2 n + 3 ) ! ! + x 2 n + 5 2 × 4 × ( 2 n + 5 ) ! ! ∓ ⋯ U n ( x ) = 1 ( 2 n + 1 ) ! ! − x 2 2 × ( 2 n + 3 ) ! ! + x 4 2 × 4 × ( 2 n + 5 ) ! ! ∓ ⋯ {\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}}} <p>and therefore we have: </p> U n ( x ) = A n ( x ) x 2 n + 1 . {\displaystyle U_{n}(x)={\frac {A_{n}(x)}{x^{2n+1}}}.\,} <p>So </p> A n + 1 ( x ) x 2 n + 3 = U n + 1 ( x ) = − U n ′ ( x ) x = − 1 x d d x ( A n ( x ) x 2 n + 1 ) = − 1 x ( A n ′ ( x ) ⋅ x 2 n + 1 − ( 2 n + 1 ) x 2 n A n ( x ) x 2 ( 2 n + 1 ) ) = ( 2 n + 1 ) A n ( x ) − x A n ′ ( x ) x 2 n + 3 {\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}}} <p>which is equivalent to </p> A n + 1 ( x ) = ( 2 n + 1 ) A n ( x ) − x 2 A n − 1 ( x ) . {\displaystyle A_{n+1}(x)=(2n+1)A_{n}(x)-x^{2}A_{n-1}(x).\,} <p>Using the definition of the sequence and employing induction we can show that </p> A n ( x ) = P n ( x 2 ) sin ( x ) + x Q n ( x 2 ) cos ( x ) , {\displaystyle A_{n}(x)=P_{n}(x^{2})\sin(x)+xQ_{n}(x^{2})\cos(x),\,} <p>where P n {\displaystyle P_{n}} and Q n {\displaystyle Q_{n}} are polynomial functions with integer coefficients and the degree of P n {\displaystyle P_{n}} is smaller than or equal to ⌊ 1 2 n ⌋ . {\displaystyle {\bigl \lfloor }{\tfrac {1}{2}}n{\bigr \rfloor }.} In particular, A n ( 1 2 π ) = P n ( 1 4 π 2 ) . {\displaystyle A_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=P_{n}{\bigl (}{\tfrac {1}{4}}\pi ^{2}{\bigr )}.} </p><p>Hermite also gave a closed expression for the function A n , {\displaystyle A_{n},} namely </p> A n ( x ) = x 2 n + 1 2 n n ! ∫ 0 1 ( 1 − z 2 ) n cos ( x z ) d z . {\displaystyle A_{n}(x)={\frac {x^{2n+1}}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z.\,} <p>He did not justify this assertion, but it can be proved easily. First of all, this assertion is equivalent to </p> 1 2 n n ! ∫ 0 1 ( 1 − z 2 ) n cos ( x z ) d z = A n ( x ) x 2 n + 1 = 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).} <p>Proceeding by induction, take n = 0. {\displaystyle n=0.} </p> ∫ 0 1 cos ( x z ) d z = sin ( x ) x = U 0 ( x ) {\displaystyle \int _{0}^{1}\cos(xz)\,\mathrm {d} z={\frac {\sin(x)}{x}}=U_{0}(x)} <p>and, for the inductive step, consider any <a href="/facts/Natural_number/u65og8JE">natural number</a> n . {\displaystyle n.} If </p> 1 2 n n ! ∫ 0 1 ( 1 − z 2 ) 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),} <p>then, using <a href="/facts/Integration_by_parts/aV6JgBNd">integration by parts</a> and <a href="/facts/Leibniz_integral_rule/bFKo8h0I">Leibniz's rule</a>, one gets </p> 1 2 n + 1 ( n + 1 ) ! ∫ 0 1 ( 1 − z 2 ) n + 1 cos ( x z ) d z = 1 2 n + 1 ( n + 1 ) ! ( ( 1 − z 2 ) n + 1 sin ( x z ) x | z = 0 z = 1 ⏞ = 0 + ∫ 0 1 2 ( n + 1 ) ( 1 − z 2 ) n z sin ( x z ) x d z ) = 1 x ⋅ 1 2 n n ! ∫ 0 1 ( 1 − z 2 ) n z sin ( x z ) d z = − 1 x ⋅ d d x ( 1 2 n n ! ∫ 0 1 ( 1 − z 2 ) n cos ( x z ) d z ) = − U n ′ ( x ) x = U n + 1 ( x ) . {\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}}} <p>If 1 4 π 2 = p / q , {\displaystyle {\tfrac {1}{4}}\pi ^{2}=p/q,} with p {\displaystyle p} and q {\displaystyle q} in N {\displaystyle \mathbb {N} } , then, since the coefficients of P n {\displaystyle P_{n}} are integers and its degree is smaller than or equal to ⌊ 1 2 n ⌋ , {\displaystyle {\bigl \lfloor }{\tfrac {1}{2}}n{\bigr \rfloor },} q ⌊ n / 2 ⌋ P n ( 1 4 π 2 ) {\displaystyle q^{\lfloor n/2\rfloor }P_{n}{\bigl (}{\tfrac {1}{4}}\pi ^{2}{\bigr )}} is some integer N . {\displaystyle N.} In other words, </p> N = q ⌊ n / 2 ⌋ A n ( 1 2 π ) = q ⌊ n / 2 ⌋ 1 2 n n ! ( p q ) n + 1 2 ∫ 0 1 ( 1 − z 2 ) n cos ( 1 2 π z ) d z . {\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.} <p>But this number is clearly greater than 0. {\displaystyle 0.} On the other hand, the limit of this quantity as n {\displaystyle n} goes to infinity is zero, and so, if n {\displaystyle n} is large enough, N < 1. {\displaystyle N<1.} Thereby, a contradiction is reached. </p><p>Hermite did not present his proof as an end in itself but as an afterthought within his search for a proof of the <a href="/facts/Transcendental_number/IrB7JppM">transcendence</a> of π . {\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 e {\displaystyle e} <a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a>). </p><p>Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, A n ( x ) {\displaystyle A_{n}(x)} is the "residue" (or "remainder") of Lambert's continued fraction for tan x . {\displaystyle \tan x.} <a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a> </p> <h2 id="cartwrights-proof">Cartwright's proof</h2> <p><a href="/facts/Harold_Jeffreys/yWPSFpPB">Harold Jeffreys</a> wrote that this proof was set as an example in an exam at <a href="/facts/Cambridge_University/0la7aX9o">Cambridge University</a> in 1945 by <a href="/facts/Mary_Cartwright/5GPGzsX2">Mary Cartwright</a>, but that she had not traced its origin.<a class="footnote-ref" id="fnref:7" href="#fn:7"><sup>7</sup></a> It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University.<a class="footnote-ref" id="fnref:8" href="#fn:8"><sup>8</sup></a> </p><p>Consider the integrals </p> I n ( x ) = ∫ − 1 1 ( 1 − z 2 ) n cos ( x z ) d z , {\displaystyle I_{n}(x)=\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz,} <p>where n {\displaystyle n} is a non-negative integer. </p><p>Two <a href="/facts/Integration_by_parts/aV6JgBNd">integrations by parts</a> give the <a href="/facts/Recurrence_relation/JolXC71M">recurrence relation</a> </p> x 2 I n ( x ) = 2 n ( 2 n − 1 ) I n − 1 ( x ) − 4 n ( n − 1 ) I n − 2 ( x ) . ( n ≥ 2 ) {\displaystyle x^{2}I_{n}(x)=2n(2n-1)I_{n-1}(x)-4n(n-1)I_{n-2}(x).\qquad (n\geq 2)} <p>If </p> J n ( x ) = x 2 n + 1 I n ( x ) , {\displaystyle J_{n}(x)=x^{2n+1}I_{n}(x),} <p>then this becomes </p> J n ( x ) = 2 n ( 2 n − 1 ) J n − 1 ( x ) − 4 n ( n − 1 ) x 2 J n − 2 ( x ) . {\displaystyle J_{n}(x)=2n(2n-1)J_{n-1}(x)-4n(n-1)x^{2}J_{n-2}(x).} <p>Furthermore, J 0 ( x ) = 2 sin x {\displaystyle J_{0}(x)=2\sin x} and J 1 ( x ) = − 4 x cos x + 4 sin x . {\displaystyle J_{1}(x)=-4x\cos x+4\sin x.} Hence for all n ∈ Z + , {\displaystyle n\in \mathbb {Z} _{+},} </p> J n ( x ) = x 2 n + 1 I n ( x ) = n ! ( P n ( x ) sin ( x ) + Q n ( 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 )},} <p>where P n ( x ) {\displaystyle P_{n}(x)} and Q n ( x ) {\displaystyle Q_{n}(x)} are <a href="/facts/Polynomial/Lzak8VVx">polynomials</a> of degree ≤ n , {\displaystyle \leq n,} and with <a href="/facts/Integer/PUwwrawx">integer</a> coefficients (depending on n {\displaystyle n} ). </p><p>Take x = 1 2 π , {\displaystyle x={\tfrac {1}{2}}\pi ,} and suppose if possible that 1 2 π = a / b {\displaystyle {\tfrac {1}{2}}\pi =a/b} where a {\displaystyle a} and b {\displaystyle b} are natural numbers (i.e., assume that π {\displaystyle \pi } is rational). Then </p> a 2 n + 1 n ! I n ( 1 2 π ) = P n ( 1 2 π ) b 2 n + 1 . {\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}.} <p>The right side is an integer. But 0 < I n ( 1 2 π ) < 2 {\displaystyle 0<I_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}<2} since the interval [ − 1 , 1 ] {\displaystyle [-1,1]} has length 2 {\displaystyle 2} and the function being integrated takes only values between 0 {\displaystyle 0} and 1. {\displaystyle 1.} On the other hand, </p> a 2 n + 1 n ! → 0 as n → ∞ . {\displaystyle {\frac {a^{2n+1}}{n!}}\to 0\quad {\text{ as }}n\to \infty .} <p>Hence, for sufficiently large n {\displaystyle n} </p> 0 < a 2 n + 1 I n ( π 2 ) n ! < 1 , {\displaystyle 0<{\frac {a^{2n+1}I_{n}\left({\frac {\pi }{2}}\right)}{n!}}<1,} <p>that is, we could find an integer between 0 {\displaystyle 0} and 1. {\displaystyle 1.} That is the contradiction that follows from the assumption that π {\displaystyle \pi } is rational. </p><p>This proof is similar to Hermite's proof. Indeed, </p> J n ( x ) = x 2 n + 1 ∫ − 1 1 ( 1 − z 2 ) n cos ( x z ) d z = 2 x 2 n + 1 ∫ 0 1 ( 1 − z 2 ) n cos ( x z ) d z = 2 n + 1 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}}} <p>However, it is clearly simpler. This is achieved by omitting the inductive definition of the functions A n {\displaystyle A_{n}} and taking as a starting point their expression as an integral. </p> <h2 id="nivens-proof">Niven's proof</h2> <p>This proof uses the characterization of π {\displaystyle \pi } as the smallest positive <a href="/facts/Zero_of_a_function/mlK3dhoT">zero</a> of the <a href="/facts/Sine/gQ6LXK8z">sine</a> function.<a class="footnote-ref" id="fnref:9" href="#fn:9"><sup>9</sup></a> </p><p>Suppose that π {\displaystyle \pi } is rational, i.e. π = a / b {\displaystyle \pi =a/b} for some integers a {\displaystyle a} and b {\displaystyle b} which may be taken <a href="/facts/Without_loss_of_generality/gVP4dmEe">without loss of generality</a> to both be positive. Given any positive integer n , {\displaystyle n,} we define the polynomial function: </p> f ( x ) = x n ( a − b x ) n n ! {\displaystyle f(x)={\frac {x^{n}(a-bx)^{n}}{n!}}} <p>and, for each x ∈ R {\displaystyle x\in \mathbb {R} } let </p> F ( x ) = f ( x ) − f ″ ( x ) + f ( 4 ) ( x ) + ⋯ + ( − 1 ) n f ( 2 n ) ( x ) . {\displaystyle F(x)=f(x)-f''(x)+f^{(4)}(x)+\cdots +(-1)^{n}f^{(2n)}(x).} <p>Claim 1: F ( 0 ) + F ( π ) {\displaystyle F(0)+F(\pi )} is an integer. </p><p>Proof: Expanding f {\displaystyle f} as a sum of monomials, the coefficient of x k {\displaystyle x^{k}} is a number of the form c k / n ! {\displaystyle c_{k}/n!} where c k {\displaystyle c_{k}} is an integer, which is 0 {\displaystyle 0} if k < n . {\displaystyle k<n.} Therefore, f ( k ) ( 0 ) {\displaystyle f^{(k)}(0)} is 0 {\displaystyle 0} when k < n {\displaystyle k<n} and it is equal to ( k ! / n ! ) c k {\displaystyle (k!/n!)c_{k}} if n ≤ k ≤ 2 n {\displaystyle n\leq k\leq 2n} ; in each case, f ( k ) ( 0 ) {\displaystyle f^{(k)}(0)} is an integer and therefore F ( 0 ) {\displaystyle F(0)} is an integer. </p><p>On the other hand, f ( π − x ) = f ( x ) {\displaystyle f(\pi -x)=f(x)} and so ( − 1 ) k f ( k ) ( π − x ) = f ( k ) ( x ) {\displaystyle (-1)^{k}f^{(k)}(\pi -x)=f^{(k)}(x)} for each non-negative integer k . {\displaystyle k.} In particular, ( − 1 ) k f ( k ) ( π ) = f ( k ) ( 0 ) . {\displaystyle (-1)^{k}f^{(k)}(\pi )=f^{(k)}(0).} Therefore, f ( k ) ( π ) {\displaystyle f^{(k)}(\pi )} is also an integer and so F ( π ) {\displaystyle F(\pi )} is an integer (in fact, it is easy to see that F ( π ) = F ( 0 ) {\displaystyle F(\pi )=F(0)} ). Since F ( 0 ) {\displaystyle F(0)} and F ( π ) {\displaystyle F(\pi )} are integers, so is their sum. </p><p>Claim 2: </p> ∫ 0 π f ( x ) sin ( x ) d x = F ( 0 ) + F ( π ) {\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx=F(0)+F(\pi )} <p>Proof: Since f ( 2 n + 2 ) {\displaystyle f^{(2n+2)}} is the zero polynomial, we have </p> F ″ + F = f . {\displaystyle F''+F=f.} <p>The <a href="/facts/Table_of_derivatives/9q1b7Ev6">derivatives</a> of the <a href="/facts/Sine/gQ6LXK8z">sine</a> and <a href="/facts/Cosine/czej7ur0">cosine</a> function are given by sin' = cos and cos' = −sin. Hence the <a href="/facts/Product_rule/MyKxsP5t">product rule</a> implies </p> ( F ′ ⋅ sin − F ⋅ cos ) ′ = f ⋅ sin {\displaystyle (F'\cdot \sin {}-F\cdot \cos {})'=f\cdot \sin } <p>By the <a href="/facts/Fundamental_theorem_of_calculus/pHeSGtUx">fundamental theorem of calculus</a> </p> ∫ 0 π f ( x ) sin ( x ) d x = ( F ′ ( x ) sin x − F ( x ) cos x ) | 0 π . {\displaystyle \left.\int _{0}^{\pi }f(x)\sin(x)\,dx={\bigl (}F'(x)\sin x-F(x)\cos x{\bigr )}\right|_{0}^{\pi }.} <p>Since sin 0 = sin π = 0 {\displaystyle \sin 0=\sin \pi =0} and cos 0 = − cos π = 1 {\displaystyle \cos 0=-\cos \pi =1} (here we use the above-mentioned characterization of π {\displaystyle \pi } as a zero of the sine function), Claim 2 follows. </p><p>Conclusion: Since f ( x ) > 0 {\displaystyle f(x)>0} and sin x > 0 {\displaystyle \sin x>0} for 0 < x < π {\displaystyle 0<x<\pi } (because π {\displaystyle \pi } is the <i>smallest</i> positive zero of the sine function), Claims 1 and 2 show that F ( 0 ) + F ( π ) {\displaystyle F(0)+F(\pi )} is a <i>positive</i> integer. Since 0 ≤ x ( a − b x ) ≤ π a {\displaystyle 0\leq x(a-bx)\leq \pi a} and 0 ≤ sin x ≤ 1 {\displaystyle 0\leq \sin x\leq 1} for 0 ≤ x ≤ π , {\displaystyle 0\leq x\leq \pi ,} we have, by the original definition of f , {\displaystyle f,} </p> ∫ 0 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!}}} <p>which is smaller than 1 {\displaystyle 1} for large n , {\displaystyle n,} hence F ( 0 ) + F ( π ) < 1 {\displaystyle F(0)+F(\pi )<1} for these n , {\displaystyle n,} by Claim 2. This is impossible for the positive integer F ( 0 ) + F ( π ) . {\displaystyle F(0)+F(\pi ).} This shows that the original assumption that π {\displaystyle \pi } is rational leads to a contradiction, which concludes the proof. </p><p>The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula </p> ∫ 0 π f ( x ) sin ( x ) d x = ∑ j = 0 n ( − 1 ) j ( f ( 2 j ) ( π ) + f ( 2 j ) ( 0 ) ) + ( − 1 ) n + 1 ∫ 0 π f ( 2 n + 2 ) ( x ) sin ( x ) d x , {\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,} <p>which is obtained by 2 n + 2 {\displaystyle 2n+2} <a href="/facts/Integration_by_parts/aV6JgBNd">integrations by parts</a>. Claim 2 essentially establishes this formula, where the use of F {\displaystyle F} hides the iterated integration by parts. The last integral vanishes because f ( 2 n + 2 ) {\displaystyle f^{(2n+2)}} is the zero polynomial. Claim 1 shows that the remaining sum is an integer. </p><p>Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight.<a class="footnote-ref" id="fnref:10" href="#fn:10"><sup>10</sup></a> In fact, </p> J n ( x ) = x 2 n + 1 ∫ − 1 1 ( 1 − z 2 ) n cos ( x z ) d z = ∫ − 1 1 ( x 2 − ( x z ) 2 ) 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}}} <p>Therefore, the <a href="/facts/Integration_by_substitution/NvXfTTaf">substitution</a> x z = y {\displaystyle xz=y} turns this integral into </p> ∫ − x x ( x 2 − y 2 ) n cos ( y ) d y . {\displaystyle \int _{-x}^{x}(x^{2}-y^{2})^{n}\cos(y)\,dy.} <p>In particular, </p> J n ( π 2 ) = ∫ − π / 2 π / 2 ( π 2 4 − y 2 ) n cos ( y ) d y = ∫ 0 π ( π 2 4 − ( y − π 2 ) 2 ) n cos ( y − π 2 ) d y = ∫ 0 π y n ( π − y ) n sin ( y ) d y = n ! b n ∫ 0 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}}} <p>Another connection between the proofs lies in the fact that Hermite already mentions<a class="footnote-ref" id="fnref:11" href="#fn:11"><sup>11</sup></a> that if f {\displaystyle f} is a polynomial function and </p> F = f − f ( 2 ) + f ( 4 ) ∓ ⋯ , {\displaystyle F=f-f^{(2)}+f^{(4)}\mp \cdots ,} <p>then </p> ∫ f ( x ) sin ( x ) d x = F ′ ( x ) sin ( x ) − F ( x ) cos ( x ) + C , {\displaystyle \int f(x)\sin(x)\,dx=F'(x)\sin(x)-F(x)\cos(x)+C,} <p>from which it follows that </p> ∫ 0 π f ( x ) sin ( x ) d x = F ( π ) + F ( 0 ) . {\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx=F(\pi )+F(0).} <h2 id="bourbakis-proof">Bourbaki's proof</h2> <p><a href="/facts/Nicolas_Bourbaki/qRJhs9nr">Bourbaki</a>'s proof is outlined as an exercise in his <a href="/facts/Calculus/MEmUTYoz">calculus</a> treatise.<a class="footnote-ref" id="fnref:12" href="#fn:12"><sup>12</sup></a> For each natural number <i>b</i> and each non-negative integer n , {\displaystyle n,} define </p> A n ( b ) = b n ∫ 0 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.} <p>Since A n ( b ) {\displaystyle A_{n}(b)} is the integral of a function defined on [ 0 , π ] {\displaystyle [0,\pi ]} that takes the value 0 {\displaystyle 0} at 0 {\displaystyle 0} and π {\displaystyle \pi } and which is greater than 0 {\displaystyle 0} otherwise, A n ( b ) > 0. {\displaystyle A_{n}(b)>0.} Besides, for each natural number b , {\displaystyle b,} A n ( b ) < 1 {\displaystyle A_{n}(b)<1} if n {\displaystyle n} is large enough, because </p> x ( π − x ) ≤ ( π 2 ) 2 {\displaystyle x(\pi -x)\leq \left({\frac {\pi }{2}}\right)^{2}} <p>and therefore </p> A n ( b ) ≤ π b n 1 n ! ( π 2 ) 2 n = π ( b π 2 / 4 ) 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!}}.} <p>On the other hand, <a href="/facts/Integration_by_parts/aV6JgBNd">repeated integration by parts</a> allows us to deduce that, if a {\displaystyle a} and b {\displaystyle b} are natural numbers such that π = a / b {\displaystyle \pi =a/b} and f {\displaystyle f} is the polynomial function from [ 0 , π ] {\displaystyle [0,\pi ]} into R {\displaystyle \mathbb {R} } defined by </p> f ( x ) = x n ( a − b x ) n n ! , {\displaystyle f(x)={\frac {x^{n}(a-bx)^{n}}{n!}},} <p>then: </p> A n ( b ) = ∫ 0 π f ( x ) sin ( x ) d x = [ − f ( x ) cos ( x ) ] x = 0 x = π − [ − f ′ ( x ) sin ( x ) ] x = 0 x = π + ⋯ ± [ f ( 2 n ) ( x ) cos ( x ) ] x = 0 x = π ± ∫ 0 π f ( 2 n + 1 ) ( 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}}} <p>This last integral is 0 , {\displaystyle 0,} since f ( 2 n + 1 ) {\displaystyle f^{(2n+1)}} is the null function (because f {\displaystyle f} is a polynomial function of degree 2 n {\displaystyle 2n} ). Since each function f ( k ) {\displaystyle f^{(k)}} (with 0 ≤ k ≤ 2 n {\displaystyle 0\leq k\leq 2n} ) takes integer values at 0 {\displaystyle 0} and π {\displaystyle \pi } and since the same thing happens with the sine and the cosine functions, this proves that A n ( b ) {\displaystyle A_{n}(b)} is an integer. Since it is also greater than 0 , {\displaystyle 0,} it must be a natural number. But it was also proved that A n ( b ) < 1 {\displaystyle A_{n}(b)<1} if n {\displaystyle n} is large enough, thereby reaching a <a href="/facts/Proof_by_contradiction/vHtFzFQ5">contradiction</a>. </p><p>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 ) {\displaystyle A_{n}(b)} are integers. </p> <h2 id="laczkovichs-proof">Laczkovich's proof</h2> <p><a href="/facts/Mikl%C3%B3s_Laczkovich/NPJAshyl">Miklós Laczkovich</a>'s proof is a simplification of Lambert's original proof.<a class="footnote-ref" id="fnref:13" href="#fn:13"><sup>13</sup></a> He considers the functions </p> f k ( x ) = 1 − x 2 k + x 4 2 ! k ( k + 1 ) − x 6 3 ! k ( k + 1 ) ( k + 2 ) + ⋯ ( k ∉ { 0 , − 1 , − 2 , … } ) . {\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 \}).} <p>These functions are clearly defined for any real number x . {\displaystyle x.} Besides </p> f 1 / 2 ( x ) = cos ( 2 x ) , {\displaystyle f_{1/2}(x)=\cos(2x),} f 3 / 2 ( x ) = sin ( 2 x ) 2 x . {\displaystyle f_{3/2}(x)={\frac {\sin(2x)}{2x}}.} <p>Claim 1: The following <a href="/facts/Recurrence_relation/JolXC71M">recurrence relation</a> holds for any real number x {\displaystyle x} : </p> x 2 k ( k + 1 ) f k + 2 ( x ) = f k + 1 ( x ) − f k ( x ) . {\displaystyle {\frac {x^{2}}{k(k+1)}}f_{k+2}(x)=f_{k+1}(x)-f_{k}(x).} <p>Proof: This can be proved by comparing the coefficients of the powers of x . {\displaystyle x.} </p><p>Claim 2: For each real number x , {\displaystyle x,} </p> lim k → + ∞ f k ( x ) = 1. {\displaystyle \lim _{k\to +\infty }f_{k}(x)=1.} <p>Proof: In fact, the sequence x 2 n / n ! {\displaystyle x^{2n}/n!} is bounded (since it converges to 0 {\displaystyle 0} ) and if C {\displaystyle C} is an upper bound and if k > 1 , {\displaystyle k>1,} then </p> | f k ( x ) − 1 | ⩽ ∑ n = 1 ∞ C k n = C 1 / k 1 − 1 / k = C k − 1 . {\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}}.} <p>Claim 3: If x ≠ 0 , {\displaystyle x\neq 0,} x 2 {\displaystyle x^{2}} is rational, and k ∈ Q ∖ { 0 , − 1 , − 2 , … } {\displaystyle k\in \mathbb {Q} \smallsetminus \{0,-1,-2,\ldots \}} then </p> f k ( x ) ≠ 0 and f k + 1 ( x ) f k ( x ) ∉ Q . {\displaystyle f_{k}(x)\neq 0\quad {\text{ and }}\quad {\frac {f_{k+1}(x)}{f_{k}(x)}}\notin \mathbb {Q} .} <p>Proof: Otherwise, there would be a number y ≠ 0 {\displaystyle y\neq 0} and integers a {\displaystyle a} and b {\displaystyle b} such that f k ( x ) = a y {\displaystyle f_{k}(x)=ay} and f k + 1 ( x ) = b y . {\displaystyle f_{k+1}(x)=by.} To see why, take y = f k + 1 ( x ) , {\displaystyle y=f_{k+1}(x),} a = 0 , {\displaystyle a=0,} and b = 1 {\displaystyle b=1} if f k ( x ) = 0 {\displaystyle f_{k}(x)=0} ; otherwise, choose integers a {\displaystyle a} and b {\displaystyle b} such that f k + 1 ( x ) / f k ( x ) = b / a {\displaystyle f_{k+1}(x)/f_{k}(x)=b/a} and define y = f k ( x ) / a = f k + 1 ( x ) / b . {\displaystyle y=f_{k}(x)/a=f_{k+1}(x)/b.} In each case, y {\displaystyle y} cannot be 0 , {\displaystyle 0,} because otherwise it would follow from claim 1 that each f k + n ( x ) {\displaystyle f_{k+n}(x)} ( n ∈ N {\displaystyle n\in \mathbb {N} } ) would be 0 , {\displaystyle 0,} which would contradict claim 2. Now, take a natural number c {\displaystyle c} such that all three numbers b c / k , {\displaystyle bc/k,} c k / x 2 , {\displaystyle ck/x^{2},} and c / x 2 {\displaystyle c/x^{2}} are integers and consider the sequence </p> g n = { f k ( x ) n = 0 c n k ( k + 1 ) ⋯ ( k + n − 1 ) f k + n ( x ) n ≠ 0 {\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}}} <p>Then </p> g 0 = f k ( x ) = a y ∈ Z y and g 1 = c k f k + 1 ( x ) = b c k y ∈ Z y . {\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.} <p>On the other hand, it follows from claim 1 that </p> g n + 2 = c n + 2 x 2 k ( k + 1 ) ⋯ ( k + n − 1 ) ⋅ x 2 ( k + n ) ( k + n + 1 ) f k + n + 2 ( x ) = c n + 2 x 2 k ( k + 1 ) ⋯ ( k + n − 1 ) f k + n + 1 ( x ) − c n + 2 x 2 k ( k + 1 ) ⋯ ( k + n − 1 ) f k + n ( x ) = c ( k + n ) x 2 g n + 1 − c 2 x 2 g n = ( c k x 2 + c x 2 n ) g n + 1 − c 2 x 2 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}}} <p>which is a linear combination of g n + 1 {\displaystyle g_{n+1}} and g n {\displaystyle g_{n}} with integer coefficients. Therefore, each g n {\displaystyle g_{n}} is an integer multiple of y . {\displaystyle y.} Besides, it follows from claim 2 that each g n {\displaystyle g_{n}} is greater than 0 {\displaystyle 0} (and therefore that g n ≥ | y | {\displaystyle g_{n}\geq |y|} ) if n {\displaystyle n} is large enough and that the sequence of all g n {\displaystyle g_{n}} converges to 0. {\displaystyle 0.} But a sequence of numbers greater than or equal to | y | {\displaystyle |y|} cannot converge to 0. {\displaystyle 0.} </p><p>Since f 1 / 2 ( 1 4 π ) = cos 1 2 π = 0 , {\displaystyle f_{1/2}({\tfrac {1}{4}}\pi )=\cos {\tfrac {1}{2}}\pi =0,} it follows from claim 3 that 1 16 π 2 {\displaystyle {\tfrac {1}{16}}\pi ^{2}} is irrational and therefore that π {\displaystyle \pi } is irrational. </p><p>On the other hand, since </p> tan x = sin x cos x = x f 3 / 2 ( x / 2 ) f 1 / 2 ( x / 2 ) , {\displaystyle \tan x={\frac {\sin x}{\cos x}}=x{\frac {f_{3/2}(x/2)}{f_{1/2}(x/2)}},} <p>another consequence of Claim 3 is that, if x ∈ Q ∖ { 0 } , {\displaystyle x\in \mathbb {Q} \smallsetminus \{0\},} then tan x {\displaystyle \tan x} is irrational. </p><p>Laczkovich's proof is really about the <a href="/facts/Hypergeometric_function/yTEC6NBJ">hypergeometric function</a>. In fact, f k ( x ) = 0 F 1 ( k − x 2 ) {\displaystyle f_{k}(x)={}_{0}F_{1}(k-x^{2})} and <a href="/facts/Carl_Friedrich_Gauss/f4eSMBD7">Gauss</a> found a continued fraction expansion of the hypergeometric function using its <a href="/facts/Functional_equation/4ovU4CAt">functional equation</a>.<a class="footnote-ref" id="fnref:14" href="#fn:14"><sup>14</sup></a> 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. </p><p>Laczkovich's result can also be expressed in <a href="/facts/Bessel_function/hSyvFPqC">Bessel functions of the first kind J ν ( x ) {\displaystyle J_{\nu }(x)} </a>. In fact, Γ ( k ) J k − 1 ( 2 x ) = x k − 1 f k ( x ) {\displaystyle \Gamma (k)J_{k-1}(2x)=x^{k-1}f_{k}(x)} (where Γ {\displaystyle \Gamma } is the <a href="/facts/Gamma_function/SjGQcnyw">gamma function</a>). So Laczkovich's result is equivalent to: If x ≠ 0 , {\displaystyle x\neq 0,} x 2 {\displaystyle x^{2}} is rational, and k ∈ Q ∖ { 0 , − 1 , − 2 , … } {\displaystyle k\in \mathbb {Q} \smallsetminus \{0,-1,-2,\ldots \}} then </p> x J k ( x ) J k − 1 ( x ) ∉ Q . {\displaystyle {\frac {xJ_{k}(x)}{J_{k-1}(x)}}\notin \mathbb {Q} .} <h2 id="see-also">See also</h2> <ul> <li>Mathematics portal</li></ul> <ul><li><a href="/facts/Proof_that_e_is_irrational/FtB9sNP5">Proof that e is irrational</a></li> <li><a href="/facts/Proof_that_%CF%80_is_transcendental/BafvlpWm">Proof that π is transcendental</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Lindemann, Ferdinand von (2004) [1882], "Ueber die Zahl π", in Berggren, Lennart; Borwein, Jonathan M.; Borwein, Peter B. (eds.), Pi, a source book (3rd ed.), New York: Springer-Verlag, pp. 194–225, ISBN 0-387-20571-3. <a href="0-387-20571-3" target="_blank">0-387-20571-3</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>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. (eds.), Pi, a source book (3rd ed.), New York: Springer-Verlag, pp. 129–140, ISBN 0-387-20571-3. <a href="0-387-20571-3" target="_blank">0-387-20571-3</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>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: 303–311. <a href="/wiki/Charles_Hermite" target="_blank">/wiki/Charles_Hermite</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>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: 342–344. <a href="/wiki/Charles_Hermite" target="_blank">/wiki/Charles_Hermite</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>Hermite, Charles (1912) [1873]. "Sur la fonction exponentielle". In Picard, Émile (ed.). Œuvres de Charles Hermite (in French). Vol. III. Gauthier-Villars. pp. 150–181. <a href="/wiki/Charles_Hermite" target="_blank">/wiki/Charles_Hermite</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>Zhou, Li (2011). "Irrationality proofs à la Hermite". The Mathematical Gazette. 95 (534): 407–413. arXiv:0911.1929. doi:10.1017/S0025557200003491. S2CID 115175505. <a href="/wiki/ArXiv_(identifier)" target="_blank">/wiki/ArXiv_(identifier)</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> <li id="fn:7"><p>Jeffreys, Harold (1973), Scientific Inference (3rd ed.), Cambridge University Press, p. 268, ISBN 0-521-08446-6 <a href="0-521-08446-6" target="_blank">0-521-08446-6</a> <a href="#fnref:7" class="footnote-back-ref">↩</a></p></li> <li id="fn:8"><p>"Department of Pure Mathematics and Mathematical Statistics". www.dpmms.cam.ac.uk. Retrieved 2022-04-19. <a href="http://www.dpmms.cam.ac.uk/study/IA/AnalysisI/" target="_blank">http://www.dpmms.cam.ac.uk/study/IA/AnalysisI/</a> <a href="#fnref:8" class="footnote-back-ref">↩</a></p></li> <li id="fn:9"><p>Niven, Ivan (1947), "A simple proof that π is irrational" (PDF), Bulletin of the American Mathematical Society, vol. 53, no. 6, p. 509, doi:10.1090/s0002-9904-1947-08821-2 <a href="/wiki/Ivan_Niven" target="_blank">/wiki/Ivan_Niven</a> <a href="#fnref:9" class="footnote-back-ref">↩</a></p></li> <li id="fn:10"><p>Zhou, Li (2011). "Irrationality proofs à la Hermite". The Mathematical Gazette. 95 (534): 407–413. arXiv:0911.1929. doi:10.1017/S0025557200003491. S2CID 115175505. <a href="/wiki/ArXiv_(identifier)" target="_blank">/wiki/ArXiv_(identifier)</a> <a href="#fnref:10" class="footnote-back-ref">↩</a></p></li> <li id="fn:11"><p>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: 303–311. <a href="/wiki/Charles_Hermite" target="_blank">/wiki/Charles_Hermite</a> <a href="#fnref:11" class="footnote-back-ref">↩</a></p></li> <li id="fn:12"><p>Bourbaki, Nicolas (1949), Fonctions d'une variable réelle, chap. I–II–III, Actualités Scientifiques et Industrielles (in French), vol. 1074, Hermann, pp. 137–138 <a href="/wiki/Nicolas_Bourbaki" target="_blank">/wiki/Nicolas_Bourbaki</a> <a href="#fnref:12" class="footnote-back-ref">↩</a></p></li> <li id="fn:13"><p>Laczkovich, Miklós (1997), "On Lambert's proof of the irrationality of π", American Mathematical Monthly, vol. 104, no. 5, pp. 439–443, doi:10.2307/2974737, JSTOR 2974737 <a href="/wiki/Mikl%C3%B3s_Laczkovich" target="_blank">/wiki/Mikl%C3%B3s_Laczkovich</a> <a href="#fnref:13" class="footnote-back-ref">↩</a></p></li> <li id="fn:14"><p>Gauss, Carl Friedrich (1811–1813), "Disquisitiones generales circa seriem infinitam", Commentationes Societatis Regiae Scientiarum Gottingensis Recentiores (in Latin), 2 <a href="/wiki/Carl_Friedrich_Gauss" target="_blank">/wiki/Carl_Friedrich_Gauss</a> <a href="#fnref:14" class="footnote-back-ref">↩</a></p></li> </ol>