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 a / b , {\displaystyle a/b,} where a {\displaystyle a} and b {\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 (pseudonym of a group of mathematicians). 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 {\displaystyle \pi } is not just irrational, but transcendental as well.
Lambert's proof
In 1768, Johann Heinrich Lambert published a proof that π {\displaystyle \pi } is irrational by first showing that this continued fraction expansion holds:
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 }}}}}}}}.}
Then, Lambert proved that if x {\displaystyle x} is non-zero and rational, then this expression must be irrational. Assuming x = a / b {\displaystyle x=a/b} is a rational number (with a , b ∈ Z + {\displaystyle a,b\in \mathbb {Z} ^{+}} ), the continued fraction for tan ( a / b ) {\displaystyle \tan(a/b)} takes the form:
tan ( a b ) = a b − a 2 3 b − a 2 5 b − … {\displaystyle \tan \left({\frac {a}{b}}\right)={\cfrac {a}{b-{\cfrac {a^{2}}{3b-{\cfrac {a^{2}}{5b-\dots }}}}}}}
Let v n = C n D n {\displaystyle v_{n}={\frac {C_{n}}{D_{n}}}} represent the n {\displaystyle n} -th tail/remainder of this continued fraction, where C n , D n ∈ Z + {\displaystyle C_{n},D_{n}\in \mathbb {Z} ^{+}} . The recursive relation between consecutive tails is given by:
v n = a 2 ( 2 n + 1 ) b − v n + 1 {\displaystyle v_{n}={\cfrac {a^{2}}{(2n+1)b-v_{n+1}}}}
Substituting v n = C n D n {\displaystyle v_{n}={\frac {C_{n}}{D_{n}}}} and v n + 1 = C n + 1 D n + 1 {\displaystyle v_{n+1}={\frac {C_{n+1}}{D_{n+1}}}} into this recurrence relation gives:
C n D n = a 2 ( 2 n + 1 ) b − C n + 1 D n + 1 = a 2 D n + 1 ( 2 n + 1 ) b D n + 1 − C n + 1 {\displaystyle {\frac {C_{n}}{D_{n}}}={\frac {a^{2}}{(2n+1)b-{\frac {C_{n+1}}{D_{n+1}}}}}={\frac {a^{2}D_{n+1}}{(2n+1)bD_{n+1}-C_{n+1}}}}
Equating the numerators on both sides yields:
C n = a 2 D n + 1 ⟹ D n + 1 = C n a 2 {\displaystyle C_{n}=a^{2}D_{n+1}\implies D_{n+1}={\frac {C_{n}}{a^{2}}}}
Deep in the expansion, since 0 < v n < 1 {\displaystyle 0<v_{n}<1} , we have C n < D n {\displaystyle C_{n}<D_{n}} . Using C n = a 2 D n + 1 {\displaystyle C_{n}=a^{2}D_{n+1}} and assuming a ≥ 1 {\displaystyle a\geq 1} , this leads to D n > D n + 1 {\displaystyle D_{n}>D_{n+1}} , setting up an infinite descent of positive integers ( D n > D n + 1 > D n + 2 > ⋯ > 0 {\displaystyle D_{n}>D_{n+1}>D_{n+2}>\dots >0} ), which is impossible because subsequent terms are always decreasing from a finite starting term yet always greater than zero. Hence, the assumption that x {\displaystyle x} is rational is false. Since tan ( π 4 ) = 1 {\displaystyle \tan \left({\frac {\pi }{4}}\right)=1} , it follows that π 4 {\displaystyle {\frac {\pi }{4}}} is irrational, and thus π {\displaystyle \pi } is also irrational. Lambert obtained his formula for the continued fraction expression of tan(x) via induction. An elementary derivation of the continued fraction for the tangent function can be constructed by matching its power series expansion to an infinite continued fraction with unknown coefficients.
Statement of Lambert's Continued Fraction for tan(x) For any real angle x {\displaystyle x} (in radians) satisfying | x | < π 2 {\displaystyle |x|<{\frac {\pi }{2}}} , the tangent function can be expressed as the infinite continued fraction:
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 }}}}}}}}}
where the denominators follow the sequence of positive odd integers 1 , 3 , 5 , 7 , … {\displaystyle 1,3,5,7,\dots }
Derivation by coefficient matching
Step 1: Power series of tan(x) Recall the standard Maclaurin series for sine and cosine:
sin x = x − x 3 6 + x 5 120 − x 7 5040 + … {\displaystyle \sin x=x-{\frac {x^{3}}{6}}+{\frac {x^{5}}{120}}-{\frac {x^{7}}{5040}}+\dots }
cos x = 1 − x 2 2 + x 4 24 − x 6 720 + … {\displaystyle \cos x=1-{\frac {x^{2}}{2}}+{\frac {x^{4}}{24}}-{\frac {x^{6}}{720}}+\dots }
Performing polynomial long division on sin x cos x {\displaystyle {\frac {\sin x}{\cos x}}} yields the power series for tan x {\displaystyle \tan x} :
tan x = x + 1 3 x 3 + 2 15 x 5 + 17 315 x 7 + … (Eq. 1) {\displaystyle \tan x=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+\dots \quad {\text{(Eq. 1)}}}
Step 2: Parameterizing the continued fraction Assume tan x {\displaystyle \tan x} can be written in the form:
tan x = x 1 − x 2 a 1 − x 2 a 2 − x 2 a 3 − ⋱ {\displaystyle \tan x={\cfrac {x}{1-{\cfrac {x^{2}}{a_{1}-{\cfrac {x^{2}}{a_{2}-{\cfrac {x^{2}}{a_{3}-\ddots }}}}}}}}}
where a 1 , a 2 , a 3 , … {\displaystyle a_{1},a_{2},a_{3},\dots } are unknown constants to be determined such that the expansion of the continued fraction matches Eq. 1 term-by-term. The geometric series formula 1 1 − u = 1 + u + u 2 + u 3 + … {\displaystyle {\frac {1}{1-u}}=1+u+u^{2}+u^{3}+\dots } is applied repeatedly to convert each convergents' fraction into a power series.
Step 3: Determining coefficients
Finding a₁ Truncating the fraction at the first unknown constant yields the second convergent C 2 ( x ) {\displaystyle C_{2}(x)} :
C 2 ( x ) = x 1 − x 2 a 1 {\displaystyle C_{2}(x)={\frac {x}{1-{\frac {x^{2}}{a_{1}}}}}}
Applying the geometric series formula with u = x 2 a 1 {\displaystyle u={\frac {x^{2}}{a_{1}}}} :
C 2 ( x ) = x ( 1 + x 2 a 1 + x 4 a 1 2 + … ) = x + x 3 a 1 + O ( x 5 ) {\displaystyle C_{2}(x)=x\left(1+{\frac {x^{2}}{a_{1}}}+{\frac {x^{4}}{a_{1}^{2}}}+\dots \right)=x+{\frac {x^{3}}{a_{1}}}+{\mathcal {O}}(x^{5})}
Comparing the coefficient of x 3 {\displaystyle x^{3}} with Eq. 1:
1 a 1 = 1 3 ⟹ a 1 = 3 {\displaystyle {\frac {1}{a_{1}}}={\frac {1}{3}}\implies a_{1}=3}
Finding a₂ Including the next layer yields C 3 ( x ) {\displaystyle C_{3}(x)} :
C 3 ( x ) = x 1 − x 2 3 − x 2 a 2 {\displaystyle C_{3}(x)={\cfrac {x}{1-{\cfrac {x^{2}}{3-{\cfrac {x^{2}}{a_{2}}}}}}}}
Expanding the inner denominator:
1 3 − x 2 a 2 = 1 3 ⋅ 1 1 − x 2 3 a 2 = 1 3 ( 1 + x 2 3 a 2 + … ) = 1 3 + x 2 9 a 2 + … {\displaystyle {\frac {1}{3-{\frac {x^{2}}{a_{2}}}}}={\frac {1}{3}}\cdot {\frac {1}{1-{\frac {x^{2}}{3a_{2}}}}}={\frac {1}{3}}\left(1+{\frac {x^{2}}{3a_{2}}}+\dots \right)={\frac {1}{3}}+{\frac {x^{2}}{9a_{2}}}+\dots }
Substituting this back into the primary denominator:
1 − x 2 ( 1 3 + x 2 9 a 2 + … ) = 1 − x 2 3 − x 4 9 a 2 − … {\displaystyle 1-x^{2}\left({\frac {1}{3}}+{\frac {x^{2}}{9a_{2}}}+\dots \right)=1-{\frac {x^{2}}{3}}-{\frac {x^{4}}{9a_{2}}}-\dots }
Expanding C 3 ( x ) {\displaystyle C_{3}(x)} using the geometric series formula where v = x 2 3 + x 4 9 a 2 {\displaystyle v={\frac {x^{2}}{3}}+{\frac {x^{4}}{9a_{2}}}} :
C 3 ( x ) = x [ 1 + (