In 1761, Johann Heinrich Lambert proved that is irrational by first showing that this continued fraction expansion holds:
Then Lambert proved that if is non-zero and rational, then this expression must be irrational. Since , it follows that is irrational, and thus is also irrational.[2] A simplification of Lambert's proof is given below.
Hermite's proofedit
Written in 1873, this proof uses the characterization of as the smallest positive number whose half is a zero of the cosine function and it actually proves that is irrational.[3][4] As in many proofs of irrationality, it is a proof by contradiction.
Consider the sequences of real functions and for defined by:
If with and in , then, since the coefficients of are integers and its degree is smaller than or equal to is some integer In other words,
But this number is clearly greater than On the other hand, the limit of this quantity as goes to infinity is zero, and so, if is large enough, 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 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 [5]).
Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, is the "residue" (or "remainder") of Lambert's continued fraction for [6]
Cartwright's proofedit
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]
where and are polynomials of degree and with integer coefficients (depending on ).
Take and suppose if possible that where and are natural numbers (i.e., assume that is rational). Then
The right side is an integer. But since the interval has length and the function being integrated takes only values between and On the other hand,
Hence, for sufficiently large
that is, we could find an integer between and That is the contradiction that follows from the assumption that is rational.
This proof is similar to Hermite's proof. Indeed,
However, it is clearly simpler. This is achieved by omitting the inductive definition of the functions and taking as a starting point their expression as an integral.
Niven's proofedit
This proof uses the characterization of as the smallest positive zero of the sine function.[9]
Suppose that is rational, i.e. for some integers and which may be taken without loss of generality to both be positive. Given any positive integer we define the polynomial function:
and, for each let
Claim 1: is an integer.
Proof: Expanding as a sum of monomials, the coefficient of is a number of the form where is an integer, which is if Therefore, is when and it is equal to if ; in each case, is an integer and therefore is an integer.
On the other hand, and so for each non-negative integer In particular, Therefore, is also an integer and so is an integer (in fact, it is easy to see that ). Since and are integers, so is their sum.
Since and (here we use the above-mentioned characterization of as a zero of the sine function), Claim 2 follows.
Conclusion: Since and for (because is the smallest positive zero of the sine function), Claims 1 and 2 show that is a positive integer. Since and for we have, by the original definition of
which is smaller than for large hence for these by Claim 2. This is impossible for the positive integer This shows that the original assumption that 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
which is obtained by integrations by parts. Claim 2 essentially establishes this formula, where the use of hides the iterated integration by parts. The last integral vanishes because 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,
Therefore, the substitution turns this integral into
In particular,
Another connection between the proofs lies in the fact that Hermite already mentions[3] that if is a polynomial function and
then
from which it follows that
Bourbaki's proofedit
Bourbaki's proof is outlined as an exercise in his calculus treatise.[10] For each natural number b and each non-negative integer define
Since is the integral of a function defined on that takes the value at and and which is greater than otherwise, Besides, for each natural number if is large enough, because
and therefore
On the other hand, repeated integration by parts allows us to deduce that, if and are natural numbers such that and is the polynomial function from into defined by
then:
This last integral is since is the null function (because is a polynomial function of degree ). Since each function (with ) takes integer values at and and since the same thing happens with the sine and the cosine functions, this proves that is an integer. Since it is also greater than it must be a natural number. But it was also proved that if 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 are integers.
Laczkovich's proofedit
Miklós Laczkovich's proof is a simplification of Lambert's original proof.[11] He considers the functions
These functions are clearly defined for any real number Besides
Proof: This can be proved by comparing the coefficients of the powers of
Claim 2: For each real number
Proof: In fact, the sequence is bounded (since it converges to ) and if is an upper bound and if then
Claim 3: If is rational, and then
Proof: Otherwise, there would be a number and integers and such that and To see why, take and if ; otherwise, choose integers and such that and define In each case, cannot be because otherwise it would follow from claim 1 that each () would be which would contradict claim 2. Now, take a natural number such that all three numbers and are integers and consider the sequence
Then
On the other hand, it follows from claim 1 that
which is a linear combination of and with integer coefficients. Therefore, each is an integer multiple of Besides, it follows from claim 2 that each is greater than (and therefore that ) if is large enough and that the sequence of all converges to But a sequence of numbers greater than or equal to cannot converge to
Since it follows from claim 3 that is irrational and therefore that is irrational.
On the other hand, since
another consequence of Claim 3 is that, if then is irrational.
Laczkovich's proof is really about the hypergeometric function. In fact, and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation.[12] 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.
^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, ISBN0-387-20571-3.
^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.
^ abZhou, Li (2011). "Irrationality proofs à la Hermite". The Mathematical Gazette. 95 (534): 407–413. arXiv:0911.1929. doi:10.1017/S0025557200003491. S2CID 115175505.
^Bourbaki, Nicolas (1949), Fonctions d'une variable réelle, chap. I–II–III, Actualités Scientifiques et Industrielles (in French), vol. 1074, Hermann, pp. 137–138
^Gauss, Carl Friedrich (1811–1813), "Disquisitiones generales circa seriem infinitam", Commentationes Societatis Regiae Scientiarum Gottingensis Recentiores (in Latin), 2
April 10, 2024
proof, that, irrational, 1760s, johann, heinrich, lambert, first, prove, that, number, irrational, meaning, cannot, expressed, fraction, displaystyle, where, displaystyle, displaystyle, both, integers, 19th, century, charles, hermite, found, proof, that, requi. In the 1760s Johann Heinrich Lambert was the first to prove that the number p 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 Another proof which is a simplification of Lambert s proof is due to Miklos Laczkovich Many of these are proofs by contradiction In 1882 Ferdinand von Lindemann proved that p displaystyle pi is not just irrational but transcendental as well 1 Contents 1 Lambert s proof 2 Hermite s proof 3 Cartwright s proof 4 Niven s proof 5 Bourbaki s proof 6 Laczkovich s proof 7 See also 8 ReferencesLambert s proof edit nbsp Scan of formula on page 288 of Lambert s Memoires sur quelques proprietes remarquables des quantites transcendantes circulaires et logarithmiques Memoires de l Academie royale des sciences de Berlin 1768 265 322In 1761 Johann Heinrich Lambert proved that p displaystyle pi nbsp is irrational by first showing that this continued fraction expansion holds tan x x1 x23 x25 x27 displaystyle tan x cfrac x 1 cfrac x 2 3 cfrac x 2 5 cfrac x 2 7 ddots nbsp Then Lambert proved that if x displaystyle x nbsp is non zero and rational then this expression must be irrational Since tan p4 1 displaystyle tan tfrac pi 4 1 nbsp it follows that p4 displaystyle tfrac pi 4 nbsp is irrational and thus p displaystyle pi nbsp is also irrational 2 A simplification of Lambert s proof is given below Hermite s proof editWritten in 1873 this proof uses the characterization of p displaystyle pi nbsp as the smallest positive number whose half is a zero of the cosine function and it actually proves that p2 displaystyle pi 2 nbsp is irrational 3 4 As in many proofs of irrationality it is a proof by contradiction Consider the sequences of real functions An displaystyle A n nbsp and Un displaystyle U n nbsp for n N0 displaystyle n in mathbb N 0 nbsp defined by A0 x sin x An 1 x 0xyAn y dyU0 x sin x x Un 1 x Un x x displaystyle begin aligned A 0 x amp sin x amp amp A n 1 x int 0 x yA n y dy 4pt U 0 x amp frac sin x x amp amp U n 1 x frac U n x x end aligned nbsp Using induction we can prove that An x x2n 1 2n 1 x2n 32 2n 3 x2n 52 4 2n 5 Un x 1 2n 1 x22 2n 3 x42 4 2n 5 displaystyle begin aligned A n x amp 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 amp frac 1 2n 1 frac x 2 2 times 2n 3 frac x 4 2 times 4 times 2n 5 mp cdots end aligned nbsp and therefore we have Un x An x x2n 1 displaystyle U n x frac A n x x 2n 1 nbsp So An 1 x x2n 3 Un 1 x Un x x 1xddx An x x2n 1 1x An x x2n 1 2n 1 x2nAn x x2 2n 1 2n 1 An x xAn x x2n 3 displaystyle begin aligned frac A n 1 x x 2n 3 amp 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 amp 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 amp frac 2n 1 A n x xA n x x 2n 3 end aligned nbsp which is equivalent to An 1 x 2n 1 An x x2An 1 x displaystyle A n 1 x 2n 1 A n x x 2 A n 1 x nbsp Using the definition of the sequence and employing induction we can show that An x Pn x2 sin x xQn x2 cos x displaystyle A n x P n x 2 sin x xQ n x 2 cos x nbsp where Pn displaystyle P n nbsp and Qn displaystyle Q n nbsp are polynomial functions with integer coefficients and the degree of Pn displaystyle P n nbsp is smaller than or equal to 12n displaystyle bigl lfloor tfrac 1 2 n bigr rfloor nbsp In particular An 12p Pn 14p2 displaystyle A n bigl tfrac 1 2 pi bigr P n bigl tfrac 1 4 pi 2 bigr nbsp Hermite also gave a closed expression for the function An displaystyle A n nbsp namely An x x2n 12nn 01 1 z2 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 nbsp He did not justify this assertion but it can be proved easily First of all this assertion is equivalent to 12nn 01 1 z2 ncos xz dz An x x2n 1 Un 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 nbsp Proceeding by induction take n 0 displaystyle n 0 nbsp 01cos xz dz sin x x U0 x displaystyle int 0 1 cos xz mathrm d z frac sin x x U 0 x nbsp and for the inductive step consider any natural number n displaystyle n nbsp If 12nn 01 1 z2 ncos xz dz Un x displaystyle frac 1 2 n n int 0 1 1 z 2 n cos xz mathrm d z U n x nbsp then using integration by parts and Leibniz s rule one gets 12n 1 n 1 01 1 z2 n 1cos xz dz 12n 1 n 1 1 z2 n 1sin xz x z 0z 1 0 012 n 1 1 z2 nzsin xz xdz 1x 12nn 01 1 z2 nzsin xz dz 1x ddx 12nn 01 1 z2 ncos xz dz Un x x Un 1 x displaystyle begin aligned amp frac 1 2 n 1 n 1 int 0 1 left 1 z 2 right n 1 cos xz mathrm d z amp 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 amp 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 amp 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 amp qquad frac U n x x 4pt amp qquad U n 1 x end aligned nbsp If 14p2 p q displaystyle tfrac 1 4 pi 2 p q nbsp with p displaystyle p nbsp and q displaystyle q nbsp in N displaystyle mathbb N nbsp then since the coefficients of Pn displaystyle P n nbsp are integers and its degree is smaller than or equal to 12n displaystyle bigl lfloor tfrac 1 2 n bigr rfloor nbsp q n 2 Pn 14p2 displaystyle q lfloor n 2 rfloor P n bigl tfrac 1 4 pi 2 bigr nbsp is some integer N displaystyle N nbsp In other words N q n 2 An 12p q n 2 12nn pq n 12 01 1 z2 ncos 12pz dz 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 nbsp But this number is clearly greater than 0 displaystyle 0 nbsp On the other hand the limit of this quantity as n displaystyle n nbsp goes to infinity is zero and so if n displaystyle n nbsp is large enough N lt 1 displaystyle N lt 1 nbsp 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 p displaystyle pi nbsp 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 nbsp 5 Moreover Hermite s proof is closer to Lambert s proof than it seems In fact An x displaystyle A n x nbsp is the residue or remainder of Lambert s continued fraction for tan x displaystyle tan x nbsp 6 Cartwright s proof editHarold 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 In x 11 1 z2 ncos xz dz displaystyle I n x int 1 1 1 z 2 n cos xz dz nbsp where n displaystyle n nbsp is a non negative integer Two integrations by parts give the recurrence relation x2In x 2n 2n 1 In 1 x 4n n 1 In 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 nbsp If Jn x x2n 1In x displaystyle J n x x 2n 1 I n x nbsp then this becomes Jn x 2n 2n 1 Jn 1 x 4n n 1 x2Jn 2 x displaystyle J n x 2n 2n 1 J n 1 x 4n n 1 x 2 J n 2 x nbsp Furthermore J0 x 2sin x displaystyle J 0 x 2 sin x nbsp and J1 x 4xcos x 4sin x displaystyle J 1 x 4x cos x 4 sin x nbsp Hence for all n Z displaystyle n in mathbb Z nbsp Jn x x2n 1In 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 nbsp where Pn x displaystyle P n x nbsp and Qn x displaystyle Q n x nbsp are polynomials of degree n displaystyle leq n nbsp and with integer coefficients depending on n displaystyle n nbsp Take x 12p displaystyle x tfrac 1 2 pi nbsp and suppose if possible that 12p a b displaystyle tfrac 1 2 pi a b nbsp where a displaystyle a nbsp and b displaystyle b nbsp are natural numbers i e assume that p displaystyle pi nbsp is rational Then a2n 1n In 12p Pn 12p b2n 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 nbsp The right side is an integer But 0 lt In 12p lt 2 displaystyle 0 lt I n bigl tfrac 1 2 pi bigr lt 2 nbsp since the interval 1 1 displaystyle 1 1 nbsp has length 2 displaystyle 2 nbsp and the function being integrated takes only values between 0 displaystyle 0 nbsp and 1 displaystyle 1 nbsp On the other hand a2n 1n 0 as n displaystyle frac a 2n 1 n to 0 quad text as n to infty nbsp Hence for sufficiently large n displaystyle n nbsp 0 lt a2n 1In p2 n lt 1 displaystyle 0 lt frac a 2n 1 I n left frac pi 2 right n lt 1 nbsp that is we could find an integer between 0 displaystyle 0 nbsp and 1 displaystyle 1 nbsp That is the contradiction that follows from the assumption that p displaystyle pi nbsp is rational This proof is similar to Hermite s proof Indeed Jn x x2n 1 11 1 z2 ncos xz dz 2x2n 1 01 1 z2 ncos xz dz 2n 1n An x displaystyle begin aligned J n x amp x 2n 1 int 1 1 1 z 2 n cos xz dz 5pt amp 2x 2n 1 int 0 1 1 z 2 n cos xz dz 5pt amp 2 n 1 n A n x end aligned nbsp However it is clearly simpler This is achieved by omitting the inductive definition of the functions An displaystyle A n nbsp and taking as a starting point their expression as an integral Niven s proof editThis proof uses the characterization of p displaystyle pi nbsp as the smallest positive zero of the sine function 9 Suppose that p displaystyle pi nbsp is rational i e p a b displaystyle pi a b nbsp for some integers a displaystyle a nbsp and b displaystyle b nbsp which may be taken without loss of generality to both be positive Given any positive integer n displaystyle n nbsp we define the polynomial function f x xn a bx nn displaystyle f x frac x n a bx n n nbsp and for each x R displaystyle x in mathbb R nbsp let F x f x f x f 4 x 1 nf 2n x displaystyle F x f x f x f 4 x cdots 1 n f 2n x nbsp Claim 1 F 0 F p displaystyle F 0 F pi nbsp is an integer Proof Expanding f displaystyle f nbsp as a sum of monomials the coefficient of xk displaystyle x k nbsp is a number of the form ck n displaystyle c k n nbsp where ck displaystyle c k nbsp is an integer which is 0 displaystyle 0 nbsp if k lt n displaystyle k lt n nbsp Therefore f k 0 displaystyle f k 0 nbsp is 0 displaystyle 0 nbsp when k lt n displaystyle k lt n nbsp and it is equal to k n ck displaystyle k n c k nbsp if n k 2n displaystyle n leq k leq 2n nbsp in each case f k 0 displaystyle f k 0 nbsp is an integer and therefore F 0 displaystyle F 0 nbsp is an integer On the other hand f p x f x displaystyle f pi x f x nbsp and so 1 kf k p x f k x displaystyle 1 k f k pi x f k x nbsp for each non negative integer k displaystyle k nbsp In particular 1 kf k p f k 0 displaystyle 1 k f k pi f k 0 nbsp Therefore f k p displaystyle f k pi nbsp is also an integer and so F p displaystyle F pi nbsp is an integer in fact it is easy to see that F p F 0 displaystyle F pi F 0 nbsp Since F 0 displaystyle F 0 nbsp and F p displaystyle F pi nbsp are integers so is their sum Claim 2 0pf x sin x dx F 0 F p displaystyle int 0 pi f x sin x dx F 0 F pi nbsp Proof Since f 2n 2 displaystyle f 2n 2 nbsp is the zero polynomial we have F F f displaystyle F F f nbsp The derivatives of the sine and cosine function are given by sin cos and cos sin Hence the product rule implies F sin F cos f sin displaystyle F cdot sin F cdot cos f cdot sin nbsp By the fundamental theorem of calculus 0pf x sin x dx F x sin x F x cos x 0p displaystyle left int 0 pi f x sin x dx bigl F x sin x F x cos x bigr right 0 pi nbsp Since sin 0 sin p 0 displaystyle sin 0 sin pi 0 nbsp and cos 0 cos p 1 displaystyle cos 0 cos pi 1 nbsp here we use the above mentioned characterization of p displaystyle pi nbsp as a zero of the sine function Claim 2 follows Conclusion Since f x gt 0 displaystyle f x gt 0 nbsp and sin x gt 0 displaystyle sin x gt 0 nbsp for 0 lt x lt p displaystyle 0 lt x lt pi nbsp because p displaystyle pi nbsp is the smallest positive zero of the sine function Claims 1 and 2 show that F 0 F p displaystyle F 0 F pi nbsp is a positive integer Since 0 x a bx pa displaystyle 0 leq x a bx leq pi a nbsp and 0 sin x 1 displaystyle 0 leq sin x leq 1 nbsp for 0 x p displaystyle 0 leq x leq pi nbsp we have by the original definition of f displaystyle f nbsp 0pf x sin x dx p pa nn displaystyle int 0 pi f x sin x dx leq pi frac pi a n n nbsp which is smaller than 1 displaystyle 1 nbsp for large n displaystyle n nbsp hence F 0 F p lt 1 displaystyle F 0 F pi lt 1 nbsp for these n displaystyle n nbsp by Claim 2 This is impossible for the positive integer F 0 F p displaystyle F 0 F pi nbsp This shows that the original assumption that p displaystyle pi nbsp 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 0pf x sin x dx j 0n 1 j f 2j p f 2j 0 1 n 1 0pf 2n 2 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 nbsp which is obtained by 2n 2 displaystyle 2n 2 nbsp integrations by parts Claim 2 essentially establishes this formula where the use of F displaystyle F nbsp hides the iterated integration by parts The last integral vanishes because f 2n 2 displaystyle f 2n 2 nbsp 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 Jn x x2n 1 11 1 z2 ncos xz dz 11 x2 xz 2 nxcos xz dz displaystyle begin aligned J n x amp x 2n 1 int 1 1 1 z 2 n cos xz dz amp int 1 1 left x 2 xz 2 right n x cos xz dz end aligned nbsp Therefore the substitution xz y displaystyle xz y nbsp turns this integral into xx x2 y2 ncos y dy displaystyle int x x x 2 y 2 n cos y dy nbsp In particular Jn p2 p 2p 2 p24 y2 ncos y dy 0p p24 y p2 2 ncos y p2 dy 0pyn p y nsin y dy n bn 0pf x sin x dx displaystyle begin aligned J n left frac pi 2 right amp int pi 2 pi 2 left frac pi 2 4 y 2 right n cos y dy 5pt amp 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 amp int 0 pi y n pi y n sin y dy 5pt amp frac n b n int 0 pi f x sin x dx end aligned nbsp Another connection between the proofs lies in the fact that Hermite already mentions 3 that if f displaystyle f nbsp is a polynomial function and F f f 2 f 4 displaystyle F f f 2 f 4 mp cdots nbsp then f x sin x dx 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 nbsp from which it follows that 0pf x sin x dx F p F 0 displaystyle int 0 pi f x sin x dx F pi F 0 nbsp Bourbaki s proof editBourbaki s proof is outlined as an exercise in his calculus treatise 10 For each natural number b and each non negative integer n displaystyle n nbsp define An b bn 0pxn p x nn sin x dx displaystyle A n b b n int 0 pi frac x n pi x n n sin x dx nbsp Since An b displaystyle A n b nbsp is the integral of a function defined on 0 p displaystyle 0 pi nbsp that takes the value 0 displaystyle 0 nbsp at 0 displaystyle 0 nbsp and p displaystyle pi nbsp and which is greater than 0 displaystyle 0 nbsp otherwise An b gt 0 displaystyle A n b gt 0 nbsp Besides for each natural number b displaystyle b nbsp An b lt 1 displaystyle A n b lt 1 nbsp if n displaystyle n nbsp is large enough because x p x p2 2 displaystyle x pi x leq left frac pi 2 right 2 nbsp and therefore An b pbn1n p2 2n p bp2 4 nn 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 nbsp On the other hand repeated integration by parts allows us to deduce that if a displaystyle a nbsp and b displaystyle b nbsp are natural numbers such that p a b displaystyle pi a b nbsp and f displaystyle f nbsp is the polynomial function from 0 p displaystyle 0 pi nbsp into R displaystyle mathbb R nbsp defined by f x xn a bx nn displaystyle f x frac x n a bx n n nbsp then An b 0pf x sin x dx f x cos x x 0x p f x sin x x 0x p f 2n x cos x x 0x p 0pf 2n 1 x cos x dx displaystyle begin aligned A n b amp int 0 pi f x sin x dx 5pt amp Big f x cos x Big x 0 x pi Big f x sin x Big x 0 x pi cdots 5pt amp 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 nbsp This last integral is 0 displaystyle 0 nbsp since f 2n 1 displaystyle f 2n 1 nbsp is the null function because f displaystyle f nbsp is a polynomial function of degree 2n displaystyle 2n nbsp Since each function f k displaystyle f k nbsp with 0 k 2n displaystyle 0 leq k leq 2n nbsp takes integer values at 0 displaystyle 0 nbsp and p displaystyle pi nbsp and since the same thing happens with the sine and the cosine functions this proves that An b displaystyle A n b nbsp is an integer Since it is also greater than 0 displaystyle 0 nbsp it must be a natural number But it was also proved that An b lt 1 displaystyle A n b lt 1 nbsp if n displaystyle n nbsp 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 displaystyle A n b nbsp are integers Laczkovich s proof editMiklos Laczkovich s proof is a simplification of Lambert s original proof 11 He considers the functions fk x 1 x2k x42 k k 1 x63 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 nbsp These functions are clearly defined for any real number x displaystyle x nbsp Besides f1 2 x cos 2x displaystyle f 1 2 x cos 2x nbsp f3 2 x sin 2x 2x displaystyle f 3 2 x frac sin 2x 2x nbsp Claim 1 The following recurrence relation holds for any real number x displaystyle x nbsp x2k k 1 fk 2 x fk 1 x fk x displaystyle frac x 2 k k 1 f k 2 x f k 1 x f k x nbsp Proof This can be proved by comparing the coefficients of the powers of x displaystyle x nbsp Claim 2 For each real number x displaystyle x nbsp limk fk x 1 displaystyle lim k to infty f k x 1 nbsp Proof In fact the sequence x2n n displaystyle x 2n n nbsp is bounded since it converges to 0 displaystyle 0 nbsp and if C displaystyle C nbsp is an upper bound and if k gt 1 displaystyle k gt 1 nbsp then fk x 1 n 1 Ckn C1 k1 1 k Ck 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 nbsp Claim 3 If x 0 displaystyle x neq 0 nbsp x2 displaystyle x 2 nbsp is rational and k Q 0 1 2 displaystyle k in mathbb Q smallsetminus 0 1 2 ldots nbsp then fk x 0 and fk 1 x fk x Q displaystyle f k x neq 0 quad text and quad frac f k 1 x f k x notin mathbb Q nbsp Proof Otherwise there would be a number y 0 displaystyle y neq 0 nbsp and integers a displaystyle a nbsp and b displaystyle b nbsp such that fk x ay displaystyle f k x ay nbsp and fk 1 x by displaystyle f k 1 x by nbsp To see why take y fk 1 x displaystyle y f k 1 x nbsp a 0 displaystyle a 0 nbsp and b 1 displaystyle b 1 nbsp if fk x 0 displaystyle f k x 0 nbsp otherwise choose integers a displaystyle a nbsp and b displaystyle b nbsp such that fk 1 x fk x b a displaystyle f k 1 x f k x b a nbsp and define y fk x a fk 1 x b displaystyle y f k x a f k 1 x b nbsp In each case y displaystyle y nbsp cannot be 0 displaystyle 0 nbsp because otherwise it would follow from claim 1 that each fk n x displaystyle f k n x nbsp n N displaystyle n in mathbb N nbsp would be 0 displaystyle 0 nbsp which would contradict claim 2 Now take a natural number c displaystyle c nbsp such that all three numbers bc k displaystyle bc k nbsp ck x2 displaystyle ck x 2 nbsp and c x2 displaystyle c x 2 nbsp are integers and consider the sequence gn fk x n 0cnk k 1 k n 1 fk n x n 0 displaystyle g n begin cases f k x amp n 0 dfrac c n k k 1 cdots k n 1 f k n x amp n neq 0 end cases nbsp Then g0 fk x ay Zy and g1 ckfk 1 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 nbsp On the other hand it follows from claim 1 that gn 2 cn 2x2k k 1 k n 1 x2 k n k n 1 fk n 2 x cn 2x2k k 1 k n 1 fk n 1 x cn 2x2k k 1 k n 1 fk n x c k n x2gn 1 c2x2gn ckx2 cx2n gn 1 c2x2gn displaystyle begin aligned g n 2 amp 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 amp 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 amp frac c k n x 2 g n 1 frac c 2 x 2 g n 5pt amp left frac ck x 2 frac c x 2 n right g n 1 frac c 2 x 2 g n end aligned nbsp which is a linear combination of gn 1 displaystyle g n 1 nbsp and gn displaystyle g n nbsp with integer coefficients Therefore each gn displaystyle g n nbsp is an integer multiple of y displaystyle y nbsp Besides it follows from claim 2 that each gn displaystyle g n nbsp is greater than 0 displaystyle 0 nbsp and therefore that gn y displaystyle g n geq y nbsp if n displaystyle n nbsp is large enough and that the sequence of all gn displaystyle g n nbsp converges to 0 displaystyle 0 nbsp But a sequence of numbers greater than or equal to y displaystyle y nbsp cannot converge to 0 displaystyle 0 nbsp Since f1 2 14p cos 12p 0 displaystyle f 1 2 tfrac 1 4 pi cos tfrac 1 2 pi 0 nbsp it follows from claim 3 that 116p2 displaystyle tfrac 1 16 pi 2 nbsp is irrational and therefore that p displaystyle pi nbsp is irrational On the other hand since tan x sin xcos x xf3 2 x 2 f1 2 x 2 displaystyle tan x frac sin x cos x x frac f 3 2 x 2 f 1 2 x 2 nbsp another consequence of Claim 3 is that if x Q 0 displaystyle x in mathbb Q smallsetminus 0 nbsp then tan x displaystyle tan x nbsp is irrational Laczkovich s proof is really about the hypergeometric function In fact fk x 0F1 k x2 displaystyle f k x 0 F 1 k x 2 nbsp and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation 12 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 Jn x displaystyle J nu x nbsp In fact G k Jk 1 2x xk 1fk x displaystyle Gamma k J k 1 2x x k 1 f k x nbsp where G displaystyle Gamma nbsp is the gamma function So Laczkovich s result is equivalent to If x 0 displaystyle x neq 0 nbsp x2 displaystyle x 2 nbsp is rational and k Q 0 1 2 displaystyle k in mathbb Q smallsetminus 0 1 2 ldots nbsp then xJk x Jk 1 x Q displaystyle frac xJ k x J k 1 x notin mathbb Q nbsp See also edit nbsp Mathematics portalProof that e is irrational Proof that p is transcendentalReferences edit Lindemann Ferdinand von 2004 1882 Ueber die Zahl p 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 Lambert Johann Heinrich 2004 1768 Memoire sur quelques proprietes remarquables des quantites 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 b Hermite Charles 1873 Extrait d une lettre de Monsieur Ch Hermite a Monsieur Paul Gordan Journal fur die reine und angewandte Mathematik in French 76 303 311 Hermite Charles 1873 Extrait d une lettre de Mr Ch Hermite a Mr Carl Borchardt Journal fur die reine und angewandte Mathematik in French 76 342 344 Hermite Charles 1912 1873 Sur la fonction exponentielle In Picard Emile ed Œuvres de Charles Hermite in French Vol III Gauthier Villars pp 150 181 a b Zhou Li 2011 Irrationality proofs a la Hermite The Mathematical Gazette 95 534 407 413 arXiv 0911 1929 doi 10 1017 S0025557200003491 S2CID 115175505 Jeffreys Harold 1973 Scientific Inference 3rd ed Cambridge University Press p 268 ISBN 0 521 08446 6 Department of Pure Mathematics and Mathematical Statistics www dpmms cam ac uk Retrieved 2022 04 19 Niven Ivan 1947 A simple proof that p is irrational PDF Bulletin of the American Mathematical Society vol 53 no 6 p 509 doi 10 1090 s0002 9904 1947 08821 2 Bourbaki Nicolas 1949 Fonctions d une variable reelle chap I II III Actualites Scientifiques et Industrielles in French vol 1074 Hermann pp 137 138 Laczkovich Miklos 1997 On Lambert s proof of the irrationality of p American Mathematical Monthly vol 104 no 5 pp 439 443 doi 10 2307 2974737 JSTOR 2974737 Gauss Carl Friedrich 1811 1813 Disquisitiones generales circa seriem infinitam Commentationes Societatis Regiae Scientiarum Gottingensis Recentiores in Latin 2 Retrieved from https en wikipedia org w index php title Proof that p is irrational amp oldid 1217172163, wikipedia, wiki, book, books, library,