Proof that π is irrational

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 , where and are both integers. In the 19th century, Charles Hermite found a proof that requires no prerequisite knowledge beyond basic calculus. Three simplifications of Hermite's proof are due to Mary Cartwright, Ivan Niven, and Nicolas Bourbaki. Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich. Many of these are proofs by contradiction.

In 1882, Ferdinand von Lindemann proved that is not just irrational, but transcendental as well.[1]

Lambert's proof

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Scan of formula on page 288 of Lambert's "Mémoires sur quelques propriétés remarquables des quantités transcendantes, circulaires et logarithmiques", Mémoires de l'Académie royale des sciences de Berlin (1768), 265–322

In 1761, 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 proof

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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:

 

Using induction we can prove that

 

and therefore we have:

 

So

 

which is equivalent to

 

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

 

where   and   are polynomial functions with integer coefficients and the degree of   is smaller than or equal to   In particular,  

Hermite also gave a closed expression for the function   namely

 

He did not justify this assertion, but it can be proved easily. First of all, this assertion is equivalent to

 

Proceeding by induction, take  

 

and, for the inductive step, consider any natural number   If

 

then, using integration by parts and Leibniz's rule, one gets

 

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 proof

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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]

Consider the integrals

 

where   is a non-negative integer.

Two integrations by parts give the recurrence relation

 

If

 

then this becomes

 

Furthermore,   and   Hence for all  

 

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 proof

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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.

Claim 2:

 

Proof: Since   is the zero polynomial, we have

 

The derivatives of the sine and cosine function are given by sin' = cos and cos' = −sin. Hence the product rule implies

 

By the fundamental theorem of calculus

 

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 proof

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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 proof

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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

 
 

Claim 1: The following recurrence relation holds for any real number  :

 

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.

Laczkovich's result can also be expressed in Bessel functions of the first kind  . In fact,   (where   is the gamma function). So Laczkovich's result is equivalent to: If     is rational, and   then

 

See also

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References

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  1. ^ 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.
  2. ^ 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.
  3. ^ a b 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.
  4. ^ 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.
  5. ^ 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.
  6. ^ a b Zhou, Li (2011). "Irrationality proofs à la Hermite". The Mathematical Gazette. 95 (534): 407–413. arXiv:0911.1929. doi:10.1017/S0025557200003491. S2CID 115175505.
  7. ^ Jeffreys, Harold (1973), Scientific Inference (3rd ed.), Cambridge University Press, p. 268, ISBN 0-521-08446-6
  8. ^ "Department of Pure Mathematics and Mathematical Statistics". www.dpmms.cam.ac.uk. Retrieved 2022-04-19.
  9. ^ 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
  10. ^ Bourbaki, Nicolas (1949), Fonctions d'une variable réelle, chap. I–II–III, Actualités Scientifiques et Industrielles (in French), vol. 1074, Hermann, pp. 137–138
  11. ^ 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
  12. ^ Gauss, Carl Friedrich (1811–1813), "Disquisitiones generales circa seriem infinitam", Commentationes Societatis Regiae Scientiarum Gottingensis Recentiores (in Latin), 2