In the 18th century, Johann Heinrich Lambert proved that the number π (pi) is irrational: that is, it cannot be expressed as a fraction *a*/*b*, where *a* is an integer and *b* is a non-zero integer. 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 Bourbaki. Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich.

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

In 1761, Lambert proved that π is irrational by first showing that this continued fraction expansion holds:

Then Lambert proved that if *x* is non-zero and rational then this expression must be irrational. Since tan(π/4) = 1, it follows that π/4 is irrational and therefore that π is irrational.^{[2]} A simplification of Lambert's proof is given below.

This proof uses the characterization of π as the smallest positive number whose half is a zero of the cosine function and it actually proves that π^{2} is irrational.^{[3]}^{[4]} As in many proofs of irrationality, the argument proceeds by reductio ad absurdum.

Consider the sequences (*A*_{n})_{n ≥ 0} and (*U*_{n})_{n ≥ 0} of functions from **R** into **R** thus defined:

It can be proved by induction that

and that

and therefore that

So

which is equivalent to

It follows by induction from this, together with the fact that *A*_{0}(*x*) = sin(*x*) and that *A*_{1}(*x*) = −*x* cos(*x*) + sin(*x*), that *A _{n}*(

Hermite also gave a closed expression for the function *A _{n}*, 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 *n* = 0.

and, for the inductive step, consider any *n* ∈ **Z**_{+}. If

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

If π^{2}/4 = *p*/*q*, with *p* and *q* in **N**, then, since the coefficients of *P _{n}* are integers and its degree is smaller than or equal to ⌊

But this number is clearly greater than 0. On the other hand, the limit of this quantity as n goes to infinity is zero, and so, if *n* is large enough, *N* < 1. 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 π. 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*^{[5]}).

Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, *A*_{n}(*x*) is the "residue" (or "remainder") of Lambert's continued fraction for tan(*x*).^{[6]}

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]}

Consider the integrals

where *n* is a non-negative integer.

Two integrations by parts give the recurrence relation

If

then this becomes

Furthermore, *J*_{0}(*x*) = 2sin(*x*) and *J*_{1}(*x*) = −4*x* cos(*x*) + 4sin(*x*). Hence for all *n* ∈ **Z**_{+},

where *P*_{n}(*x*) and *Q*_{n}(*x*) are polynomials of degree ≤ 2*n*, and with integer coefficients (depending on *n*).

Take *x* = π/2, and suppose if possible that π/2 = *a*/*b*, where *a* and *b* are natural numbers (i.e., assume that π is rational). Then

The right side is an integer. But 0 < *I*_{n}(π/2) < 2 since the interval [−1, 1] has length 2 and the function that is being integrated takes only values between 0 and 1. On the other hand,

Hence, for sufficiently large *n*

that is, we could find an integer between 0 and 1. That is the contradiction that follows from the assumption that π is rational.

This proof is similar to Hermite's proof. Indeed,

However, it is clearly simpler. This is achieved by passing the inductive definition of the functions *A _{n}* and taking as a starting point their expression as an integral.

This proof uses the characterization of π as the smallest positive zero of the sine function.^{[8]}

Suppose that π is rational, i.e. π = *a* /*b* for some integers *a* and *b* ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer *n*, we define the polynomial function *f* from **R** into **R** defined by

and, for each *x* ∈ **R** denote by

the alternating sum of *f* and its first *n* derivatives of even order.

**Claim 1:** *F*(0) + *F*(π) is an integer.

**Proof:** Expanding *f* as a sum of monomials, the coefficient of *x ^{k}* is a number of the form

On the other hand, *f*(π – *x*) = *f*(*x*) and so (–1)^{k}*f*^{ (k)}(π – *x*) = *f*^{ (k)}(*x*) for each non-negative integer *k*. In particular, (–1)^{k}*f*^{ (k)}(π) = *f*^{ (k)}(0). Therefore, *f*^{ (k)}(π) is also an integer and so *F*(π) is an integer (in fact, it is easy to see that *F*(π) = *F*(0), but that is not relevant to the proof). Since *F*(0) and *F*(π) are integers, so is their sum.

**Claim 2:**

**Proof:** Since *f*^{ (2n + 2)} 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 sin 0 = sin π = 0 and cos 0 = – cos π = 1 (here we use the above-mentioned characterization of π as a zero of the sine function), Claim 2 follows.

**Conclusion:** Since *f*(*x*) > 0 and sin *x* > 0 for 0 < *x* < π (because π is the *smallest* positive zero of the sine function), Claims 1 and 2 show that *F*(0) + *F*(π) is a *positive* integer. Since 0 ≤ *x*(*a* – *bx*) ≤ π*a* and 0 ≤ sin *x* ≤ 1 for 0 ≤ *x* ≤ π, we have, by the original definition of *f*,

which is smaller than 1 for large *n*, hence *F*(0) + *F*(π) < 1 for these *n*, by Claim 2. This is impossible for the positive integer *F*(0) + *F*(π).

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 2*n* + 2 integrations by parts. Claim 2 essentially establishes this formula, where the use of *F* hides the iterated integration by parts. The last integral vanishes because *f*^{ (2n + 2)} 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 *xz* = *y* turns this integral into

In particular,

Another connection between the proofs lies in the fact that Hermite already mentions^{[3]} that if *f* is a polynomial function and

then

from which it follows that

Bourbaki's proof is outlined as an exercise in his Calculus treatise.^{[9]} For each natural number *b* and each non-negative integer *n*, define

Since *A _{n}*(

and therefore

On the other hand, recursive integration by parts allows us to deduce that, if *a* and *b* are natural number such that π = *a*/*b* and *f* is the polynomial function from [0,π] into **R** defined by

then:

This last integral is 0, since *f*^{(2n + 1)} is the null function (because *f* is a polynomial function whose degree is 2*n*). Since each function *f*^{(k)} (with 0 ≤ *k* ≤ 2*n*) takes integer values on 0 and on π (see Claim 1 from Niven's proof) and since the same thing happens with the sine and the cosine functions, this proves that *A*_{n}(*b*) is an integer. Since it is also greater than 0, it must be a natural number. But it was also proved that *A _{n}*(

This proof is quite close to Niven's proof, the main difference between them being the way of proving that the numbers *A _{n}*(

Miklós Laczkovich's proof is a simplification of Lambert's original proof.^{[10]} He considers the functions

These functions are clearly defined for all *x* ∈ **R**. Besides

**Claim 1:** The following recurrence relation holds:

**Proof:** This can be proved by comparing the coefficients of the powers of *x*.

**Claim 2:** For each *x* ∈ **R**,

**Proof:** In fact, the sequence *x*^{2n}/*n*! is bounded (since it converges to 0) and if *C* is an upper bound and if *k* > 1, then

**Claim 3:** If *x* ≠ 0 and if *x*^{2} is rational, then

**Proof:** Otherwise, there would be a number *y* ≠ 0 and integers *a* and *b* such that *f _{k}*(

Then

On the other hand, it follows from claim 1 that

which is a linear combination of *g*_{n + 1} and *g _{n}* with integer coefficients. Therefore, each

Since *f*_{1/2}(π/4) = cos(π/2) = 0, it follows from claim 3 that π^{2}/16 is irrational and therefore that π is irrational.

On the other hand, since

another consequence of claim 3 is that, if *x* ∈ **Q** \ {0}, then tan *x* is irrational.

Laczkovich's proof is really about the hypergeometric function. In fact, *f _{k}*(

Laczkovich's result can also be expressed in Bessel functions of the first kind *J*_{ν}(*x*). In fact, Γ(*k*)*J*_{k − 1}(2*x*) = *x*^{k − 1}*f _{k}*(

**^**Lindemann, Ferdinand von (2004) [1882], "Ueber die Zahl π", in Berggren, Lennart; Borwein, Jonathan M.; Borwein, Peter B.,*Pi, a source book*(3rd ed.), New York: Springer-Verlag, pp. 194–225, ISBN 0-387-20571-3**^**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.,*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 à Monsieur Paul Gordan",*Journal für die reine und angewandte Mathematik*(in French),**76**, pp. 303–311 **^**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**, pp. 342–344**^**Hermite, Charles (1912) [1873], "Sur la fonction exponentielle", in Picard, Émile,*Œuvres de Charles Hermite*(in French),**III**, Gauthier-Villars, pp. 150–181- ^
^{a}^{b}Zhou, Li (2011), "Irrationality proofs à la Hermite",*Math. Gazette*(November), arXiv:0911.1929 **^**Jeffreys, Harold (1973),*Scientific Inference*(3rd ed.), Cambridge University Press, p. 268, ISBN 0-521-08446-6**^**Niven, Ivan (1947), "A simple proof that π is irrational" (PDF),*Bulletin of the American Mathematical Society*,**53**(6), p. 509, doi:10.1090/s0002-9904-1947-08821-2**^**Bourbaki, Nicolas (1949),*Fonctions d'une variable réelle, chap. I–II–III*, Actualités Scientifiques et Industrielles (in French),**1074**, Hermann, pp. 137–138**^**Laczkovich, Miklós (1997), "On Lambert's proof of the irrationality of π",*American Mathematical Monthly*,**104**(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**

This page is based on a Wikipedia article written by authors
(here).

Text is available under the CC BY-SA 3.0 license; additional terms may apply.

Images, videos and audio are available under their respective licenses.