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

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.

is irrational by first showing that this continued fraction expansion holds: Then Lambert proved that if

as the smallest positive number whose half is a zero of the cosine function and it actually proves that

are polynomial functions with integer coefficients and the degree of

Hermite also gave a closed expression for the function

First of all, this assertion is equivalent to Proceeding by induction, take

and, for the inductive step, consider any natural number

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

is the "residue" (or "remainder") of Lambert's continued fraction for

[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] It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University.

Two integrations by parts give the recurrence relation If then this becomes Furthermore,

and taking as a starting point their expression as an integral.

Hence the product rule implies By the fundamental theorem of calculus Since

is the smallest positive zero of the sine function), Claims 1 and 2 show 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

Claim 1 shows that the remaining sum is an integer.

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 is outlined as an exercise in his calculus treatise.

is large enough, because and therefore On the other hand, repeated integration by parts allows us to deduce that, if

and since the same thing happens with the sine and the cosine functions, this proves that

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

are integers and consider the sequence Then On the other hand, it follows from claim 1 that which is a linear combination of

[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