Proof that π is irrational


In the 1760s, Johann Heinrich Lambert proved that the number 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 Nicolas Bourbaki. Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich.
In 1882, Ferdinand von Lindemann proved that is not just irrational, but transcendental as well.

Lambert's proof

In 1761, Lambert proved that 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 = 1, it follows that /4 is irrational and therefore that is irrational. A simplification of Lambert's proof is given below.

Hermite's proof

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 2 is irrational. As in many proofs of irrationality, it is a proof by contradiction.
Consider the sequences of functions An and Un from into 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 Pn and Qn are polynomial functions with integer coefficients and the degree of Pn is smaller than or equal to ⌊n/2⌋. In particular, An = Pn.
Hermite also gave a closed expression for the function An, 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. If
then, using integration by parts and Leibniz's rule, one gets
If 2/4 = p/q, with p and q in, then, since the coefficients of Pn are integers and its degree is smaller than or equal to ⌊n/2⌋, q⌊n/2⌋Pn is some integer N. In other words,
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. 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, which Hermite could easily see.
Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, An is the "residue" of Lambert's continued fraction for tan.

Cartwright's proof

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.
Consider the integrals
where n is a non-negative integer.
Two integrations by parts give the recurrence relation
If
then this becomes
Furthermore, J0 = 2sin and J1 = −4x cos + 4sin. Hence for all n ∈ Z+,
where Pn and Qn are polynomials of degree ≤ n, and with integer coefficients.
Take x = /2, and suppose if possible that /2 = a/b, where a and b are natural numbers. Then
The right side is an integer. But 0 < In < 2 since the interval 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 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 An and taking as a starting point their expression as an integral.

Niven's proof

This proof uses the characterization of as the smallest positive zero of the sine function.
Suppose that is rational, i.e. for some integers a and, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function:
and, for each x ∈ ℝ let
Claim 1: is an integer.
Proof:
Expanding f as a sum of monomials, the coefficient of xk is a number of the form where ck is an integer, which is 0 if. Therefore, is 0 when and it is equal to if ; in each case, is an integer and therefore F is an integer.
On the other hand, = f and so = for each non-negative integer k. In particular, = Therefore, is also an integer and so F is an integer = F. Since F and F 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 , Claim 2 follows.
Conclusion: Since and for , Claims 1 and 2 show that is a positive integer. Since and for, we have, by the original definition of f,
which is smaller than 1 for large n, hence for these n, by Claim 2. This is impossible for the positive integer.
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 F 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 proof than it appears at first sight. 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 that if f is a polynomial function and
then
from which it follows that

Bourbaki's proof

's proof is outlined as an exercise in his calculus treatise. For each natural number b and each non-negative integer n, define
Since An is the integral of a function which defined on that takes the value 0 on 0 and on and which is greater than 0 otherwise, An > 0. Besides, for each natural number b, An < 1 if n is large enough, because
and therefore
On the other hand, recursive integration by parts allows us to deduce that, if a and b are natural number such that = a/b and f is the polynomial function from into R defined by
then:
This last integral is 0, since f is the null function. Since each function f takes integer values on 0 and on and since the same thing happens with the sine and the cosine functions, this proves that An is an integer. Since it is also greater than 0, it must be a natural number. But it was also proved that An < 1 if n 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 are integers.

Laczkovich's proof

's proof is a simplification of Lambert's original proof. 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 x2n/n! is bounded and if C is an upper bound and if k > 1, then
Claim 3: If x ≠ 0 and if x2 is rational, then
Proof: Otherwise, there would be a number y ≠ 0 and integers a and b such that fk = ay and fk + 1 = by. In order to see why, take y = fk + 1, a = 0 and b = 1 if fk = 0; otherwise, choose integers a and b such that fk + 1/fk = b/a and define y = fk/a = fk + 1/b. In each case, y cannot be 0, because otherwise it would follow from claim 1 that each fk + n would be 0, which would contradict claim 2. Now, take a natural number c such that all three numbers bc/k, ck/x2 and c/x2 are integers and consider the sequence
Then
On the other hand, it follows from claim 1 that
which is a linear combination of gn + 1 and gn with integer coefficients. Therefore, each gn is an integer multiple of y. Besides, it follows from claim 2 that each gn is greater than 0 if n is large enough and that the sequence of all gn converges to 0. But a sequence of numbers greater than or equal to |y| cannot converge to 0.
Since f1/2 = cos = 0, it follows from claim 3 that 2/16 is irrational and therefore that is irrational.
On the other hand, since
another consequence of Claim 3 is that, if x ∈ Q \ , then tan x is irrational.
Laczkovich's proof is really about the hypergeometric function. In fact, fk = 0F1 and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation. 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. In fact, ΓJk − 1 = xk − 1fk. So Laczkovich's result is equivalent to: If x ≠ 0 and if x2 is rational, then