Proof that π is irrational

Proof that π is irrational
In the 1760s, Johann Heinrich Lambert was the first to prove that the number 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.
Lambert's proof
In 1761, 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. A simplification of Lambert's proof is given below.
Hermite's proof
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. 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 ).
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
Cartwright's proof
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. It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University.
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
This proof uses the characterization of as the smallest positive zero of the sine function.
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. In fact,
Therefore, the substitution turns this integral into
In particular,
Another connection between the proofs lies in the fact that Hermite already mentions that if is a polynomial function and
then
from which it follows that
Bourbaki's proof
Bourbaki's proof is outlined as an exercise in his calculus treatise. 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
Miklós Laczkovich's proof is a simplification of Lambert's original proof. 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. 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
Proof that e is irrational
Proof that is transcendental
References
Pi
Article proofs
Irrational numbers