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