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Our aim is to prove that π is irrational.  The proof presented here is due to a mathematician named Ivan Niven, and was developed and reported in 1947. 

In this case, the statement we need to prove is: 

                    S = The number π is irrational. 

The converse of the statement S is 

                    T = The number π is rational. 

If statement T is true, then our original assertion in statement S is false.  So, let us assume that statement T is true, and see if we can find a contradiction.  If π is rational, then we can define integers a and b such that

where a and b are integers with no common factors.  Now, let us define a function f(x) as follows:

In this expression, n is another integer that we won’t specify at the present time – we’ll choose it later on.  Now, if we make the variable transformation

we can see that our function f(x) has the following property:

                                 (1)

Now let us define another function in terms of the derivatives of f(x):

Differentiating this function twice and adding, we can see that.

since derivatives of f(x) of order higher than 2n are zero.   In order to proceed, we want to prove an important property about the function f(x) and its derivatives.  Let’s begin by noting that we can write the function f(x) as follows:

where the c coefficients take integral values.  Therefore, any derivative of f(x) lower than the n^th derivative is zero at x = 0.  The (n+J) derivative (J ≤ n), at x = 0, is simply equal to

We therefore conclude that the derivatives of f(x), evaluated at x = 0, are either zero or take integral values.  This follows by noting that the quotient of the two factorials in the above expression is an integer.  Further, from (1), we have that

We can therefore further conclude that the derivatives of f(x) evaluated at x = π are also either zero or take integral values.

With all this in mind, we deduce that

             take integral values.

Now by using very simple differential calculus, it is easy to show that

and hence by integrating both sides of this expression, and noting that

we find that

                             (A)

This has an integral value, as we have just shown that the two terms on the right-hand side of the above expression take integral values.

But, let us go back to our original definition of f(x).  By inspection of the definition of f(x), it can be seen that

and hence

                                (B)

This provides us with the contradiction that we seek.  Look back at the integral (A).  We’ve just shown that it takes an integral value.  And yet, result (B) shows that the integrand can be made as small as we like, by choosing a sufficiently large value of n.  Therefore, on the basis of (B), we can make the integral (A) as small as we like.

The point is that we cannot have both situations at once, namely that the integral (A) is both an integer and arbitrarily small.  This is an impossibility, and so the statement T is false. 

Hence, statement S is true, and π is an irrational number.

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