From Euclid to Spigotry #15
Niven Irrationality
Last modified: September 14, 2004
Reference, Ivan Niven, "A Simple Proof that p is Irrational" (Pi Source Book, p. 276)
Lecture topic 1: Niven's proof strategy:
This is bizarre, but it has become the standard way of showing that p and other numbers are irrational or transcendental.
Step 1: Assume that p is rational: that  where a and b are integers.  Famous approximations
to p are    and  , but Niven is assuming that p is exactly of the form , with the intent of developing a contradiction.
Step 2: Define a family of polynomials, one for each positive integer n.
Niven's formula is  where n! ("n factorial") means the product .
All you need to notice about this initially is that it is zero for x = 0 and for x = p and positive in between.
Step 3: Look at the integral In =  , which is some positive quantity since the function being integrated is positive everywhere between 0 and p.
Step 4: Prove that for any n, In is an integer (a positive integer, since In is positive).  This is the hard part of the proof.
Step 5: Prove that n can be chosen to be so large that In < 1.  It is the n! in the denominator of f(x) that makes this happen.
Step 6: Note that the assumption that  has led to the absurd conclusion that there exists a positive integer that is smaller than 1.  Therefore this assumption is wrong, and p is irrational.
Lecture topic 2: The function Fn(x).
For reasons that will become apparent later, Niven wants a function Fn(x) with the property that
F''n(x) + Fn(x) = fn(x).  Here F''(x) means the second derivative of x: the derivative of the derivative.
He creates this function by a technique that works for any polynomial.
First show that for any polynomial f(x), if you keep taking the derivative over and over, you will eventually end up with zero.  Note that for f(x) = sin x, this would not be the case.

So Niven defines
Fn(x) = fn(x) - f''n(x) + f''''n(x) - f''''''n(x) +....
This sum has only a finite number of terms.
Show that F''n(x) + Fn(x) = fn(x).

Lecture topic 3: Evaluating the integral

Since Niven wants to evaluate In = , he needs to find a function Gn(x) whose derivative is  Show that Gn(x) = Fn'(x) sin x - F(x) cos x   does the job.
Recall that the derivative of sin x is cos x and that the derivative of cos x is  - sin x.
You will need the product rule for differentiation and the formula from topic 2.
Conclude that  In =  = G(p) - G(0) = F(p) + F(0)
You must use the fact that sin p = 0 and that cos p = -1.  This is really the only place that properties of p are invoked.

Lecture topic 4: Properties of fn (x) and of its derivatives
Do a binomial expansion of fn(x):
= 
Now point out the following:
a. If you differentiate the function fewer than n times and then set x = 0, you get zero.
b. If you differentiate the function between n and 2n times, you end up with precisely one term
that does not involve x.  But in the process of taking the first n derivatives, you kill off the n! in the denominator, so the result is an integer.
Since Fn(x) is the sum of fn(x) and its derivatives, Fn(0) is an integer.
Now show that fn(a/b - x) = fn(x). By the chain rule,
fn''(a/b - x) = fn''(x), and a similar result holds when you take any even number of derivatives.
So fn(p) = fn(a/b - p) = fn(0) 
fn''(p) = fn''(a/b - p) = fn''(0)
and so on for any even number of derivatives.  Conclude that Fn(p) is an integer.
Refer to topic 3 to conclude that In =  = F(p) + F(0) is an integer.
Lecture topic 5: How large is the integral?
First show that if x lies between 0 and p, 
fn(x) <  and sin x < 1. Now, with that aid of a sketch and the concept of integral as area
conclude that In =  < 
Now set c = pa and consider what happens if you choose K >2c.
IK may be greater than 1, it may be quite large, but there is some power of 2 that is even larger, say IK < 2m. 
Now set n = K + m.
You multiply IK by , then by , then by , and so on. Each of these m factors is less than 1/2, and you have m of them.  When you reach n = K + m, you will have divided IK by 2 m times, and so In < 1. 
An alternative crude but effective approach is simply to note that for n > c = pa, the effect of increasing n by 1 is to multiply by a factor that is less than 1. Just choose n so large that ,<1 and In must be less than 1.
But In was earlier shown to be a positive integer, so you have achieved the contradiction that completes the proof!