Why π is irrational (part 5 of 5)

In this post we investigate the size of the integral. In particular, we look at how we may tweak parameters to arbitrarily lower its magnitude.

Table of Contents
1. Outline
2. A polynomial and its derivatives
3. A new function
4. The integral is an integer
5. The size of the integral

Note: in this post, we will restrict our discussion to the interval [0, π].

We are concerned with the definite integral


We now show that the integral is strictly greater than 0 and, for suitably large n, strictly less than 1.

A lower bound

Note that xn and n! are positive when x is positive. We also note that, since (by assumption) π=a/b, (a-bx)n is positive when x < a/b. Thus, f(x) is positive between 0 and π. Since the sine is also positive on this interval, f(x)sin(x) must be positive, so

An upper bound

Note that, since x < π and a-bx < a on the interval,

Since the sine is never greater than 1, f(x) sin(x) can only be less than this:

Thus, we may consider another integral: the integral of the right hand of the inequality.

The integrand on the right is a constant, so the right-hand integral evaluates to that constant times the width of the interval:

And here comes the climax: we can choose n such that it is as large as we want. What does that do to our integral (the one on the right, that is)? It shrinks it! If we increase n by 1, then the numerator increases by a factor of πa, while the denominator increases by a factor of (n+1). Thus, considered as a function, (πa)n/n! increases until n overtakes πa, at which point the function begins to decrease. Once it starts to decrease, it will continue to decrease to a value as small as we want–increasing n indefinitely will get us to 0. In our case, “as small as we want” means “one over π.” Once we’ve gotten it to less than that value, the right hand side of the inequality will be less than 1. Thus, there exists an n such that

The contradiction

Note that there are no integers strictly greater than 0 and strictly less than 1; thus, the integral cannot be an integer. But in the previous part of this proof, we showed that the integral must be an integer! From this contradiction, we conclude that our initial assumption–that we can write π as the ratio of two integers–is false; thus, π is irrational.



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