 Irrationality Of Pi

Problem

Using "school level" mathematics prove that $\pi \approx 3.1415926535...$ is irrational.

Solution

This problem is extremely difficult and "five stars" would have perhaps been a more appropriate difficulty rating. So for that reason the solution should be used as a learning experience. In fact, using theorems from higher mathematics it is trivial to prove the irrationality of π, as it becomes a direct consequence of those theorems. But as those results are far from trivial to prove we shall utilise elementary concepts throughout, albeit pushing the boundaries on "school level" mathematics. We shall prove this by contradiction.

Let us suppose that π is rational and can be written as the ratio of two positive integers: $\pi = \dfrac{a}{b}$.

We now define the function, $f(x) = \dfrac{x^n(a-bx)^n}{n!}$, of which we shall make extensive use.

As $a = b\pi$ we can write $f(x) = \dfrac{x^n(b\pi - bx)^n}{n!} = \dfrac{(bx)^n(\pi - x)^n}{n!}$

$\begin{eqnarray}\therefore f(\pi - x) & = & \dfrac{(b(\pi - x))^n(\pi - (\pi - x))^n}{n!}\\& = & \dfrac{(b(\frac{a}{b} - x))^n x^n}{n!}\\& = & \dfrac{(a - bx)^n x^n}{n!}\\& = & f(x)\end{eqnarray}$

So when $x = 0$ we get $f(0) = f(\pi) = 0,$  which is integer.

Now let us define $G(x) = f(x) - f^{(2)}(x) + f^{(4)}(x) - ... + (-1)^n f^{(2n)}(x),$  where $f^{(k)}(x)$ represents the $k^{th}$ derivative of $f(x)$.

$\begin{eqnarray}f(x) & = & \dfrac{x^n(a - bx)^n}{n!}\\& = & \dfrac{x^n \left({n \choose 0}a^n + {n \choose 1}a^{n-1}(-bx) + {n \choose 2}a^{n-2}(-bx)^2 + ... + {n \choose n}(-bx)^n \right)}{n!}\\& = & \dfrac{{n \choose 0}a^n x^n - {n \choose 1}a^{n-1} bx^{n+1} + {n \choose 2}a^{n-2} b^2x^{n+2} - ... + (-1)^n{n \choose n}b^n x^{2n}}{n!}\\& = & \dfrac{m_n x^n + m_{n+1}x^{n+1} + m_{n+2}x^{n+2} + ... + m_{2n} x^{2n}}{n!}\\\end{eqnarray}$

In other words, $f(x)$ is a degree $2n$ polynomial where the coefficient of $x^k$ is given by $\dfrac{m_k}{n!}$, and the lowest power is $x^n$.

It should be clear that if we differentiate $f(x)$ less than $n$ times we will have a polynomial with no constant term. Hence $f^{(k)}(0) = 0$ for $k \lt n$. Similarly if we differentiate more than $2n$ times we will have no terms left and so $f^{(k)}(0) = 0$ for $k \lt 2n$.

For $n \lt k \lt 2n$ there will be a single constant term, and as the remaining terms will contain an "$x$", it follows that $f^{(k)}(0)$ will be equal to this constant term. If $x^k$ is differentiated $k$ times then it becomes $k \times (k-1) \times (k-2) \times ... \times 2 \times 1 = k!$.

Therefore the constant term will be $\dfrac{k! m_k}{n!}$ and as $k \gt n$ this will be integer.

Thus $f^{(k)}(0)$ is integer for all $k$.

As we have already shown that $f(x) = f(\pi - x)$ we get the following by repeated use of the chain rule.

$\begin{eqnarray} f^{(1)}(x) & = & -f^{(1)}(\pi - x)\\f^{(2)}(x) & = & f^{(2)}(\pi - x)\\f^{(3)}(x) & = & -f^{(3)}(\pi -x)\\... & = & ...\\f^{(k)}(x) & = & (-1)^k f^{(k)}(\pi - x)\end{eqnarray}$

Therefore $f^{(k)}(0) = (-1)^k f^{(k)}(\pi)$ and we can see that $f^{(k)}(\pi)$ must also be integer for all $k$.

Hence $G(0)$ and $G(\pi)$ by definition are integer.

From,

$G(x) = f(x) - f^{(2)}(x) + f^{(4)}(x) - ... + (-1)^n f^{(2n)}(x)$

we differentiate twice to get,

$G^{(2)} = f^{(2)}(x) - f^{(4)}(x) + f^{{6}}(x) - ... + (-1)^n f^{(2n+2)}(x)$

As we have already noted, $f^{(2n+2)}(x) = 0$.

$\therefore G(x) + G^{(2)} = f(x$)

Let $z = G^{(1)}(x)sin(x) - G(x)cos(x)$ and if we differentiate, using the product rule, we get:

$\begin{eqnarray}\dfrac{dz}{dx} & = & \left(G^{(2)}(x)sin(x) + G^{(1)}(x)cos(x)\right) - \left(G^{(1)}(x)cos(x) - G(x)sin(x)\right)\\& = & G^{(2)}(x)sinx(x) + G(x)sin(x)\\& = & \left(G^{(2)}(x) + G(x))sin(x\right)\\& = & f(x)sin(x)\end{eqnarray}$

Hence by using the Fundamental Theorem of Calculus (if the differential of X is Y then the integral of Y is X) it follows that the integral of $f(x)sin(x)$ must be $z = G^{(1)}(x)sin(x)-G(x)cos(x)$.

$\begin{eqnarray}\therefore \int_{0}^{\pi}f(x)sin(x) \,dx & = & \left(G^{(1)}(\pi)sin(\pi)-G(\pi)cos(\pi)\right) - \left(G^{(1)}(0)sin(0) - G(0)cos(0)\right)\\& = & G(0) - G(\pi)\end{eqnarray}$

And we have already established that $G(0)$ and $G(\pi)$ are integer.

Recall that $f(x) = \dfrac{x^n(a - bx)^n}{n!},$ so for $0 \lt x \lt \pi, 0 \lt sin(x) \lt 1, 0 \lt x^n \lt \pi^n.$

As $x \gt 0, bx \gt 0 \Rightarrow 0 \lt a - bx \lt a$.

$\therefore 0 \lt (a - bx)^n \lt a^n$

$\therefore 0 \lt f(x)sin(x) \lt \dfrac{\pi^n a^n}{n!} = \dfrac{(a\pi)^n}{n!}$

In other words, no point on the curve $y = f(x)sin(x)$ between $0$ and $\pi$ is is higher than $\dfrac{(a\pi)^n}{n!},$ so the area below the curve must be less than the rectangular region given by $\pi \times \dfrac{(a\pi)^n}{n!}$.

$\therefore 0 \lt \displaystyle{\int_{0}^{\pi}f(x)sin(x) \,dx} \lt \dfrac{a^n \pi^{n+1}}{n!}$

Consider the sequence given by $u_n = \dfrac{a^n}{n!}$.
The next term, $u_{n+1} = \dfrac{a^{n+1}}{(n+1)!} = \dfrac{a \cdot a^n}{(n+1)n!} = \dfrac{a}{n+1}u_n.$

However, it can be see that as $n \to \infty, \dfrac{a}{n+1} \to 0$.

$\therefore \displaystyle{\lim_{n \to \infty}\left(\dfrac{a^n}{n!}\right)} = 0,$ and for sufficiently large $n, \dfrac{a^n\pi^{n+1}}{n!} \lt 1$.

$\therefore 0 \lt \displaystyle{\int_{0}^{\pi} f(x)sin(x) \,dx} \lt 1$

But we have already shown that the value of the integral is integer. Thus by reductio ad absurdum we prove that our initial assumption that $\pi$ could be written as the ratio of two integers is false and $\pi$ must be irrational. Q.E.D.

Problem ID: 371 (24 Dec 2009)     Difficulty: 4 Star

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