Proof that Pi is Irrational

Theorem 1: The number $\pi$ is irrational.

There are many proofs to show that $\pi$ is irrational. The proof below is due to Ivan Niven.

• Proof: Suppose instead that $\pi$ is rational. Then there exists integers $a$ and $b$ with $b \neq 0$ such that $\displaystyle{\pi = \frac{a}{b}}$. Define:

(1)

\begin{align} \quad f(x) = \frac{x^n (a - bx)^n}{n!} \end{align}

(2)

\begin{align} \quad F(x) = f(x) - f^{(2)}(x) + f^{(4)}(x) - ... (-1)^n f^{(2n)}(x) \end{align}

• Note that $f^{(k)}(0)$ and $f^{(k)}(\pi)$ are integers for every nonnegative integer $k$.

• We now compute the derivative of $F'(x) \sin x - F(x) \cos x$ by the product rule for differentiation to get:

(3)

\begin{align} \quad [F'(x) \sin x - F(x) \cos x]' &= F''(x) \sin x + F'(x) \cos x - F'(x) \cos x + F(x) \sin x \\ &= F''(x) \sin x + F(x) \sin x \\ &= [F''(x) + F(x)]\sin x \end{align}

• Observe that:

(4)

\begin{align} \quad F''(x) + F(x) &= \left [ f(x) - f^{(2)}(x) + f^{(4)}(x) - ... (-1)^n f^{(2n)}(x) \right ]'' + [f(x) - f^{(2)}(x) + f^{(4)}(x) - ... (-1)^n f^{(2n)}(x)] \\ &= \left [f^{(2)}(x) - f^{(4)}(x) + ... + (-1)^n f^{(2n + 2)}(x) \right ] + [f(x) - f^{(2)}(x) + f^{(4)}(x) - ... (-1)^n f^{(2n)}(x)] \\ &= f(x) + (-1)^n f^{(2n + 2)}(x) \end{align}

• But the $(2n + 2)^{\mathrm{th}}$ derivative of $f$ is zero since $f(x)$ is a polynomial of degree $2n$. So $F''(x) + F(x) = f(x)$ and hence:

(5)

\begin{align} \quad [F'(x) \sin x - F(x) \cos x]' = f(x) \sin x \end{align}

• We integrate $f(x) \sin x$ over $[0, \pi]$ by using the Fundamental theorem of Calculus to get:

(6)

\begin{align} \quad \int_0^{\pi} f(x) \sin x \: dx &= [F'(x) \sin x - F(x) \cos x]_0^{\pi} \\ &= [F'(\pi) \sin \pi - F(\pi) \cos \pi] - [F'(0) \sin 0 - F(0) \cos 0 ] \\ &= F(\pi) + F(0) \quad (*) \end{align}

• The values of $F(\pi)$ and $F(0)$ are completely determined by sums of the values of $f^{(k)}(0)$ and $f^{(k)}(\pi)$ which we have already noted to be integer values. So $F(\pi) + F(0)$ is an integer.

• We now show that $F(\pi) + F(0)$ is a positive integer. Observe that on the interval $(0, \pi)$ we have that $f(x) > 0$ and $\sin x > 0$. So $f(x) \sin x > 0$ on this interval. So from $(*)$ we must have that $F(\pi) + F(0) > 0$.

• Now note that for every $x \in (0, \pi)$ we have that $0 < \sin x < 1$. So on this interval, $0 < f(x) \sin x < f(x)$. Also on this interval, $a - bx < a$. So $(a - bx)^n < a^n$. But also $x^n < \pi^n$ on this interval. So $\displaystyle{f(x) < \frac{\pi^n a^n}{n!}}$. Putting this all together tells us that for every $x \in (0, \pi)$ we have that:

(7)

\begin{align} \quad 0 < f(x) \sin x < \frac{\pi^n a^n}{n!} \end{align}

• Integrating both sides of this equation on $[0, \pi]$ gives us:

(8)

\begin{align} \quad 0 < F(\pi) + F(0) \overset{(*)} = \int_0^{\pi} f(x) \sin x \: dx < \int_0^{\pi} \frac{\pi^n a^n}{n!} \: dx = \frac{\pi^{n+1} a^n}{n!} \end{align}

• But as $n \to \infty$, the righthand side of the inequality above goes to $0$ - which is a contradiction, since $F(\pi) + F(0)$ is supposed to be a positive integer. Therefore the assumption that $\pi$ was rational was false. So $\pi$ is irrational. $\blacksquare$