Ivan Niven's Proof that \(\pi\) is Irrational

Ivan Niven (1915-1999)

Ivan Niven gave a one page proof that \(\pi\) is irrational in 1946.^[1] I had to work a bit to understand it, so thought a write-up was in order. The proof is by contradiction. Niven starts by assuming that \(\pi = a/b\) for positive integers \(a, b\). Then define:
\begin{align*}
f(x) = f_n(x) = \frac{x^n(a-bx)^n}{n!},
\end{align*}
for some positive integer \(n\). I'm generally going to stick to the notation \(f(x)\) in what follows, otherwise things get a bit top-heavy. Just keep in mind that \(f(x)\) depends on \(n\).

The first step is to show that \(f(x)\) and all its derivatives have integral values at \(x=0\). Consider when \(n=3\) for example:
\begin{align*}
f(x) &= \frac{x^3(a-bx)^3}{3!}\\[0.7em]
g(x) &= 6 \cdot f(x) = x^3(a-bx)^3\\[0.3em]
&= a_3x^3 + a_4x^4 + a_5x^5 + a_6x^6,
\end{align*}
where the \(a_i\) are nonzero integers because they are monomials of the form \(\pm ca^ib^j\), where \(c\) is a binomial coefficient arising from the expansion of \((a-bx)^3\). Therefore:
\begin{equation*}
\begin{split}
g'(x) &= 3 \cdot a_3x^2 + 4 \cdot a_4x^3 + 5 \cdot a_5x^4 + 6 \cdot a_6x^5\\[0.3em]
g''(x) &= 2 \cdot 3 \cdot a_3x + 3 \cdot 4 \cdot a_4x^2 + 4 \cdot 5 \cdot a_5x^3 + 5 \cdot 6 \cdot a_6x^4\\\\
\hline\\
g'''(x) &= 1 \cdot 2 \cdot 3 \cdot a_3 + 2 \cdot 3 \cdot 4 \cdot a_4x + 3 \cdot 4 \cdot 5 \cdot a_5x^2 + 4 \cdot 5 \cdot 6 \cdot a_6x^3\\[0.3em]
g^{(4)}(x) &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot a_4 + 2 \cdot 3 \cdot 4 \cdot 5 \cdot a_5x + 3 \cdot 4 \cdot 5 \cdot 6 \cdot a_6x^2\\[0.3em]
g^{(5)}(x) &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot a_5 + 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot a_6x\\[0.3em]
g^{(6)}(x) &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot a_6.\\[0.3em]
\end{split}
\end{equation*}
\(g(x)\) and its first two derivatives (above the line) have \(x\) as a factor, so they evaluate to 0 when \(x = 0\). The seventh and subsequent derivatives further along in this list are uniformly 0, so they too evaluate to 0 when \(x=0\). The same is true of \(f(x) = g(x)/6\). Note that each of the numerical coefficients of the third derivative (just underneath the line) is a product of three consecutive integers, so each of those coefficients is a multiple of both 2 and 3 and therefore of \(2 \cdot 3 = 6\), and so the derivative is too. That is equally true of the fourth, fifth, and sixth derivatives, so they too are multiples of 6. Consequently, whenever \(x\) is an integer, \(f^{(m)}(x) = g^{(m)}(x)/6\) is also an integer when \(3 \leq m \leq 6\). In particular, \(f^{(m)}(0)/6\) is an integer when \(3 \leq m \leq 6\). This establishes that \(f(x)\) and all its derivatives have integral values at \(x=0\) when \(n=3\).

The proof proceeds similarly for any whole number \(n : \; f^{(m)}(0) = 0\) when \(m < n\) or \(m > 2n\). And for \(n \leq m \leq 2n, \; g^{(m)}(x)\) is a multiple of \(n!\), so \(f^{(m)}(x) = g^{(m)}(x)/n!\) is an integer and in particular, \(f^{(0)}\) is an integer. What is true for \(n = 3\) is true for all \(n\): \(f(x)\) and all its derivatives have integral values at \(x=0\).

Note that:
\begin{align*}
f(a/b-x) &= \frac{(a/b-x)^n \cdot \big{(}a-b(a/b-x)\big{)}^n}{n!}\\[0.7em]
&= \frac{(a/b-x)^n \cdot (a-a+bx)^n}{n!}\\[0.7em]
&= \frac{(a/b-x)^n \cdot (bx)^n}{n!}\\[0.7em]
&= \frac{(a-bx)^n \cdot x^n}{n!}\\[0.7em]
&= f(x).\tag{1}
\end{align*}
Plugging in \(x=0\) into (1) results in \(f(\pi)=f(a/b) = f(0)\). Put \(h(x)=a/b-x\). Then:
\begin{align*}
f(x) &= f(a/b-x) = f(h(x))\\
\therefore f'(x) &= f(h(x))'\\
&=f'(h(x)) \cdot h'(x)\\
&= f'(h(x)) \cdot (-1) = -f'(g(x)).\tag{2}
\end{align*}
Plugging \(x=0\) into (2) results in \(f'(a/b) = -f'(0)\). The sign reverses again when going to the second derivative, and continuing in this fashion, \(f^{(m)}(a/b) = \pm f^{(m)}(0)\). In short, \(f^{(m)}(\pi)=f^{(m)}(a/b)\) is an integer just as \(f^{(m)}(0)\) is.

String together every other derivative like this:
\begin{align*}
F(x) = f(x) -f''(x) + f^{(4)}(x) - \cdots + (-1)^nf^{(2n)}(x).
\end{align*}
Then by applying the product rule for derivatives and then canceling opposite-signed values in \(F''(x)+F(x)\):
\begin{align*}
\{F'(x)\sin{x} - F(x)\cos{x}\}' &= F''(x)\sin{x} + F'(x)\cos{x}\\
&\hspace{6pt}-F'(x)\cos{x} + F(x)\sin{x}\\
&= F''(x)\sin{x} + F(x)\sin{x}\\
&= (F''(x) + F(x))\sin{x}\\
&= f(x)\sin{x}.
\end{align*}
Taking the definite integral:
\begin{align*}
\int_0^\pi f(x)\sin{x}dx &= \Big{[}F'(x)\sin{x} - F(x)\cos{x}\Big{]}_0^\pi\tag{3}\\[0.7em]
&= F'(\pi)\sin{\pi} - F(\pi)\cos{\pi} - F'(0)\sin{0} + F(0)\cos{0}\\[0.3em]
&= F(\pi) + F(0).
\end{align*}
\(F(x)\) is a sum of derivatives \(f^{(j)}(x)\) and the latter all evaluate to a integer at \(x=\pi\), so \(F(\pi)\) is an integer; and similarly \(F(0)\) is an integer and their sum is as well. \(f(x)\) and \(\sin(x)\) are positive and bounded between 0 and \(\pi = a/b\), so the integral in (3) is a positive finite value (not 0!). Between 0 and \(\pi\), \(0 < x^n < \pi^n\) and \(0 < (a-bx)^n < a^n\), so:
\begin{align*}
0 < f(x)\sin{x} < \frac{\pi^na^n}{n!} = \frac{(\pi a)^n}{n!}.\tag{4}
\end{align*}
\(\pi a\) is just a constant, so the expression on the right side of (4) goes to 0 with \(n\). Therefore the integrand in (3) can be made as small as desired by choosing \(n\) large enough and the same holds for the integral as well. It follows that \(F(\pi) + F(0)\) is a positive integer less than 1 for large \(n\). Contradiction!

The entire development depends on assuming that \(\pi = a/b\) is rational, and since that assumption leads to a contradiction, \(\pi\) must be irrational. QED.

Mike Bertrand

October 4, 2024

^ 1. "A Simple Proof that \(\pi\) is Irrational", by Ivan Niven, Bulletin of the American Mathematical Society, 53 (1947), p.509.