Simple proofs: The irrationality of pi

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Mankind has been fascinated with $\pi$, the ratio between the circumference of a circle and its diameter, for at least 2500 years. Ancient Hebrews used the approximation 3 (see 1 Kings 7:23 and 2 Chron. 4:2). Babylonians used the approximation 3 1/8. Archimedes, in the first rigorous analysis of $\pi$, proved that 3 10/71 < $\pi$ < 3 1/7, by means of a sequence of inscribed and circumscribed triangles. Later scholars in India (where decimal arithmetic was first developed, at least by 300 CE), China and the Middle East computed $\pi$ ever more accurately. In 1665, Newton computed 16 digits, but, as he later confessed, “I am ashamed to tell you to how many figures I carried these computations, having no other business at the time.” In 1844 the computing prodigy Johan Dase produced 200 digits. In 1874, Shanks published 707 digits, although later it was found that only the first 527 were correct.

In the 20th century, $\pi$ was computed to thousands, then to millions, then to billions of digits, in part due to some remarkable new formulas and algorithms for $\pi$, and in part due to clever computational techniques, all accelerated by the relentless advance of Moore’s Law. The most recent computation, as far as the present author is aware, produced 22.4 trillion digits. One may download up to the first trillion digits here. For additional details on the history and computation of $\pi$, see The quest for $\pi$ and The computation of previously inaccessible digits of pi^2 and Catalan’s constant.

For many years, part of the motivation for computing $\pi$ was to answer the question of whether $\pi$ was a rational number — if $\pi$ was rational, the decimal expansion would eventually repeat. But given that no repetitions were found in early computations, many leading mathematicians in the 17th and 18th century concluded that $\pi$ must be irrational. In 1761, Johann Heinrich Lambert settled the question by proving that $\pi$ is irrational. Then in 1882 Ferdinand von Lindemann proved that $\pi$ is transcendental, which proved once and for all that the ancient Greek problem of squaring the circle is impossible (because ruler-and-compass constructions can only produce power-of-two degree algebraic numbers).

We present here what we believe to be the simplest proof that $\pi$ is irrational. It was first published in the 1930s as an exercise in the Bourbaki treatise on calculus. It requires only a familiarity with integration, including integration by parts, which is a staple of any high school or first-year college calculus course. It is similar to, but simpler than (in the present author’s opinion), a related proof due to Ivan Niven.

Gist of the proof:
The basic idea is to define a function $A_n(b)$, based on an integral from $0$ to $\pi$. This function has the property that for each positive integer $b$ and for all sufficiently large integers $n$, $A_n(b)$ lies strictly between 0 and 1. Yet another line of reasoning, assuming $\pi$ is a rational number, and applying integration by parts, concludes that $A_n(b)$ must be an integer. This contradiction shows that $\pi$ must be irrational.

THEOREM: $\pi$ is irrational.

Proof:
For each positive integer $b$ and non-negative integer $n$, define $$A_n(b) = b^n \int_0^\pi \frac{x^n(\pi – x)^n \sin(x)}{n!} \, {\rm d}x.$$ Note that the integrand function of $A_n(b)$ is zero at $x = 0$ and $x = \pi$, but is strictly positive elsewhere in the integration interval. Thus $A_n(b) > 0$. In addition, since $x (\pi – x) \le (\pi/2)^2$, we can write $$A_n(b) \le \frac{\pi b^n}{n!}\left(\frac{\pi}{2}\right)^{2n} = \frac{\pi (b \pi^2 / 4)^n}{n!},$$ which is less than one for large $n$, since, by Stirling’s formula, the $n!$ in the denominator increases faster than the $n$-th power in the numerator. Thus we have established, for any integer $b \ge 1$ and all sufficiently large $n$, that $0 < A_n(b) < 1.$

Now let us assume that $\pi = a/b$ for relatively prime integers $a$ and $b$. Define $f(x) = x^n (a – bx)^n / n!$. Then we can write, using integration by parts, $$A_n(b) = \int_0^\pi f(x) \sin(x) \, {\rm d}x = \left[-f(x) \cos(x)\right]_0^\pi – \left[-f'(x) \sin(x)\right]_0^\pi + \cdots$$ $$ \cdots \pm \left[f^{(2n)} (x) \cos(x)\right]_0^\pi \pm \int_0^\pi f^{(2n+1)} (x) \cos(x) \, {\rm d}x.$$ Now note that for $k = 1$ to $k = n$, the derivative $f^{(k)} (x)$ is zero at $x = 0$ and $x = \pi$. For $k = n + 1$ to $2n$, $f^{(k)} (x)$ takes various integer values at $x = 0$ and $x = \pi$; and for $k = 2n + 1$, the derivative is zero. Also, the function $\sin(x) $ is $0$ at $x = 0$ and $x = \pi$, while the function $\cos(x)$ is $1$ at $x = 0$ and $-1$ at $x = \pi$. Combining these facts, we conclude that $A_n (b)$ must be an integer. This contradiction proves that $\pi$ is irrational.

For other proofs in this series see the listing at Simple proofs of great theorems.

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