A motivated and simple proof that pi is irrational

Today is 3/14/15 — Super Pi Day — so was I telling my 7-year-old son all about the number $\pi$ this afternoon. When I told him that $\pi$ keeps on going forever and ever he asked “How do you know that?” Although I don’t know a proof that I could explain to a 7-year-old, I wanted to record the following proof which uses only basic calculus. It is essentially Niven’s famous proof, as generalized by Alan Parks, but I have tried to write it in a way which is more motivated than the usual treatments. As a bonus, the proof also shows that e is irrational.

Let $c$ be a positive real number. Our goal is to assume that $c$ is rational and use this assumption to construct a continuous function $f(x)$ and a sequence of polynomials $g_k(x)$ for $k=1,2,\ldots$ with the following properties:

(1) $f(x)$ and all $g_k(x)$ are positive on the interval $(0,c)$ .

(2) There is a sequence of real numbers $M_k$ with $M_k \to 0$ as $k \to \infty$ such that for each $k$ , $|g_k(x)| \leq M_k$ for all $x \in [0,c].$

(3) The integral $\int_0^c f(x) g_k(x) dx$ is an integer for all $k$ .

To see why this gives a contradiction, note that (1) and (3) imply that $\int_0^c f(x) g_k(x) dx \geq 1$ for all $k$ , but if $L$ is the maximum value of $f(x)$ on $[0,c]$ then (2) implies that

$\int_0^c f(x) g_k(x) dx \leq \int_0^c L M_k dx \leq c \cdot L \cdot M_k$

which is less than 1 for $k$ sufficiently large.

Let $R$ be the set of all continuous functions $g(x)$ on $[0,c]$ with the property that both $g(0)$ and $g(c)$ are integers. Note that $R$ is closed under addition, subtraction, and multiplication (it is what mathematicians call a ring).

We define $R_*$ to be the set of all $g(x)$ in $R$ whose $k^{\rm th}$ derivative belongs to $R$ for all $k=1,2,\ldots$ . The product rule for derivatives shows that $R_*$ is closed under addition, subtraction, and multiplication, and by definition it is closed under taking derivatives.

Key Observation #1: If $g(x) \in R_*$ and $g(0) = g(c) = 0$ , then $\frac{g(x)^k}{k!} \in R_*$ for all $k=1,2,\ldots$

Proof: The proof is by induction on $k$ . The case $k=1$ is trivial. Assume that $\frac{g(x)^{k-1}}{(k-1)!} \in R_*$ . We have $\frac{g(x)^k}{k!} \in R$ by assumption, and by the chain rule we have $(\frac{g(x)^k}{k!})' = \frac{g(x)^{k-1}}{(k-1)!} g'(x)$ . Since $g'(x) \in R_*$ and $R_*$ is closed under multiplication, the result follows. Q.E.D.

We define $R^*$ to be the set of all $g(x)$ in $R$ with a $k^{\rm th}$ antiderivative in $R$ for all $k=1,2,\ldots$ More formally, $R^*$ consists of all functions $g(x)$ in $R$ such that for each $k=1,2,\ldots$ , there exists $G_k(x)$ in $R$ whose $k^{\rm th}$ derivative is $g(x)$ . Note that by definition we in fact have $G_k(x) \in R^*$ .

Key Observation #2: If $f(x) \in R^*$ and $g(x) \in R_*$ is a polynomial, then $\int_0^c f(x) g(x) dx$ is an integer.

Proof: The proof is by induction on the degree $d$ of $g(x)$ . If $d=0$ then $g(x)$ is constant, and the result follows easily from the fundamental theorem of calculus. Assume the result is true for polynomials of degree $d-1$ . Letting $F(x)$ be an antiderivative of $f(x)$ in $R^*(x)$ and applying integration by parts, we obtain

$\int_0^c f(x) g(x) dx = F(c) g(c) - F(0)g(0) - \int_0^c F(x) g'(x) dx$

which is an integer by the inductive hypothesis. Q.E.D.

We can now prove our main theorem, which as we will see implies that both $\pi$ and $e$ are irrational.

Main Theorem: Suppose there is a function $f(x) \in R^*$ with $f(x)>0$ for $x \in (0,c)$ . Then $c$ is irrational.

Proof: Assume for the sake of contradiction that $c = \frac{m}{n}$ with $m,n$ positive integers. Then the function $g(x)=x(m-nx)$ belongs to $R_*$ and satisfies $g(0) = g(c) = 0$ . (The derivative of $g(x)$ is $(m-nx) - nx$ and its second derivative is $-2n$ .) It follows from Key Observation #1 that $g_k(x) :=\frac{g(x)^k}{k!}$ belongs to $R_*$ for all $k=1,2,\ldots$ By Key Observation #2, $\int_0^c f(x) g_k(x) dx$ is an integer for all $k$ . On the other hand, both $f(x)$ and all $g_k(x)$ are positive on $(0,c)$ . Thus conditions (1) and (3) above are satisfied. To see that (2) is satisfied as well, let $M$ be the maximum value of $g(x)$ on $[0,c]$ . Then $g_k(x) \leq M_k := \frac{M^k}{k!}$ , which goes to zero as $k \to \infty$ . As discussed above, conditions (1),(2),(3) together now lead to a contradiction. Q.E.D.

Corollary 1: $\pi$ is irrational.

Proof: Take $f(x) = \sin(x)$ and $c = \pi$ . Q.E.D.

Corollary 2: $e$ is irrational.

Proof: Assume $e =\frac{m}{n}$ with $m,n$ positive integers and take $f(x) = n e^x$ and $c = 1$ . Q.E.D.

Concluding remarks

1. As mentioned in the introduction, this post is a reorganization and repackaging of the proof given in this article by Alan E. Parks, which originally appeared in the American Math Monthly. I first came across Parks’s proof in the book “Biscuits of Number Theory”, edited by Arthur Benjamin and Ezra Brown.

2. As Parks notes in his article, the main theorem easily implies the following generalizations of Corollary 1 and Corollary 2:

If $0 < |r| \leq \pi$ and if $\cos(r)$ and $\sin(r)$ are both rational, then $r$ is irrational.
If $r > 1$ is a positive rational number then $\ln(r)$ is irrational.

3. The strategy of proof here is a very natural one. Every irrationality proof that I know of for $\pi$ or $e$ involves reaching a contradiction by constructing a positive integer less than 1. And the argument used to bound the integral in (3) appears frequently (albeit in a different context) in analysis, especially elementary complex analysis. And the Key Observations, while clever, are simple consequences of the product rule for derivatives, the chain rule, and the fundamental theorem of calculus. So unlike other irrationality proofs that I’ve seen, I find this argument quite motivated. It is also accessible to anyone who has studied calculus and mathematical induction. And finally, it’s nice that the same argument proves irrationality of $\pi$ and $e$ at the same time.

4 thoughts on “A motivated and simple proof that pi is irrational”

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ianagol says:

September 11, 2019 at 6:36 pm

I know that this is a slick way to prove that π is irrational. However, I’ve always been fond of the original concept behind Lambert’s proof, the idea being that tan(r) is irrational for r rational. This follows from the continued fraction expansion tan(r) = r/(3-r^2/(5-r^2/(7-r^2… .

The point here is that if r is rational, the the partial convergents of this continued fraction give rational numbers which converge too quickly to tan(r) for tan(r) to be irrational. Of course the proof requires some care, both to prove convergence of the continued fractions and to show the limit is irrational. But I like this approach because of the analogy with other irrationality proofs using continued fractions. A complete argument is given here: http://www.maths.unsw.edu.au/~angell/5535/chapter7.ps

- Matt Baker says:
  
  September 11, 2019 at 8:17 pm
  
  Yes, Lambert’s continued fraction argument is also great! (And a similar analysis with tanh instead of tan proves that e is irrational, though of course this is a much easier fact than the irrationality of pi.)

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A motivated and simple proof that pi is irrational

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