The Irrationality of e and Pi

Some irrational numbers, most notably $ e $ and $ \pi $, are not defined as roots of polynomials but rather as limits of infinite processes. For such numbers, the Pythagorean divisibility approach does not apply directly. We need techniques that exploit series structure and integral representations.

Euler proved the irrationality of $ e $ in 1737 using continued fractions. Fourier gave a simpler proof around 1815 using the series definition. Lambert proved $ \pi $ irrational in 1761 via continued fractions for the tangent function, and Niven produced a remarkably short calculus-based proof in 1947.

The golden ratio rectangle

Fourier’s Proof That e Is Irrational

Euler’s Number

Definition (Euler’s Number). Euler’s number is defined by the infinite series:

$$ e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \cdots $$

This series converges very rapidly because the factorials in the denominators grow extremely fast.

The Proof

Important: Theorem (Fourier, c. 1815). Euler’s number $ e $ is irrational.

Proof. Assume $ e = a/b $ where $ a, b $ are positive integers.

Choose any integer $ n > \max(b, 1) $. Define:

$$ N = n!\left(e – \sum_{k=0}^{n} \frac{1}{k!}\right). $$

This is $ n! $ times the “tail” of the series for $ e $ (all terms after the $ (n+1) $-th).

Step 1: $ N $ is a positive integer.

$ N > 0 $ because the tail of the convergent series is positive.

We show $ N $ is an integer by examining each piece:

$$ N = n!\cdot e – n!\sum_{k=0}^{n}\frac{1}{k!} = n!\cdot\frac{a}{b} – \sum_{k=0}^{n}\frac{n!}{k!}. $$

Since $ n > b $, the factor $ b $ divides $ n! $, so $ n! \cdot a/b $ is an integer. Each term $ n!/k! $ is an integer for $ k \leq n $. Therefore $ N $ is an integer.

Step 2: $ N < 1 $.

Expand the tail:

$$ \begin{aligned} N &= n!\left(\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots\right) \\ &= \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \cdots \end{aligned} $$

Each denominator is at least as large as the corresponding power of $ (n+1) $:

$$ N < \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \cdots = \frac{1/(n+1)}{1 – 1/(n+1)} = \frac{1}{n}. $$

Since $ n \geq 2 $, we have $ N < 1/n \leq 1/2 < 1 $.

Step 3: Contradiction.

$ N $ is a positive integer less than 1. But there are no positive integers less than 1.

Therefore $ e $ is irrational. $ \blacksquare $

This proof is often attributed to Fourier (c. 1815), though Euler had already established the irrationality of $ e $ in 1737 using continued fractions. Fourier’s argument is notable for its elegance and accessibility: it uses nothing beyond the definition of $ e $ and basic arithmetic.

A Criterion for Irrationality via Series

The structure of Fourier’s proof can be abstracted into a general criterion.

Theorem (Irrationality Criterion). Let $ x = \sum_{n=0}^{\infty} a_n $ be a convergent series of rationals. If for every positive integer $ N $, the quantity

$$ T_N = N! \sum_{n=N+1}^{\infty} a_n $$

is a positive non-integer, then $ x $ is irrational.

Fourier’s proof of the irrationality of $ e $ is essentially an application of this criterion with $ a_n = 1/n! $.

The Irrationality of $ e^2 $

Theorem. $ e^2 $ is irrational.

Proof (sketch). We use the series:

$$ e^2 = \sum_{n=0}^{\infty} \frac{2^n}{n!}. $$

The approach mirrors Fourier’s proof, but a direct application runs into difficulty: the bound on the tail involves $ 2^{n+1}/(n-1) $, which does not go to zero fast enough. Instead, one uses the fact that $ e^2 $ and $ e^{-2} $ are linked through

$$ e^{-2} = \sum_{n=0}^{\infty} \frac{(-2)^n}{n!}. $$

By working with both $ e^2 $ and $ e^{-2} $ simultaneously, one can squeeze out the required contradiction. If $ e^2 = a/b $, then $ e^{-2} = b/a $, and the alternating series for $ e^{-2} $ provides the tighter bounds needed to force a quantity to be simultaneously a nonzero integer and less than 1 in absolute value.

The full details can be found in Hardy and Wright’s An Introduction to the Theory of Numbers, Theorem 45. $ \blacksquare $

The Continued Fraction of e

Euler discovered a beautiful pattern in the continued fraction expansion of $ e $.

Theorem (Euler, 1737).

$$ e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, \ldots] $$

The pattern is: $ [2; 1, 2k, 1] $ repeating for $ k = 1, 2, 3, \ldots $. More precisely, the continued fraction coefficients satisfy:

$$ a_{3k-1} = 2k \quad (k = 1, 2, 3, \ldots), \qquad a_n = 1 \text{ otherwise (for } n \geq 1\text{)}. $$

Since this continued fraction is infinite, $ e $ is irrational. Furthermore, the beautiful pattern in its coefficients reflects the deep structure of $ e $.

Euler also established generalized continued fractions for $ e $:

$$ e = 1 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{2}{3 + \cfrac{3}{4 + \cfrac{4}{5 + \ddots}}}}} $$

and

$$ e = 1 + \cfrac{2}{1 + \cfrac{1}{6 + \cfrac{1}{10 + \cfrac{1}{14 + \cfrac{1}{18 + \ddots}}}}} $$

These are not simple continued fractions (the numerators are not all 1), but they too are infinite.

Lambert’s Proof That Pi Is Irrational

Proving that $ \pi $ is irrational is substantially harder than proving the irrationality of $ \sqrt{2} $ or $ e $. The number $ \pi $ is not defined by a simple algebraic equation or a rapidly converging series with factorial denominators. The first proof was given by Johann Heinrich Lambert in 1761.

Lambert’s Approach

Lambert proved that $ \tan(x) $ is irrational whenever $ x $ is a nonzero rational number. Since $ \tan(\pi/4) = 1 $ is rational, it follows that $ \pi/4 $, and hence $ \pi $, must be irrational.

Lambert’s proof uses the continued fraction expansion:

$$ \tan(x) = \cfrac{x}{1 – \cfrac{x^2}{3 – \cfrac{x^2}{5 – \cfrac{x^2}{7 – \ddots}}}} $$

He showed that if $ x = p/q $ is rational and nonzero, then this continued fraction is irrational, contradicting $ \tan(\pi/4) = 1 $ if $ \pi $ were rational.

The Continued Fraction of $ \pi $

Theorem.

$$ \pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, \ldots]. $$

Unlike $ \sqrt{2} $ or $ e $, the continued fraction of $ \pi $ has no known pattern. Its coefficients appear essentially random (though they are, of course, completely determined). The continued fraction is infinite, confirming that $ \pi $ is irrational.

There are also beautiful generalized continued fractions for $ \pi $:

$$ \pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \cfrac{7^2}{2 + \ddots}}}}} $$

and

$$ \pi = 3 + \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \cfrac{7^2}{6 + \ddots}}}} $$

Niven’s Proof That Pi Is Irrational

Ivan Niven gave a remarkably short proof of the irrationality of $ \pi $ in 1947 using only basic calculus.

Important: Theorem (Niven, 1947). $ \pi $ is irrational.

Proof. Assume $ \pi = a/b $ with $ a, b \in \mathbb{Z}^+ $. For a positive integer $ n $ (to be chosen later), define the polynomial:

$$ f(x) = \frac{x^n(a – bx)^n}{n!}. $$

Note that $ f(x) = f(\pi – x) $ (by the substitution $ a – bx = b(\pi – x) $, using $ a = b\pi $).

Properties of $ f $:

(i) For $ 0 < x < \pi $, we have $ 0 < x(\pi – x) \leq \pi^2/4 $. Hence:

$$ 0 < f(x) = \frac{b^n x^n(\pi – x)^n}{n!} \leq \frac{(b\pi^2/4)^n}{n!}. $$

This tends to 0 as $ n \to \infty $.

(ii) $ f(0) = 0 $ and $ f(\pi) = 0 $.

(iii) The $ k $-th derivative $ f^{(k)}(0) $ is an integer for every $ k \geq 0 $. This is because $ f(x) = \frac{1}{n!}\sum_{j=0}^{n}\binom{n}{j}a^{n-j}(-b)^j x^{n+j} $, so the $ k $-th derivative at 0 is an integer when $ k \geq n $, and 0 when $ k < n $.

(iv) By symmetry $ f(x) = f(\pi – x) $, we have $ f^{(k)}(\pi) = (-1)^k f^{(k)}(0) $, which is also an integer.

Now define:

$$ F(x) = f(x) – f''(x) + f^{(4)}(x) – f^{(6)}(x) + \cdots + (-1)^n f^{(2n)}(x). $$

One verifies that:

$$ \frac{d}{dx}\left[F'(x)\sin x – F(x)\cos x\right] = \left[F''(x) + F(x)\right]\sin x = f(x)\sin x. $$

The last equality holds because $ F”(x) + F(x) = f(x) $ by the telescoping construction of $ F $.

Therefore:

$$ \int_0^{\pi} f(x)\sin x\,dx = \left[F'(x)\sin x – F(x)\cos x\right]_0^{\pi} = F(\pi) + F(0). $$

By properties (iii) and (iv), $ F(0) $ and $ F(\pi) $ are both integers. So the integral is a positive integer.

But by property (i), for large $ n $:

$$ 0 < \int_0^{\pi} f(x)\sin x\,dx \leq \pi \cdot \frac{(b\pi^2/4)^n}{n!} \to 0. $$

For $ n $ sufficiently large, the integral is strictly between 0 and 1. But it must be a positive integer. Contradiction.

Therefore $ \pi $ is irrational. $ \blacksquare $

Niven’s proof is celebrated for requiring nothing beyond basic calculus and properties of polynomials. It avoids the continued fraction machinery of Lambert’s proof entirely.

Why Irrationality Matters

Irrationality proofs are more than mathematical curiosities. They reveal fundamental truths about the structure of the real number line.

Geometry

The Pythagorean discovery of $ \sqrt{2} $ showed that geometry and arithmetic are not perfectly aligned: the diagonal of a unit square has no “common measure” with the sides. This required a complete rethinking of the Greek theory of proportions.

Calculus

The number $ e $ is the base of the natural logarithm, the unique function equal to its own derivative, and the cornerstone of analysis. Its irrationality is essential to the structure of exponential growth and decay.

Physics

Both $ e $ and $ \pi $ appear throughout physics, in wave equations, quantum mechanics, thermodynamics, and general relativity. Their irrationality means that physical constants often cannot be expressed as simple fractions.

Music

The equal-tempered musical scale uses frequency ratios of $ 2^{k/12} $ for $ k = 0, 1, \ldots, 11 $. These are all irrational (except $ k = 0 $ and $ k = 12 $), which means no interval in equal temperament is a “pure” ratio.

Number Theory

Techniques from irrationality proofs, including divisibility, unique factorization, Bezout’s identity, and continued fractions, are foundational tools used throughout modern number theory.

Open Questions

The study of irrationality is far from complete. Open questions remain about the irrationality of numbers such as:

$ \pi + e $, is this irrational? Almost certainly, but unproven.
$ \pi \cdot e $, same situation.
$ \pi^e $, completely unknown.
Euler’s constant $ \gamma \approx 0.5772\ldots $, we do not even know if this is irrational.
Many values of the Riemann zeta function at odd integers.

These problems continue to drive research at the frontier of mathematics.

Summary: Choosing the Right Method

Number: $ \sqrt{n} $ (non-perfect square) | Best Method: Pythagorean | Key Idea: Parity / prime divisibility
Number: $ \log_p q $ ($ p,q $ distinct primes) | Best Method: Pythagorean | Key Idea: Unique factorization
Number: $ \sqrt{2} + \sqrt{3} $ | Best Method: Algebraic manipulation | Key Idea: Reduce to known irrational
Number: $ e $ | Best Method: Power series (Fourier) | Key Idea: Factorial tail argument
Number: $ \pi $ | Best Method: Niven’s integral | Key Idea: Polynomial integral trick
Number: Quadratic irrationals | Best Method: Continued fractions | Key Idea: Periodicity
Number: General algebraic irrationals | Best Method: Continued fractions | Key Idea: Non-termination
Exercises

Fill in the details of Fourier’s proof for $ e $ with $ n = 3 $. Compute $ N = 3!(e – \sum_{k=0}^{3} 1/k!) $ numerically and verify that $ 0 < N < 1 $.
The number $ e $ has the continued fraction $ [2; 1, 2, 1, 1, 4, 1, 1, 6, \ldots] $. Compute the first six convergents $ p_n/q_n $ and verify they approach $ e \approx 2.71828 $.
Explain why Fourier’s proof does not directly work for $ \pi $. What property of $ e $ does the proof exploit that $ \pi $ lacks?
In Niven’s proof, verify the identity $ F”(x) + F(x) = f(x) $ for the case $ n = 1 $, where $ f(x) = x(a – bx) $.
Lambert proved that $ \tan(x) $ is irrational for nonzero rational $ x $. Use this to prove that $ \pi/6 $ is irrational. (Hint: what is $ \tan(\pi/6) $?)
Prove that $ e $ is not a quadratic irrational, that is, $ e $ does not satisfy any equation $ ae^2 + be + c = 0 $ with integer coefficients. (Hint: use the irrationality of $ e^2 $.)
The number $ \pi $ has a continued fraction that begins $ [3; 7, 15, 1, 292, \ldots] $. The convergent $ [3; 7] = 22/7 $ is the famous approximation. Compute $ [3; 7, 15] $ and $ [3; 7, 15, 1] $ and determine which is closer to $ \pi $.
Prove that if $ e $ were rational, then $ e^{-1} $ would also be rational. Then adapt Fourier’s proof to show directly that $ e^{-1} $ is irrational (using the series $ e^{-1} = \sum_{n=0}^{\infty} (-1)^n/n! $).