The Number e and Its Series Expansion

Rational, irrational, transcendental

What really needs to be proved

A real number xx is rational if it can be written as a quotient of two integers:

x∈Q  ⟺  ∃(p,q)∈Z×N∗,x=pqx \in \mathbb{Q} \iff \exists (p, q) \in \mathbb{Z} \times \mathbb{N}^*,\quad x = \frac{p}{q}

It is irrational otherwise. Proving that a number is irrational therefore means proving a negative property: no fraction, however complicated, will do. That is why almost all such proofs proceed by contradiction — we assume the fraction is given, and derive an arithmetic contradiction from it.

Numerical computation proves nothing

The first decimal digits of ee betray no structure:

e=2.718 281 828 459 045 235 360…e = 2.718\,281\,828\,459\,045\,235\,360\ldots

The pattern 18281828 appears twice in a row, which is pure coincidence. Above all, no finite computation can settle the matter: every rational number with a very large denominator has a periodic decimal expansion, but its period may be gigantic.

Number Decimal expansion Nature
13\dfrac{1}{3} 0.333…0.333\ldots, period 11 rational
17\dfrac{1}{7} 0.142857‾0.\overline{142857}, period 66 rational
197\dfrac{1}{97} period of 9696 digits rational
2\sqrt{2} infinite, non-periodic algebraic irrational
ee infinite, non-periodic transcendental irrational

Irrational is not transcendental

Two notions not to be confused.

A real number is algebraic if it is a root of a nonzero polynomial with integer coefficients, and transcendental otherwise. Thus 2\sqrt{2} is irrational, but algebraic, since it is a root of X2−2X^2 - 2. The number ee, on the other hand, is irrational and transcendental: Hermite proved it in 1873, and Lindemann did the same for π\pi in 1882.

Q⊂{algebraic numbers}⊂R\mathbb{Q} \subset \left\{\text{algebraic numbers}\right\} \subset \mathbb{R}

The irrationality of ee is therefore a strictly weaker result than its transcendence — but its proof fits in half a page, whereas Hermite's requires several pages of analysis.

Classic pitfall

Writing "ee is irrational because its decimal expansion is infinite". That argument is wrong: 13\dfrac{1}{3} also has an infinite decimal expansion. The correct criterion is the absence of periodicity, and it is useless in practice, since one can never verify the absence of a period from a finite number of decimal digits.