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The Hermite theorem states that the base $e$ of the natural logarithm is a transcendental number.

We use these countability facts: the integers are countable; a finite Cartesian product of countable sets is countable; a countable union of countable sets is countable; and every subset of a countable set is countable. Finite products can be enumerated by listing index tuples in increasing maximum coordinate, and countable unions by diagonal enumeration of pairs of indices.

Fix $d\ge 1$. For each degree bound $D$, there are finitely many monomials $X_1^{{\alpha }_1}\cdots X_d^{{\alpha }_d}$ with ${\alpha }_1+\cdots +{\alpha }_d\le D$. An integer polynomial of total degree at most $D$ is specified by an integer coefficient for each of them, so these polynomials form a finite Cartesian product of copies of $\mathbb{Z}$ and are countable. Every polynomial has some finite degree bound. Taking the union over $D=0,1,2,\dots$ proves the countability of integer polynomial rings:It is infinite because it contains all integer constant polynomials.

For $d=1$, each nonzero polynomial has only finitely many real roots: repeated application of the factor theorem bounds their number by its degree. The real algebraic numbers are the union of these finite root sets over the countably many nonzero integer polynomials. Thus they are countable. If the real transcendental numbers were countable too, their union with the algebraic numbers would make $\mathbb{R}$ countable, contrary to the allowed uncountability assumption. Hence there are uncountably many real transcendental numbers.

To construct an algebraically independent real sequence, choose $x_1$ outside the algebraic numbers. Suppose $x_1,\dots ,x_d$ already satisfy the required nonvanishing condition. For each nonzero $P\in \mathbb{Z}[X_1,\dots ,X_d,T]$, writeAt least one coefficient polynomial $P_j$ is nonzero, and the inductive hypothesis makes its value at $(x_1,\dots ,x_d)$ nonzero. Consequently $P(x_1,\dots ,x_d,T)$ is a nonzero one-variable polynomial and has only finitely many real roots. There are countably many choices of $P$, so the union of all these forbidden roots is countable. Choose $x_{d+1}$ outside it, which is possible because $\mathbb{R}$ is uncountable. This extends the nonvanishing condition to $d+1$. Recursion therefore givesIn particular the chosen numbers are distinct, since a relation $X_i-X_j$ is forbidden. Clearing rational denominators shows that every finite initial segment consists of algebraically independent elements over $\mathbb{Q}$.

The first series is telescoping, sinceIts $N$th partial sum is $1-1/(N+1)$, soa rational number.

The second series converges absolutely, for example by the ratio test, and is the odd part of the exponential series:This value is irrational. Indeed, if $a=\sinh 1$ were algebraic, then $e$ would satisfymaking $e$ algebraic over the algebraic numbers and hence algebraic, contrary to the Hermite theorem on the transcendence of e. Thus the second sum is in fact transcendental.

LetThe tail satisfiesFor every fixed positive integer $d$, this is smaller than $q_N^{-d}$ for all sufficiently large $N$.

The binary expansion has $1$ digits at factorial positions and arbitrarily long blocks of zeros, but it is neither eventually zero nor eventually periodic. Since a rational number has an eventually periodic binary expansion, $\alpha$ is irrational. If it were an algebraic number of degree $d$, part (b), or equivalently the Liouville approximation theorem, would give a fixed positive multiple of $q^{-d}$ as a lower bound for all rational approximations. The fractions $p_N/q_N$ violate that bound. Thus $\alpha$ is a transcendental number.

This can be false. For any nonzero transcendental number $\beta$, the number $\gamma =1/\beta$ is also transcendental, since otherwise its reciprocal would be algebraic. Yet $\beta \gamma =1$ is algebraic.

LetIf $x$ were rational, thenwould make $\sqrt[3]{6}=(x^3-5)/(3x)$ rational. But if $\sqrt[3]{6}=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by unique prime factorization: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore

Use the given convergent seriesSuppose this number were $a/b\in \mathbb{Q}$. Choose $N$ large enough that $b\mid (2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their differencewould therefore be an integer. It is positive, whilefor sufficiently large $N$, a contradiction. Hence

A transcendental number is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. LetIf $a\ne 0$ and $y$ were an algebraic number, then $e$ would satisfyover the algebraic extension $\mathbb{Q}(y)$. By transitivity of algebraic extensions, $e$ would be algebraic over $\mathbb{Q}$, contradicting its transcendence. If $a=0$, then $b\ne 0$ and $e=b/y$ would again be algebraic. Thus