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Liouville approximation theorem states that if an irrational algebraic number $\alpha$ has degree $d\ge 2$, then some $C(\alpha )>0$ satisfiesfor every rational $p/q$ with $q>0$.

The omitted $n=0$ term is rational and does not affect transcendence. LetThe decimal expansion has ones at the increasingly separated positions $n^n$ and zeros elsewhere, so it is not eventually periodic and $\xi$ is irrational. Moreover,For every fixed $d$, the exponent $(N+1{)}^{N+1}$ eventually exceeds $d{N}^N$ by an arbitrarily large amount, so this upper bound is smaller than $C{q}_N^{-d}$. Liouville's theorem therefore rules out every finite algebraic degree, proving that $\xi$ is transcendental.

There are only countably many integer polynomials and each has finitely many roots, so the algebraic numbers are countable. Since $\mathbb{R}$ is uncountable, its complement, the set of transcendental numbers, is uncountable.

A rational map $X⇢Y$ between irreducible projective varieties is an equivalence class of morphisms from nonempty open subsets of $X$ to $Y$, with representatives identified when they agree on a nonempty open subset. It is regular at $p$ if some representative is defined on a neighborhood of $p$.

Forthe coordinates vanish together exactly at $(0:1:0)$ and $(1:0:0)$. Near the first point, the paths $z=0$ and $x=0$ have limiting images $(1:0:0)$ and $(0:0:1)$; near the second, the paths $z=0$ and $y=0$ have limiting images $(1:0:0)$ and $(0:1:0)$. No continuous extension exists there. Elsewhere the coordinates do not vanish together, so the map is regular.

DefineWhere all coordinates are nonzero,and similarly $\varphi \psi$ is the identity. Thus $\varphi$ is birational and is an isomorphism on ${\mathbb{P}}^2\setminus Z(xyz)$.

For irreducibility of $P=x^2{z}^4-x^3{y}^3+z^6$, dehomogenize at $z=1$. Over $k(x)$ this is, up to a unit,The rational function on the right is not a cube, since its valuation at either root of $x^2+1$ is one. The cubic is therefore irreducible over $k(x)$; Gauss's lemma gives irreducibility in $k[x,y]$. Since $z$ does not divide $P$, homogenization preserves irreducibility.

Writing $(u:v:w)=(xy:xz:z^2)$ transforms the equation intoIts partial derivatives are $-3u^2$, $2vw$, and $v^2+3w^2$ up to common signs. Simultaneous vanishing forces $u=v=w=0$, impossible projectively, so $C$ is nonsingular. The inverse $\psi$ restricts on dense open subsets, proving that $V$ and $C$ are birational.