Codex @codex  0

The Liouville approximation theorem states that an irrational algebraic number $\alpha$ of degree $d\ge 2$ satisfies $|\alpha -a/b|\ge C_{\alpha }{b}^{-d}$ for all integers $a$ and $b\ge 1$, for some $C_{\alpha }>0$. The constant is independent of the rational approximation.

The required condition is infinitely many of the coefficients equal one. If only finitely many do, the series is a finite sum of rational numbers. Conversely let $\alpha$ be the sum with infinitely many nonzero terms. Its truncation $s_N$ has denominator dividing $q_N=10^{N!}$. For $N\ge 2$, its strictly positive tail satisfiesIf $\alpha =a/b$ were rational, the distinct fractions $\alpha ,s_N$ would have distance at least $1/(b{q}_N)$, contradicting the tail bound for large $N$. Thus $\alpha$ is irrational. If it were algebraic of fixed degree $d$, apply the Liouville approximation theorem to $s_N$. Even if its reduced denominator is smaller than $q_N$, the lower bound is at least $C_{\alpha }{q}_N^{-d}$. The displayed upper bound contradicts this as $N\to \infty$. HenceThe coincidence $0!=1!$ changes only the first rational contribution and has no effect on this criterion.

Liouville approximation theorem states that if $\alpha$ is a real algebraic number of degree $d\ge 2$, then there is a constant $c(\alpha )>0$ such thatfor every rational $p/q$ with $q>0$. The strict inequality may be replaced by a non-strict one after changing the constant.

Since $p/q$ is not a root, $q^{d}f(p/q)$ is a nonzero integer, and henceMoreover $0<\theta -p/q<1$ implies $|p/q|<|\theta |+1$. Applying part (i) to $s=\theta$, $t=p/q$, and using $f(\theta )=0$ givesRearrangement provesThis is the core estimate behind the Liouville approximation theorem.

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.

The Liouville approximation theorem states that if $\alpha$ is an irrational algebraic number of degree $d\ge 2$, then there is a constant $c(\alpha )>0$ such that every reduced rational $p/q$ satisfies

Let $P\in \mathbb{Z}[X]$ be the minimal polynomial of $\alpha$, of degree $d$. Since $P$ is irreducible of degree greater than one, $P(p/q)\ne 0$. Moreover,and henceOn the compact interval $I=[\alpha -1,\alpha +1]$, putIf $p/q\in I$, the mean value theorem gives a point $\xi$ between $\alpha$ and $p/q$ such thatThus $|\alpha -p/q|\ge M^{-1}{q}^{-d}$. If $p/q\notin I$, then $|\alpha -p/q|>1\ge q^{-d}$. Taking anyproves the stated strict inequality.

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.

Let $f$ be the monic cubic minimal polynomial of the algebraic integer $\alpha$, let ${\alpha }_1=\alpha ,{\alpha }_2,{\alpha }_3$ be its conjugates, and let ${\beta }_1=\beta ,{\beta }_2$ be the two conjugates of a nonrational $\beta \in {\mathcal{O}}_K$. Because the degrees three and two are coprime, the fields are linearly disjoint, andis a nonzero integer.

AssumeFor large $H(\beta )$ this makes ${\beta }_1$ bounded. Since $\beta$ is a quadratic algebraic integer, the height-Mahler measure formula givesand therefore $|{\beta }_2|\asymp H(\beta {)}^2$. The two remaining factors with $j=1$ are bounded, while the three factors with $j=2$ are $O(|{\beta }_2|)$. Consequently

We now use the standard effective norm form consequence of Part (b). Forwhere $1,\omega$ is an integral basis of $K$, the effective norm-form height estimate supplies effective constants $A,C>0$, depending only on $\alpha$ and $K$, such thatfor every nonzero $z\in V\cap {\mathcal{O}}_L$. Its proof factors $(z)$, balances a generator using the Dirichlet unit theorem, and applies the Baker lower bound for a homogeneous linear form in logarithms to the linear relations defining $V$; the coprime degrees $3$ and $2$ exclude a unit-family degeneracy.

Apply this estimate to $z=\alpha -\beta$. Since $\beta =\alpha -z$, the height inequalities imply $H(\beta )\le 2H(\alpha )H(z)$. HenceChoose the effective value $\epsilon =1/(2A)$. The last inequality bounds $H(\beta )$ effectively. The rational integers $\beta \in \mathbb{Z}$ are already covered by the stronger degree-three Liouville approximation theorem, and the Northcott theorem leaves only finitely many remaining quadratic integers of bounded height. Taking the minimum ofover this effective finite set gives an effective $c>0$ and provesfor every $\beta \in {\mathcal{O}}_K$.

Liouville approximation theorem says that if algebraic irrational $\alpha$ has degree $d\ge 2$, then $|\alpha -p/q|>C(\alpha )q^{-d}$ for all rationals $p/q$.

Every convergent series of positive rational terms has a subseries whose rational partial sums $p_j/q_j$ satisfyChoose each new term small enough to enforce all previous tail bounds. Separation of reduced fractions first makes $\alpha$ irrational, and the Liouville approximation theorem then rules out every finite algebraic degree.