Algebraic number

The complex algebraic numbers form a field, the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$. To see closure under the field operations, if $\alpha ,\beta$ are algebraic numbers, the tower law makes $\mathbb{Q}(\alpha ,\beta )$ a finite-dimensional field extension of $\mathbb{Q}$. Every element of this extension is an algebraic element, because its successive powers are linearly dependent. In particular it contains $\alpha \pm \beta$, $\alpha \beta$ and $\alpha /\beta$ when $\beta \ne 0$.

It is an infinite-degree algebraic extension. A primitive $p$th root of unity has minimal polynomial of an algebraic element $1+X+\cdots +X^{p-1}$ of degree $p-1$ for every prime $p$, so the degrees of its finite subextensions are unbounded. Finally a complex root of a polynomial whose coefficients are algebraic numbers is itself algebraic over their finite-degree coefficient field and hence over $\mathbb{Q}$, again by the tower law. Thus every such polynomial splits in this field, establishing the algebraic closure assertion.

A quadratic irrational number is a real irrational root of a quadratic polynomial with integer coefficients.

For a positive nonsquare integer $D$ and positive integers $a,q$, the integer $D{q}^2-a^2$ is nonzero. Therefore $|\sqrt{D}-a/q|=|D{q}^2-a^2|/(q^2(\sqrt{D}+a/q))$. If $a/q\le \sqrt{D}+1$, the denominator factor is at most $2\sqrt{D}+1$; otherwise the approximation error already exceeds one. This proves that a quadratic irrational of this form is a badly approximable number. For $D=2$, splitting instead at $a/q=2$ gives the strict bound $|\sqrt{2}-a/q|>1/(4q^2)$.

// Target: analytic-number-theory.bigb

The golden ratio is the positive numberwhich satisfies ${\phi }^2=\phi +1$ and $1/\phi =\phi -1$.

Integer polynomials form a countable set and each has finitely many roots, so the algebraic numbers are countable.