In mathematics, an algebraic number is a number that is a root of a non-zero polynomial in one variable with integer (or, equivalently, rational) coefficients. For example, the golden ratio ${\displaystyle (1+{\sqrt {5}})/2}$ is an algebraic number, because it is a root of the polynomial ${\displaystyle X^{2}-X-1}$, i.e., a solution of the equation ${\displaystyle x^{2}-x-1=0}$, and the complex number ${\displaystyle 1+i}$ is algebraic as a root of ${\displaystyle X^{4}+4}$. Algebraic numbers include all integers, rational numbers, and n-th roots of integers.

Algebraic complex numbers are closed under addition, subtraction, multiplication and division, and hence form a field, denoted ${\displaystyle {\overline {\mathbb {Q} }}}$. The set of algebraic real numbers ${\displaystyle {\overline {\mathbb {Q} }}\cap \mathbb {R} }$ is also a field.

Numbers which are not algebraic are called transcendental and include π and e. There are countably infinite algebraic numbers, hence almost all real (or complex) numbers (in the sense of Lebesgue measure) are transcendental.

For any ⁠${\displaystyle \alpha }$⁠, the simple extension of the rationals by ⁠${\displaystyle \alpha }$⁠, denoted by ${\displaystyle \mathbb {Q} (\alpha )}$ (whose elements are the ${\displaystyle f(\alpha )}$ for ${\displaystyle f}$ a rational function with rational coefficients which is defined at ${\displaystyle \alpha }$), is of finite degree if and only if ⁠${\displaystyle \alpha }$⁠ is an algebraic number.

The condition of finite degree means that there is a fixed set of numbers ${\displaystyle \{a_{i}\}}$ of finite cardinality ⁠${\displaystyle k}$⁠ with elements in ${\displaystyle \mathbb {Q} (\alpha )}$ such that ${\displaystyle \textstyle \mathbb {Q} (\alpha )=\sum _{i=1}^{k}a_{i}\mathbb {Q} }$; that is, each element of ${\displaystyle \mathbb {Q} (\alpha )}$ can be written as a sum ${\displaystyle \textstyle \sum _{i=1}^{k}a_{i}q_{i}}$ for some rational coefficients ${\displaystyle \{q_{i}\}}$.

Since the ${\displaystyle a_{i}}$ are themselves members of ${\displaystyle \mathbb {Q} (\alpha )}$, each can be expressed as sums of products of rational numbers and powers of ⁠${\displaystyle \alpha }$⁠, and therefore this condition is equivalent to the requirement that for some finite ${\displaystyle n}$, $${\displaystyle \mathbb {Q} (\alpha )={\biggl \lbrace }\sum _{i=-n}^{n}\alpha ^{i}q_{i}\mathbin {\bigg |} q_{i}\in \mathbb {Q} {\biggr \rbrace }.}$$

The latter condition is equivalent to ${\displaystyle \alpha ^{n+1}}$, itself a member of ${\displaystyle \mathbb {Q} (\alpha )}$, being expressible as ${\displaystyle \textstyle \sum _{i=-n}^{n}\alpha ^{i}q_{i}}$ for some rationals ${\displaystyle \{q_{i}\}}$, so ${\displaystyle \textstyle \alpha ^{2n+1}=\sum _{i=0}^{2n}\alpha ^{i}q_{i-n}}$ or, equivalently, ⁠${\displaystyle \alpha }$⁠ is a root of ${\displaystyle \textstyle x^{2n+1}-\sum _{i=0}^{2n}x^{i}q_{i-n}}$; that is, an algebraic number with a minimal polynomial of degree not larger than ${\displaystyle 2n+1}$.

It can similarly be proven that for any finite set of algebraic numbers ${\displaystyle \alpha _{1}}$, ${\displaystyle \alpha _{2}}$... ${\displaystyle \alpha _{n}}$, the field extension ${\displaystyle \mathbb {Q} (\alpha _{1},\alpha _{2},...\alpha _{n})}$ has a finite degree.