A real or complex number is said to be transcendental if it is not the root of any (nonzero) integer-coefficient polynomial. ("Transcendental" means "not algebraic".)

Examples and non-examples

Many of the most interesting numbers are not transcendental.

Every integer is not transcendental (i.e. is algebraic): the integer is the root of the integer-coefficient polynomial $x - n$ .

Every rational is algebraic: the rational $\frac{p}{q}$ is the root of the integer-coefficient polynomial $q x - p$ .

$\sqrt{2}$ is algebraic: it is a root of $x^{2} - 2$ .

$i$ is algebraic: it is a root of $x^{2} + 1$ .

$e^{i π / 2}$ (or $\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$ ) is algebraic: it is a root of $x^{4} + 1$ .

However, $π$ and $e$ are both transcendental. (Both of these are difficult to prove.)

Proof that there is a transcendental number

There is a very sneaky proof that there is some transcendental real number, though this proof doesn't give us an example. In fact, the proof will tell us that "almost all" real numbers are transcendental. (The same proof can be used to demonstrate the existence of irrational numbers.)

It is a fairly easy fact that the non-transcendental numbers (that is, the algebraic numbers) form a countable subset of the real numbers. Indeed, the Fundamental Theorem of Algebra states that every polynomial of degree $n$ has exactly $n$ complex roots (if we count them with multiplicity, so that $x^{2} + 2 x + 1$ has the "two" roots $x = - 1$ and $x = - 1$ ). There are only countably many integer-coefficient polynomials

spell out why

, and each has only finitely many complex roots (and therefore only finitely many—possibly

0

—real roots), so there can only be countably many numbers which are roots of any integer-coefficient polynomial.

But there are uncountably many reals (proof), so there must be some real (indeed, uncountably many!) which is not algebraic. That is, there are uncountably many transcendental numbers.

Explicit construction of a transcendental number

Liouville's constant