[summary(Technical): A real or complex number $~$z$~$ is said to be transcendental if there is no (nonzero) Integer-coefficient [-polynomial] which has $~$z$~$ as a [root_of_polynomial root].]

[summary: A transcendental number $~$z$~$ is one such that there is no (nonzero) [-polynomial] function which outputs $~$0$~$ when given $~$z$~$ as input. $~$\frac{1}{2}$~$, $~$\sqrt{6}$~$, $~$i$~$ and $~$e^{i \pi/2}$~$ are not transcendental; $~$\pi$~$ and $~$e$~$ are both transcendental.]

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_number 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 $~$n$~$ 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 $~$qx - 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 \pi/2}$~$ (or $~$\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$~$) is algebraic: it is a root of $~$x^4+1$~$.

However, $~$\pi$~$ 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_every 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 Fundamental Theorem of Algebra] states that every polynomial of degree $~$n$~$ has exactly $~$n$~$ complex roots (if we count them with [multiplicity multiplicity], so that $~$x^2+2x+1$~$ has the "two" roots $~$x=-1$~$ and $~$x=-1$~$). There are only countably many integer-coefficient polynomials [todo: 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 ([reals_are_uncountable 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

[todo: Liouville's constant]