# Transcendental number

https://arbital.com/p/transcendental_number

by Patrick Stevens Aug 17 2016 updated Aug 20 2016

A transcendental number is one which is not the root of any integer-coefficient polynomial.

[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]