# Algebraic Number

Most numbers we use every day are Algebraic Numbers.

But some are **not**, such as π (pi) and * e* (Euler's number).

## Algebraic Number

To be algebraic, a number must be a root of a non-zero polynomial with rational coefficients.

**Put more simply**, when we have a polynomial like (for example):

### 2x^{3} − 5x + 39

Then **x**is algebraic.

Because all conditions are met:

- 2x
^{3}− 5x + 39 is a non-zero polynomial (a polynomial which is not just "0") **x**is a root (i.e.**x**gives the result of**zero**for the function 2x^{3}− 5x + 39)- the coefficients (the numbers 2, −5 and 39) are rational numbers

Let's find an algebraic number:

### Example: 2x^{3} − 5x + 39

We need to find the value of x **where 2x ^{3} − 5x + 39 is equal to 0**

Well **x = −3** works, because 2(−3)^{3} − 5(−3) + 39 = −54+15+39 = 0

so **−3 is an Algebraic Number**

Let's try another polynomial (remember: the coefficients must be rational).

### Example: 2x^{3} − ¼ = 0

The coefficients are 2 and −¼, both rational numbers.

And **x = 0.5**, because 2(0.5)^{3} − ¼ = 0

so **0.5 is an Algebraic Number**

In fact most numbers we use daily are algebraic.

## Not Algebraic? Then Transcendental!

When a number is not algebraic, it is called transcendental.

It is known that π (pi) and * e* (Euler's number) are

**not algebraic**, and so are transcendental.

What about the square root of 2?

### Example: is √2 (the square root of 2) algebraic or transcendental?

√2 is a solution to x^{2} − 2 = 0, so it is **algebraic** (and not transcendental).

**not**algebraic.

## Properties

All algebraic numbers are computable and so they are definable.

The set of algebraic numbers is **countable**. Put simply, the list of whole numbers is "countable", and you can arrange the algebraic numbers in a 1-to-1 manner with whole numbers, so they are also countable.

The imaginary number **i** is algebraic (it is the solution to x^{2} + 1 = 0).

All rational numbers are algebraic, but an irrational number may or may not be algebraic.