# 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 equation with rational coefficients.

Put more simply, when we have an equation like (for example):

### 2x3 − 5x + 39 = 0

Then x is algebraic.

Because all conditions are met:

• 2x3 − 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 2x3 − 5x + 39)
• the coefficients (the numbers 2, −5 and 39) are rational numbers

Let's find an algebraic number:

### Example: 2x3 − 5x + 39 = 0

We need to find the value of x where 2x3 − 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: 2x3 − 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:

All integers and rational numbers are algebraic,
but an irrational number may or may not be 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 they are transcendental.

But it can actually be very hard to prove that a number is transcendental.

## More

Let's investigate a few more numbers

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

√2 is a solution to x2 − 2 = 0, so it is algebraic (and not transcendental).

### Example: the unit imaginary numberi

Well, we know that i2 = −1, so i is the solution to:

x2 + 1 = 0

When x = i we get −1 + 1 = 0

So the imaginary number i is an Algebraic Number

### Example: φ (the Golden Ratio)

φ is the solution to x2 − x − 1 = 0

So φ is an Algebraic Number

### Example: x2 + 2x + 10 = 0

The solutions to this quadratic equation are complex numbers :

• x = −1 + 3i
• x = −1 − 3i

(Try putting them into the equation, and remember that i2 = −1)

So they are both Algebraic Numbers

It seems like every number is algebraic:

• We know all integers are
• And all rational numbers are
• And √2 is
• And the imaginary number i is, and so is φ

BUT, strangely enough, almost all real and complex numbers are not algebraic, and so are transcendental.

Why? Well, imagine some random real number where each digit is randomly chosen, and you get something like 7.17493614485672... (on for infinity). It is almost certain to be transcendental.

But in every day life we use carefully chosen numbers like 6 or 3.5 or 0.001, so most numbers we deal with (except π and e) are algebraic, but any truly randomly chosen real or complex number is almost certain to be transcendental.

## 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.