Algebraic Number

Most numbers we use every day are Algebraic Numbers
But some are not, such as π and e

Algebraic Number

Put simply, when we have a polynomial equation like (for example)

2x2 + 4x − 7 = 0

whose coefficients (the numbers 2, 4 and −7) are rational numbers (whole numbers or simple fractions) ...

... then x is Algebraic.

We can imagine all kinds of polynomials:

In each case x is algebraic.

In fact all integers, all rational numbers, some irrational numbers (such as √2) are Algebraic.

Testing Game

We can make a game of it!

Start with our number, and we have to get it to zero using only:

  • whole numbers
  • add, subtract, multiply (but not divide)
  • whole number exponents

Here are some examples:

NumberGetting it to Zero
55 − 5 = 0
−7−7 + 7 = 0
½2(½) − 1 = 0
√2(√2)2 − 2 = 0
3√2(3√2)2 − 18 = 0

All those numbers are algebraic!

Try some yourself and see how you go.

Now try π (pi) and see if you have any success.

More Formally

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

So x is algebraic in this example:

2x3 − 5x + 39 = 0

Because all conditions are met:

So we know x is algebraic

But let's see its value anyway:

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

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

Example: 2x3 − ¼ = 0

The coefficients are 2 and −¼, both rational numbers. So x is an Algebraic Number

We can also discover that x = 0.5, because 2(0.5)3 − ¼ = 0

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.

Back in 1844 Joseph Liouville created a number:


(it has a 1 in every factorial numbered position such as 1, 2, 6, 24, etc)

And he showed it is not algebraic ... he had broken mathematics!

Just kidding. But it did transcend algebraic numbers and was

the first known transcendental number

Then in 1873 Charles Hermite proved that e (Euler's number) is transcendental, and in 1882 Ferdinand von Lindemann proved that π (pi) is transcendental.

It is actually hard to prove that a number is transcendental.


Let's investigate a few more numbers

Example: the unit imaginary number i

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

x2 + 1 = 0

So the imaginary number i is an Algebraic Number

Note: using the "testing game":  i2 + 1 = 0

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)

They are both Algebraic Numbers

Which is More Common?

It might seem that transcendental numbers are rare, but:

almost all real and complex numbers 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.


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.