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):

2x3 − 5x + 39

Then x is algebraic.

Because all conditions are met:

Let's find an algebraic number:

Example: 2x3 − 5x + 39

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

It can actually be very hard to prove that a number is not algebraic.


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 x2 + 1 = 0).

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