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Algebraic Numbers

Algebraic Numbers

Real numbers that can occur as roots of polynomial equations that have integer coefficients. For example, all rational numbers are algebraic. So are all surds such as \(\sqrt 7 \), as well as numbers built from surds such as the number below.

\[\frac{{42 + \sqrt[3]{{15.2}}}}{{\sqrt {4 - \sqrt 3 } }}\].

Note: Real numbers which are not algebraic are known as transcendental numbers.

 

Nested diagram showing number sets: natural, whole, integers, rationals, algebraic, and real numbers, alongside complex and...

 

 

See also

Irrational numbers, natural numbers, whole numbers, imaginary numbers, complex numbers

Key Formula

anxn+an−1xn−1+⋯+a1x+a0=0a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0
Where:
  • xx = The algebraic number (the root of the polynomial)
  • a0,a1,…,ana_0, a_1, \ldots, a_n = Integer coefficients, with $a_n \neq 0$
  • nn = The degree of the polynomial (a positive integer)

Worked Example

Problem: Show that 2\sqrt{2} is an algebraic number by finding a polynomial equation with integer coefficients that it satisfies.
Step 1: Let x=2x = \sqrt{2}.
x=2x = \sqrt{2}
Step 2: Square both sides to eliminate the square root.
x2=2x^2 = 2
Step 3: Rearrange into standard polynomial form with integer coefficients.
x2−2=0x^2 - 2 = 0
Step 4: Verify: the polynomial x2−2x^2 - 2 has integer coefficients (a2=1a_2 = 1, a1=0a_1 = 0, a0=−2a_0 = -2), and 2\sqrt{2} is indeed a root since (2)2−2=0(\sqrt{2})^2 - 2 = 0.
1⋅(2)2+0⋅(2)+(−2)=2−2=0  ✓1 \cdot (\sqrt{2})^2 + 0 \cdot (\sqrt{2}) + (-2) = 2 - 2 = 0 \; \checkmark
Answer: 2\sqrt{2} is algebraic because it is a root of x2−2=0x^2 - 2 = 0, a polynomial with integer coefficients.

Another Example

This example shows the simplest case: a rational number is always algebraic via a degree-1 (linear) polynomial. The first example involved an irrational surd requiring a degree-2 polynomial.

Problem: Show that the rational number 35\frac{3}{5} is algebraic.
Step 1: Let x=35x = \frac{3}{5}.
x=35x = \frac{3}{5}
Step 2: Multiply both sides by 5 to clear the fraction.
5x=35x = 3
Step 3: Rearrange into polynomial form.
5x−3=05x - 3 = 0
Step 4: This is a degree-1 polynomial with integer coefficients (a1=5a_1 = 5, a0=−3a_0 = -3). In fact, every rational number pq\frac{p}{q} satisfies qx−p=0qx - p = 0, so every rational number is algebraic.
5⋅35−3=3−3=0  ✓5 \cdot \frac{3}{5} - 3 = 3 - 3 = 0 \; \checkmark
Answer: 35\frac{3}{5} is algebraic because it is a root of 5x−3=05x - 3 = 0.

Frequently Asked Questions

What is the difference between algebraic and transcendental numbers?
An algebraic number is a root of some polynomial with integer coefficients; a transcendental number is a real (or complex) number that is not the root of any such polynomial. For example, 5\sqrt{5} is algebraic (root of x2−5=0x^2 - 5 = 0), while π\pi and ee are transcendental — no polynomial with integer coefficients has π\pi or ee as a root. Proving a number is transcendental is generally much harder than proving it is algebraic.
Is every irrational number algebraic?
No. Some irrational numbers are algebraic (like 2\sqrt{2}), and some are transcendental (like π\pi). Being irrational simply means the number cannot be written as a fraction of two integers. Algebraic irrationals satisfy polynomial equations with integer coefficients, while transcendental numbers do not.
Are there more algebraic numbers or transcendental numbers?
Perhaps surprisingly, there are far more transcendental numbers. The set of algebraic numbers is countable (meaning they can be listed in a sequence), while the set of transcendental numbers is uncountable. In a precise mathematical sense, "almost all" real numbers are transcendental.

Algebraic Numbers vs. Transcendental Numbers

Algebraic NumbersTranscendental Numbers
DefinitionRoot of a polynomial with integer coefficientsNot the root of any polynomial with integer coefficients
Examples13\frac{1}{3}, 2\sqrt{2}, 53\sqrt[3]{5}, the golden ratio ϕ\phiπ\pi, ee, ln⁡2\ln 2, 222^{\sqrt{2}}
Rational or irrational?Can be either rational or irrationalAlways irrational
How many?Countably infiniteUncountably infinite
Ease of proofJust exhibit a polynomial it satisfiesTypically requires advanced proof techniques

Why It Matters

Algebraic numbers appear throughout algebra courses whenever you solve polynomial equations — every solution you find (rational or irrational) is an algebraic number. The distinction between algebraic and transcendental numbers is fundamental in higher mathematics and is closely tied to classic impossibility results, such as the proof that you cannot square the circle with a compass and straightedge (because π\pi is transcendental). Understanding this classification also helps clarify the structure of the real number line: it is split into algebraic and transcendental parts, with the transcendental part being vastly larger.

Common Mistakes

Mistake: Assuming all irrational numbers are transcendental.
Correction: Many irrational numbers are algebraic. For instance, 2\sqrt{2} is irrational but algebraic (it satisfies x2−2=0x^2 - 2 = 0). Only irrational numbers that satisfy no polynomial with integer coefficients are transcendental.
Mistake: Thinking the polynomial must have rational (not integer) coefficients.
Correction: The definition uses integer coefficients, but this is equivalent to allowing rational coefficients — you can always multiply through by the least common denominator to clear fractions. For example, 12x2−1=0\frac{1}{2}x^2 - 1 = 0 becomes x2−2=0x^2 - 2 = 0 after multiplying by 2.

Related Terms