Transcendental Numbers — Definition, Examples & Properties

Transcendental Numbers

Real numbers that are not algebraic. That is, real numbers that cannot be a root of a polynomial equation with integer coefficients. e and π are transcendental.

Nested diagram showing number sets: natural, whole, integers, rationals, algebraic, reals, and complex/imaginary numbers with...

Key Formula

a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0

Where:

$x$ = A real number being tested — if no polynomial of this form has x as a root, then x is transcendental
$a_0, a_1, \ldots, a_n$ = Integer coefficients (not all zero)
$n$ = A positive integer representing the degree of the polynomial

Example

Problem: Show that √2 is algebraic (not transcendental) by finding a polynomial equation with integer coefficients that has √2 as a root.

Step 1: Let x = √2. We want to eliminate the square root by forming a polynomial equation.

x = \sqrt{2}

Step 2: Square both sides to remove the radical.

x^2 = 2

Step 3: Rearrange into standard polynomial form with integer coefficients.

x^2 - 2 = 0

Step 4: Check: the coefficients are 1, 0, and −2 — all integers. Substituting x = √2 gives (√2)² − 2 = 2 − 2 = 0. ✓

1 \cdot x^2 + 0 \cdot x + (-2) = 0

Answer: Since √2 satisfies x² − 2 = 0, it is algebraic, not transcendental.

Another Example

Unlike the first example (which showed a number IS algebraic), this example shows why a famous number is NOT algebraic. Proving transcendence is far harder — you must rule out every possible polynomial, not just find one.

Problem: Explain why π is transcendental and cannot satisfy any polynomial equation with integer coefficients.

Step 1: Suppose, for contradiction, that π is algebraic. Then there exists some polynomial with integer coefficients such that π is a root.

a_n \pi^n + a_{n-1} \pi^{n-1} + \cdots + a_0 = 0

Step 2: In 1882, Ferdinand von Lindemann proved that no such polynomial can exist. His proof used deep results from complex analysis connecting exponential functions and algebraic numbers (the Lindemann–Weierstrass theorem).

Step 3: Because no polynomial equation with integer coefficients has π as a root, π is transcendental by definition. This proof also settled the ancient problem of 'squaring the circle' — constructing a square with the same area as a given circle using only compass and straightedge is impossible.

Answer: π is transcendental because Lindemann proved in 1882 that it cannot be a root of any polynomial with integer coefficients.

Frequently Asked Questions

What is the difference between irrational numbers and transcendental numbers?

Every transcendental number is irrational, but not every irrational number is transcendental. An irrational number simply cannot be written as a fraction of two integers. A transcendental number has the stronger property that it cannot be a root of any polynomial with integer coefficients. For example, √2 is irrational but algebraic (it satisfies x² − 2 = 0), while π is both irrational and transcendental.

Are there more transcendental numbers or algebraic numbers?

There are vastly more transcendental numbers. The set of algebraic numbers is countable (you can list them), while the set of transcendental numbers is uncountable. In a precise mathematical sense, almost every real number is transcendental, even though it is usually very difficult to prove any specific number is transcendental.

How do you prove a number is transcendental?

Proving transcendence is notoriously difficult. You must show that no polynomial equation with integer coefficients — of any degree — has that number as a root. This typically requires advanced techniques from number theory and analysis. Liouville constructed the first known transcendental number in 1844, and later Hermite (1873) proved e is transcendental and Lindemann (1882) proved π is transcendental.

Transcendental Numbers vs. Algebraic Numbers

	Transcendental Numbers	Algebraic Numbers
Definition	Cannot be a root of any polynomial with integer coefficients	Can be a root of at least one polynomial with integer coefficients
Examples	π, e, 2^(√2)	√2, ∛5, (1+√5)/2, all rationals
Rational or irrational?	Always irrational	Can be rational or irrational
How many exist?	Uncountably many (almost all real numbers)	Countably many
Ease of identification	Very hard to prove for a given number	Often easy — just find a polynomial it satisfies

Why It Matters

Transcendental numbers appear throughout calculus, physics, and engineering — every time you use

\pi

in a circle formula or

e

in exponential growth, you are working with transcendental numbers. The concept also resolved famous ancient problems: proving

\pi

is transcendental showed that squaring the circle with compass and straightedge is impossible. Understanding the distinction between algebraic and transcendental numbers deepens your grasp of how the real number line is structured.

Common Mistakes

Mistake: Assuming all irrational numbers are transcendental.

Correction: Many irrational numbers are algebraic. For instance, √2, √3, and the golden ratio (1+√5)/2 are all irrational but satisfy simple polynomial equations with integer coefficients. Transcendental is a strictly stronger condition than irrational.

Mistake: Thinking transcendental numbers are rare or exotic.

Correction: In fact, almost every real number is transcendental. The algebraic numbers form a countable set, while the transcendental numbers are uncountable. They only seem rare because it is hard to prove specific numbers are transcendental.

Related Terms

Algebraic Numbers — The complement — numbers that ARE polynomial roots
Real Numbers — The set containing all algebraic and transcendental numbers
Irrational Numbers — All transcendental numbers are irrational
Rational Numbers — All rationals are algebraic, never transcendental
Polynomial — Central to the definition of transcendental
e — The most famous transcendental number alongside π
Pi — Proven transcendental by Lindemann in 1882
Root — Transcendental numbers are never roots of integer polynomials