From Wikipedia, we have the following definitions:

  • A transcendental number is a real or complex number that is not algebraic

  • A transcendental function is an analytic function that does not satisfy a polynomial equation

However these definitions are arguably rather cryptic to those who are not familiar with the literature of higher mathematics.

So in layman's terms, what exactly does it mean to be transcendental? How would a transcendental number be different from an ordinary number, say $5.$ And respectively, how would a transcendental function be different from an ordinary function, say $f(x) = x^2$

  • 7
    $\begingroup$ Well, it would be equivalent to get a handle on what it means to be an algebraic number. What makes this hard is that there are polynomials which aren't solvable in terms of radicals, so it's not easy to get a more intuitive equivalent definition. $\endgroup$
    – Ian
    Commented Mar 6, 2016 at 22:50
  • 48
    $\begingroup$ Beware of the phrase "ordinary number", because transcendentals outnumber other types, so in some sense they are the "ordinary" ones. Similar to what Georg Cantor said, "The rationals are spotted in the line like stars in a black sky while the dense blackness is the firmament of the irrationals". $\endgroup$ Commented Mar 7, 2016 at 1:25
  • 6
    $\begingroup$ @DanielR.Collins The tricky part about that being that the rationals are still dense in the reals, which is what confuses a lot of students starting analysis I think $\endgroup$
    – MT_
    Commented Mar 7, 2016 at 18:01
  • 2
    $\begingroup$ Obligatory reference: pi and e on a blind date (at the very beginning, about 8 seconds in). $\endgroup$ Commented Mar 9, 2016 at 21:44
  • 4
    $\begingroup$ Wikipedia has a feature called "Simple English", the description there is very short but reasonable: simple.wikipedia.org/wiki/Transcendental_number $\endgroup$
    – Bitwise
    Commented Mar 10, 2016 at 0:24

11 Answers 11


We will play a game. Suppose you have some number $x$. You start with $x$ and then you can add, subtract, multiply, or divide by any integer, except zero. You can also multiply by $x$. You can do these things as many times as you want. If the total becomes zero, you win.

For example, suppose $x$ is $\frac23$. Multiply by $3$, then subtract $2$. The result is zero. You win!

Suppose $x$ is $\sqrt[3] 7$. Multiply by $x$, then by $x$ again, then subtract $7$. You win!

Suppose $x$ is $\sqrt2 +\sqrt3$. Here it's not easy to see how to win. But it turns out that if you multiply by $x$, subtract 10, multiply by $x$ twice, and add $1$, then you win. (This is not supposed to be obvious; you can try it with your calculator.)

But if you start with $x=\pi$, you cannot win. There is no way to get from $\pi$ to $0$ if you add, subtract, multiply, or divide by integers, or multiply by $\pi$, no matter how many steps you take. (This is also not supposed to be obvious. It is a very tricky thing!)

Numbers like $\sqrt 2+ \sqrt 3$ from which you can win are called algebraic. Numbers like $\pi$ with which you can't win are called transcendental.

Why is this interesting? Each algebraic number is related arithmetically to the integers, and the winning moves in the game show you how so. The path to zero might be long and complicated, but each step is simple and there is a path. But transcendental numbers are fundamentally different: they are not arithmetically related to the integers via simple steps.

  • 64
    $\begingroup$ This is the best answer because it is accurate and uses only the requested "layman's terms" in the explanation. Someone not smarter than a 5th-grader should still be able to follow this. $\endgroup$
    – Jed Schaaf
    Commented Mar 7, 2016 at 3:55
  • 188
    $\begingroup$ Small addendum: If you can reach zero without using the “multiply by $x$” move, then we say that the number is rational. $\endgroup$
    – MJD
    Commented Mar 7, 2016 at 4:03
  • 33
    $\begingroup$ @fleablood $(((((x+1)\cdot x)\cdot x)\cdot x)\cdot x)-7 = 0$. See Horner's method. $\endgroup$
    – MJD
    Commented Mar 7, 2016 at 19:24
  • 32
    $\begingroup$ @fleablood: Any 6th-grader could understand the above rules. How many laymen understand "polynomial" or "coefficients"? This is the beauty of this answer, you don't need any mathematical education to understand it $\endgroup$ Commented Mar 7, 2016 at 21:15
  • 31
    $\begingroup$ @fleablood it is if in your vocabulary there are words like "polynomial", "integer coefficients", and "plug into that". I would bet a lot of money that the average person will have no clue what you're talking about without thinking about it a lot. On the other hand, people love games, these are clear rules, and everybody is happy $\endgroup$
    – Ant
    Commented Mar 7, 2016 at 21:16

$\sqrt2$ satisfies the equation: $$x^2-2=0$$ Similarly, $\sqrt[\Large3]3$ satisfies the equation: $$x^3-3=0$$ Numbers like this, that satisfy polynomial equations, are called algebraic numbers. (Specifically, the coefficients of these polynomials need to be integers.)

Another algebraic number is $\frac12$, since it satisfies: $$2x-1=0$$ In fact, all rational numbers are algebraic. But, as the first two examples show, not every algebraic number is rational.

Now, it's not obvious, but if you add up or multiply together two algebraic numbers, you get another algebraic number. For example, $\sqrt2+\sqrt[\Large3]3$ satisfies the equation: $$x^6-6x^4-6x^3+12x^2-36x+1=0$$

(In case you're wondering, complex numbers can also be algebraic. In fact, it's not hard to show that a complex number is algebraic if and only if its real and imaginary parts are algebraic.)

A real (or complex) number that's not algebraic is called transcendental. In 1873, the number $e\approx2.71828$ was proven transcendental. In 1882, $\pi\approx3.14159$ was, too. It is unknown if $e+\pi$ is transcendental. In fact, we're not even sure if it's irrational! Same goes for similar numbers such as $\pi^\pi$ and $e\pi$. ($e^\pi$, however, is transcendental.)

  • 3
    $\begingroup$ "In fact, we're not even sure if it's rational!" .. would that be irrational? $\endgroup$
    – muru
    Commented Mar 7, 2016 at 15:30
  • 18
    $\begingroup$ @muru Fixed. (Though, logically speaking, they're the same thing; we don't know if it's rational or irrational.) $\endgroup$ Commented Mar 7, 2016 at 15:31
  • 4
    $\begingroup$ This doesn't explain what a transcendental function is though. @AkivaWeinberger $\endgroup$
    – user311559
    Commented Mar 8, 2016 at 0:54
  • 1
    $\begingroup$ what of $\pi^e$ $\endgroup$
    – tox123
    Commented Mar 13, 2016 at 1:58
  • $\begingroup$ @tox123 Also unknown. $e^\pi$ is easier because of its relation to the equation $e^{\pi i}=-1$. $\endgroup$ Commented Mar 13, 2016 at 2:01

Among the real numbers, some are integer.

Others are rational, i.e. they are solutions of a linear equation such as

$$px=q$$ where $p,q$ are integer. The numbers not falling in this scheme are called irrational.

A rather obvious generalization of this principle are numbers that are solutions of a polynomial equation such as

$$px^3+qx^2+rx+s=0$$ where $p,q,r,s$ are integer (any other degree can do). These numbers are called algebraic, which is the converse of transcendental.

The algebraic numbers enjoy a special property: even though there is an infinity of them, they can be numbered (they are said to be countable). By contrast, the transcendental numbers cannot, there is a "larger" infinity of them.

You easily understand that all integers are rational and all rationals are algebraic.

Among the functions of the real variable, some are polynomials.

A rational fraction is the quotient of two polynomials, i.e. a function $y=\dfrac{Q(x)}{P(x)}$, that verifies an equation like


More generally, an algebraic function $y=f(x)$ is such that it can be expressed as the root of a polynomial with coefficients that are themselves polynomials in $x$:


A function that is not algebraic is called transcendental.

Looking closer, one can observe that algebraic items are defined from equations that use a finite number of additions and multiplications. Transcendental items require "stronger" tools (such as an infinite number of terms).

  • 2
    $\begingroup$ This is a very good answer as it shows the connection between these numbers and functions and algebra in general. +1 $\endgroup$
    – Allawonder
    Commented May 23, 2018 at 15:43

The only thing cryptic I see in the quoted definition of "transcendental number" is that you haven't first defined what an algebraic number is. An algebraic number is a number that is a root of a polynomial with rational coefficients. That is equivalent to saying it's a root of a polynomial with integer coefficients. Thus the roots of $$ \frac 5 8 x^3 - \frac{21}2 x^2 + \frac{17}{12} x + 19 = 0 $$ are algebraic numbers. The common denominator of these coefficients is $24$, and multiplying both sides by that we get $$ 15x^3 - 252 x^2 + 34 x + 456 = 0 $$ and that equation has the same roots but has integer coefficients.

Rational numbers are algebraic numbers. For example $\dfrac{17}{12}$ is a root of $$ x - \frac {17}{12} = 0 $$ or of $$ 12x - 17 = 0. $$

The function $x\mapsto \sqrt[3] x = f(x)$ is an algebraic function by the given definition, since it satisfies the polynomial equation $$ f(x)^3 - x = u^3 - x = 0 $$ in the variable $u=f(x)$. In other words, it is not only polynomial functions that satisfy polynomial equations.


A transcendental number is a number that is not a root of a nonzero polynomial with integer coefficients. An example of a transcendental number is $\pi$. On the other hand, $5$ is not transcendental because it is a root of the polynomial $x - 5$.

Similarly, a transcendental function is a function $f(x)$ that does not satisfy a nontrivial polynomial equation $P(x, f(x)) = 0$ (nontrivial meaning that at least one coefficient is nonzero). An example of a transcendental function is $\sin(x)$. On the other hand, $f(x) = \sqrt{x}$ is not transcendental because it satisfies the equation $f(x)^2 - x = 0$.

  • 3
    $\begingroup$ The explanation for a transcendental function needs more explanation. The polynomial equation involves a polynomial in terms of $f(x)$ equal to $0$, with coefficients being polynomials in terms of $x$ with (usually) integer coefficients. $\endgroup$
    – user236182
    Commented Mar 6, 2016 at 22:23
  • 1
    $\begingroup$ You should include the condition of being non-zero to the polynomial in the opening definition. $\endgroup$ Commented Mar 7, 2016 at 1:44
  • $\begingroup$ Shouldn't you also mention ...a polynomial equation "having a finite number of terms"... Because I really don't see how your definition prevents me from proving that sin(x) is Not transcendental by taking the Taylor series of sine function as the polynomial in question( figuratively). $\endgroup$ Commented Oct 19, 2017 at 16:18
  • 2
    $\begingroup$ If there are infinitely many terms, it's not a polynomial. $\endgroup$ Commented Oct 19, 2017 at 22:30

The historical origin of the term is with René Descartes and Gottfried Wilhelm Leibniz in the 17th Century.

In his La Géométrie (1637): LIVRE SECOND De la nature des lignes courbes, Descartes discuss the traditional classification of curves.

Descartes called a curve "geometrical" if it could be described by a polynomial equation in two variables. Thus, a geometrical curve is what would be called "algebraic" in modern terminology.

A curve that is not geometrical was called "mechanical" by Descartes [see the Quadratrix of Dinostratus for an example: it offers a "mechanical solution" to the squaring of the circle, based on the coordinate motion of two lines].

In his paper: DE VERA PROPORTIONE CIRCULI AD QUADRATUM CIRCUMSCRIPTUM IN NUMERIS RATIONALIBUS EXPRESSA, Act.Erudit.Lips.1682 (reprinted into: Gottfried Wilhelm von Leibniz, Leibnizens mathematische Schriften, herausgegeben von C.I.Gerhardt (1858), page 118-on), Leibniz introduces the distinction between algebraic curves and transcendental ones, where these are the curves not definable with a polynomial (like the trigonometric functions).

See page 120:

transcendens inter alia habetur per aequationes gradus indefiniti.

See Part II of Henk J.M. Bos, Redefining Geometrical Exactness: Descartes’ Transformation of the Early Modern Concept of Construction (2001).

  • 3
    $\begingroup$ Since when Latin, French, and German are considered "Simple English"? :-) $\endgroup$
    – guga
    Commented Mar 14, 2016 at 19:19
  • $\begingroup$ @guga - Historical documentation on the issue at hand is always valuable, even if it doesn't completely meet the OP's request. Thank you very much, Mauro ALLEGRANZA, for posting this. $\endgroup$ Commented Jun 18, 2022 at 13:38

In my simpledt words: transcendental is what cannot be expressed in a sum of powers (a finite number thereof).

There are different definitions for algebra. Let us take a simple one (that does not covers all the acceptions for "algebraic"), possibly the elementary algebra used at school. You are allowed a finite number of simple operations: additions, subtractions, multiplications, divisions of rationals and unknowns. You are algebraic (a number or a function) is a finite number of these operations reaches $0$. For instance the $x$ in $x\times x +3= 0$. If not, if reaching zero requires an infinite number of elementary operations, you are transcendental.

Leibniz apparently introduced the name transcendal for $\sin(x)$ which is not an algebraic function of $x$. Indeed, one can instead write (in certain contexts): $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$ as an infinite quantity of sums, product, etc. And $\sin \pi = 0$, from which you might suspect (not a proof) that $\pi$ might be transcendental.

Funnily, it is easier to approximate some transcendental numbers that some algebraic ones with sequences of rationals, which is the essence of many proofs of transcendance (using for instance Diophantine approximation).

Euler apparently introduced the modern notion of algebraic numbers. I wonder if in the original text the related notions of transcendance and infinity where inspired by religious type of concepts.

  • 2
    $\begingroup$ " it is easier to approximate a transcendental number that an algebraic one with sequences of rationals" is not true. It is easier for some transcendental numbers and algebraic numbers, but certainly not for all. $\endgroup$ Commented Mar 7, 2016 at 19:31
  • $\begingroup$ Corrected accordingly $\endgroup$ Commented Mar 7, 2016 at 19:38

A transcendental number is one which is not the root of a polynomial with integer (or equivalently rational) coefficients.

From a different Wikipedia article: a transcendental function is one which "cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction."

  • 2
    $\begingroup$ That second description is confusing, since "root extraction" could easily be confused with taking radicals, which is too weak. But if it means finding a root of some nonzero polynomial, then the operations of addition and multiplication are no longer necessary (though they don't spoil anything either). $\endgroup$ Commented Mar 7, 2016 at 12:44
  • 1
    $\begingroup$ Since the question asked for 'simple english' this is the closest one. The number is transcendental because it transcends the simple representaiton through addition, multiplication and root extraction. $\endgroup$
    – Lighthart
    Commented Mar 9, 2016 at 0:09

The quoted definition isn't very precise. Without symbols, here's how I'd say it:

Definition. A real number $x_0$ is said to be transcendental if and only if, for every polynomial function $P$ in one argument, if $P$ has integer coefficients distinct from $0$, then the result of applying $P$ to $x_0$ is distinct from $0$.

This is a bit hard to understand without symbols, though, so lets introduce some notation. We write $\mathbb{Z}[x]$ for the set of polynomials with coefficients in $\mathbb{Z}$ in the symbol $x$. For example, the polynomial $x^2-1$ belongs to $\mathbb{Z}[x]$, while the polynomial $\sqrt{2}x^2-$ does not.

Now given $P \in \mathbb{Z}[x]$ and a real number $x_0$, lets write $[x \mapsto x_0]P$ for the real number obtained by replacing each instance of the symbol $x$ with the real number $x_0$. For example: $$[x \mapsto \sqrt{2}](x^2-2) = (\sqrt{2})^2-2 = 2-2 = 0$$

In this notation, we can write:

Definition. A real number $x_0$ is said to be transcendental if and only if, for every $P \in \mathbb{Z}[x]$, if $P$ is distinct from $0$, then the real number $[x \mapsto x_0]P$ is distinct from $0$.

For example, $\sqrt{2}$ fails to be transcendental (despite that it is irrational, because as we saw previously $$[x \mapsto \sqrt{2}](x^2-2) = 0.$$

On the other hand, to say that $\pi$ is transcendental, is to say that we can never get $0$ in this way. For example, all the following are non-zero real numbers:

$$[x \mapsto \pi](x^2-2), \qquad [x \mapsto \pi](x^3-2x+1), \qquad [x \mapsto \pi](x^5-3x^4+2x+1)$$

In other words, all the following are non-zero real numbers: $$\pi^2-2, \qquad \pi^3-2\pi+1, \qquad \pi^5-3\pi^4+2\pi+1$$

  • 8
    $\begingroup$ That definition, while symbol free, seems to be the opposite of "simple English". $\endgroup$ Commented Mar 6, 2016 at 22:37
  • 1
    $\begingroup$ Apparently I don't know what a symbol is. $\endgroup$
    – djechlin
    Commented Mar 10, 2016 at 22:24
  • $\begingroup$ @djechlin, what I probably should have said is: "Without symbols, except for certain placeholder symbols whose meaning I introduce at the moment I introduce them, and except for the symbol $0$ whose use is so widespread even outside of mathematics that there's really no reason to hesitate to use it, here's how I'd say it:" But, this is a little long-winded. $\endgroup$ Commented Mar 11, 2016 at 1:29
  • $\begingroup$ How can we be sure that the last three numbers are nonzero ? $\endgroup$
    – user65203
    Commented May 23, 2018 at 19:11

If $E/K$ is an extension of fields, with $e \in E$ transcendental over $K$, then $K(e) \simeq K(X) \simeq \operatorname{Frac}(K[X]) \simeq \operatorname{Frac}(K[e])$.

What this means is, if you took the formal field of fractions of all polynomials with coefficients in $K$ and evaluated them at $e$, nothing would simplify (up to multiplication by a unit). This is because $e$ satisfies no algebraic relations with $K$; in that sense, it's "free". On the other hand, evaluating an algebraic element makes many things either vanish or blow up. The evaluations that make sense constitute the field generated around $K \cup \{e\}$, which won't be isomorphic to $K(X)$.

  • 21
    $\begingroup$ I'm not sure this is exactly what the OP meant by "layman's terms". $\endgroup$
    – bof
    Commented Mar 7, 2016 at 10:50
  • 4
    $\begingroup$ If this is layman's terms, then I'm definitely a layman. I have my undergraduate degree in math (albeit not from a top school) and the isomorphism stuff with $\text{Frac}$ is all gibberish to me. $\endgroup$ Commented Mar 7, 2016 at 14:50

Well, in simple english which isn't rigorous, "transcendental" means it can not be expressed in terms of whole numbers and roots. $\sqrt[3]{2 + 1/\sqrt{3}}, \sqrt[7]{43} + 5\sqrt[3]{49},$ etc (or simply $\sqrt[4]{3}$ or even $39/47$ or $5$) can be and so are not transcendental. $\pi$ can not be expressed in any such way (take my word for it) so it is transcendental.

But you can see that is a terrible, poorly defined, ambiguous definition that is useless in any formal mathematical sense.

More formally a real number $m$ is algebraic if there is some polynomial $P(x) = a_nx^n + a_{n-1}x^{n-1} + .... + a_2x^2 + a_1x + a_0$ where each of the $a_i$s are integers and each of the $x^i$ are raised to a non-negative integer power-- if there is such a polynomial where if you plug in your number $m$ and get $P(m) = 0$-- if that is possible, the number is algebraic. If it is not possible for there to be any such polynomial, the number is *transcendental.

So take my number $m =\sqrt[3]{2 + 1/\sqrt{3}}$ it is the solution to ... $P(x) = 3x^6-12x^3 + 11 = 0$[*]. There is no finite polynomial with integer coefficients where $P(\pi) = 0$. (Take my word for that).


$3(\sqrt[3]{2 + 1/\sqrt{3}})^6 - 12 (\sqrt[3]{2 + 1/\sqrt{3}})^3 + 11 =$

$3({2 + 1/\sqrt{3}})^2 - 12({2 + 1/\sqrt{3}}) + 11 =$

$3(4 + 4/\sqrt{3} + 1/3) - 12(2 + 1/\sqrt{3}) + 11 =$

$12 + 12/\sqrt{3} + 1 - 24 - 12/\sqrt{3} + 11 = 0$

My other number $\sqrt[7]{43} + 5\sqrt[3]{49}$ is also a solution to polynomials but it'd be a real pain for me to figure out which ones. (It'd be a 21 degree polynomial. If I set $P(x) = a_{21}x^{21} + ... + a_1x + a_0$, plug in $ \sqrt[7]{43} + 5\sqrt[3]{49}$, set equal to 0, I will get 21 equations to solve for 22 of the $a_i$s in terms of the 22nd $a_i$ which I would choose just to make them all integers. Very tedious but doable. Always doable for any finite expression in terms of roots and whole numbers.)

$\sqrt[4]{3}$ is the solution to $x^4 - 3 = 0$.

$39/47$ is the solution to $47x - 39 = 0$ (which is a single degree polynomial) and $5$ is the solution to $x - 5 = 0$.

  • 9
    $\begingroup$ You probably know this already fleablood, but its worth pointing out that what Galois famously proved is that certain quintic polynomials have solutions that cannot be expressed in terms of whole numbers and roots; in other words, that your opening definition isn't quite right. $\endgroup$ Commented Mar 8, 2016 at 2:24
  • $\begingroup$ @goblin I think you are thinking of Abel. $\endgroup$
    – MJD
    Commented Mar 9, 2016 at 2:06
  • $\begingroup$ @MJD, oh, I didn't realize it was Abel who proved that. In that case, what is Galois famous for? $\endgroup$ Commented Mar 9, 2016 at 4:19
  • $\begingroup$ @goblin You were right. Galois proved it also. $\endgroup$
    – MJD
    Commented Mar 9, 2016 at 8:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .