From Wikipedia

  • 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 $f(x) = x^2$

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    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. – Ian Mar 06 '16 at 22:50
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    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". – Daniel R. Collins Mar 07 '16 at 01:25
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    @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 – MCT Mar 07 '16 at 18:01
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    Obligatory reference: [pi and e on a blind date](https://www.youtube.com/watch?v=nKq6_vjrxMo) (at the very beginning, about 8 seconds in). – Peter Mortensen Mar 09 '16 at 21:44
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    Wikipedia has a feature called "Simple English", the description there is very short but reasonable: https://simple.wikipedia.org/wiki/Transcendental_number – Bitwise Mar 10 '16 at 00:24
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    Ask them to watch [this Numberphile episode](https://www.youtube.com/watch?v=seUU2bZtfgM). – Steinar Lima Mar 11 '16 at 22:36
  • When writing about transcendental issues, be transcendentally clear. Rene Descartes (1596-1650) – Aditya ultra Mar 15 '16 at 06:39

11 Answers11


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.

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    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. – Jed Schaaf Mar 07 '16 at 03:55
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    Small addendum: If you can reach zero *without* using the “multiply by $x$” move, then we say that the number is _rational_. – MJD Mar 07 '16 at 04:03
  • I don't think the rules are clear. You can't get $\sqrt[3]7$ to zero by the stated rules because the rules only state you can manipulate x by whole numbers. You can't multiply x by itself if x isn't a whole number. If you *can* include x and the derived values from x then $\pi$ (and *any* number) *is* algebraic because $\pi - \pi = 0$. – fleablood Mar 07 '16 at 19:02
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    @fleablood What's not clear about “You can also multiply by $x$”? – MJD Mar 07 '16 at 19:07
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    _no matter how many steps you take._ I think that this is false. The number of steps in the game have to be finite. See [here](http://math.stackexchange.com/questions/1330746/wrong-proof-but-where-is-the-mistake) – Eminem Mar 07 '16 at 19:21
  • I didn't see that rule. Okay, Let x = the solution to $x^5 + x^4 = 7$. You can get x^5 and you can get x^4 but you can't, so far as I see, get x^5 + x^4. At least it's not apparent. If you add a rule we can add and subtract derived numbers we get x - x. We have to somehow declare a rule that if we derive a power of x we can not derive another another to the same power or something equivalent. – fleablood Mar 07 '16 at 19:21
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    @fleablood $(((((x+1)\cdot x)\cdot x)\cdot x)\cdot x)-7 = 0$. See [Horner's method](https://en.wikipedia.org/wiki/Horner's_method). – MJD Mar 07 '16 at 19:24
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    @Eminem "no matter how many steps you take" can very legitimately mean how many steps you can take. As you can't take infinite steps it's legit I think. After all it's an acceptable to say "you can never count all the numbers no matter how long you take". Well, if I'm allowed infinite time I can. I can even do it in under 2 seconds if I'm allowed infinite speed in counting. But I'm not allowed either of those things... so I can't. – fleablood Mar 07 '16 at 19:25
  • Horner's method is good. But... what's wrong with simply saying "plug it into any polynomial with integer coefficients"? That just seems ... much more straightforward. – fleablood Mar 07 '16 at 19:32
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    @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 – Guntram Blohm Mar 07 '16 at 21:15
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    @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 – Ant Mar 07 '16 at 21:16
  • I didn't get the idea that the OP was either a 6th grader or an average man on the street. I don't think a 6th grader or the average man on the street would really get or believe that that are some numbers where you can't get to zero. In fact I think the average person would assume it's obvious you can get from any number to any number by the rules and would be very perplexed that you can't with pi. – fleablood Mar 07 '16 at 21:28
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    @fleablood Then tell him to try it. The game above is easy enough for the average layman to understand, and its purpose is to show which numbers are transcendental and which are algebraic, that's all. – Simply Beautiful Art Mar 07 '16 at 22:04
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    @fleablood No, he/she gave the impression that he/she was a layman, and this is the easiest way a layman could probably understand it. If you could show me your above as a solution to some polynomial, then show me that, but it most certainly isn't easier if it isn't harder. – Simply Beautiful Art Mar 07 '16 at 22:25
  • what about x = π then substract by π? – Alex Mar 08 '16 at 08:40
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    @Alex You can only add/subtract **integers**. The only other operation you are allowed is to multiply by the number itself, so $x = \pi$ you can do $x \cdot \pi = \pi^2$. – Bakuriu Mar 08 '16 at 08:53
  • Anyway regarding the finite vs infinite number of steps: isn't there the difference between *computable* trascendental numbers and *non computable* trascendental numbers? – Bakuriu Mar 08 '16 at 08:56
  • @Bakuriu aha.. I doubt I get the magic behind it. So I couldnt do: x = π, devide by x - 1? – Alex Mar 08 '16 at 15:06
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    @Alex No, you can only *multiply* by either the number in question or an integer different from $0$. No divisions allowed. – Bakuriu Mar 08 '16 at 16:03
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    @fleablood I agree with you that this game is deceptively simple such that a layman may confuse "I can't do it" with "it can't be done." An example like sqrt(2) + sqrt(3) should help illustrate the difference. But I bet that if you gave your definition to a typical person and asked them "is 7 transcendental?" they wouldn't be sure, and in that respect, this game is a great explanation. – yshavit Mar 09 '16 at 00:10
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    I really like this, but it doesn’t touch transcendental *functions*. – KRyan Mar 10 '16 at 13:25
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    @KRyan, hopefully MJD comes back and changes it up a bit to include functions too! – AlanSTACK Mar 10 '16 at 23:32
  • @KRyan Well, if we describe the game that determines if a number is algebraic as "algebra" (add integers, divide integers, multiply by x), then an algebraic function is just a function whose action is some set of algebra. Gussy that up to make action and function make sense? – Yakk Mar 11 '16 at 19:09
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    Great explanation indeed, but doesn't give any intuition behind why you would come up with a random definition like this and consider it useful. – user541686 Mar 12 '16 at 07:16
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    another point, if you can reach 0 without the multiplication, division, or multiplication by x, then it is an integer. – tox123 Mar 13 '16 at 01:57
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    Whether or not this proposed "game" is mathematically a perfect explanation or not is ,as far as I am concerned, besides the point. If you truly want to capture all aspects as to define what algebraic and transcendental numbers are, then for somebody who is mathematically inept or who is a sixth grader, you won't be able to explain. But this answer comes very close to a proper explanation without (and that's remarkable!) heavy mathematical concepts. I think it is awesome!. – imranfat Mar 13 '16 at 04:05
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    by the way, if you can reach 0 without adding, subtracting and multiplication, then the number is actually 0. – Cthulhu Mar 14 '16 at 10:46
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    @Cthulhu, stronger statement: if you can reach 0 without adding or subtracting, the number is 0. – Wildcard Dec 10 '16 at 22:41
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    Are there *any* numbers for which division is *necessary*? I can't think of any. Likewise, subtracting integers is redundant; I believe all you need is "you can add any integer; you can multiply by any non-zero integer; you can multiply by $x$." – Wildcard Aug 09 '17 at 23:56
  • Wow. I never would have thought of it this way. Do you, by any chance, have a Combinatorial Game Theory background? This reminds me quite a bit of some of math games people create, and the CGT approach to mathematics in general. Games are useful! – DukeZhou Aug 26 '17 at 05:41
  • This one and the second most upvoted answer do not address the second part of OP's question. – HeWhoMustBeNamed Oct 19 '17 at 15:50
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    The way I got to this explanation was: several people had already jumped in to say the thing about algebraic numbers being zeroes of polynomials. But this seemed unsatisfying, because it's just explaining one obscure mathematical abstraction (algebraic numbers) in terms of another (polynomials). Why do we care about polynomials, what is important about them? And when I asked that question, the answer that came to mind was that the polynomials are the free ring over $x$. Aha! That _is_ important, and it's easy to explain, so that's what I did. – MJD May 26 '18 at 23:29
  • And if you never need to multiply or divide by anything except $x$, it's an algebraic integer. – NoLongerBreathedIn Jun 25 '20 at 15:34
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    @wildcard I could have left out subtraction and division, but I put them in because it makes the explanation more natural. The target audience is people with no significant mathematical training. To a mathematician, it's obvious that addition of integers includes subtraction as a special case. Non-mathematicians do not always think this way. They have learned that there are four arithmetic operations. If I had said "addition and multiplication" many people would have been distracted and wondered "why those two? Why not some other two?". Including all four avoids this distraction. – MJD Nov 19 '21 at 14:39
  • Is there a good way to find these steps if you have a number of the form $\sqrt{2} + \sqrt{3}$? You can convert a polynomial into the steps of the game with Hormerʼs method mentioned above, but is there an easy way to take $\sqrt{2} + \sqrt{3}$ and find $x^4 - 10x^2 + 1 = 0$? (I know you canʼt go from arbitrary polynomial where the degree can be ≥ 5 to its roots easily, but somehow I never thought of the reverse question until now.) – Daniel H Nov 20 '21 at 05:49
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    @DanielH [This post explains the general method](https://math.stackexchange.com/a/359073/25554), with $\sqrt2+\sqrt 3$ as an example. The jargon term you're looking for is “minimal polynomial”. – MJD Nov 20 '21 at 21:07
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    It's interesting to think about how this answer is the "inverse" of the standard definition given in many other answers. In the standard definition, the algebraic numbers are the numbers obtainable from the integers using a certain tool (polynomial roots). In your answer, the algebraic numbers are the numbers _from which the integers are obtainable_ using a certain tool (the moves you define). – Akiva Weinberger Dec 14 '21 at 05:41
  • (Cont'd) The nontrivial fact that the sum of two algebraic numbers is algebraic has an interesting interpretation in this framework: instead of finding a set of moves that take you to zero, it's (nonobviously) good enough to _split_ the number into two pieces, each of which has a path to zero. – Akiva Weinberger Dec 14 '21 at 05:44
  • It may seem different, but it's the exact same thing. All that the "steps" in the game are doing is evaluating a polynomial to see if $x$ is a root. The only difference is that we're using a shorter sequence of operations (Horner's method) to evaluate the polynomial. – MJD Dec 14 '21 at 07:43

$\sqrt2$ 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.)

Akiva Weinberger
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    "In fact, we're not even sure if it's rational!" .. would that be *ir*rational? – muru Mar 07 '16 at 15:30
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    @muru Fixed. (Though, logically speaking, they're the same thing; we don't know if it's rational or irrational.) – Akiva Weinberger Mar 07 '16 at 15:31
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    This doesn't explain what a transcendental function is though. @AkivaWeinberger –  Mar 08 '16 at 00:54
  • what of $\pi^e$ – tox123 Mar 13 '16 at 01:58
  • @tox123 Also unknown. $e^\pi$ is easier because of its relation to the equation $e^{\pi i}=-1$. – Akiva Weinberger Mar 13 '16 at 02:01
  • "*Specifically, the coefficients of these polynomials need to be* **integers**"$-$ Would it make any difference if the coefficients are instead *rational*? (I don't think so since in that case it'd just take a couple of steps to transform the equation to have integer coefficients). – HeWhoMustBeNamed Oct 19 '17 at 13:28
  • @MrReality No, it would not make any difference. --In fact, it wouldn't even make a difference if the coefficients were allowed to be algebraic! (Clearly every algebraic number $\alpha$ would satisfy the equation $x-\alpha=0$; the surprising fact is that we don't get any _more_ numbers when we use algebraic coefficients.) – Akiva Weinberger Oct 19 '17 at 14:32

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).

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    This is a very good answer as it shows the connection between these numbers and functions and algebra in general. +1 – Allawonder May 23 '18 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.

Michael Hardy
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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 $f(x)^2 - x = 0$.

Robert Israel
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    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. – user236182 Mar 06 '16 at 22:23
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    You should include the condition of being non-zero to the polynomial in the opening definition. – goblin GONE Mar 07 '16 at 01:44
  • 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). – HeWhoMustBeNamed Oct 19 '17 at 16:18
  • If there are infinitely many terms, it's not a polynomial. – Robert Israel Oct 19 '17 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).

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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.

Laurent Duval
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    " 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. – Robert Israel Mar 07 '16 at 19:31
  • Corrected accordingly – Laurent Duval Mar 07 '16 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."

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    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). – Marc van Leeuwen Mar 07 '16 at 12:44
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    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. – Lighthart Mar 09 '16 at 00:09

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: $$\pi^2-2, \qquad \pi^3-2\pi+1, \qquad \pi^5-3\pi^4+2\pi+1$$

goblin GONE
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    That definition, while symbol free, seems to be the opposite of "simple English". – siegehalver Mar 06 '16 at 22:37
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    Apparently I don't know what a symbol is. – djechlin Mar 10 '16 at 22:24
  • @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. – goblin GONE Mar 11 '16 at 01:29
  • How can we be sure that the last three numbers are nonzero ? –  May 23 '18 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 $K(X)$.

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    I'm not sure this is exactly what the OP meant by "layman's terms". – bof Mar 07 '16 at 10:50
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    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. – Clarinetist Mar 07 '16 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 $x - 5 = 0$.

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    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. – goblin GONE Mar 08 '16 at 02:24
  • @goblin I think you are thinking of Abel. – MJD Mar 09 '16 at 02:06
  • @MJD, oh, I didn't realize it was Abel who proved that. In that case, what is Galois famous for? – goblin GONE Mar 09 '16 at 04:19
  • @goblin You were right. Galois proved it also. – MJD Mar 09 '16 at 08:15