To be honest, I never really understood the importance of algebraic numbers. If we lived in an universe where $\pi$ was algebraic, would there be a palpable difference between that universe and ours?

My choice of $\pi$ for this question isn't really that important, any other ''famous'' transcendental number (i.e. $e$) could work. I'm aware there are a lot of open problems about deciding whether some number is transcendental or algebraic (for an example Apery's constant, Euler-Mascheroni constant and even $\pi + e$).

Are those problems important only because they are hard to tackle? Are they important at all? If tomorrow was published a proof of algebraicity of those numbers, what would we gain from it?

EDIT: OK, maybe I took too much ''artistic freedom'' with the title of my question. I wasn't really curious about alternate universes. Bottom line was: why are those proofs important? Why is ''being a transcendental number'' important property of a number?

Rodrigo de Azevedo
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    This is not much different from "what if $\sqrt 2$ was rational"? – Pedro Aug 16 '13 at 01:50
  • @PeterTamaroff: No, as outlined in the Math-Overflow response referenced in my answer below. However, any universe in which pi is non-transcendental is probably so much simpler than ours, that intelligent life could not exist. – Pieter Geerkens Aug 16 '13 at 03:58
  • @PieterGeerkens "I wouldn't claim that $πT/T$ satisfies all reasonable definitions of $π$." makes it debatable. The OP seems to be endowing $\pi$ with some "mystical" property. Rationals are solutions to $ax+b=0$ with $a,b$ integers. Irrationals might solve higher degree polynomials, algebraics solve them all. We're left with transcendental numbers. But yes, I guess my knowledge might be too narrow to judge this question correctly. – Pedro Aug 16 '13 at 04:01
  • @PeterTamaroff: Then I will defer to your greater understanding of **Dedekind cuts**. – Pieter Geerkens Aug 16 '13 at 04:03
  • @PieterGeerkens Could you rephrase that? – Pedro Aug 16 '13 at 04:05
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    I think the moral here for you is that if you want to ask an question, you should ask the question you actually intend. If what you really wanted to know was the significance of algebraic numbers, but asked instead about alternate universes, you'll get answers about alternate universes. – ShreevatsaR Aug 16 '13 at 04:07
  • @PeterTamaroff: If you are going to dispute a professional mathematician's interpretation of **Dedekind cuts** you must have a much deeper understanding of them than I have. – Pieter Geerkens Aug 16 '13 at 04:16
  • @PieterGeerkens I am not saying what is written on MO is flawed or anything. Of course, I cannot even judge its correctness! I am just saying that the analogy with "our" $\pi$ is questionable if this "other" $\pi$ hasn't the defining properties of "our" $\pi$. When I say "makes it debatable" I meant your "No" in response to my first comment. Algebraic numbers satisfy polynomial equations. Rational numbers satisfy special polynomial equations, the linear ones, and irrational numbers are those who don't satisfy any linear equation. That is what I thought when I wrote that. – Pedro Aug 16 '13 at 04:35
  • "*If we lived in an universe where π was algebraic, would there be a palpable difference between that universe and ours?*" This question doesn't express what you want to ask straight forwardly. The answer is a tautological no. Interpreted such questions otherwise, if you'd say "*If I had a tiger, what color would it's fur be?*", you don't want the answer to be "*You have no tiger!*" – Nikolaj-K Aug 16 '13 at 08:31
  • @ShreevatsaR Yes, you are right. Lesson definitely learned! – ante.ceperic Aug 16 '13 at 08:33
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    I upvoted your question because it attracted such a crop of unusually putrid answers. – MJD Aug 21 '13 at 08:09
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    Well, at least, in that universe, this question would ask what if it wasn't an algebraic number instead. – CODE Aug 24 '13 at 17:54
  • @MJD : I posted an answer for somewhat similar reasons. – zyx Oct 07 '13 at 17:53

11 Answers11


No such universe is possible, it would be a universe in which $1$ is equal to $2$.

That said, a rational approximation to $\pi$ with error $\lt 10^{-200}$ is undoubtedly good enough for all practical purposes.

Lindemann's proof that $\pi$ is transcendental was a great achievement, but knowing the result has no consequences outside mathematics.

André Nicolas
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  • Now I am thinking about pink elephants! – ncmathsadist Aug 16 '13 at 00:41
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    Can you elaborate on why $\pi$ being algebraic would lead to $1$ equaling $2$? – Jonathan Aug 16 '13 at 01:03
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    There are various *proofs* that $\pi$ is transcendental. So if as a matter of fact $pi$ were algebraic, there would be a contradiction within mathematics. And from $A$ and not $A$, you can prove anything.One has to remember that $\pi$ is a *mathematical* construct. There are in the world objects whose properties can be modelled with great accuracy by the abstract mathematical notion of circle. However, in principle such objects are not mathematical circles. – André Nicolas Aug 16 '13 at 01:08
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    This is not true in **all** mathematical structures. By choosing a suitably reduced set of starting axioms, as *Francois G Dorais* notes in the post referenced in my answer, one can construct systems in which π is rational. – Pieter Geerkens Aug 16 '13 at 04:21
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    "No such universe is possible." Could you give requirements for something to be *possible*? It appears you use that term synonymously with *non-contradictory*. – Nikolaj-K Aug 16 '13 at 08:40
  • A necessary condition, certainly not a sufficient one. The area has been the subject of philosophical study for a long time. No answers, but many careful analyses.. – André Nicolas Aug 16 '13 at 13:48

You have to understand that although $\pi$ is a real number, it's not actually a real number. That is, it's in $\mathbb{R}$, but that set does not exist in the physical universe. It's an abstraction, just like the imaginary number $i$ is an abstraction, and one that has found significant use in physics (quantum mechanics and electrical engineering among others). Just like the idea of a number at all is an abstraction: the abstraction of assigning the same description to different quantities that are not directly related.

My point is that the quality of the universe that allows such abstractions to be imagined by intelligent creatures seems not to be separable from the quality that allows intelligent, imaginative creatures to exist at all. It requires only a sufficiently descriptive language, such as the kind considered in mathematical logic, to write down the formal definition of $\pi$ and indeed, of all of our mathematics, which implies all the algebraic and analytic properties of $\pi$ that we have proven because we wrote the proofs in that language!

So no such alternate physical universe can exist. On the other hand, one could imagine basing the definition of $\pi$ on alternate axioms, such as those specifying a particular non-Euclidean geometry, in some of which one does have $\pi = 3$, say. At least for some circles.

Ryan Reich
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  • And for other circles, you might have $\pi = 2.71828...$ or $\pi = 1.618033...$. But here's a question: can you have circles where $\pi$ is both less than and greater than $3.14159...$ in the same geometry? – Joe Z. Aug 16 '13 at 12:51
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    Well, it can happen on a Riemannian manifold with variable curvature. – Ryan Reich Aug 16 '13 at 20:36
  • I don't know enough about manifolds to understand that, but thanks for your answer. – Joe Z. Aug 17 '13 at 01:35
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    Oh, sorry, I'll expand. A Riemannian manifold is a generalized geometry (think of a surface like what you'd see in multivariable calculus). Around a point where it curves like a sphere, $\pi \leq 3.14\dots$; around a saddle point, $\pi \geq 3.14\dots$. – Ryan Reich Aug 17 '13 at 02:44
  • It's not your fault, it's mine. :P But thanks for the explanation again. – Joe Z. Aug 17 '13 at 03:20

Interesting that nobody has mentioned: A practical consequence is that you cannot construct $\pi$ using a compass and a straightedge. This has saved so-so many man-hours; if Lindemann hasn't proved $\pi$ were transcendental we wouldn't have, e.g. caramel macchiato (or more significantly, aircraft).

Lord Soth
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    I think this answer mixes something very valuable -- i.e., the observation that the transcendence of $\pi$ solves one of the outstanding mathematical problems left over from antiquity -- with some rather silly assertions. Most people working in applied fields (at any point in time) were not spending serious time trying to square the circle. Some people with a lot of time on their hands are still trying to square the circle to this day. – Pete L. Clark Aug 16 '13 at 02:11
  • @PeteL.Clark Sorry, I did not get what you meant by "silly assertions." – Lord Soth Aug 16 '13 at 02:14
  • If only we could unfold circles... – Jean-Sébastien Aug 16 '13 at 02:23
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    $@$Lord Soth: I mean the assertions that if Lindemann had not proved $\pi$ is transcendental we wouldn't have caramel macchiato or aircraft. Asserting such a strong connection between a 19th century mathematical theorem and a 20th century beverage seems "silly" (to me). – Pete L. Clark Aug 16 '13 at 02:34
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    @PeteL.Clark Well, you never know, maybe the person who invented caramel macchiato was trying to square the circle before his invention; when Lindemann showed it impossible, he gave up and spent his time on unconventional coffee experiments. In any case, my latter assertions (that you have called silly and thus broke my heart) relied on an older-than-antiquity subject called "Joke theory." – Lord Soth Aug 16 '13 at 02:40
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    @LordSoth: PeteL.Clark's comment was that those who are practically minded (those who invented caramel macchiato or aircraft) would almost surely not be spending time trying to square the circle without Lindemann's proof. We know this because *even before the proof,* hardly anyone was seriously spending time on squaring the circle. Meanwhile, the kind of people who spend time on such things are mostly still doing so despite the proof, including those who every year publish "disproofs" of Cantor's or Godel's theorems, P=NP, etc. – ShreevatsaR Aug 16 '13 at 07:17
  • @PeteL.Clark, true. I'm mentally filtering out the silly parts, and upvoting for the clever part of the answer. – Paul Draper May 17 '15 at 03:19

Just to expand a bit on trb456's answer.

If $\pi$ was algebraic then that could mean there is some reasonably simple polynomial $p(x)$ for which it was a zero. This would be really nice because then all I'd have to do with a mystery number $\xi$ to check if it was $\pi$ is to check if $p(\xi)=0$ and then perhaps do a bit more book-keeping to verify that $\xi$ was really the real $\pi$.

For example, imagine we had some mystery number $\eta$ and we want to check that $\eta = \sqrt{2}$. How to do this? (for the purposes of this hypothetical, suppose calculators are all controlled by evil, self-aware, robot masters, they cannot be trusted, we have to do calculations by pencil and paper to be safe) $p(x)=x^2-2$ has $p(\sqrt{2}) = 0$, but $p(-\sqrt{2}) = 0$ so as a check on the number being $\sqrt{2}$ I'd also need to check on the sign of the number by some method.

So, perhaps you can see the utility of a number being algebraic. There is some finite number of algebraic operators we can perform on a potential candidate to ascertain if it is in fact the algebraic number in question.

In contrast, to show some potential number is $\pi$ we'll have to resort to a deeper mathematics. Some analysis, series, etc... We have convenient notations to hide the sophsitication, but $\sqrt{2}$ is much easier to define than $\pi$.

In any event, it probably should be agreed that there is some sufficiently precise rational number which captures the concept of $\pi$ for the need of any physical problem which involves $\pi$, so the absence of the polynomial check it of little concern. Most of the time $p(x)=x-22/7$ will do just fine.

James S. Cook
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  • Why do you say the polynomial would be "reasonably simple"? Supposing the discussion makes sense at all, why couldn't the minimal polynomial for $\pi$ have degree $17^{23^{119}}$ and extremely large coefficients? – MJD Aug 21 '13 at 08:08
  • @MJD I suppose it could, but then we'd at least know it exactly after a stupidly large finite computation. Unless of course the computation is so stupendous that all the computation power in the accessible universe failed to complete it. Then, pragmatically, I suppose we'd be back to approximate methods. But, if the computational powers were sufficient then we'd have something. I will agree, a better thing to say would be "polynomial could be reasonably simple in the sense of allowing physically plausible computation" then the spirit of the answer lives on. – James S. Cook Aug 22 '13 at 01:53
  • I'm not following you. Do we know $\sqrt 2$ exactly after a finite computation? If so, then in what sense to we know it, and why wouldn't we know the value of a transcendental number exactly in the same sense? – MJD Aug 22 '13 at 16:58
  • @MJD is $a>0$ the $\sqrt{2}$? All we need to do is check, does $a^2-2=0$?. On the other hand, what finite algebraic formula can we use to ascertain if a given number $a$ was in fact $\pi$? At the moment, given the fact $\pi$ is of course not algebraic, what we can do is check $|a-22/7|< \epsilon$ (or some more sophisticated approximation of $\pi$) and we'd know $a$ was close to $\pi$. But, at best, we know it's close. If there was a formula, then we could check that $a$ was indeed the exact value $\pi$. As it stands, so far as I know, some analytical slight of hand is needed to obtain $\pi$. – James S. Cook Aug 22 '13 at 17:10
  • @MJD for example, I can say $\pi = \sqrt{6(1+\frac{1}{4}+ \frac{1}{9} + \cdots)} $ but this does not given me a finite algebraic check on $\pi$. – James S. Cook Aug 22 '13 at 17:12

This probably doesn't answer what you really want to know. It seems like you are more interested in knowing about the significance of algebraic numbers.

That said, This paper by Ivan Niven provides a proof that $\pi$ is transcendental. The proof is a proof by contradiction. That is, Niven assumes that $\pi$ is algebraic and derives a contradiction. So when you ask what would happen if $\pi$ was algebraic, then Ivan Niven actually has something very concrete. You could try to take a look at the paper to figure out what contradiction he derives.

And the question now becomes: what other statements can you derive from this? The fact is that you can prove anything from a false statement. So, as mentioned in the other answers, you can prove that any statement is true. And this is (at least one place) where the "significance" is.

(I found the link to the article in this answer.)

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Being an algebraic number is just a property, like being an even integer. Not all integers are even, and not all real numbers are algebraic. No big deal. The algebraic numbers happen to be the zero of some polynomial in one variable over the integers. That's all. But it's a nice property, and it's easier to identify algebraic numbers versus non-algebraic (transcendental) numbers. We know very little about individual transcendentals, but we know lots about algebraics!


In his answer to this question on Math-Overflow, Francois G Dorais considers such a possibility, in a weaker system of arithmetic:

Shepherdson presented a simple method for constructing such models, I will present such a model where π is rational!

I admit that I cannot follow the entire proof, but his closing remarks can be understood by those with only essential maths.

Pieter Geerkens
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$\pi$ is a pretty important real number which has been studied since antiquity. It is natural to ask every question you can about such numbers to see what you can find out; that's how mathematics progresses. First, is $\pi$ an integer? The answer, of course, is no, and this has been known probably as long as the concept of $\pi$ existed. Next, you might as if $\pi$ is the ratio of two integers; that is, if it's the solution to an equation of the form $p \pi + q = 0$ for integers $p,q$. That was answered by Lambert negatively in the 18th century.

Now that you've considered that, it's a natural extension to ask whether $\pi$ is a root of any polynomial equation, rather than just linear equations. If it is, we can probably say a lot about it with number theoretic methods. The fact that this is impossible was established in 1882 by Lindemann, who proved that $\pi$ is transcendental. So in that sense, it's more of a no-go sort of theorem which says that $\pi$ can't be described purely algebraically based on $\mathbb Q$.

So this is just an example of a natural question about a natural object. The answer turned out to be the less interesting of two possibilities for the purpose of practical work. The fact that $\pi$ is transcendental isn't really that important from a modern perspective, and certainly not for anything as applied as physics or engineering. Ultimately, the more important thing is the proof itself. Proving basically anything in transcendence theory is hard because it requires both algebraic and analytic tools to be used in conjunction, and combining them is always tricky.

Also, it's worth pointing out that the Lindemann–Weierstrass theorem proves more than just the transcendence of $\pi$. It's actually quite a far-reaching result with a number of applications and significant importance. $\pi$ being transcendental just happens to be one of them. This continues to be one of the few nontrivial facts with proofs in transcendence theory, and a lot of modern research is related to various generalizations and related conjectures.

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If $\pi$ were algebraic, it could be the limit of the ratio of consecutive terms in an integer sequence satisfying a linear recurrence! A famous property of $\phi$, the golden ratio, is that it is the limit of the ratio of consecutive Fibonacci numbers. Since $\pi$ is transcendental, there is no similar corresponding sequence. Here are some details:

Suppose a sequence of integers $\{a_n\}$ satisfies a linear recurrence

$$a_n + c_{n-1} a_{n-1} + c_{n-2}a_{n-2} + \cdots + c_{n-d}a_{n-d} = 0,$$

where each $c_i$ is rational. Such sequences often (*) behave like a geometric sequence in the long run:

$$a_n \sim c r^n,$$

where $r$ is an algebraic number. The root $r$ with maximum modulus of the polynomial

$$p(x) = x^d + c_{n-1} x^{d-1} + c_{n-2} x^{d-2} + \cdots + c_{n-d}$$

is the limit of the ratios of consecutive terms of the sequence (*) provided there is only one root with that maximum modulus.

For example, the Fibonacci sequence satisfies

$$a_n - a_{n-1} - a_{n-2} = 0,$$

though this is usually written as $a_n = a_{n-1} + a_{n-2}$. The associated polynomial is $p(x) = x^2 - x - 1$. Its largest root is $\phi$, so the ratio of consecutive terms tends to $\phi = \frac{1+\sqrt{5}}{2}$ as $n$ gets large. We will never see a similar result for $\pi$ or $e$.

Linear recurrences like this arise frequently, for example, in counting paths in a digraph or counting words of size $n$ in a regular language.

Hugh Denoncourt
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Lots of good answers up there, but what strikes me is that we do have to have transcendental numbers. The algebraic numbers (roots of polynomials with rational coefficients) are countable; yet we know the set of real numbers is uncountable. We thus need a lot more numbers than the algebraics to make up the real line.

That in itself doesn't make any particular number transcendental. The fact that π is transcendental has to do with the metric in which we are measuring. If we use the supremum norm where the "length" of a vector x is |xi| where xi is the component having largest absolute value, the unit circle winds up looking to our eyes like a square, and the ratio of its diameter to its circumference is 4. That is a nice integer number and avoids the inconvenience of having to approximate π for any practical applications.

So why do we use the Euclidean metric where |x| = $\sqrt{\sum x_i^2}$ leading to a transcendental π? Well, because it is everywhere differentiable, indeed analytic, and so we can apply the power of calculus to any questions that we might have.

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Betty Mock
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If tomorrow was published a proof of algebraicity of those numbers, what would we gain from it?

New theory that applies where those numbers appear, which for $\pi$ is most of mathematics (physics, engineering, ...).

When a transcendentally constructed number is algebraic or rational, it can be a sign of additional structure and a theory waiting to be built. In some cases, a very substantial theory, such as "arithmetic geometry" coming from observations about numerical values of zeta- and L-functions. If a number as ubiquitous as $\pi$ had previously undetected algebraic structure, it would probably be so for a systematic reason, such as hidden symmetries that once discovered can apply to all kinds of problems where $\pi$ appears.

If $\pi$ were rational this would potentially be more dramatic. For example, $\pi = 22/7$ might imply some secret $11$-fold symmetry of the circle, or the existence of a previously unsuspected category of spaces in which the circle participates in a $22$- or $7$-fold covering with some interesting new structures. And the notion of geometry would be updated to reflect these new possibilities.

This is not that far from what really happened for numbers that generalize $\pi$ (periods of algebraic varieties): there is a compelling point of view that such numbers are just the numerical manifestation of more structured "upgraded" objects, and statements about transcendence and irrationality of the numbers then acquire the meaningful and practical interpretation that all

relations between the numbers are accounted for by relations between the structures.

The highlighted sentence (stripped of "the"'s) is the real answer to the question.

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