If x is a fraction, x^x and x^x^x are irrational. But can x^x^x^x be rational? | Hacker News Reader

If x is a fraction, x^x and x^x^x are irrational. But can x^x^x^x be rational?

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68 pointsxSwag42 comments

Let me try to explain Schanuel's conjecture a little bit, if you're lost on words like transcendence degree. This turned out longer than I hoped so I hope it's not just a wall of text.

sqrt(2), π, π/2, π^2, and e are all irrational. Now think about π and π/2: these are very closely related. What I mean is:

π - 2(π/2) = 0

So we can add some combination of the two numbers, and get 0.

Now what about π and e? These feel different. After working with them one gets the feeling that aπ + be will never be zero, if a and b are rational[1]. If that's true, then π and e are called "linearly independent over the rationals". We do not know whether π and e are linearly independent over the rationals. (In particular, we don't know whether e+pi is rational.) It turns out that stuff like that is hard.

Now, there's an even stronger thing than being "linearly independent", and that's being "algebraicly independent". When talking about linear indepence we only thought about multiplying by rationals and adding (linear relations). For example, π and π/2 are NOT algebraicly independent. What about π and π^2? Well: (π)^2 - (π^2) = 0 which is rational. This shows that π and π^2 are also not algebraicly independent. (The equation is obvious, but think about what it means: on the left of the minus, I have π, and I'm raising it to a power of 2. on the right of the minus, I have π^2, and I'm raising it to a power of 1). We do not know whether π and e are mutually transcendental.

So you should think: linear independence of rationals is a weak way of saying "these numbers aren't that related." Algebraic independence means there even less related. You can think of both as ways of proving huge amounts of numbers are irrational. If we can prove that e and pi are algebraicly independent then for example we would know e + 5e^3 + e^7 + pi + 4pi^4 is irrational.

Now, you can understand what this conjecture says. Take z_1, z_2, ..., z_n which are complex numbers. If you want, you can just imagine them to be irrational numbers. Suppose they are linearly independent over the rationals. The conjecture is that: at least n of z_1, z_2, ..., z_n, e^(z_1), e^(z_2), ..., e^(z_n) are algebraicly independent.

[1] And not both 0 ;)

Related: If you know some linear algebra, think about this. sqrt(2) is irrational. Think of 1 and sqrt(2) has "vectors" which you can multiply by scalars, but the only scalars you're allowed to multiply by are rational numbers. In this way you get a "vector space over the rationals" which includes things like 3+sqrt(2), 5, and 7sqrt(2). This vector space is also a field: we can multiply things together, and we get another element of the vector space: (3+sqrt(2))(7sqrt(2)) = 3x7sqrt(2) + 7x2 = 14 + 21sqrt(2).

Because sqrt(2) is algebraic over the rationals, this vector space-field thing is finite. But now do it with pi instead of sqrt(2), and you'll get an infinite dimensional vector space with the vectors 1, pi, pi^2, pi^3.... Elements of this field are things like 4+pi+7pi^3.

Why are you involving the rationals? If ax + by is zero for some non-zero rationals a and b, there also are non-zero integers i and j such that ix + jy is zero (multiply a and b by the product of their denumerators to get one such pair)

> Why are you involving the rationals?

Number systems with a well behaved division operation (fields) are heavily studied and have lots of wonderful properties. If we allow for division, thus turning the integers into the rationals (which form a field), then we have access to all this.

So, while you are correct, to a mathematician it might be simpler to start with a field and then be able to wave our hands at centuries of work in field theory and just say, "All that stuff applies here."

In particular, the notion of linear independence is generally defined only for subsets of a vector space. And being a vector space requires that the scalars (allowable coefficients) form a field.

This is the kind of comments that I love. Sorry for the meta-comment, but I thought it deserved it. An easy read, and still informative and most importantly perks your imagination if you're into this kind of thing.

In the first equation involving pi - how can the sum of two positive numbers be zero?

I believe he meant minus. For the purposes of linear independence, there is no significance of positive vs negative coefficients.

EDIT: Wow, go team.

It cannot be. What the OP meant is π - 2(π/2) = 0

By the magic of typos.

Thanks, edited.

The proof given is fascinating, but...translucent.

Can we get a simple, and correct, "yes" or "no" here?

The answer is no, assuming Schanuel's conjecture.

Any set of algebraically independent numbers over a field are going to be transcendental over that field. In this case, being transcendental means they're not rational. And the number in question is in that set, so it's not rational.

But Schanuel's conjecture hasn't been proven. So would you say this reduces to Schanuel's conjecture?

The Schanuel's conjecture is widely believed to be true. A friend of mine actually did his Ph.D. on it, and the tools used to attack the problem are truly mind-blowing.

Basically, Zilber used model-theory to construct a field that "behaves like C" (the field of complex numbers), and where the conjecture is forced to be true. Not only this field exists, but it is unique. It would be no surprise then if this field was exactly C, and this would prove the conjecture for complex numbers.

Note that to prove that the two fields are isomorphic it would suffice to prove that the "artificial" field has a continuous automorphism of order two (that would basically be the conjugation in C); then its fixed set would be the real line, and it would be easy to construct the rest of the isomorphism.

The existence of such an automorphism is even less surprising, yet, nobody has managed to prove it.

That sounds so baller. But one question.

You say that he constructed a field that "behaves like C." Granted, you put that phrase in quotes for a reason, but it seems to me as though if you pinned down the behavior of multiplication in this artificial field, then conjugation wouldn't be too far away. So what does it mean to "behave like C"? Or better yet, do you have a link to you friend's thesis or other papers?

Of course the details are very tricky :) "behaves like C" means that you construct a first-order theory that all the properties of C that you want, plus the conjecture.

I don't know much about this, but I think that there is no conjugation in the language of exponential fields, and if you try to add it everything becomes intractable (for one, every proof would first require to prove Schanuel's conjecture).

The Wikipedia has a few references: https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber...

No; for that to be the case, equivalence would have to be proven (i.e. construct a proof of Shanuel that works when you assume this theory to be true.)

Right now, it could be that there exists a proof that is valid and does not depend on Shanuel's conjecture.

Not necessarily. The proof presented depends Schanuel's conjecture, but there may be another proof that doesn't depend on that conjecture, directly or indirectly.

The question you're asking is one of metalogic. It's like asking, what is the necessary structure of the proof of some proposition? Mathematicians definitely ask questions like that, but those are tough nuts to crack.

The given answer is "no"[1], all powers are irrational. I'm not sure how this simple "no" would be helpful to you without understanding any part of the proof.

[1] assuming https://en.wikipedia.org/wiki/Schanuel%27s_conjecture

A simple "no" is helpful because it answers the title question without having to grok the proof. Some of us appreciate math geekery without being one.

I don't think I've seen this notation before:

    (a,b) = 1

I guess it means gcd(a, b) = 1

Yes, this is a short notation of gcd, and it is quite common among mathematicians.

Is there any good reason for this short notation other than saving three letters for the well-initiated? Does it make sense in the context of other mathematical notation conventions (ie could you guess that it meant gdc(a,b) without a lot of incriminating context pointing towards it)? Mathematics is hard enough for laymen as it is.

As I learned recently, there is actually a good reason why gcd(a, b) is often abbreviated as (a, b). Roughly speaking, in ring theory, an ideal I is some subset of a ring (which you can think of as a number system, sorta) for which ai is in I (if a is any element of the ring and i is in I). Given any element x, we can generate an ideal from x by just taking every other y in the ring and computing yx - if every yx is in the ideal, then multiplying by some other z to get zyx is the same as just having (zy)x ,which is just a different multiple of x. This ideal - the ideal generated by x - is written as (x); if it's generated by multiple elements, it's written as (x1, x2, ...).

If you consider the ring of the integers, then the ideals are multiples of some integer. For instance, the multiples of three are an ideal, because multiplying a multiple of three by any integer yields a multiple of three (so if you multiply the set of multiples of three by any integer, you just get back something that's in the ideal).

The final point is this: the ideal generated by two integers is actually just the set of multiples of their gcd. Therefore, if a and b are integers, (a, b) is the set of multiples of their gcd - just like (3) is the set of multiples of three. This is why the gcd is often written in this way.

The cool thing is that this works in rings in general, not just integers. You can extend the concepts of gcd, primality, divisibility, etc to rings in general, and operate on things besides just integers, such as matrices, polynomials, or rotations of a cube.

For more info:

  http://en.wikipedia.org/wiki/Ring_theory
  http://en.wikipedia.org/wiki/Ideal_(ring_theory)

> You can extend the concepts of gcd, primality, divisibility, etc to rings in general

Thats true, but you have a "real" GCD only in special types of rings (principal ideal domains). In other rings, you only have Ideals as rough generalization of GCDs.

> Therefore, if a and b are integers, (a, b) is the set of multiples of their gcd - just like (3) is the set of multiples of three.

To make this more clear, in the notation of Ideals, you can write this:

  (15,6) = (3)

That is, the Ideal generated by 15 and 6 is same as the Ideal generated by 3. And for nonnegative integers, this essentially means the same as saying that 3 is the GCD of 15 and 6:

  gcd(15,6) = 3

There is still some "unclean" step involved here (that is, identifying numbers by their principal ideal, i.e. treating 3 and (3) as if these were equal), but I think this justifies the notation nevertheless.

You're is exactly right. To add to what you were saying and expand a little bit:

A principal ideal domain is a ring in which every ideal is generated by only one element, so whenever we see (a, b), we know there is some element c such that (a, b) = (c). I think this is what vog meant by having a "real" GCD - only in a principal ideal domain is your gcd unique. Without uniqueness, we can still define a greatest common divisor such that if gcd(a, b) = g, we know that there is nothing we can multiply by g to get a divisor of both a and b; that is, there's no extra factor we can add to g in order to get another factor of both a and b. That is enough to call g a GCD - but it's not necessarily unique! It turns out that in rings which aren't principal ideal domains, you can have more than one GCD! It's bizarre to think exactly what "greatest" means in this context, but you can also just think of it as "can't add any more factor while still dividing both a and b".

Ring theory is fun! (And practical, sometimes - you can describe some algorithms very elegantly via embedding the things you're working with in unusual rings.)

More info, with some examples and counterexamples:

  https://en.wikipedia.org/wiki/Principal_ideal_domain

This is a great counterexample to those that decry mathematical notation. Ambiguity may not be introduced for raw efficiency, but instead to indicate powerf techniques through punning.

The same holds not just for math symbols but also for math names. Many things have similar/same names because there is a deep mathematical connection that justifies the conflation.

And just as often the names are totally arbitrary.

There is some precedent for the notation (a, b) or <a, b> or [a, b] meaning "some function applied to a and b". For example, inner products or commutators (for a group, the commutator [a, b] of a and b is equal to a^-1 b^-1 ab).

The compactness really can matter sometimes... you might have a gcd appearing in the initializer of a summation symbol. http://en.wikipedia.org/wiki/Mobius_inversion gives you a good example of the "d divides n" notation appearing in such a position... pretty sure there exist similar constructions but with gcds.

Also, when stretching analogies far enough, one could argue that the GCD serves a similar purpose in number theory like the inner product in linear algebra. :-)

However, I think the real reason is the notation of GCD as an Ideal generated over two elements of a Ring, as described in the other reply: https://news.ycombinator.com/item?id=6090688

Yes. This is standard notation in many textbooks and university classes.

It's actually standard notation for the highest common factor.

This is nit-picky, but why does the HN title use "fraction" when the linked question uses the more-correct "rational"? It doesn't replace the other instance of "rational".

"Fraction" is shorthand (at least to a layman, which the vast majority of HN is in mathematics) for "rational non-integer." x must not be an integer for the stated property to hold, but the result might be.

This is at the very least highly confusing, because "fraction" doesn't actually imply "non-integer". But maybe that's because I also know what rational numbers are, which according to you makes me an expert mathematician.

I knew all of those things and it wasn't highly confusing to me and any concerns were checked by RTFA

x<1>=x and by induction x<n+1>=x^x<n>

How does one induct (induce?) this?

Gro-Tsen is just defining x<n> recursively. "by induction" there doesn't mean "proof by induction".

Prove it for the base case (x<1> = x), then assume it works for n-1, and show it works for the nth case.

That is a beautiful proof. Thanks for sharing this xSwag.

Mathematics is a fascinating and interesting subject, but I just wonder if this is the right place to post? If this is, are we expecting to see similar stuff from biology, physics, etc?

Yes. eg. https://news.ycombinator.com/item?id=5878664

Reply on news.ycombinator.com