7 ms·

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

• ctdonath 13y ago
  The proof given is fascinating, but...translucent. Can we get a simple, and correct, "yes" or "no" here?
  • seliopou 13y ago
    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.
    • cookingrobot 13y ago
      But Schanuel's conjecture hasn't been proven. So would you say this reduces to Schanuel's conjecture?
      • seliopou 13y ago
        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.
      • thomasbk 13y ago
        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.
      • ot 13y ago
        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.
        • seliopou 13y ago
          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?
          • ot 13y ago
            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.27s_pseudo-exponentiation https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber...
  • thomasbk 13y ago
    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 https://en.wikipedia.org/wiki/Schanuel%27s_conjecture
    • ctdonath 13y ago
      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.
  • deleted 13y ago
    [deleted]
• aidos 13y ago
  I don't think I've seen this notation before: (a,b) = 1 I guess it means gcd(a, b) = 1
  • acchow 13y ago
    Yes. This is standard notation in many textbooks and university classes.
  • vog 13y ago
    Yes, this is a short notation of gcd, and it is quite common among mathematicians.
    • Dewie 13y ago
      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.
      • wging 13y ago
        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 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.
        • vog 13y ago
          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 https://news.ycombinator.com/item?id=6090688
• tomrod 13y ago
  That is a beautiful proof. Thanks for sharing this xSwag.
• snake_plissken 13y ago
  x<1>=x and by induction x<n+1>=x^x<n> How does one induct (induce?) this?
  • tomrod 13y ago
    Prove it for the base case (x<1> = x), then assume it works for n-1, and show it works for the nth case.
  • qiemem 13y ago
    Gro-Tsen is just defining x<n> recursively. "by induction" there doesn't mean "proof by induction".
• happywolf 13y ago
  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?
  • JonnieCache 13y ago
    Yes. eg. https://news.ycombinator.com/item?id=5878664 https://news.ycombinator.com/item?id=5878664
• carstimon 13y ago
  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.
• bmm6o 13y ago
  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".
  • GeneralMayhem 13y ago
    "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.
    • codeflo 13y ago
      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.
      • vehementi 13y ago
        I knew all of those things and it wasn't highly confusing to me and any concerns were checked by RTFA