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I can think of three cases where "irrational-looking" numbers turn out to be rational. However, they're actually mathematically trivial, not "long suspected irr

I can think of three cases where "irrational-looking" numbers turn out to be rational. However, they're actually mathematically trivial, not "long suspected irrational". 1. e^(i * pi) "looks" like nonsense. If you think of exponentiation as iterated multiplication, use algebra (roots) to compute rational exponents, use continuity to fill in the non-integer real values... then, still, how do you "do something" an imaginary number of times? Of course, it's actually -1. This doesn't qualify (IMO) because as soon as mathematicians understood complex exponents, this result was probably fairly obvious. 2. The infinite tower sqrt(2)^sqrt(2)^sqrt(2)^... has a finite value, which is 2. (It's the fixed point, sqrt(2)^x = x). Yet again, we have a simple (integer) value coming out of something that seems infinitely complex. However, yet again, once you make the conceptual leap necessary to evaluate such "towers", the answer is trivial. 3. Binet's formula "looks" like it should be producing irrational numbers for integer inputs, but actually produces the Fibonacci sequence. But this is just high school algebra: the problematic terms cancel and you're left with something reducing to an integer. I suspect that all of the "long suspected irrationals" like e^pi and pi^e are irrational. Almost all mathematicians, if it were an even-odds bet, would bet on them being so. (Of course, intuition can and does fail in all sorts of ways.) It's just that, for all but the simplest numbers, we don't have the tools to prove irrationality when, in fact, they are. That requires quantifying over all the integers, and we already know that some first-order statements are undecidable. The "scary" (and, in my mind, likely) possibility is that, for some of these simple-to-express real numbers, it's formally undecidable whether they are rational. More weirdly, no specific real number will ever be proven to be in the "unprovably irrational" category (since that would, by definition, constitute proof of irrationality).

tel 12y ago
I think a much better example is of the style of the highest rated answer in the thread. If you look at the wiki page [0] you see that the constant has a non-trivial relationship with a major mathematical question ("what is the behavior of the prime numbers?") and, as you compute estimations of it within your computational ability it waves above and below its true value. For things like e^i you almost cannot define the idea of complex powers without knowing the right concepts. The sqrt(2) example clearly converges toward 2 as you stack the tower. Binet's formula was arrived at via derivation from the Fibonacci recursion formula... so anyone who had it would obviously know its significance. So, to be clear, the only reason why those examples would look strange is if someone showed it to you (armed with only high school mathematics) without explanation. Or if you don't know how to estimate the solution to f^n(1), f(x) = sqrt(2)^x for n, say, in [1,10]. "Legendre's Constant" has none of those features. That said, with today's computational power deriving a sufficiently strong estimate that it "might be 1" isn't so hard. One can imagine that other slower to converge examples exist today [0] http://en.wikipedia.org/wiki/Legendre%27s_constant http://en.wikipedia.org/wiki/Legendre%27s_constant
3JPLW 12y ago
Given that the set of rationals is countable, and that the set of irrationals is uncountable, nearly "all" real numbers are irrational. In our experience, however, we mostly interact with rational numbers, and so irrational numbers seem to be the "strange" ones. But there are infinitely more of them! Were one able to devise a method to choose a real number at random, it'd be nearly impossible to ever get a rational number. As such, it would indeed be extraordinary to have some non-trivial function transform an irrational number (or two, like e^pi) in an unexpected way into a rational number. I imagine most mathematicians would take a very lopsided bet that they're irrational.
- Dove 12y ago
  This statement can actually be made stronger. Not only are the vast majority of real numbers irrational, they are transcendental[1] -- indescribable in algebraic terms. The numbers we can actually talk about and name are the countable algebraic numbers -- like the square root of two -- and a few transcendentals we know about. Not many. You could count them on your fingers and toes. The rest of the reals -- and it is most of that vast sea -- are completely inaccessible, unnameable, forever anonymous. [1] http://en.wikipedia.org/wiki/Transcendental_number http://en.wikipedia.org/wiki/Transcendental_number
  - jsbgir 12y ago
    The set of transcendental but computable numbers is also countable, however: there is a countable number of algorithms. Therefore, the majority of the reals are uncomputable, like Chaintin's constant (the probability that a random algorithm halts). This can in turn be made stronger by noticing that the definition of Chaitin's constant is finite, and by the same argument, definable real numbers are countable. It follows that the majority of the reals are indefinable.
  - bashinator 12y ago
    Seems nicely analogous to "regular matter" vs dark matter and dark energy in the Universe.
- gizmo686 12y ago
S4M 12y ago
There is also gamma (lim_{n -> infinity} (sum_{k=1..n} 1/k)-ln n), that, as far as I know, hasn't been proved to be either rational, either irrational, that is "long suspected irrational", and probably is.
codeflo 12y ago
> 2. The infinite tower sqrt(2)^sqrt(2)^sqrt(2)^... has a finite value, which is 2. (It's the fixed point, sqrt(2)^x = x). Yet again, we have a simple (integer) value coming out of something that seems infinitely complex. However, yet again, once you make the conceptual leap necessary to evaluate such "towers", the answer is trivial. Is it really that simple? sqrt(2)^4 = 4, so the answer should also 4 by the same argument. Which makes the whole expression not well-defined IMHO.
- dragonwriter 12y ago
  The expression is well-defined, the statement that the infinite tower is equal to the fixed point is just slightly oversimplified. That's only true for attracting fixed points, but its possible to have repelling fixed points, as well.
  - codeflo 12y ago
    Fair enough, I'm starting to read up on this. Obviously, you can define sqrt(2)^sqrt(2)^... to mean the (unique?) attracting fixed point of x -> sqrt(2)^x. But that makes your explanation a bit circular (it is so because it's defined to be so). Probably that's the only useful definition, but that fact is certainly not trivial or self-evident.
    - dragonwriter 12y ago
      Well, the thing is that you can test whether a fixed point is an attracting fixed point; per Wikipedia (which I am relying on for the details of the test because I haven't used this for years, but the test looks correct), any fixed point x of a function f which has an open neighborhood where f is continuously differentiable and f'(x) < 1 is attracting.
  - michaelochurch 12y ago
    Very good point. Thanks for stating this.
  - Chinjut 12y ago

tel 12y ago

I think a much better example is of the style of the highest rated answer in the thread. If you look at the wiki page [0] you see that the constant has a non-trivial relationship with a major mathematical question ("what is the behavior of the prime numbers?") and, as you compute estimations of it within your computational ability it waves above and below its true value. For things like e^i you almost cannot define the idea of complex powers without knowing the right concepts. The sqrt(2) example clearly converges toward 2 as you stack the tower. Binet's formula was arrived at via derivation from the Fibonacci recursion formula... so anyone who had it would obviously know its significance. So, to be clear, the only reason why those examples would look strange is if someone showed it to you (armed with only high school mathematics) without explanation. Or if you don't know how to estimate the solution to f^n(1), f(x) = sqrt(2)^x for n, say, in [1,10]. "Legendre's Constant" has none of those features. That said, with today's computational power deriving a sufficiently strong estimate that it "might be 1" isn't so hard. One can imagine that other slower to converge examples exist today [0] http://en.wikipedia.org/wiki/Legendre%27s_constant http://en.wikipedia.org/wiki/Legendre%27s_constant

3JPLW 12y ago

Given that the set of rationals is countable, and that the set of irrationals is uncountable, nearly "all" real numbers are irrational. In our experience, however, we mostly interact with rational numbers, and so irrational numbers seem to be the "strange" ones. But there are infinitely more of them! Were one able to devise a method to choose a real number at random, it'd be nearly impossible to ever get a rational number. As such, it would indeed be extraordinary to have some non-trivial function transform an irrational number (or two, like e^pi) in an unexpected way into a rational number. I imagine most mathematicians would take a very lopsided bet that they're irrational.

Dove 12y ago
This statement can actually be made stronger. Not only are the vast majority of real numbers irrational, they are transcendental[1] -- indescribable in algebraic terms. The numbers we can actually talk about and name are the countable algebraic numbers -- like the square root of two -- and a few transcendentals we know about. Not many. You could count them on your fingers and toes. The rest of the reals -- and it is most of that vast sea -- are completely inaccessible, unnameable, forever anonymous. [1] http://en.wikipedia.org/wiki/Transcendental_number http://en.wikipedia.org/wiki/Transcendental_number
- jsbgir 12y ago
  The set of transcendental but computable numbers is also countable, however: there is a countable number of algorithms. Therefore, the majority of the reals are uncomputable, like Chaintin's constant (the probability that a random algorithm halts). This can in turn be made stronger by noticing that the definition of Chaitin's constant is finite, and by the same argument, definable real numbers are countable. It follows that the majority of the reals are indefinable.
- bashinator 12y ago
  Seems nicely analogous to "regular matter" vs dark matter and dark energy in the Universe.
gizmo686 12y ago

S4M 12y ago

There is also gamma (lim_{n -> infinity} (sum_{k=1..n} 1/k)-ln n), that, as far as I know, hasn't been proved to be either rational, either irrational, that is "long suspected irrational", and probably is.

codeflo 12y ago

> 2. The infinite tower sqrt(2)^sqrt(2)^sqrt(2)^... has a finite value, which is 2. (It's the fixed point, sqrt(2)^x = x). Yet again, we have a simple (integer) value coming out of something that seems infinitely complex. However, yet again, once you make the conceptual leap necessary to evaluate such "towers", the answer is trivial. Is it really that simple? sqrt(2)^4 = 4, so the answer should also 4 by the same argument. Which makes the whole expression not well-defined IMHO.

dragonwriter 12y ago
The expression is well-defined, the statement that the infinite tower is equal to the fixed point is just slightly oversimplified. That's only true for attracting fixed points, but its possible to have repelling fixed points, as well.
- codeflo 12y ago
  Fair enough, I'm starting to read up on this. Obviously, you can define sqrt(2)^sqrt(2)^... to mean the (unique?) attracting fixed point of x -> sqrt(2)^x. But that makes your explanation a bit circular (it is so because it's defined to be so). Probably that's the only useful definition, but that fact is certainly not trivial or self-evident.
  - dragonwriter 12y ago
    Well, the thing is that you can test whether a fixed point is an attracting fixed point; per Wikipedia (which I am relying on for the details of the test because I haven't used this for years, but the test looks correct), any fixed point x of a function f which has an open neighborhood where f is continuously differentiable and f'(x) < 1 is attracting.
- michaelochurch 12y ago
  Very good point. Thanks for stating this.
- Chinjut 12y ago