
Have any long-suspected irrational numbers turned out to be rational? - colinprince
http://mathoverflow.net/questions/32967/have-any-long-suspected-irrational-numbers-turned-out-to-be-rational
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x0x0
In the age of computers, there will be fewer fun surprises like that.

What surprises me is the euler-mascheroni constant, which shows up all over
number theory and also in probability (coupon collector's problem, entropy of
weibull and levy) is so difficult to work with that not only has no-one proven
it transcendental, it hasn't even been proven irrational.

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mcherm
> In the age of computers, there will be fewer fun surprises like that.

Really? I can't think of any way in which computers make it easier to
distinguish rational from irrational numbers. What is it that I am missing?

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GFK_of_xmaspast
Go read Kuperberg's answer where he mentions Borwein&Borwein (the one with 51
upvotes)

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mcherm
Oh, that's REALLY interesting!

I admit I was surprised that "Most (but not all) interesting numbers admit a
polynomial-time algorithm to compute their digits." But this is a really good
reason why computers would make a difference.

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murbard2
One of the reason such proofs are hard is that there is no reason for these
numbers to be rational, it would be a huge coincidence. Proofs typically
happen when there is a good reason for things to be a certain way.

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tel
And "almost all" numbers are irrational, so you sort of do need a reason to be
rational.

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michaelochurch
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).

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3JPLW
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.

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Dove
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)

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jsbgir
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.

