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.
If you can compute a number's digits (i.e. approximate it), then you can see if they start repeating or not. If you compute a billion digits and they don't repeat yet, and you have no reason to expect the number to be a fraction with a huge denominator, then a good conjecture is that the number is irrational and vice versa when there is repetition. You'll be right more often than not.
(Of course, like many heuristics this one is easily exploitable by an adversary trying to make you give the wrong answer. And clearly this is not a proof, just a rule of thumb where you expect to be wrong now and then but not usually.)
The only thing that computers add is the ability to prove that a number is not rational to a larger number of decimal places than a human working on their own could or any other technique that requires brute force calculation on a large scale.
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.
It's not much easier to prove it. But when you can quickly compute the first few billion digits, you can just look for repetitions and conjecture that a number is rational.
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.
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).
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.
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.
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.
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
> 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.
There's no obvious, singular definition of "is" for something like that tower. You have to describe it via a generating process or an equation. If you ask for the values "x" such that "x = sqrt(2)^x" then you need to consider the chance there are multiple such "x"es. If you consider it the limit result of continually applying f(x) = sqrt(2)^x then you'll get the least fixed point.
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.
But a function could just as well have multiple attracting fixed points (consider the cube root function, with attracting fixed points at both +1 and -1).
Instead, we can look at it like this: If f is a continuous, order-preserving function from a closed interval to itself, and x <= f(x), then x <= f(x) <= f(f(x)) <= ..., with this sequence converging to the least fixed point >= x.
In our case, we can think of f(x) = sqrt(2)^x as a continuous function from [-infinity, +infinity] to itself, and take our starting point to be x = sqrt(2) [or perhaps 1, or 0, or -infinity, each one step back from the previous]; thus, the sequence sqrt(2), sqrt(2)^sqrt(2), sqrt(2)^(sqrt(2)^sqrt(2)), ..., converges to the least fixed point of f, which will be 2 rather than 4.
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.
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.
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.
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.