Also, lest I be rude,
Aw, shucks. Thanks for the praise!
For example, in the .999=1 thread (post #1059), he just described the infinitely many points of a line segment using the intuitively clear terminology “shitload”. Mathematical exposition at its finest! 
Subsequent posts by Trinopus (#1061) and myself (#1064) delved the meaning of that. 
ETA: Link.
The integers with sqrt(-5) adjoined does not have unique factorization. The example is 37 = (1+2sqrt(-5))(1-2sqrt(-5)) and none of the factors has a proper divisor. Whether this leads directly to a counter-example to the OP is not clear to me.
I have a solution but it won’t fit in the margin of this post.
There’s no counter-example in the context of the integers with sqrt(-5) adjoined, as even though this is not a unique factorization domain, it is still integrally closed (as -5 is squarefree and not 1 mod 4).
[Sorry everyone else; I’ll work on making that plain-spoken later…]
(Well, I can make it plain-spoken easily. Notice that the OP’s theorem can actually be quite strongly generalized: whenever r is a rational root of a polynomial p(x) with integer coefficients, the lowest-terms denominator of r divides the leading coefficient of p. [Proof: Let L be the leading coefficient of p, let r = n/d in lowest terms, and consider p(n/d)d[sup]deg§[/sup] = 0. Distributing out the left hand side, we see that its leading term is Ln[sup]deg§[/sup], and all the rest of its terms are multiples of d. But this means Ln[sup]deg§[/sup] is also divisible by d, and since d is coprime to n, we must have that d divides L.].
In particular, if r is a rational (i.e., ratio of integers) root of a polynomial with integer coefficients and leading coefficient 1, then r is an integer. This property is called “integral closure”, and we might discuss whether the same property holds when “integer” is replaced throughout by something else.
Dedekind, et al, studied this problem when “integer” was replaced by “number of the form a + b * sqrt(K), for integers a and b and a fixed integer K” way back when, and discovered, among other things, that integral closure continues to hold just in case either K is a square (in which case we haven’t actually changed anything), or the only square factor of K is 1 and K - 1 is not a multiple of 4.
But the actual argument for this, well, I’ll leave that for anyone interested to look up…)
You deserve it. If you’ve ever published anything, it might be useful to use it for Wikipedia sources. Too many math pages get too technical without coming up for air once in a while.
BTW, I feel the same way about Chronos’ posts in physics.
There are a number of amazing posters in dopeworld. I agree on Chronos. And while we are dishing out the praise, there is also Stranger On A Train. I always feel, “woohoo! I’m talking with a genuine rocket scientist.”
What Indistinguishable said. About integrally closed rings, I mean. You might try numbers of the form a + bsqrt(-3) or a + bsqrt(5), neither of which is integrally closed in its field of fractions, but in each case there is a fraction that satisfies and integral quadratic, but I couldn’t readily see a fraction whose square is in the ring.
By leaving rings of numbers, here is an easy example. Take all polynomials of the form a_0 + a_2x^2 + a_3x^3 + …, i.e. whose degree 1 term is 0. You can see that neither multiplication of such polynomials will take you outside of that set, so it is a ring in the usual way. In that ring, x^2 lacks a square root, but it has a fractional square root: x^3/x^2.
Oh, here is a simpler example. Look at all numbers of the form a + bsqrt(8) (with integer coefficients). There is no sqrt(2) in that ring, but the fraction sqrt(8)/2 is a square root of 2. This comes from the preceding example by letting x = sqrt(2). Then x^2 = 2, x^3 = sqrt(8), x^4 = 4, x^5 = 2sqrt(8), etc.
And Pasta. Check him out on Hamiltonians. Now I can fling that word around:).
What, technically, is “proof-making,” the goal of the OP, called? “Mathematics?”
A real question. (This place being lousy with Ninjas, I sometimes feel compelled to add that I am not one myself here.)
Yes. (I’m a bit surprised by your hesitance. Why would you not think this was mathematics?)