> in the end we have no idea what it looks like -- is it >0.5? No clue. No, that isn't true. The proof Colin gives is quite explicit. There's an explicit listi

> in the end we have no idea what it looks like -- is it >0.5? No clue. No, that isn't true. The proof Colin gives is quite explicit. There's an explicit listing of the algebraic numbers (except that he leaves the order of various things unstated; for present purposes I'll just work with the order of appearance in his writeup). Then there's a procedure that considers these one by one and chooses intervals of exponentially decreasing size to concentrate on. You can get as many decimal places as you like of the transcendental number Colin's constructing, just by doing enough steps. Let's be more explicit. The first algebraic numbers in Colin's list are 0, 0, -1, 1, 0. (I assume we aren't counting repeated roots of a single polynomial, but also aren't eliminating duplicates between different polynomials in the list.) So, first we take the intervals [0,1/3], [1/3,2/3], [2/3,1] and the algebraic number 0. We take the leftmost interval not containing 0, namely [1/3,2/3]. Now we subdivide into three again: [1/3,4/9], [4/9,5/9], [5/9,2/3]. And again our algebraic number is 0. The leftmost interval not containing 0 is [1/3,4/9]. (So we've already established that the number is < 0.5.) In the next three steps we never see an algebraic number inside our interval so we'll just take the leftmost third each time; so we get [1/3,1/3+1/243]. We've now found the first two decimal places of the transcendental number we're constructing, namely 0.33. It would (at least in principle) be easy to computerize the process and find many more decimal places.