5 ms·
The linked paper on the arXiv is pretty bad. The idea as it stands appears to be very mundane -- that we define some new number, which he calls ① and I'll call
by drostie 15y ago
The linked paper on the arXiv is pretty bad. The idea as it stands appears to be very mundane -- that we define some new number, which he calls ① and I'll call m, to denote the size of the natural numbers N = {0, 1, 2, ...}. Then m² is the cardinality of N × N, you see. And we could say things like m + 1 > m, surely, and everything would be peachy keen.
Well, not so fast. That sort of mathematics is fraught with difficulty. So, for example, the set of pairs of natural numbers is called N² = {(0, 0), (1, 0), (0, 1), (2, 0), ...}. How many of those are there? The arXiv paper very distinctly says that there are m² of these numbers. In doing so he's apparently unaware of the huge breakthrough that Hilbert made: that in a deep sense, there are only m of them. For every pair, you can associate a number. It's not even hard, the function is:
def n(x, y):
return (x + y + 1)*(x + y)/2 + y
It's only a little tougher to invert this mapping, which is not strictly necessary, but is always nice:
from math import sqrt
def pair(n):
row = int(0.5 * (sqrt(1 + 8*n) - 1))
col = n - row * (row + 1)/2
return (row - col, col)
Mind you, that's just with examples that he uses. The whole mathematics is poorly defined. Supposing that you remove one element, like the number 0, from the natural numbers, he wants to say that there are m - 1 numbers in the set {1, 2, 3, 4, ...}. And if I take away 2, then presumably there are m - 2. And if I remove 2n for each n in N, I have performed m 'remove' operations, and you would presumably say that the set {1, 3, 5, 7, ...} has m - m elements, no? And even if you say that the resulting set of odd numbers has a size "something like" m/2, how do you know it's not m/2 + 1 or m/2 - 1? Suppose we define the sum of all the odd numbers up to the n'th one as {1, 1+3, 1+3+5, 1+3+5+7, ...} -- this surely has m/2 elements, correct? No -- those sums are {1, 4, 9, 16, ...} and this has sqrt(m) - 1 elements. Wha...? Does sqrt(m) even exist? His other publications suggest that he usually expands calculations in terms of (positive/negative) integer powers of m; there is little discussion about fractional powers, but presumably they must exist.
In short, we must accept that the mathematics of m has no correspondence (or only ad-hoc after-the-fact correspondence) to the natural numbers M. When you do this and uncouple the two, you'll probably discover that his "Infinity computer" is just a normal computer which can represent polynomials, and his ① is just the polynomial's 'x'.