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by throwaway080383 2780 days ago
To back up my last claim, let me prove that 100% of positive integers are even in the same spirit:

Fix any odd positive integer x, and consider its "twists" x{d}, which for positive d I define to be x*2^d. Every integer is of the form x{d} for a unique choice of x and d. Now, for fixed N, if I consider all "twists" for which d<N, the proportion which are even is (N-1)/N. Thus, as N tends to infinity, the proportion tends to 1.

1 comments

How does a twist x{d} being even equate to the integer d being even?
The twist x{d} is even as soon as d>0.
Yes, but that doesn't mean the integer d is even.
I never say anything about d being even. Rather, as soon as d > 0, x{d} is even, so "almost all" x{d} are even. As you vary x and d, you cover all integers exactly once, so from this perspective, "almost all" integers are even.

My point is that this is very similar to how elliptic curves are being counted here.