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I edited my post while you wrote this reply. Edit: And by the way, the immediate direction I'm walking you into with that edit is that, if for some a < b, (a,b

I edited my post while you wrote this reply. Edit: And by the way, the immediate direction I'm walking you into with that edit is that, if for some a < b, (a,b) is one of your pixels, then there are no pixel boundaries in that interval (a,b), unless your pixels overlap.

Dylan16807 12y ago
Well it's the non-overlapping pixels that are the interesting case, so let's stick with those. So no a and b, just a. Then I suppose my first attempt to define a pixel would be the zero-width area stating at a, non-inclusive. So the zero width area starting at 3/8, non-inclusive. Is that fundamentally flawed? (Probably. Edit 2: I mean seriously I have no idea how to handle ranges that are infinitely small. But they clearly exist in some form, I can look at the set of irrational numbers...) > one rational (I think you mean one number of the form k/2^n) Yes, in this context. It doesn't really matter if it's k/2^n or any rational, but for consistency's sake let's stick with k/2^n dividers. Edit: How about this. Is it meaningful to talk about rounding reals to the nearest rational? If not, why not?
- SamReidHughes 12y ago
  > So the zero width area starting at 3/8, non-inclusive. So yeah, is that just the interval (3/8, 3/8), i.e. the empty set, or what? If it were inclusive, you'd have [3/8, 3/8], which means the same thing as {3/8} -- it's the set that contains just the value 3/8. > Is it meaningful to talk about rounding reals to the nearest rational? If not, why not? It is not. The reason is, there is no nearest rational (if it's not already rational). Suppose x is an irrational number, and p is some rational number. Then p != x. That means |x-p| > 0. That means there's some integer n such that 1/n < |x-p|. If p > x, then x < (p - 1/n) < p. If p < x, then p < (p + 1/n) < x. In each case we've constructed a rational number (p-1/n or p+1/n) that is closer to x than p was. In general, between any two irrational numbers, there's infinitely many rational numbers, and between any two rational numbers, there's (uncountably) infinitely many irrational numbers.
  - Dylan16807 12y ago
    > is that the empty set Pretty much. It was not the most accurate first attempt. I was hoping to find a way to fix it. > no rounding Well that argument is the obvious answer yes, but it requires a certain framework. Kind of like arguing that there is no biggest number, but there is ω... Anyway I accept that with normal semantics you can't define non-point non-overlapping ranges between dense rationals. I was wrong. Thank you for your time.

Dylan16807 12y ago

Well it's the non-overlapping pixels that are the interesting case, so let's stick with those. So no a and b, just a. Then I suppose my first attempt to define a pixel would be the zero-width area stating at a, non-inclusive. So the zero width area starting at 3/8, non-inclusive. Is that fundamentally flawed? (Probably. Edit 2: I mean seriously I have no idea how to handle ranges that are infinitely small. But they clearly exist in some form, I can look at the set of irrational numbers...) > one rational (I think you mean one number of the form k/2^n) Yes, in this context. It doesn't really matter if it's k/2^n or any rational, but for consistency's sake let's stick with k/2^n dividers. Edit: How about this. Is it meaningful to talk about rounding reals to the nearest rational? If not, why not?

SamReidHughes 12y ago
> So the zero width area starting at 3/8, non-inclusive. So yeah, is that just the interval (3/8, 3/8), i.e. the empty set, or what? If it were inclusive, you'd have [3/8, 3/8], which means the same thing as {3/8} -- it's the set that contains just the value 3/8. > Is it meaningful to talk about rounding reals to the nearest rational? If not, why not? It is not. The reason is, there is no nearest rational (if it's not already rational). Suppose x is an irrational number, and p is some rational number. Then p != x. That means |x-p| > 0. That means there's some integer n such that 1/n < |x-p|. If p > x, then x < (p - 1/n) < p. If p < x, then p < (p + 1/n) < x. In each case we've constructed a rational number (p-1/n or p+1/n) that is closer to x than p was. In general, between any two irrational numbers, there's infinitely many rational numbers, and between any two rational numbers, there's (uncountably) infinitely many irrational numbers.
- Dylan16807 12y ago
  > is that the empty set Pretty much. It was not the most accurate first attempt. I was hoping to find a way to fix it. > no rounding Well that argument is the obvious answer yes, but it requires a certain framework. Kind of like arguing that there is no biggest number, but there is ω... Anyway I accept that with normal semantics you can't define non-point non-overlapping ranges between dense rationals. I was wrong. Thank you for your time.