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This goes against generally accepted mathematical jargon (http://en.wikipedia.org/wiki/List_of_mathematical_jargon#Descriptive_informalities http://en.wikipedia

This goes against generally accepted mathematical jargon (http://en.wikipedia.org/wiki/List_of_mathematical_jargon#Descriptive_informalities http://en.wikipedia.org/wiki/List_of_mathematical_jargon#Des...). For countable sets like integers, "almost all" has a meaning "all except a _finite_ set". Obviously the set of numbers that do _not_ contain 3 (or any other digit) is an infinite countable subset of all integers. Frequency has nothing to do with it.

musername 12y ago
okay, but what do you call it if the frequency is diverging to zero for infinity?
- mynegation 12y ago
  "Asymptotic distribution" or "density" probably. Check out this, for an example: http://en.wikipedia.org/wiki/Prime_number_theorem http://en.wikipedia.org/wiki/Prime_number_theorem
jerf 12y ago
I don't think I'm going to ding what is clearly intended as a general-audience presentation for failing to precisely conform to an uncited definition on Wikipedia, under a section with the word "Informality" in its name.
ColinWright 12y ago
That turns out not to be the case: > almost all > A shorthand term for "all except for a set of measure zero", > when there is a measure to speak of. In this case there is no measure to speak of, so it is normal to talk about taking the ration in a finite initial segment and then letting the size go to infinity and seeing what happens to the ratio. And a set of measure zero does not have to be finite, either. Almost all numbers are transcendental in the sense given above, and yet there are infinitely many non-transcendental numbers. So I think you are mistaken, although I would be interested to hear a more complete statement of what you mean, in case I have misunderstood you.
- mynegation 12y ago
  The same excerpt, starting from "One can also speak" describes how that "almost all" terminology is applied to integers (or, by extension, to any countable set, that is already measure zero). The example statement is "almost all prime numbers are odd", because there is a finite number of primes that are even (only one number 2). Put another way, if we stick to the definition of "almost all" as elements not having the property forming measure zero set, then you do not need the video to prove the title: all integers are already measure zero. So, for subsets of integers, "almost all" means something else in jargon. I could not watch the video, but I found (I believe) equivalent text. And "almost all" means that as we extend the reach the share of numbers not containing 3 goes to zero. It is mildly entertaining (albeit a bit too obvious) result, but my point was that usage of "almost all" is confusing because _I_ am used to a very specific meaning of it.
  - drostie 12y ago
    It's also a little confusing because there are sets which do not have a well-defined natural density among the integers, like: leastpow2 x = head $ dropWhile (\k -> 2*k <= x) $ map (2^) [0..] p k = k < 3 * (leastpow2 k `div` 2) limitless = [k | k <- [0..], k == 0 || p k] which oscillates between attaining 1/2 and then 2/3 for its partial densities.
- JadeNB 12y ago
  > In this case there is no measure to speak of, so it is normal to talk about taking the ration in a finite initial segment and then letting the size go to infinity and seeing what happens to the ratio. Indeed it is normal to talk about it, but this condition is usually called having "density 1" (https://en.wikipedia.org/wiki/Natural_density https://en.wikipedia.org/wiki/Natural_density). I think that "almost all" is not terribly common in this connection, except possibly in a community where this is the only kind of nearly-true statement that one would want to discuss.
- redxaxder 12y ago
  Number Theorists use a different definition of "almost all" from the one you see in measure theory. They use the limit version shown in the video.
moomin 12y ago
Wikipedia actually supports his usage on a more definitive page http://en.wikipedia.org/wiki/Almost_all http://en.wikipedia.org/wiki/Almost_all. Speaking as an ex-mathematician, I started laughing as soon as I read the title, because it pretty much tells you how to demonstrate it.

musername 12y ago

okay, but what do you call it if the frequency is diverging to zero for infinity?

mynegation 12y ago
"Asymptotic distribution" or "density" probably. Check out this, for an example: http://en.wikipedia.org/wiki/Prime_number_theorem http://en.wikipedia.org/wiki/Prime_number_theorem

jerf 12y ago

I don't think I'm going to ding what is clearly intended as a general-audience presentation for failing to precisely conform to an uncited definition on Wikipedia, under a section with the word "Informality" in its name.

ColinWright 12y ago

That turns out not to be the case: > almost all > A shorthand term for "all except for a set of measure zero", > when there is a measure to speak of. In this case there is no measure to speak of, so it is normal to talk about taking the ration in a finite initial segment and then letting the size go to infinity and seeing what happens to the ratio. And a set of measure zero does not have to be finite, either. Almost all numbers are transcendental in the sense given above, and yet there are infinitely many non-transcendental numbers. So I think you are mistaken, although I would be interested to hear a more complete statement of what you mean, in case I have misunderstood you.

mynegation 12y ago
The same excerpt, starting from "One can also speak" describes how that "almost all" terminology is applied to integers (or, by extension, to any countable set, that is already measure zero). The example statement is "almost all prime numbers are odd", because there is a finite number of primes that are even (only one number 2). Put another way, if we stick to the definition of "almost all" as elements not having the property forming measure zero set, then you do not need the video to prove the title: all integers are already measure zero. So, for subsets of integers, "almost all" means something else in jargon. I could not watch the video, but I found (I believe) equivalent text. And "almost all" means that as we extend the reach the share of numbers not containing 3 goes to zero. It is mildly entertaining (albeit a bit too obvious) result, but my point was that usage of "almost all" is confusing because _I_ am used to a very specific meaning of it.
- drostie 12y ago
  It's also a little confusing because there are sets which do not have a well-defined natural density among the integers, like: leastpow2 x = head $ dropWhile (\k -> 2*k <= x) $ map (2^) [0..] p k = k < 3 * (leastpow2 k `div` 2) limitless = [k | k <- [0..], k == 0 || p k] which oscillates between attaining 1/2 and then 2/3 for its partial densities.
JadeNB 12y ago
> In this case there is no measure to speak of, so it is normal to talk about taking the ration in a finite initial segment and then letting the size go to infinity and seeing what happens to the ratio. Indeed it is normal to talk about it, but this condition is usually called having "density 1" (https://en.wikipedia.org/wiki/Natural_density https://en.wikipedia.org/wiki/Natural_density). I think that "almost all" is not terribly common in this connection, except possibly in a community where this is the only kind of nearly-true statement that one would want to discuss.
redxaxder 12y ago
Number Theorists use a different definition of "almost all" from the one you see in measure theory. They use the limit version shown in the video.

moomin 12y ago

Wikipedia actually supports his usage on a more definitive page http://en.wikipedia.org/wiki/Almost_all http://en.wikipedia.org/wiki/Almost_all. Speaking as an ex-mathematician, I started laughing as soon as I read the title, because it pretty much tells you how to demonstrate it.