6 ms·

Yep, that's right. Its cardinality is the same as rationals, since it's countable.

by xtacy 7y ago

perl4ever 7y ago
How disappointing. Unlike most of the time, I read the article first and now that I'm here, that was the question I had - clearly it's smaller than reals, but how and why is this field larger than rational numbers? Guess it's not.
- klodolph 7y ago
  It’s larger than the rational numbers in the sense that it is a strict superset. Cardinality is what a lot of people reach for when they are talking about “larger” or “smaller”, but there are lots of other useful concepts which we can translate to “larger” and “smaller”. So when someone says “larger” or “smaller”, your first step might be to try and translate that relationship into a more precise mathematical concept, like cardinality or measure. Casual terminology also leads to weird discussions. Like when someone asks whether some function is “close” to another, and these functions are defined in terms of vector spaces. Unfortunately, “closeness” does not necessarily exist in a vector space. So the answer may be that the question does not make sense.
  - effie 7y ago
    > “closeness” does not necessarily exist in a vector space. The asker will give a definition. For example, two vectors are close if sqrt of dot product of difference of the two vectors is smaller than some number delta.
    - deleted 7y ago
      [deleted]
  - perl4ever 7y ago
    "It’s larger than the rational numbers in the sense that it is a strict superset" Ok, did I miss the explanation of that? Or is it something in "part 2" which I didn't see a link to?
    - shawnz 7y ago
      Clearly every single rational number is "useful", plus others that are not rational
      - perl4ever 7y ago
        So I guess what I lack is an understanding of why that doesn't affect cardinality.
        shawnz 7y ago
        There is a 1-to-1 mapping between the sets, so by definition that means they have the same cardinality. Let me give a more common example. Consider these two sets: N, the set of non-negative integers, and Z, the set of all integers. Clearly, everything in N is also in Z, and then some. But N and Z still have the same cardinality, because there are 1-to-1 mappings between the two sets. Here is one example of such a mapping: N | Z --------- 0 -> 0 1 -> 1 2 -> -1 3 -> 2 4 -> -2 5 -> 3 6 -> -3 ..etc. The formula for this mapping would be floor(n/2)*(-1)^(n%2). Clearly everything in the left set has exactly one corresponding item in the right set and vice versa, so they must be the same "size", even though the right set contains every item in the left set and then some.
- ur-whale 7y ago
  > Guess it's not. Isn't there a theorem that speaks of the existence or non-existence of a set whose cardinality is strictly larger than Q and strictly smaller than R. And a conjecture that says this theorem might well be unprovable?
  - pdonis 7y ago
    A "theorem" is a proved proposition, so there is no such thing as a theorem that might well be unprovable. The proposition you refer to (which is not a theorem since no proof is known, and in fact it has been shown that this proposition is logically independent of the usual foundations of set theory, so it cannot be proved in that framework) is called the Continuum Hypothesis: https://en.wikipedia.org/wiki/Continuum_hypothesis https://en.wikipedia.org/wiki/Continuum_hypothesis
    - lonelappde 7y ago
      The theorem is that the continuum hypothesis is independent of ZFC.
  - BeetleB 7y ago
    In this particular case, though, that is irrelevant. This set is strictly of the same cardinality as the rationals.
- thaumasiotes 7y ago
  > that was the question I had - clearly it's smaller than reals, but how and why is this field larger than rational numbers? As pdonis points out sidethread, this isn't really a valid question. (Or rather, the question is fine, but the answer to all questions of this form is already well-known, so there's no point in asking this specific question.) It is not possible to prove that a set is both smaller than the reals and larger than the rationals, because such a set would disprove the continuum hypothesis. (And symmetrically, it isn't possible to prove that no such set exists, because that would be a proof of the continuum hypothesis.)
  - perl4ever 7y ago
    "It is not possible to prove that a set is both smaller than the reals and larger than the rationals, because such a set would disprove the continuum hypothesis." Sure, even without much of a mathematical background, people generally take it for granted. Which is why it's disappointing that a suggestion of overturning it isn't fulfilled.
    - thaumasiotes 7y ago
      Suggestion of overturning it? It's a proof. That article would be headlined "disproving the independence of the continuum hypothesis" or some such; it would be huge news, not somebody's fun blog post.
      - perl4ever 7y ago
        So, you know better than to be fooled by this clickbait. Noted. All I'm saying is that it was very effective on me, unlike a lot of things.
  - lonelappde 7y ago
    Aside, it's a shape that hypothesis in math (axiom) is the opposite of hypothesis in science (claim to test)
    - 7y ago
- naniwaduni 7y ago
  Given that continuum hypothesis is independent of ZF, it would be very interesting to find a set of interest to a general audience whose cardinality lies between the rationals and the reals.

Yep, that's right. Its cardinality is the same as rationals, since it's countable.

by xtacy 7y ago

Yep, that's right. Its cardinality is the same as rationals, since it's countable.

How disappointing. Unlike most of the time, I read the article first and now that I'm here, that was the question I had - clearly it's smaller than reals, but how and why is this field larger than rational numbers? Guess it's not.

klodolph 7y ago
It’s larger than the rational numbers in the sense that it is a strict superset. Cardinality is what a lot of people reach for when they are talking about “larger” or “smaller”, but there are lots of other useful concepts which we can translate to “larger” and “smaller”. So when someone says “larger” or “smaller”, your first step might be to try and translate that relationship into a more precise mathematical concept, like cardinality or measure. Casual terminology also leads to weird discussions. Like when someone asks whether some function is “close” to another, and these functions are defined in terms of vector spaces. Unfortunately, “closeness” does not necessarily exist in a vector space. So the answer may be that the question does not make sense.
- effie 7y ago
  > “closeness” does not necessarily exist in a vector space. The asker will give a definition. For example, two vectors are close if sqrt of dot product of difference of the two vectors is smaller than some number delta.
  - deleted 7y ago
    [deleted]
- perl4ever 7y ago
  "It’s larger than the rational numbers in the sense that it is a strict superset" Ok, did I miss the explanation of that? Or is it something in "part 2" which I didn't see a link to?
  - shawnz 7y ago
    Clearly every single rational number is "useful", plus others that are not rational
    - perl4ever 7y ago
      So I guess what I lack is an understanding of why that doesn't affect cardinality.
      - shawnz 7y ago
        There is a 1-to-1 mapping between the sets, so by definition that means they have the same cardinality. Let me give a more common example. Consider these two sets: N, the set of non-negative integers, and Z, the set of all integers. Clearly, everything in N is also in Z, and then some. But N and Z still have the same cardinality, because there are 1-to-1 mappings between the two sets. Here is one example of such a mapping: N | Z --------- 0 -> 0 1 -> 1 2 -> -1 3 -> 2 4 -> -2 5 -> 3 6 -> -3 ..etc. The formula for this mapping would be floor(n/2)*(-1)^(n%2). Clearly everything in the left set has exactly one corresponding item in the right set and vice versa, so they must be the same "size", even though the right set contains every item in the left set and then some.
ur-whale 7y ago
> Guess it's not. Isn't there a theorem that speaks of the existence or non-existence of a set whose cardinality is strictly larger than Q and strictly smaller than R. And a conjecture that says this theorem might well be unprovable?
- pdonis 7y ago
  A "theorem" is a proved proposition, so there is no such thing as a theorem that might well be unprovable. The proposition you refer to (which is not a theorem since no proof is known, and in fact it has been shown that this proposition is logically independent of the usual foundations of set theory, so it cannot be proved in that framework) is called the Continuum Hypothesis: https://en.wikipedia.org/wiki/Continuum_hypothesis https://en.wikipedia.org/wiki/Continuum_hypothesis
  - lonelappde 7y ago
    The theorem is that the continuum hypothesis is independent of ZFC.
- BeetleB 7y ago
  In this particular case, though, that is irrelevant. This set is strictly of the same cardinality as the rationals.
thaumasiotes 7y ago
> that was the question I had - clearly it's smaller than reals, but how and why is this field larger than rational numbers? As pdonis points out sidethread, this isn't really a valid question. (Or rather, the question is fine, but the answer to all questions of this form is already well-known, so there's no point in asking this specific question.) It is not possible to prove that a set is both smaller than the reals and larger than the rationals, because such a set would disprove the continuum hypothesis. (And symmetrically, it isn't possible to prove that no such set exists, because that would be a proof of the continuum hypothesis.)
- perl4ever 7y ago
  "It is not possible to prove that a set is both smaller than the reals and larger than the rationals, because such a set would disprove the continuum hypothesis." Sure, even without much of a mathematical background, people generally take it for granted. Which is why it's disappointing that a suggestion of overturning it isn't fulfilled.
  - thaumasiotes 7y ago
    Suggestion of overturning it? It's a proof. That article would be headlined "disproving the independence of the continuum hypothesis" or some such; it would be huge news, not somebody's fun blog post.
    - perl4ever 7y ago
      So, you know better than to be fooled by this clickbait. Noted. All I'm saying is that it was very effective on me, unlike a lot of things.
- lonelappde 7y ago
  Aside, it's a shape that hypothesis in math (axiom) is the opposite of hypothesis in science (claim to test)
  - 7y ago
naniwaduni 7y ago
Given that continuum hypothesis is independent of ZF, it would be very interesting to find a set of interest to a general audience whose cardinality lies between the rationals and the reals.