3 ms·

Any source? Curious as to what is meant by "used directly"

joe__f 4y ago
Probably just that, all physically measurable quantities are rational, and the rational numbers have measure zero in the reals
- ChrisLomont 4y ago
  >all physically measurable quantities are rational That's not even clear, unless you also assume you can measure exactly a unit length, which is not physically possible. Assuming you can measure something as a perfect rational value implies infinite precision, which is not possible. All physically measurable quantities have uncertainty is what I think you mean, but that doesn't say anything about possible cardinalities.
  - joe__f 4y ago
    I think you're saying that all physically measurable quantities are rational numbers with terminating decimal expansions. The set of rational numbers with terminating decimal expansions is a subset of the set of rational numbers, so my statement that all physically measurable quantities are rational is true. I also agree that there is some uncertainty in all measurements, I don't think this is at odds with what I said.
    - ChrisLomont 4y ago
      > I think you're saying that all physically measurable quantities are rational numbers with terminating decimal expansions. No, this most certainly isn't true. For example, take a meter defined (as SI does) as a fraction of the speed of light. Now, for every length in the universe to be a rational fraction of this length is most certainly not true, because lengths, under relativity, form a continuum. Any speed is possible, so any length (via contraction) is possible. There'd be no physical reason all physical processes would be constrained to a subset of rational grid points - things can move freely. Heck, if all physical quantities are rational, are you claiming velocities only occur as rational numbers? That time is only rational numbers? It's far more likely that they'd be irrational, and our man made units are the weird things. Suppose two diagonal corners of a square with rational sides are somehow physically exact.... What is the physical diagonal length? Oh yeah, irrational. So no, physics isn't formed from rational numbers.
      - joe__f 4y ago
        I'm making a statement about all possible experimentally measurable quantities, which as you pointed out have a finite precision and so can be represented as having terminating decimal expansions. This as far as I'm aware this is not a controversial statement. I'm not making any kind of statement about whether space-time is continuous or not. You are modelling spacetime as continuous and seem to be saying this is the only possible option. That's not true; there are a number of different models of how space-time behave at the smallest length scales. Try reading about loop quantum gravity for example; in that model spacetime is taken to be a discrete lattice at the smallest scales. In your model with continuous space-time then it's true that a square of side length 1 would have irrational diagonal of length root 2. In reality we don't currently have experimental evidence to say either way, and it could be that your square of side length 1 has a rational diagonal which approximates root 2.
bobbylarrybobby 4y ago
I'm assuming they mean that for a number to be used, it must be computable.
chowells 4y ago
Source: |R| > |N| = |any string of symbols in any notation that has existed in the past, exists now, or will exist in the future|

Probably just that, all physically measurable quantities are rational, and the rational numbers have measure zero in the reals

ChrisLomont 4y ago
>all physically measurable quantities are rational That's not even clear, unless you also assume you can measure exactly a unit length, which is not physically possible. Assuming you can measure something as a perfect rational value implies infinite precision, which is not possible. All physically measurable quantities have uncertainty is what I think you mean, but that doesn't say anything about possible cardinalities.
- joe__f 4y ago
  I think you're saying that all physically measurable quantities are rational numbers with terminating decimal expansions. The set of rational numbers with terminating decimal expansions is a subset of the set of rational numbers, so my statement that all physically measurable quantities are rational is true. I also agree that there is some uncertainty in all measurements, I don't think this is at odds with what I said.
  - ChrisLomont 4y ago
    > I think you're saying that all physically measurable quantities are rational numbers with terminating decimal expansions. No, this most certainly isn't true. For example, take a meter defined (as SI does) as a fraction of the speed of light. Now, for every length in the universe to be a rational fraction of this length is most certainly not true, because lengths, under relativity, form a continuum. Any speed is possible, so any length (via contraction) is possible. There'd be no physical reason all physical processes would be constrained to a subset of rational grid points - things can move freely. Heck, if all physical quantities are rational, are you claiming velocities only occur as rational numbers? That time is only rational numbers? It's far more likely that they'd be irrational, and our man made units are the weird things. Suppose two diagonal corners of a square with rational sides are somehow physically exact.... What is the physical diagonal length? Oh yeah, irrational. So no, physics isn't formed from rational numbers.
    - joe__f 4y ago
      I'm making a statement about all possible experimentally measurable quantities, which as you pointed out have a finite precision and so can be represented as having terminating decimal expansions. This as far as I'm aware this is not a controversial statement. I'm not making any kind of statement about whether space-time is continuous or not. You are modelling spacetime as continuous and seem to be saying this is the only possible option. That's not true; there are a number of different models of how space-time behave at the smallest length scales. Try reading about loop quantum gravity for example; in that model spacetime is taken to be a discrete lattice at the smallest scales. In your model with continuous space-time then it's true that a square of side length 1 would have irrational diagonal of length root 2. In reality we don't currently have experimental evidence to say either way, and it could be that your square of side length 1 has a rational diagonal which approximates root 2.

bobbylarrybobby 4y ago

I'm assuming they mean that for a number to be used, it must be computable.

chowells 4y ago

Source: |R| > |N| = |any string of symbols in any notation that has existed in the past, exists now, or will exist in the future|