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How real are real numbers? (2004)
dhosek
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andrewla
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This isn't fair for what quantization means in reality, but I'm just pointing out that you don't have to introduce the real numbers to get physically meaningless quantities.
thaumasiotes
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You want to say that the concept of half an inch lacks physical representation, but that isn't true. You can easily demonstrate it as the ratio between one inch and two feet, compared to the reference ratio between one inch and one foot.
dhosek is saying that in a quantized space, there are reals that cannot be demonstrated this way, and he is right, but the same thing is untrue of rationals.
("In a quantized space", by the way, just means that all measured quantities are necessarily integers. That causes all kinds of problems, but "lacking examples of arbitrary rational numbers" isn't one of them.)
tromp
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You seem to have your units confused. Half an inch is a distance, while ratios, or comparisons of ratios, are all dimensionless scalars.
andrewla
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But when we say things like "the rationals are discrete" or "the computable numbers are discrete" these are very specific claims in the domain of measure theory, a theory which yields almost nothing of value except endless paradoxes and naval-gazing nonsense. Similarly when people say "the rationals are countable" and "the computable numbers are countable" this is taking for granted the Cantor notion of measuring cardinality by bijective correspondence, once again, a theory that yields nothing of value except endless paradoxes and naval-gazing nonsense.
In the practical sense the rational numbers are quite continuous -- between any two rational numbers there are an infinite (unbounded) number of rational numbers -- there's no notion of a "leap" the way there is with the integers. And any useful number can be approximated arbitrarily closely by rationals.
And for computable numbers there's even less of a gap. With rationals you can only approximate. But you can have a computable number that is exactly equal to the square root of 2, because a computable number is the algorithm by which you form arbitrarily close approximations. The square of that computable number is itself computable and is exactly equal to 2.
What do "real" numbers buy you? That is, what do you get for the trouble of building your formalism around numbers that are not computable? Basically nothing.
And if you're worried that sticking to the rationals and the computable numbers is too much of a concession to "physical reality", rest assured -- the rationals are just as unphysical as the real numbers because they are continuous already, and physics does not give us the power to measure the difference between two sufficiently precise rational numbers just as it barfs when you throw "real" numbers at it.
mathgradthrow
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BeetleB
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It buys you the rigor of doing calculus, which buys you a lot of results that, while could be computed without calculus, would also be very difficult without it.
andrewla
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Doing calculus with computable numbers is totally possible and you get all the continuity you need. You need to drop the Lebesgue formulation of the integeral and either use a Reimann integral or the gauge integral (Henstock–Kurzweil) if you need a well-behaved integral in the face of very poorly-behaved functions, but in physical reality these don't exist and in abstract mathematics they are rarely of interest and the gauge integral is as robust as Lebesgue without all the measure theory nonsense.
Intuitionalist analysis and calculus are very well established; the only thing you can't do with them is nonsense like showing that integrating over the characteristic function of the rationals is zero (who cares) or showing that you can break a three dimensional sphere up into three pieces are reassemble them after translations and rotations into a larger sphere (obviously not true).
dmfdmf
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Approximate relative to what? All actual measurement is implicitly or explicitly approximate such as L = x meters +/- epsilon. There is no infinite precision by which to discount rational measures as "approximate" and thus "invalid" in any way.
>What do "real" numbers buy you? That is, what do you get for the trouble of building your formalism around numbers that are not computable? Basically nothing.
You "buy" all of mathematics which operates on the assumption of "infinite" precision. It is an abstraction necessary to prove theorems and relationships of math. Abstracting from precision isn't a denial that it exists, it is the assumption that I can ignore it or leave it undefined. This is the assumption that distinguishes math from physics/engineering. Mathematicians deal with abstract e or pi but in the real world pi=3.14 if you are tiling your patio and 3.14159265... or whatever is necessary to get to the moon.
andrewla
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pi and e and sqrt(2) are real numbers and not rational, to be sure. But they are computable! Computable just means that they are arbitrarily approximable. "approximate relative to what" is that whatever criteria defines the number. You can't represent the "true" value of a non-rational number in the rationals, but you can prove that the error of an approximation is (rationally) bounded above and below, and you can have another approximation with a tighter bound.
Rational numbers are already infinitely precise relative to other representations -- finite decimals are another representation that is functionally equivalent to the rationals, but even a simple rational like 1/3 does not have a finite decimal value.
You can prove all the interesting theorems with computable numbers and rational/decimal numbers. You don't need the real numbers because you can't name a real number that exists and is not computable, BY DEFINITION! No mathematical construction can define a real number that is not constructible. These numbers are useless and there's no reason to continue even in abstract mathematics to pretend that they are useful because we have the formalisms to ignore them.