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I have studied the definition of real numbers as V.A.Zorich explains it in his first book Mathematical Analysis I. Basically, he says that any set of objects that respects a certain list of properties (the axioms of reals) is a "materialization" (a model) of reals and consequently, for those objects, all the theorems of analysis that are proved starting from the axioms of reals are valid. So far so good. I then ask myself: why when we face a physical problem, for example when we talk about the mass of bodies, do we use real numbers? To do this I have to imagine this:

  1. I imagine that a hypothetical label is attached to every physical body, which describes the property we call "mass";
  2. At this point, to understand if I can identify these labels as real numbers, I must define what it means to add and multiply two labels, and then verify that the axioms relating to these two operations are satisfied in the set of these labels;
  3. Regarding the sum, I define it as that label that would be attached to the body obtained as the monolithic union of two bodies;
  4. Regarding the product, what does it mean to multiply two masses?

I've never thought deeply about the point 4, but I would say that if an answer to this question cannot be found, the use of real numbers to identify the masses of bodies is logically unjustified, or am I wrong?

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    $\begingroup$ It’s worth reading Terence Tao’s description of how mathematical modeling works. You just boldly introduce mathematical quantities (such as real numbers) that you think of as being associated with various quantities, then you assume that certain equations or mathematical relationships are satisfied by those quantities, then you work out the consequences of those equations and see what predictions you obtain about the physical world. You just cross your fingers and hope for the best. Experiment is needed to see if the model makes good predictions. math.stackexchange.com/a/4785107/40119 $\endgroup$
    – littleO
    Commented Jul 3 at 19:03
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    $\begingroup$ multiplying two masses may not have a physical meaning but I can surely use multiplication of reals to find a mass of 5 electrons, for example. A physical property has a unit of measurement besides a number itself representing that property. $\endgroup$
    – Vasili
    Commented Jul 3 at 19:08
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    $\begingroup$ Short answer is possibly frustratingly inelegant: Because they seem to work. When the Greeks were measuring lengths, they would initially have thought rational numbers were sufficient, until it was shown that (for example) $\sqrt2$ is irrational. Anyone pre-foundational crisis (twiddle turn of the $20$th century) would not really have thought about it too hard. Even today it is not really in the forefront of most people's thoughts. I suspect that strictly speaking, the full force of real numbers is not needed—just computable numbers. $\endgroup$
    – Brian Tung
    Commented Jul 3 at 19:09
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    $\begingroup$ Multiplying two masses is done, for example, when computing the attractive force in Newtonian gravity. I don't know if there is an interpretation that can be disentangled from that formula and made to stand on its own two feet, as it were. $\endgroup$
    – Brian Tung
    Commented Jul 3 at 19:10
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    $\begingroup$ Some people who study quantum phenomena take the quantum serious and therefore do not use real numbers, since there is no smallest real number >0. $\endgroup$
    – Steen82
    Commented Jul 3 at 19:23

1 Answer 1

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  1. First, it's not really necessary to think of multiplication as fundamental. When we construct the real numbers using Dedekind cuts, for example, the aspect of the real numbers we take to be fundamental is the ordering $a \le b$. And this has a natural physical interpretation for masses, in terms of scales and so forth.

  2. Second, we do not assign real numbers to masses. There is no such thing as a mass of $2$! Mass is a unital quantity: we only ever measure masses by comparing them to other masses (and length, time, etc. all work the same way). So the real numbers appear as ratios of masses, not as masses. And there's a completely straightforward interpretation of what it means to multiply ratios: I have three masses $M_1, M_2, M_3$, and if $M_2$ is $r$ times as heavy as $M_1$ (as I've established using a fulcrum or whatever else), and $M_3$ is $s$ times as heavy as $M_2$, then $M_3$ is $rs$ times as heavy as $M_1$. There is no need to multiply two unital quantities here.

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  • $\begingroup$ Why do you say that the product is not necessary in the definition of real numbers? On the Zorich book I read that only when you precisely define the meaning of "+" and "x" in the set of objects you are working with, then you can call that objects real numbers. In this case I still can't understand the meaning of multiply two "ratio of masses" in physical terms. $\endgroup$
    – Nameless
    Commented Jul 3 at 19:47
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    $\begingroup$ @Nameless: that is one approach and it isn't the standard approach. In the standard approach to the real numbers we don't define them axiomatically at all and instead we just construct them from $\mathbb{Q}$ using either Cauchy sequences or Dedekind cuts. And I just explained what it means to multiply two ratios, it is a completely straightforward operation that occurs when you compare three masses. Would it help to use specific numbers? Do you agree that if $M_2$ is $2$ times as heavy as $M_1$, and $M_3$ is $3$ times as heavy as $M_2$, then $M_3$ is $6$ times as heavy as $M_1$? $\endgroup$
    – Qiaochu Yuan
    Commented Jul 3 at 19:51
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    $\begingroup$ @Nameless It might also be useful to remember that 1 kg used to literally mean "the mass of the International Prototype of the Kilogram" (before SI redefined it in terms of Planck's constant). So a mass of 2.5 kg literally would mean "having a ratio of 2.5 relative to the prototype mass". $\endgroup$
    – Tob Ernack
    Commented Jul 3 at 19:58
  • $\begingroup$ Ok, I understood. Sorry if it took me a while to understand what you meant, thank you very much. $\endgroup$
    – Nameless
    Commented Jul 3 at 20:09

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