Imagine you being an ancient person, developing a theory of rational numbers from scratch, and suppose, you've discovered all known properties of integers (you've extended, previously found by you, natural numbers for operating with a notion of debt, which is presented in your culture as well). So, you decide to define rationals as they are mostly defined in books with some notion of "splitting" or "dividing" a shape, or an object to n equal shapes/forms. You also may show that this definition of rational numbers is equivalent to them, being defined as solutions for some abstract equations of this sort: $nx = m,n \in N, m\in Z$. After discovering all properties for addition, will you actually define multiplication of rationals like this: $\frac{a}{b} * \frac{c}{d} \equiv \frac{a*c}{b*d}$? If yes, can you explain why?
There is a bit similar post Rule for multiplying rational numbers on ME, but I am not satisfied with answers because they mostly use the commutative property of this operation, but how do they know this while defining this operation?
There are people who say that "we assume that this operation is commutative", but isn't that lame? This approach looks like constructing chess rules for a first time and not something that can be found separately by a lot of people making independent discoveries.
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2The answers to the linked question only assume commutativity for multiplication of integers. They don't assume commutativity for rationals. – Jacob Jun 24 '21 at 15:47
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2+1 to your question, taking it's score from $(-1)$ back to $(0)$. Astonished that anyone would downvote your question. You clearly put a lot of thought and effort into your question, including examining a previous mathSE post. Whether your analysis is (or is not) flawed or incomplete is (in my opinion) irrelevant to the issue of whether your question deserves to be downvoted. Unfortunately, there is no quality control on downvoting/upvoting questions. – user2661923 Jun 24 '21 at 16:11
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@jacer21, I am probably missing some of them, but cannot find any answers or comments, which use only commutative property of integers without assuming that it works for rationals. Can you, please, point explicitly on them? – Maksym Khomiak Jun 24 '21 at 16:15
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Re: @user2661923's comment, definitely don't let random downvotes get to you - some people are going to downvote for no good reason. – Noah Schweber Jun 24 '21 at 16:24
2 Answers
Definitions in mathematics aren't made randomly; there's always some idea we're trying to capture or set of desired properties we're trying to achieve. So we need to start by asking, "What do we want multiplication of rationals to do?"
For example, it's easy to see that the rectangle area interpretation of multiplication, which we already have from the context of natural numbers, gives us the usual definition. The area of a ${1\over 2}$-by-${1\over 3}$ rectangle, for example, has to be $1\over 6$ since we can tile a $1$-by-$1$ rectangle with six ${1\over 2}$-by-${1\over 3}$ rectangles, and (via natural number multiplication) we've already committed to the former having area $1$.
Alternatively, maybe we go more abstractly and say that we want multiplication of rational numbers to have the same "basic properties" as multiplication of natural numbers - namely, commutativity, associativity, and distributivity over addition (of rational numbers). In this case again we can show that the usual multiplication is our only option.
Are there other intuitions or desiderata which could lead us to defining different binary operations? Sure! But this doesn't change the fact that the usual notion of multiplication of rational numbers is extremely well-motivated.
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Given the integers and addition and multiplication of integers, we can define the rational numbers as the set of "equivalence classes" of the set of ordered pairs, (m, n), n not equal to 0, with (a, b) equivalent to (m, n) if and only if an= bm.
Then we can define the product [(a, b)]*[(m, n)] to be [(am, bn)] and the sum [(a, b)]+ [(m, n)] to be [(an+ bm, bn)].
([(a, b)] is the equivalence class containing the pair (a, b)]
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If I am not mistaken, this is exactly the way of formally defining rational numbers as people do this in Real Analysis course. But, again, your answer doesn't give the reason of defining this operation in that way. See, my friend, this way of doing math is a bit uncomfortable for me. You use criteria for two pairs to be equivalent, but again, your answer doesn't have explanation of doing so. Noah's answer is something, I find explanatory, but my problem is exatcly in finding reason's of defining this operation in this way – Maksym Khomiak Jun 24 '21 at 16:40