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Let me quote from Terence Tao "Analysis 1":

Historically, the realization that numbers could be treated axiomatically is very recent, not much more than a hundred years old.

Then, how could the people who lived before the axiomatization of real numbers be sure that, for example, $a$ times $b$ always equals $b$ times $a$? Because, back then they did have a set of axioms from which they could prove things, and thus they also didn't have a notion of rigorous proof. Does this mean that they just observed the pattern that if they took two concrete numbers, it didn't matter if they said "the first times the second" or "the second times the first"; and because of this observation they assumed $a \times b = b \times a$ without a proof?

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    $\begingroup$ my guess is that his observation hinges on the fact that the concept of number changed radically in the 19th century, at some Pont -I won't even try to say when- the concept of number was shorn of the notions of quantity and magnitude. before that, a number was a number of something. you do not need purely mathematical axioms to know that 3 cows plus 2 pigs is the same as 2 pigs plus 3 cows. $\endgroup$
    – mobileink
    Commented Aug 30, 2016 at 19:41
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    $\begingroup$ You're asking how they knew [an algebraeic expression] was true about mathematicians who significantly predate algebra. Commutivity as a property was likely an implicit assumption of arithmetic derived from the grouping of physical objects (like how we teach arithmetic to children) and not even formally recognized until the advent of symbolic algebra. $\endgroup$ Commented Aug 30, 2016 at 19:51
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    $\begingroup$ ps. one more thing. The same applies to Euclid an geometry. you can add angles to angles, and you can add lengths to lenghts, but you cannot add angles to lenghts. $\endgroup$
    – mobileink
    Commented Aug 30, 2016 at 20:04
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    $\begingroup$ So we ask: how could they be sure that "area" makes sense? Compute the area of a rectangle in two different ways, $ab$ and $ba$, might we perhaps get different answers? $\endgroup$ Commented Aug 30, 2016 at 21:46
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    $\begingroup$ How can people today be sure that $ab=ba$ without proof? If it is an axiom then there can be no proof, and we are no better than our ancestors. There was no modern notion of real numbers until well into 19th century, which is about the same time as axiomatic notions were developed, so the question is moot, see hsm.stackexchange.com/questions/2740/… But even for positive integers, why would people need "proofs" to know how to use them? They didn't have a proof that water flows downhill either. $\endgroup$
    – Conifold
    Commented Aug 31, 2016 at 0:42

3 Answers 3

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Multiplication, before the invention of modern (axiomatic) algebra, was defined as the operation giving the area of a rectangle with sides of a particular length.1 Commutativity of multiplication then follows from two axioms:

  • Congruent geometrical figures have equal areas
  • Any geometrical figure reflected through a line is congruent to the original figure

and the observation that reflecting a rectangle through a line through a corner at a 45 degree angle from the two sides flips the roles of the sides of the rectangle, so the side that formerly corresponded to $a$ in the figure now corresponds to $b$ and vice-versa. Since the first rectangle corresponds to $a*b$ and the second rectangle corresponds to $b*a$, and they have the same area, $a*b = b*a$.

Note: the geometrical intuition behind this proof is still used today, in the proof in set theory that $|A\times B| = |B \times A|$. (The regular inductive proof you may have seen for integers would work for finite sets, but for infinite sets it's easier to do a direct 'geometrical' proof).

1 For example, Euclid states the theorem which today we would state as "the area of a triangle with height $h$ and base $b$ is $\frac{1}{2}bh$ as "If a parallelogram and a triangle are on same base and in the same parallels, the parallelogram is double the triangle", i.e., "the area of a triangle with height $h$ and base $b$ is half the area of a parallelogram with the same height and same base". Archimedes goes a step further and states the theorem "the area of a circle with radius $r$ is $\pi r^2$" (which he was the first to prove) as "The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference, of the circle", i.e., "the area of a circle is $\frac{1}{2}rC$". Note that $\frac{1}{2}rC = \frac{1}{2}r\pi D = \frac{1}{2}r\pi 2r = \pi r^2$", but Archimedes evidently lacks the language to express his result in that form.

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  • $\begingroup$ While I believe this answer to be correct (based on my limited knowledge of mathematical history), it would be helpful if you could cite a supporting reference. $\endgroup$
    – njuffa
    Commented Aug 31, 2016 at 16:20
  • $\begingroup$ @njuffa: I tried to find some historical documentation; I'll try to find more when I get off work $\endgroup$ Commented Aug 31, 2016 at 19:52
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    $\begingroup$ No, it was not "defined", and there were no arithmetical "axioms" until 19th century. There was a common sense notion refined by mathematicians, and various calculational techniques. Restating Euclid's theorems in modern notation significantly alters there meanings. For example, to Euclid the "product" was the rectangle built on a and b, so in his terms the artificial question of whether $ab=ba$ does not even arise, he certainly did not flip rectangles to "prove" it. $\endgroup$
    – Conifold
    Commented Sep 1, 2016 at 3:04
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    $\begingroup$ @Conifold: +1. Euclid certainly had something like our "axioms", but they were geometric (e.g. two points make a line), not arithmetic. $\endgroup$
    – mobileink
    Commented Sep 8, 2016 at 23:49
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The guesses that are ventured in this question are entirely wrong.

(1) That our reason to think multiplication is commutative is that we have an axiom that says so is wrong.

(2) That those without modern standards of logical rigor could not easily understand why multiplication is commutative is wrong.

Suppose $3\times5$ means $\overbrace{5+5+5}^\text{3 terms}.$

and $5\times3$ means $\overbrace{3+3+3+3+3}^\text{5 terms}.$

(Remember that "terms" are things that you add or subtract, and "factors" are things that you multiply.)

Here is $3\times 5{:}$

$$ \begin{array}{ccl} 1 & \bullet \bullet\bullet\bullet\bullet & \longleftarrow \text{5 things} \\ 2 & \bullet \bullet\bullet\bullet\bullet & \longleftarrow \text{5 things} \\ 3 & \bullet \bullet\bullet\bullet\bullet & \longleftarrow \text{5 things} \end{array} $$ Take one of those things from each of the three sets of five and set them down here: $$ \begin{array}{c} \bullet \\ \bullet \\ \bullet \end{array} $$ Now take another one of those things from each of the three sets of five and set them down here: $$ \begin{array}{c} \bullet \\ \bullet \\ \bullet \end{array} $$ Now take another one of those things from each of the three sets of five and set them down here: $$ \begin{array}{c} \bullet \\ \bullet \\ \bullet \end{array} $$ Now take another one of those things from each of the three sets of five and set them down here: $$ \begin{array}{c} \bullet \\ \bullet \\ \bullet \end{array} $$ Now take another one of those things from each of the three sets of five and set them down here: $$ \begin{array}{c} \bullet \\ \bullet \\ \bullet \end{array} $$ There you have $5$ sets of $3$ things, thus $$ 3+3+3+3+3 $$ or $5\times 3.$ It is readily apparent why this must work for other numbers than $5$ and $3.$

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    $\begingroup$ Being a site on history of science and math, I'd expect some references to historical works. You are just giving a plausible explanation. $\endgroup$ Commented Aug 4, 2023 at 9:10
  • $\begingroup$ @MichaelBächtold : I was giving an obvious explanation. I was pointing out that this could not have failed to be obvious to anyone who had the concept of multiplication. $\endgroup$ Commented Aug 4, 2023 at 17:17
  • $\begingroup$ What is obvious to us must not necessarily agree with what was obvious to people in the past. I believe there are quite a few examples of this. $\endgroup$ Commented Aug 6, 2023 at 5:16
  • $\begingroup$ @MichaelBächtold : Recall Euclid's proof of the infinitude of primes: if $S$ is any finite set of primes then the prime factors of $1+\prod S$ are not members of $S,$ so there are always more primes. But Euclid did not write about multiplying the members of $S,$ but rather about finding their smallest common multiple. For distinct primes, the result is the same as that of multiplying them. But if someone lacks the concept of multiplication, then simple propositions about multiplication would not be obvious to that person. However$,\,\ldots\qquad$ $\endgroup$ Commented Aug 7, 2023 at 3:44
  • $\begingroup$ However, as soon as someone did have the concept of multiplication as described in my answer above, I think this argument would be seen quickly. $\endgroup$ Commented Aug 7, 2023 at 3:45
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Form a series of conferences I heard about "arabic/islamic" mathematics (especially Al Khawarizmi), a practical motivation for completing the square (that yielded the algorithm to solve 2nd degree equation) could have been (very putative) the estimation of the quantity of tiles needed to extend a palace with a certain length of walls.

At least for integers, placing square tiles on a rectangular floor provide a natural intuition than the product commutes. You can place tiles from East to West, or South to North, and get the same floor.

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