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A kid asks you, “why is area given by multiplication?” For example, why does a rectangle of length 3 and width 2 have an area of 6? You might answer “because we can break it down into six $1 \times 1$ squares, each of area 1”.

But then we’ve got to ask: why is the area of the unit square 1? Perhaps we’ve just defined it this way. But there doesn’t seem to be any obvious reason for doing so, if there should even be one. From the point of view of physical intuition/naivete, why should length be “measured” in the same way (i.e. with numbers) as area? The idea of “units are not the same as numbers” seems unsatisfactory to me because it appears to be dodging the question and assuming the hypothesis (that the same numbers can be used to represent dimensionally different objects).

Perhaps once we’ve defined how multiplication should correspond to area, we could define an addition more naturally. Say if we’ve defined the area of a unit square to be 1, then we might “reasonably” define the sum of the areas of two unit squares to be 2.

My questions are:

  1. Would we ever want $1 \times 1 \neq 1$?

  2. Perhaps more meaningfully: Would we ever want to think of multiplication as an operation taking two objects in a space and producing an object in a different space? A binary operation on a set $X$ is by definition a map from $X \times X$ to $X$; but, as in the case of area above, it doesn’t seem “obvious” why the codomain should be $X$. It seems that something more general like $X \times X \to Y$ is more appropriate.

  3. In this sense, is addition different from multiplication? I.e. could or should these two operations be meaningfully defined on objects belonging to different spaces? In particular, if we would like our notion of multiplication to take two objects in $X$ and output an object in a different space $Y$, then we may want to think of addition as taking two objects in $Y$ and outputting an object in $Y$ as well. E.g. in the area example above, we multiply two lengths to get an area, whereas we add two areas to get another area.

  4. How does all this reconcile with the desire to have a geometric argument for things like why area is given by multiplication? In particular, if it is just a matter of definition, shouldn't we admit that instead of reducing it to seemingly circular arguments like "a unit square has area 1"?

twosigma
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  • A simple example of a "multiplication" that takes values in a different space it started with is the inner product: if $a$ and $b$ are vectors, then their inner product $\langle a | b \rangle$ is an element of the field underlying the vector space. – Jannik Pitt May 12 '20 at 18:12
  • @JannikPitt Yes, but that's sort of my question. In some cases it seems like a kind of "dot product" (or in general a map $X \times X$ to $Y$) is more natural than what we have instead (a binary operation). – twosigma May 12 '20 at 18:14
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    "If it is just a matter of definition, shouldn't we admit that?" What makes you think that we don't? – JMoravitz May 12 '20 at 18:16
  • The obvious way to define an area to a , lets say , square with side $1cm$ , is $1cm^2$. What else should it be ? – Peter May 12 '20 at 18:16
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    Can you please distingush between $1$ unit and $1$ square unit? In this case $1u\times1u=1sq.u\ne 1u$, more, I can name $1$ square unit any name I want, e.g. 'trufamalgamma'. Or it doesn't make sense?.. – Alexey Burdin May 12 '20 at 18:17
  • @JMoravitz Because the argument usually given is that a unit square has area 1, and this seems to be accepted as "geometrically obvious". – twosigma May 12 '20 at 18:17
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    Does this answer your question? Formal Proof that area of a rectangle is $ab$ – JMoravitz May 12 '20 at 18:21
  • @Peter That's my question -- why "should" that be the case? Also, back to my question about using the same set of objects (in this case numbers) to represent intuitively/naively different objects (length and area). Now I can accept that it is a natural definition, but I'd like to know more generally when it is justified to use a binary operation versus something more general. – twosigma May 12 '20 at 18:21
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    @twosigma Correct me if I'm wrong but you seem to implicitly say that there are situations where one has the option of modelling something as either (a) a multiplication from $X$ to itself or (b) a multiplication from $X$ to $Y$. Then your question is why (b) is choosen over (a). Do you have such a specific situation in mind? I cannot think of one. – Jannik Pitt May 12 '20 at 18:21
  • A unit square having area $1$ can be considered as an axiom or as part of the definition of area. – JMoravitz May 12 '20 at 18:22
  • @twosigma Do you have any other defitnition that makes at least a little sense ? Since $1\cdot 1=1$ can barely be denied, we have not many choices. – Peter May 12 '20 at 18:24
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    It is exactly a matter of definition. M-W defines area as "the number of unit squares equal in measure to the surface". Do you have a different definition in mind? – Vasili May 12 '20 at 18:25
  • @JMoravitz Ok, I'm aware of it being an axiom. I guess I just thought that this wasn't the "naive" or apparently "geometrically obvious" way to reason about it. In particular, if I'm not mistaken, the Greeks didn't really think of areas as numbers. So I suppose we can (and do) define our algebra and concept of area to align with all that, but the question then comes to: what axioms do we ultimately boil down to? Do we need an axiom of area for every "dimensionally different" object? E.g. volumes of boxes in Euclidean space are usually defined as the product of intervals. – twosigma May 12 '20 at 18:26
  • In fact most definitions are somewhat "arbitary" , but in this case, there is not really an alternative. – Peter May 12 '20 at 18:28
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    A reminder that the problems we want to solve come first. The definitions come after and are written in such a way as to be useful for solving the problems we want to solve. – JMoravitz May 12 '20 at 18:29
  • @JMoravitz Very good point ! – Peter May 12 '20 at 18:30
  • @JannikPitt As I explained in my question, if we would like to think of area as something fundamentally different than length, then naively we might prefer option (b). – twosigma May 12 '20 at 18:35
  • @AlexeyBurdin Yes, but that's my question. It doesn't matter what name you give something, but what matters is that if we feel that the objects under consideration are fundamentally different, then we might naively want to question the use of a binary operation. I don't have a problem with accepting this as an axiom/definition, but it's a question worth asking in general: the simplistic/naive interpretation of a function from $X \times X \to X$ is that which takes two similar objects and outputs a similar object. In some contexts, one may not think "length" and "area" are similar objects. – twosigma May 12 '20 at 18:39
  • @AlexeyBurdin Though, I do somewhat recognize the difference between modelling something (e.g. representing both lengths and areas by numbers) versus asserting or questioning "what things really are", which can get philosophical. I also recognize that I appear to be boiling everything down to the notion of binary operation, and I recognize that this could be a misleading/inaccurate/incomplete approach. – twosigma May 12 '20 at 18:41
  • @JMoravitz Thanks, I appreciate the link. I also like the accepted answer in this post. – twosigma May 12 '20 at 19:18
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    $(\text{Part I.})$ You are mixing up "notation", "axioms", and etc... So; Take a step back. $-$ Problem to solve: "how much area does $X$ have?" Where $X$ is a "rectangle" object distinguished by its width $w$ and height $h$. It is natural we would want to define area as a function $A(X)=A(w,h)$ of its properties. $-$ Clearly, by asking "how much" of something there is, we are referring to some "quantity". To measure a "quantity", we need to define the smallest part of it, a "unit" $u$. Then, the "quantity" is a sum of its "units" $u$. Therefore, naturally $A(X)=\sum u$, for $u\in X$. – Vepir May 12 '20 at 20:23
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    $(\text{Part II.})$ We now define (an axiom) that a $1\text { by } 1$ object $X$ is to be the "unit" $u$. That is, $A(X)=A(1,1)=1u$ (Because we decided every rectangle object is naturally broken down into unit rectangles). Now, how how many units are there in a "$w$ by $h$ rectangle object"? We sum all units by definition: $A(w,h)=\sum_w \sum_h u = \underbrace{(w+w+\dots+w)}_{h\text{ times}}u=(w\times h)u$. That is, we found out that "$w$ by $h$ rectangle" has "$w$ times $h$" units. In other words: $$w\text{ by }h \text{ (rectangle) "consists of" } (w\text{ times } h)u \text{ (units) }$$ – Vepir May 12 '20 at 20:23
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    $(\text{Part III.})$ To save ink and paper, we invent the notation for "by" to be "times" ($\times$), for "consists of" to be "$=$", and call "unit $u$" a "square unit" now notated as "$\square^2$". Now, for example we can write "a one by one centimeter rectangle consists of (has area of) one square centimeter" as: $$ 1\text{cm} \times 1\text{cm} = 1\text{cm}^{2}$$ And because we are lazy we write: $$1\times 1 = 1$$ Instead of $$A(1,1)=1u$$ (Where $u$ stands for "one unit squared", and $A(1,1)$ is "area of (one by one rectangle)".) – Vepir May 12 '20 at 20:23
  • Hello Vepir, thank you for your detailed comments, they are helpful. I admit that my framing of it was incomplete/inaccurate. – twosigma May 12 '20 at 22:08
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    I think you might be interested in the formalism of Geometric Algebra: https://en.wikipedia.org/wiki/Geometric_algebra. This makes in some sense the notion of a two-dimensional object like a rectangle being composed of two one-dimensional objects (its sides) precise. Then you have precicely an example of a multiplication which lands you in a different space: multiply two one-dimensional objects to get a two-dimensional object. – Jannik Pitt May 13 '20 at 20:49

1 Answers1

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  1. No, because $m\times n$ means $\sum_{k=1}^mn$ for $m,\,n\in\Bbb N$, which implies $1\times1=1$.

  2. What happens when I want to multiply together arbitrary numbers of values? One famous option with just two spaces involves $c$-numbers and Grassmann numbers, and a product of these is a $c$-number (Grassmann number) if an even (odd) number of the original factors are Grassmann numbers. Another, with infinitely many spaces, is encountered in dimensional analysis.

  3. Going back to dimensional analysis, we pretty much need addition to "keep us where we are" in a way multiplication needn't.

  4. I'd prefer to take my definition of multiplication on $\Bbb N$ as fundamental, then note this implies a rectangle with integer sides can be rearranged to give the obvious total area as a single width-$1$ strip. Bigger number systems aren't a problem: with non-integer sides, a rectangle motivates how we can generalize all this. Flipping edges motivates $(-a)b=a(-b)=-(ab),\,(-a)(-b)=ab$. Using smaller units naturally gives us $\times$ on $\Bbb Q$, after which we handle $\Bbb R$ by continuity.

J.G.
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  • "we pretty much need addition to "keep us where we are" in a way multiplication needn't" -- well, that's what I think is interesting. Yet they are often both defined as a binary operation on the same space (not to say that this isn't a straw man of an objection). Do you know if this relates in any way to how the distributive law (e.g. in a field) connects the two operations? Multiplication distributes over addition, but not vice versa. – twosigma May 12 '20 at 18:59
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    @twosigma Distributivity is part of it, yes. The issue is that we have "bare" versions of addition and multiplication, which can be extended in very different ways while honouring their existing axioms. For example, multiplying dimensionful quantities is regular multiplication with bells on. So for $(a+b)c$ to make sense, we need $a+b$ to have a specific dimension; to achieve $(a+b)c=ac+bc$, we need $a,,b$ to have the same dimensions. As a result, the addition of dimensionful quantities is a lot more boring. – J.G. May 12 '20 at 19:18
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    @twosigma Also, multiplication is something of a sweet spot, because $x^n$ requires dimensionless $n$, while operators from tetration onwards require everything to be dimensionless. Meanwhile, more exotic mathematical functions such as $\sin x$ need to map the dimensionless to the dimensionless. All these truths result from the various ways multiplications interact. – J.G. May 12 '20 at 19:20