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How did we transition from meters, as a unit of distance, to $\text{m}^2$, as the unit of area, I do understand that for example if we define a square that is 1meter long and 1meter wide and use it as our basic unit for measuring areas, then the area of a rectangle (expressed as the number of my meter-squares)would simply be its length times its wide.

But that aside, what justifies $\text{m}\times \text{m}$ as a unit for measuring the area, in other words, why a distance "times" another distance yields an area?

MachineLearner
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Hilbert
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    Could you clarify the question based on the comment to the answer? What you are asking for is essentially a definition. Likewise why is length measured in metres? You could frame an area as a multiple of a number of square patches lying in some museum in Paris for example. And the patches could have such a representation. – user1952500 Apr 27 '19 at 17:11
  • I do understand that there is some reference unit area (i.e., a "standard") that we use in order to measure "bigger" and "smaller" areas, but my question is essentially, why is this standard, the square meter, defined as meter "times" meter, meaning why distance "times" distance = area ? – Hilbert Apr 27 '19 at 17:27
  • I believe if you dig deep enough there won’t be a convincing enough reason. You should just assume it to be a measure of 2d space. – user1952500 Apr 27 '19 at 17:55
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    There is a mathematical points of view according to which geometric objects don’t have units. We can say a rectangle has height $20$ and width $30$ without saying whether that’s $30$ meters, $30$ Angstroms, or $30$ parsecs. Units are something we need in order to deal with the physical world. Do you question why the area of that rectangle is $600,$ or is it really just about attaching units to that number? – David K Apr 27 '19 at 17:58
  • @DavidK I see, hopefully this question will make it clearer : Why does unit distance $\times$ unit distance = unit area ? Regardless of the unit. – Hilbert Apr 27 '19 at 18:12
  • You mean, why do we prefix the word "square" to any unit to make a unit of area equal to that of a square whose side equals the same unit, for example, "square centimeters" or "square kilometers"? Note that it doesn't always work the other way; not all units of area are the areas of squares with edges equal to some other units. But what if you don't have any units at all? You just have a square of side $s,$ where $s$ is a number. Do we have to explain why we say the area of the square is $s^2,$ or is it only a question when there are units involved? – David K Apr 27 '19 at 18:58
  • @DavidK No, actually by "unit" I meant "unitary" like the unit circle that has a radius equal to 1. – Hilbert Apr 27 '19 at 20:32
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    Because some people thought that area = distance $\times$ distance was intuitive. Assuming you don't have a problem with the concept of distance itself, then expressing the concept of "area" using distance is fairly "easy", because it is "easy" to make a "unit square". From there, it's "easy" to use that "unit square" as the basic measurement unit for areas. So I'd say it's a mixture of "simplicity", convenience, and reproducibility.

    Since you either disagree with that choice, or don't understand it, my suggestion is to try to come up with your own way of measuring areas, and compare it...

     – N.Bach Apr 27 '19 at 22:36
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    ... to the existing system. I doubt it can get any simpler than just deciding to express any area as a multiple of the unit area measurement (a unit square). And it just so happens that the maths on counting squares in a grid boils down to multiplying the number of rows and columns as MachineLearner explained. In fact, as Rakibul hints at, one way to handle arbitrary shapes is to use the Lebesgue measure, which boils down to splitting your shape in small rectangles so you can apply the logic for rectangles on them. – N.Bach Apr 27 '19 at 22:41
  • I think I understand the question better, finally. Our language for talking about this is a bit problematic, because the difference between a "unit length" and a "unit of length" is a very big difference in meaning to be represented by the little word "of". – David K Apr 28 '19 at 02:18
  • @N.Bach In addition to what's been said, I think that one primary reason for defining the area of a square that has a 1 meter-long side as $1m \times1m$ is because it keeps the conversions neat and more importantly true, if we convert from $1m\times 1m$ to $cm\times cm$ we'll find : $1m\times1m= 10cm\times 10cm= 100 cm\times cm$, which is indeed true, if we patch our $1m \times1m$ by little $1cm \times 1cm$ squares we're gonna find that we'll need $100$ of them, had it not been multiplication, conversions wouldn't hold anymore. – Hilbert Apr 28 '19 at 11:54

5 Answers5

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All the theorems are just mere consequences of axioms which have no proof and must be assumed(they are not silly). Now the area of a square of $1m×1m$ is defined to be $1m^2$. So now in any figure, you may fit all these squares to get the actual "area". So you actually multiply with numbers $>1$, which signifies you are fitting that many unit squares in your figure.

  • Is there at least any motivation behind this definition ? – Hilbert Apr 27 '19 at 17:41
  • It is the start... there can be no definition because that would require a diffrent term, which we again need to define,... this is the reason why 'set' and 'point' have no definitions as they are the beginning of that specific field of study... –  Apr 27 '19 at 17:44
  • @Hilbert +1 your question, but does this really answers the relation between multiplication and area. I added a question about division https://math.stackexchange.com/questions/3816054/is-there-a-geometric-intuition-for-division-of-area-by-length – Quade Sep 06 '20 at 10:27
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I was also thinking about the reason why we can do this. Then I thought about a world in which we have discrete lengths. The following image is constructed of such sticks. In the $x$-direction we have three sticks and in the $y$-direction we have 2 sticks. We can see that the enclosed area is consisting of $2 \times 3$ squares. Hence, if we define the area of one square as $\text{m} \times \text{m}$ the total area is equal to $6 \text{ m}^2$.

enter image description here

MachineLearner
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You can divide any surface by infinitesimally small rectangles.now in these small rectangles for infinitesimally small length there is their respective width. So, if you sum up all the areas of those rectangles then you get the total area of that surface no matter how it looks like.

Rakibul Islam Prince
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The concept of measuring area in the way we do goes back to the very early days of mathematics, when lengths were measured relative to the length of a reference segment and areas were (usually) measured relative to the area of a square built on the same segment.

Why a square and not some other geometric figure, such as a circle? Possibly the reason is that squares provide a relatively easy way to measure areas: they're highly symmetric, they can be measured by the length of just one side (unlike general rectangles), and the plane can be tessellated by congruent copies of a single square (unlike circles). Moreover, humans thousands of years ago had a tendency to lay out their settlements in patterns that were either rectangular or round, so choosing equilateral triangles probably did not seem as natural a choice for a "unit area" figure.

Perhaps if our mathematics had been invented by bees, area might have been measured by regular hexagons (in the pattern of a honeycomb) rather than in the pattern of plowed fields! But we are humans, so we use human mathematics.

If we had chosen a different way to measure area, most likely it would just have introduced a constant conversion factor into our distance-to-area formulas. After all, you can take a region such as the inside of a triangle or a parallelogram, double it in size, and then cut it into four regions all congruent to the original reason. Triple it in size, then you can cut it into nine regions congruent to the original region. If the rule for the area of a triangle or parallelogram were something other than "multiply a length times a length times a constant," you would quickly run into situations where the area of the whole was clearly not the sum of areas of its parts. Back in the real world, that sort of thinking would lead to all kinds of complications with land deals.

David K
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  • I do acknoweldge that it's more convenient to define a little square as our reference, and then use many of them to " cover " other areas and then express it in terms of (i.e., the number) of squares, I wasn't really concerned about the square as being the choice for our reference, but rather, why was the area of this little square ,that has a side of 1 unit of distance, was defined to be the unit of distance square, what I've been told till now was that it's more about convenience than anything else... – Hilbert Apr 28 '19 at 10:01
  • ..., meaning we could've defined unit area = complicated_function(unit length), but why complicated ? so we just went with multiplication. – Hilbert Apr 28 '19 at 10:01
  • @Hilbert Not really, if you use an arbitrary "complicated_function" you're very likely to break several required properties for computing areas. If you really want to know, you'd have to go take a look at measure theory. As a non-native English speaker, I can't really summarize the parts I'm thinking of without looking up proper terminology first, so hopefully someone can take over for me... ? – N.Bach Apr 28 '19 at 10:11
  • @N.Bach, I wasn't really trying to view the problem from an advanced mathematical standpoint, I believe that units of distances squared way used before the advent of measure theory, btw, where you refering to thesecriteria ? – Hilbert Apr 28 '19 at 10:29
  • @Hilbert It's precisely because measure theory is "new" that I mentioned it, since it's the "modern/accepted" way to look at these things. You butchered your link, but those looks like the axioms defining a measure, and the term I was looking for is apparently a "delta-ring". – N.Bach Apr 28 '19 at 10:42
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A justification for why areas are understood as products of lengths requires a treatment based on measure theory.

In measure theory, we work with measurable spaces $(X, \Sigma)$, where $\Sigma$ satisfies:

  1. $X\in\Sigma$
  2. $S\in\Sigma\implies{X\setminus{S}}\in\Sigma$
  3. $\forall{n\in\mathbb{N}},S_n\in\Sigma\implies\bigcup_{n\in\mathbb{N}}S_n\in\Sigma$.

These spaces can be equipped with functions $\mu:\Sigma\rightarrow[0,\infty]$, which satisfy:

  1. $\mu(\emptyset)=0$
  2. $\forall{i,j\in\mathbb{N}},E_i\cap{E_j}=\emptyset\implies\mu\left(\bigcup_{n\in\mathbb{N}}E_n\right)=\sum_{n\in\mathbb{N}}E_n$.

These functions are called measures, and these are the functions that generalize the concepts of length, area, and volume, to arbitrary measurable spaces.

Imagine that you have measure spaces $(X_0,\Sigma_0,\mu_0)$ and $(X_1,\Sigma_1,\mu_1)$, where we imagine that $X_0$ and $X_1$ are spaces that have dimension $1$. For $S_0\in\Sigma_0$ and $S_1\in\Sigma_1$, $\mu_0(S_0)$ and $\mu_1(S_1)$ can be interpreted to be the "lengths" of the sets $S_0$ and $S_1$ respectively, in the context of these spaces. A very natural idea is to consider the Cartesian product of these spaces, since Cartesian products are just universally important in mathematics. The product space of these two measure spaces is given by $(X_0\times{X_1},\sigma(\Sigma_0\times{\Sigma_1}),\mu)$, where $X_0\times{X_1}$ is the Cartesian product of $X_0$ and $X_1$, $\sigma(\Sigma_0\times{\Sigma_1})$ is the $\sigma$-algebra generated by $\Sigma_0\times{\Sigma_1}$, and $\mu$ is the measure defined by $\mu(S_0\times{S_1})=\mu_0(S_0)\mu_1(S_1)$. Since $X_0$ and $X_1$ are spaces of dimension 1, $X_0\times{X_1}$ is necessarily a space of dimension 2, and since $\mu_0$ and $\mu_1$ are lengths, it must be the case that $\mu$ is a product of lengths. Since $\mu$ is also an area, it follows that area must be a product of lengths.

Angel
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