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How area and volume emerge from lower dimensions

This might sound as a silly question but I think there is some hidden complexity behind it. Esentially, my question is how is it possible that:

  1. multiplication of something with essentially no area grants you an object with an area.

  2. multiplication of something with essentially no volume grants you an object with a volume.

The thing is, even if I acknowledge that line segment could have some nonzero but really small area (for example a wooden toothpick), why its according-times multiplication would give me an area of a square? If I put these toothpicks together, they still won't fill the appropriate square (see illustration):

illustration

I know, that if I discretize it and assign some fundamental "area value" it would make sense, for example having 4 cabbages in a row and 4 rows in a total gives me 16 cabbages... BUT how to make a sense of it for the case with toothpicks etc.?

Thank you very much!

Athaeneus
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  • Well, the question is flawed because it's not that the object has essentially no area, it's that it has exactly no area. The analogy of toothpicks does not hold up to the mathematical idea because toothpicks have some small width to them whereas a line does not. So, if you stack a million toothpicks side by side you may fill up a square of some unspecified width; if you stack a million lines 'side by side' (this is not even a well defined notion) the resulting width of this configuration is still zero. Same reasoning applies to volume – Alborz Jan 06 '23 at 14:10
  • Well, but then the thing is... Why 4-times a line gives you an area of a square? How does the area emerge from something without it? Maybe I am overthinking it but I search for some fundamental understanding of it. – Athaeneus Jan 06 '23 at 14:14
  • Well.. it doesn't. If your line is length $2$, '$4$ times a line' here would give you $8$. The area of the square constructed by taking lines of length $2$ is $2^2=4$. I think you have misunderstood something here. You might be thinking of the perimeter of a square. – Alborz Jan 06 '23 at 14:17
  • 4 times a line segment gives you four lines that still have 0 area. And four times the length of a 4m line segment gives you a length measurement of 16m, which still doesn't represent any area. Now, (4m)×(4m) gives an area 16m², but there I multiplied by the length 4m, not the counting number 4. – Mark S. Jan 06 '23 at 14:19
  • @MarkS. that example may be confusing the matter further as you chose the only case where '$4$ times the line' coincides with the numerical value of the area of the square with side length of that line – Alborz Jan 06 '23 at 14:21
  • Alborz: Oh I see my mistake here, I worked with side of a square being 4. I will adjust the question properly, thanks. Now, the main thing is how even any multiplication of something leads to an area...........

    Mark S: Alright, but 4m on itself still has no area, right? So I have to multiply it 4 times and then multiply by some elementary unit (m) in which I would like to express it? Do I get it correctly or am I lost once again?

     – Athaeneus Jan 06 '23 at 14:24
  • "4 times and then...(m)" It would be a little weird to separate the 4m into the 4 and the m, but I suppose you can. "Do I get it correctly?" If you find that a helpful way to think about the situation, then yes. I make no claims that it's the only way, and there's probably a lot of philosophy about this I don't know. – Mark S. Jan 06 '23 at 14:29
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    You seem to be working from a concept of multiplication in which you multiply by adding together some number of copies of an object: $4 \times 4 = 4 + 4 + 4 + 4.$ That is not a good model for multiplication in general: how does it answer $\frac13 \times \frac13$ or $\sqrt2 \times \sqrt2$? And it is not the way we use multiplication for area. We want to measure area and we have chosen the rectangle as the prototype for an object with area, and we define its area to be the product of the lengths of two adjacent sides. – David K Jan 06 '23 at 14:32
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    @DavidK +1 Nice comment! I don't think this was necessarily your intent, but your comment may make it sound like there is no reason to define a 2m×3m rectangle to have area 6m² other than convention. I'd prefer to say we define some unit square to have area 1, and then build up other shapes out of area addition. The $\frac13\times\frac23$ rectangle has the area of $\frac29$ unit squares primarily because when you put 9 of them together you can form a rectangle that is perfectly covered by 2 unit squares. For things like $\sqrt 2\times\sqrt 2$, we can approximately cover it by 2 squares' worth – Mark S. Jan 06 '23 at 14:41
  • @DavidK This model explanation makes sense to me... BUT I still would have one counter-argument: I use only geometrical model, imagining toothpicks. I visualize this exponatiation as follows: One member of the multiplication gives me a length of toothpick measured in some general length units and the other gives me how many toothpicks I have. The $\frac{1}{3} * \frac{1}{3}$ means I have toothpick of the length $1/3$ and I have only $1/3$ of these toothpicks... I don't see a problem of model interpretation there. Could you provide some material for other multiplication models, please? – Athaeneus Jan 06 '23 at 14:42
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    @Athaeneus: How should area behave? If nothing else, a figure formed from finitely-many copies of another figure should have an area equal to the corresponding multiple of the original area. Thus —at least for integer $a$ and $b$— the area of an $a$-by-$b$ rectangle should be $a$-times the area of a $1$-by-$b$, which in turn should be $b$-times the area of a $1$-by-$1$. Calling the last of these $K$, then an $a$-by-$b$ rectangle should have area $abK$. (We easily extend this to rational $a$ and $b$; "continuity" gets us to reals.) By convention, we declare $K$ to be $1$. – Blue Jan 06 '23 at 14:58
  • @Athaeneus: See also this answer of mine. – Blue Jan 06 '23 at 14:58
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    Finite contiguous $n$-dimensional regions have $n$-dimensional "measures" (lengths for $n=1$, areas for $n=2$, volumes for $n=3$) proportional to $a^n$. Any attempt to get a measure of larger $n$ that is applicable gives $0$. A cube of side length $a$ has edges of total length $12a$ but area $0$, faces of total area $6a^2$ these edges enclose but volume $0$, volume $a^3$ these faces enclose etc. – J.G. Jan 06 '23 at 15:06
  • @Blue I think I see where my problem lies. So the answer, if I understand it correctly, esentially is that we define, by convention, some fundamental unit of area $K$ to be equal to 1, give some elementary area to a line segment in this case $b$ (that $1$-by-$b$ area) and then we use it as common comparison? So the area in a square comes from the fact that we assign some area to a line segment? Otherwise, where would the area of 1-by-b come from? Is that right? – Athaeneus Jan 06 '23 at 15:13
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    @Athaeneus: We do not assign an area to a line segment; we assign only length to a segment. We assign area to a rectangle determined by two line segments. Speaking a bit philosophically, area is something of an "emergent" property, possessed by neither segment individually but coming into being by a combination of those segments. (It's not uncommon to say that a segment has "area zero", but that's really just an informal statement about a rectangle whose "other side" has length zero.) – Blue Jan 06 '23 at 17:25
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    A similar point of confusion arises when someone learns that a line is composed of points. How can the length of the line be defined? For example, see If a point has no dimension and no area how can there be space? – David K Jan 06 '23 at 18:17

5 Answers5

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This is an intuitive answer.

The area of a square of side length $a$ is $a\cdot a$ because the square area may be subdivided into a number of exactly $a\cdot a$ smaller squares that are neither overlapping nor gapped and each small square has a sidelength of 1 and an undetermined area, which we choose to call “unit of area”.

Undetermined as it is, all smaller squares have the same area because they are congruent.

The division of a segment of length a into a number of “a” segments of length $\frac{1}{a}$ is not exactly possible in the real world, however in mathematical terms is possible to be done exactly.

The fact that we call the undetermined area as “unit of area” might seem unrigurous but is not. We simply count in terms of that small area which we regard as a reference.

Same logic applies to volumes.

WindSoul
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  • Ummmmmm.... Does this mean that the setup was never 4 lengths 4 times but rather 4 units of area 4 times? Meaning that we have actually never operated with length itself but area instead, while the use of the notation for a length was only a crude simplification of the underlying problem? – Athaeneus Jan 06 '23 at 17:34
  • Not sure what you mean by “4 lengths 4 times”. A length multiplies into another length and an area multiplies into another area. You operate with length when you subdivide the area of the square by subdividing two adjacent sides of the square, but then you get the original square subdivided into a number of smaller squares. – WindSoul Jan 06 '23 at 19:52
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You can start by simply defining length on open intervals $(a,b)$ to be $b-a$. Then, in general for cubes $(a_i,b_i)^n\subseteq \mathbb R^n$ define volume to be $\prod (b_i-a_i).$ The Lebesgue theory will then capture the intuition we have for these ideas, with the added benefit that it will also treat non-intuitive situations that arise. Another approach, for example here, uses linear functionals to develop the theory, which is perhaps more direct; and it leads to the same results.

Matematleta
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If we consider infinitesimal strips of width $x$ from one side $ dA= x \cdot dx $, so to start with area did exist in its differential form. Area did not arise from nowhere.

Integrating,

$$ A = x^2 $$

and similarly for the volume.

Narasimham
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  • Well, but this just shifts the question... Now we could ask why $x * dx$ has any area at all, which is esentially the same problem as before... The question practically being: "Why any multiplication of sole length gives you an area" – Athaeneus Jan 06 '23 at 17:21
  • Integrating, $\int x,dx = \frac12x^2 + C,$ and as a definite integral, $\int_0^a x,dx = \frac12 a^2,$ so I think this answer is actually wrong. If I were to do this via integration I would integrate $\int_0^a a,dx.$ – David K Jan 06 '23 at 20:36
  • Ok, I tried to model from OP's sketch of a square window with very thin bars/toothpicks filling up a full square. – Narasimham Jan 06 '23 at 23:54
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...multiplication of something with essentially no area grants you an object with an area...multiplication of something with essentially no volume grants you an object with a volume...

Maybe you're having trouble because your intuitive understanding of "dimension" so naturally connects to common physical length/area/volume examples that it is hard to realize that there is actually a conceptual abstraction happening.

To show how completely abstract "dimensions" really are I'm going to give an example that is entirely unrelated to your length/area/volume examples... the working-world concept of a man-hour.

If you think about it, a man-hour (or rather a worker-hour) is a 2-dimensional concept that combines the two very distinct 1-dimensional concepts of "worker(s)" and "time" into a single new concept. Can 2-workers fold 100 paper airplanes in an hour? Well then 6-workers could do the same job in 20 minutes, or 1-worker alone could do the job in two hours.

It's a pretty convenient concept, but it's also nice here in that we can understand how it is clearly an abstraction. We can intuit that workers and time are completely different categories that each contribute to the concept; and we can also understand that a question like "How many 'workers' are in an 'hour'?" shows a muddied understanding of how those two categories are distinct.

Getting back to the length/area/volume examples that you were originally asking about we could maybe decide to think of them in that same kind of abstract way as "worker-hours": is it an "area" or just a "width-length", a "volume" or a "width-length-height"?

There is something strange about how width, length, and height can all be measured with the exact same kinds of basic "distance" units, and it is also a bit amazing that we can just "rotate" things and say "I'm calling this the direction of length and that the direction of width now." and have it work out. But none of that changes the fact that the abstract concept of a "width-length-height" is something distinct from the concepts of a "height" or "width-length".

And lastly, the reason why it goes $a$, $a^2$, $a^3$ is simply because we have special words for shapes when the sides are the same. Any old rectangle could be $a \cdot b$ but if we say that $b=a$ we have $a \cdot b \to a^2$ and call it a square; same thing for 3D box shapes $a \cdot b \cdot c$ if all sides are the same $a=b=c$ just gives $a \cdot b \cdot c \to a^3$ and we call it a cube.

DotCounter
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In Euclid’s geometry you get a square only by constructing it on a given straight line segment (Elements I, 46), not by multiplying a line. Numbers are not involved at all. By the time you get to Descartes’ geometry, which makes use of algebra, “squaring” a line yields not a square (area) but another line, a third proportional, with an arbitrary line playing the role of “unit”. Similarly, “multiplication” of a line by a line yields a fourth proportional, i.e. another line, not a rectangle. For Descartes as for Euclid, only magnitudes of the same can have a ratio to one another.

So it seems a square is not produced by numerically “squaring” a line, nor a rectangle by multiplying two lines, nor a cube by numerically “cubing” a line. Squares and rectangles and cubes arise only by geometric construction, traditionally with straight-edge and compass. But if we are given a numerical measure of the line(s),we can calculate the measure of the square, rectangle, or cube by taking the “square” or product or “cube” of the number(s). E.g. if line $a$ is two unit-lines in length, then the square built on that line is $2^2=4$ unit-squares in area, and the cube is $2^3=8$ unit-cubes in volume.

Edward Porcella
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