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Throughout our known history of geometry were the units representing areas and volumes always in terms of squares and cubes respectively? Take ancient Egyptian formulas as an example, the fact that their formulas are very close to ours must mean that the units representing areas and volumes were also unit squares and unit cubes, correct? I mean we can safely stamp their formulas with for example $cm$ and $cm^2$ and get the same result. And does that mean we can safely assume that they didn't measure e.g. volume by counting the number of times you can fill up an object by throwing identical fistfuls of sand into it?

What I'm saying is basically this: If the formulas from ancient times are the same as ours (say for rectangle area as length $×$ width) then that must mean that their unit for measuring area must have also been in the shape of a square and equal to $1×1$ of whatever their units of length were?

Michael Munta
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    I don't know about the history, but your reasoning doesn't really hold up. Even if you measure in e.g. circles/spheres or equilateral triangles/tetrahedrons, you'll get the same formulas. The dimensionality of a quantity doesn't depend on the shapes we're measuring or counting. – Karl Oct 27 '21 at 18:25
  • Not really. For example, a pint isn't defined as the cube of some length. – J.G. Oct 27 '21 at 18:29
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    In order to have any meaningful discussion on the topic we must have agreed upon meanings of words used. How do you think the word "area" is defined in modern times? How do you think the word "area" (or at least the corresponding word) was defined in ancient times? – JMoravitz Oct 27 '21 at 18:29
  • @JMoravitz Well I'm thinking it is defined as a count of identical figures that fit inside an object we are measuring area for. Currently, I suppose, all of our area formulas count unit squares (not rectangles, circles, triangles). This means that say for the area of a rectangle the formula length $*$ width only holds if it counts unit squares. That formula will not give us the correct number of unit triangles for example, i.e. that number of unit triangles won't cover the area completely/correctly. – Michael Munta Oct 27 '21 at 19:41
  • @Karl I get what you are saying. But you can't get the number of say unit triangles by using our formulas because they are "pre-tuned" to count unit squares. You would have to derive a different formula that counts unit triangles, right? This is the point I am trying to get across. If the formulas are the same then they must count unit squares, not triangles, circles etc. Correct? – Michael Munta Oct 28 '21 at 07:12
  • @MichaelMunta Well I'm thinking it is defined as a count of identical figures that fit inside an object we are measuring area for. Currently, I suppose, all of our area formulas count unit squares (not rectangles, circles, triangles). No. Some civilizations in the past defined area as a count of identical figures covering a surface, but certainly not all did, and this is certainly not the modern definition of area. As already said elsewhere, the modern definition of area is in terms of measure theory. So, no, you are wrong when you say the formula is pre-tuned to count unit squares. – Angel Nov 05 '21 at 13:31
  • Why square units? – ryang Nov 26 '21 at 06:47
  • @ryang isn't the accepted answer basically stating how the formula $m×n$ concerns unit squares? The whole point of my question is to see if it is based on that simple fact. If you take a rectangle/triangle/circle as your standardized shape representing $1$ unit of area and tesselate different sized rectangles to try and derive a formula, forget calculus for a moment. You claim that if we pick any of these non-square shapes that we will get length times width formula? – Michael Munta Nov 26 '21 at 11:00

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Given a $4$m $\times$ $4$m square, sure, I can determine its area by counting unit squares.

However, a 16m$^2$ circle's area is not measured by counting little squares, nor does its derivation pertain to little squares.

The fact that area always has dimension length$^2$ (m$^2$) does not imply that area is generally determined by counting little squares (or even rectangles).

But you can't get the number of say unit triangles by using our formulas because they are "pre-tuned" to count unit squares. You would have to derive a different formula that counts unit triangles, right?

Area formulae are not pre-tuned to count squares (or even rectangles).

Similarly, distance has length (m) as dimension, yet (1) it is valid to measure the length of a curved road by connecting short pieces of non-straight rope; (2) I can use a watch (time) to determine that I've driven 100 miles.

enter image description here

ryang
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    What are you trying to show with the triangle image? – Michael Munta Oct 31 '21 at 12:41
  • @MichaelMunta It's just a supplemental point. The three points in the second half of the Answer are collectively debunking the notion of "(cm$^2$) area<---many little squares; (cm) length<---many short straight lines". Analogously, while $16$ is indeed a perfect square, that's not the only way to factorise it ($16=2\times8$). – ryang Oct 31 '21 at 12:50
  • How do we define the area then if it's not the number of certain standardized figures that cover it? What I'm trying to do is make sense of the abstract notion that is the area formula. – Michael Munta Oct 31 '21 at 13:08
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    @MichaelMunta Your posted Question is premised on measuring and deriving area by counting little squares (and doing likewise for volume by counting little cubes), and your misconception seemed to arise from the fact that the area has *dimension* [Length$^2$]. My Answer attempts to debunk both these notions. – ryang Oct 31 '21 at 13:24
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    @MichaelMunta You have now revised your position, and are asking an orthogonal question: now you're asking (having changed two parameters), "Is area not *defined* (instead of measured/derived) using little *standard shapes* (instead of squares)?" The rigorous definition of area (measure theory is beyond the scope of this discussion; as for derivation using standard shapes...yes perhaps, but it's not about simply tesselating them; I suggest clicking on the first link above about circle area. – ryang Oct 31 '21 at 13:29
  • I went with shape counting as a way to measure area because I highly doubt that people 3000 years ago had a different understanding, though we can't know for sure. Like you have drawn a triangle and measured its area using identical triangles. Is there not a way to derive a formula simply from that? Also when you say to check out circle's area regarding derivation what point on that page are you trying to emphasize? – Michael Munta Nov 01 '21 at 13:37
  • @MichaelMunta None of the techniques (formula derivations nor direct measurements) on that circle-area page involve merely tessellating shapes, in spite of your certainty (actually, just your hunch). My response to this claim/certainty/hunch is contained in the first two sentences of my Answer. – ryang Nov 01 '21 at 13:47
  • I get what you are trying to say, even if the way of arriving at the formula doesn't involve such a way of thinking, you will still in the end get the result in the "number of standardized shapes". If area of a circle is $9cm^2$ you can take $9$ such unit squares and cover the circle perfectly (of course one also has to make them fit the circle by giving them different shapes, but $9$ of them will be perfectly enough). – Michael Munta Nov 05 '21 at 10:15
  • And let's stay in the realm of rectangles just for simplicity. How is the area formula not pre-tuned to count $11$ unit squares? It perfectly gives you that amount. Multiplying length$$width will not give me the exact number of unit rectangles/triangles/circles. I am just trying to understand the first comment on my question. How can I get the same formulas if I'm counting triangles. What then makes that formula universally exact and what does it depend on if not the shapes that cover it? – Michael Munta Nov 05 '21 at 10:31
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    How is the area formula not pre-tuned to count 1∗1 unit squares? It perfectly gives you that amount. Multiplying length∗width will not give me the exact number of unit rectangles/triangles/circles. This is merely coincidence. Squares are a special case of rectangles, and the formula is a essentially a definition for rectangles, though it may not be based on rectangles either. – Angel Nov 05 '21 at 13:33
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Throughout our known history of geometry were the units representing areas and volumes always in terms of squares and cubes respectively?

No. Take a look at the Japanese units of area. The base unit is the tsubo, which is based on the size of tatami mats, and which thus has no direct relationship to the lengths of the sides. The unit itself is not easily expressible in terms of length units being multiplied together, and this idea was exported to Japan, not an idea that the Japanese discovered on their own.

Take ancient Egyptian formulas as an example, the fact that their formulas are very close to ours must mean that the units representing areas and volumes were also unit squares and unit cubes, correct? I mean we can safely stamp their formulas with for example cm and cm2 and get the same result.

I am not sure I understand this question. But dimensionalized formulae and nondimensionalized formulae tend to look differently. And while Egyptian units for area and volumes may look similar to hours, this is by no means true for every ancient civilization.

And does that mean we can safely assume that they didn't measure e.g. volume by counting the number of times you can fill up an object by throwing identical fistfuls of sand into it?

Take a look at the Japanese units for volume. For a very long time, the base unit was the koku. 1 koku is equivalent to the volume of rice needed to feed a person for an entire year, and the economy of ancient Japan was even based on this unit of volume: in fact, the koku served not only as a unit of volume, but effectively, as a unit of currency. This is not quite "throwing identical fistfuls of sand into a sack", but it comes quite close. Anyway, the point is that this unit of volume has nothing to do with the side lengths of cubes or rectangular prisms. The relationship between the koku and the Japanese unit of lengths is even more non-existing than the relationship between that of the tsubo and the Japanese units of length.

What I'm saying is basically this: If the formulas from ancient times are the same as ours (say for rectangle area as length × width) then that must mean that their unit for measuring area must have also been in the shape of a square and equal to 1×1 of whatever their units of length were?

I just provided a counterexample of this.

Angel
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  • Take a look at the rectangle formula. If Egyptians calculated it using the length times width and we calculate it the same does it mean that the unit of measure is a unit square? You provided a counterexample with Japan where we obviously can't use the same formula. I want to know for formulas that are the same. – Michael Munta Nov 05 '21 at 13:01
  • If Egyptians calculated it using the length times width and we calculate it the same does it mean that the unit of measure is a unit square? It does not mean that. The formula $A=lw$ is consistent with $1 \text{unit of area} = \text{p unit of length}\times \text{q unit of length}$ as long as $1=pq$ is true. So the standardized shape could have been a rectangle and not a unit square. Or they may not have been a standardized shape at all. There may be a variety of reasons one may care about the product of lengths for area without focusing on a standardized shape. – Angel Nov 05 '21 at 13:21
  • You provided a counterexample with Japan where we obviously can't use the same formula. I want to know for formulas that are the same. This is inaccurate. We can use the formula for Japanese units. The point of my example is that the units are not based on the formula. That does not mean the formula cannot be used for those units. The size of a tsubo can be found in units of square meters. The point is that there is nothing intrinsic about squares that makes them the basis for the formula for area. – Angel Nov 05 '21 at 13:26
  • Well, what does that formula represent then? Why is it length times width? How to explain that? – Michael Munta Nov 05 '21 at 13:46
  • @MichaelMunta Because this $\text{length}\times\text{width}$ quantity was discovered to be a useful tracker of surface area covered by those societies in the contexts in which it was used. There is no particular inherent mathematical reason for it. In ancient civilizations, mathematical definitions were not driven by abstract theories, they were driven by practice and convenience within an application. Mathematical rigor did not even exist in those times. As for an abstract mathematical reason behind why we continue to use this formula today in modern times, this again requires measure theory – Angel Nov 05 '21 at 14:20
  • For a measure-theoretic overview of why areas turn out to naturally be algebraic products of lengths, see my answer in https://math.stackexchange.com/questions/3204669/why-is-textmeter-times-textmeter-a-legitimate-unit-for-measuring-area/4297607#4297607 – Angel Nov 05 '21 at 14:21
  • if someone told you to derive the formula for rectangle by using rectangular shapes how would you do it then? – Michael Munta Nov 13 '21 at 12:45
  • Are you sure that is the question you meant to ask? As presented, your question is essentially trivial. – Angel Nov 15 '21 at 12:55
  • Here is an example using right isosceles triangles, on a $9$ by $3$ recrangle: the length times width gives us the area in unit squares, same numbers as above would be $27$ unit squares. I am not refuting the fact that we can convert these unit squares into an area represented by isosceles triangles after it has been first calculated for unit squares. Then we have $54$ unit triangles as a result. But, $9×3$ will never give $54$, it is tuned to give a precise result in square units. So I am trying to understand the formula before it has been calculated and converted to triangles. – Michael Munta Nov 18 '21 at 09:39
  • If we wanted to get unit isosceles triangles from the start we would have to do length times width times $2$(since two such rectangles make up a unit square) and then we would have a formula tuned to count unit triangles. This is what I'm trying to say that when we look at the formula from that angle then the formula does not work for any shape other than unit squares. I'm asking to convince me otherwise. – Michael Munta Nov 18 '21 at 09:40
  • No, the formula is not tuned to give precise results in unit squares. Rather, it is the other way around. We essentially decided that computing the area by multiplication is a good idea, which is why rectangles have area units of unit squares. There is nothing particularly special about this. – Angel Nov 18 '21 at 13:49