I recently realized the area formula of all polygons, and most basic figures can be proven from the areas of square and rectangle. For example if we know the area of rectangle, we can the area formula of parallelogram, then triangle and so on. That brings us to the question how to prove that the area of the square with side $a$ is $a^2$.

I just came upon ProofWiki article (http://www.proofwiki.org/wiki/Area_of_Parallelogram/Rectangle), which showed me to get the area of rectangle from that of the square, which completes the rest of the link. But the square?

I see that for such a basic figure we need some axioms to get our foot down. One that I think is absolutely essential that that of a square of $1$ units has an area of $1$ unit square, from which we can generalize to bigger squares. Another I think which we used in the rectangle proof that, if a figure $A$ is divided into different pieces then the area of $A$ is the sum of the area of the different pieces.

So my first question: what set of axioms are needed that best and unambiguously describe the area of figures?

Anyway, the one informal proof we are show in primary schools, that divided for example a $5 \cdot5 $ square into $25$ pieces of $1$ cm$^2$ (intuitively very plausible) fails to convince me. Somehow, we may generalize it to rational numbers, but what if the side is $\pi$? We cannot keep dividing it infinitely many times.

So the second question: How do we using the axioms find the area of the square? And is there any gap, when we move forward, for example with the rectangle? Thanks.

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    $\begingroup$ Have you looked at proofwiki.org/wiki/Area_of_Square and proofwiki.org/wiki/Area_of_Rectangle ? $\endgroup$ – David H Jan 27 '14 at 13:55
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    $\begingroup$ As you point out, we need to deal with irrational sides. Here is one way. Let $A(s)$ be the area of a square of side $s$. Then $A$ is monotone (if $s\le t$ then $A(s)\le A(t)$). $\endgroup$ – André Nicolas Jan 27 '14 at 16:54
  • $\begingroup$ There is a similar question (but not a duplicate!) here. $\endgroup$ – TonyK Feb 18 '14 at 13:16

The question is where do you start with the definition of "area". As far as I know, there are two options:

  1. Develop a general notion of a measure in the plane, say, in the spirit of the Lebesgue integral. Then the formula for the measure of a rectangle comes as a part of the definition of the measure. The hard part of the theory is to construct a well-defined set-function on a certain algebra of subset of the plane (or $n$-space, there is no fundamental difference between the two constructions).

  2. You can also start with axioms of Euclidean geometry (which do not include anything about areas!) and then develop a concept of area for general (not only convex) polygonal regions in the plane. If your objective is to present something that high school students or typical undergraduate students (in a Euclidean geometry class) can comprehend, then this is a way to go. Take a look at the book

E. Moise, "Elementary geometry from advanced standpoint", especially section 13.5, where he shows that area of any square (with side $a$) is $a^2$, starting merely with the normalization that area of the unit square is 1. Along the way, Moise develops the notion of area of planar polygons from scratch. The most difficult part of his proof (that area exists) is to show that area of a polygon does not depend on triangulation.

To conclude: all you need is the set of (Hilbert's) axioms of Euclidean geometry (since, as Hilbert observed, Euclid's set of axioms is incomplete). You also need a notion of real numbers (which Euclid did not have either), namely, ordered field axioms and the completeness axiom.

  • $\begingroup$ What is the basic idea he uses to show the area of the square is $a^2$? $\endgroup$ – Sawarnik Feb 15 '14 at 16:04
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    $\begingroup$ @Sawarnik: he first proves it for rational side-lengths and then uses a form of exhaustion argument to do it in general (he attributes the argument to one of his students). I suggest you check your library, they might have the book or use the inter-library loan. $\endgroup$ – Moishe Kohan Feb 15 '14 at 16:13

One very practical and natural way of approaching area is to give the following definition:

The area of a plane figure is the number of unit shapes required to cover it up completely with no overlapping or going over the edges.

We usually take the unit shape to be a square of side length $1$. Note that we have to assume that area is well-defined, in other words that we couldn't cover a figure exactly with $9$ squares, and then cover the same figure exactly with $12$ squares in a different arrangement. Intuitively this seems absurd, and we have to accept it as an axiom here. From the definition we can immediately see that the area of two combined but disjoint figures is the sum of the areas of the individual figures.

We can already prove the formula for the area of a square whose sides are of integer length by covering the square in the obvious way, but what about for fractional lengths? Actually, for fractional lengths, the area might well be undefined. Consider a square of side length $1/2$. No matter how hard you try, you won't be able to cover that up exactly with unit squares, so according to the above definition, it doesn't really have an area.

For simplicity, consider a rectangle $R$ of dimensions $1/2\times1$. What is its area? $1/2$? Why? Well, surely it's one-half... after all, $R$ is simply a unit square, cut in half... but wait, what do you mean by "cut in half"? We're going in circles.

The solution is to go to a smaller unit of measurement, extending the definition:

If it takes $n$ copies of a shape to cover a unit square, the shape has area $1/n$. If a shape can be covered exactly by any shapes at all, its area is that sum of the areas of those shapes.

So again, what is the area of a square of side length $a/b$ ? Well, such a square can be broken up into $a^2$ little squares of side length $1/b$, and it's obvious that you can arrange $b^2$ such little squares to cover a unit square exactly, thus by the above definition such a square has side length $1/b^2$. Therefore the area of the original square is $a^2\cdot1/b^2$.

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    $\begingroup$ But for irrational lengths? $\endgroup$ – Sawarnik Feb 18 '14 at 18:02
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    $\begingroup$ @Sawarnik In reality it's sufficient to think of all lengths as rational anyway, and since by its nature this approach is really only suitable for more practical, intuitive forms of mathematics, I'm not sure it would be useful to extend the definition to irrational lengths. I suppose you could use some sort of exhaustion based argument (note, of course, that you first need to define what irrational multiplication means). $\endgroup$ – Jack M Feb 18 '14 at 18:16

I don't think it is about axioms. It is about definitions. Usually you define area in terms of the area of a square. This means that you should take "side squared" as the definition of the area of a square.

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    $\begingroup$ I thought of this but I think we can prove it by assuming that 1*1 square has area 1 and some other axioms. I feel axioms are needed for example the second one. And I do not think we describe area in terms of squares, is it not the space bounded by the figure or some formalization of it. Because that way defining the area of some figures like circle would become extremely difficult. $\endgroup$ – Sawarnik Jan 27 '14 at 13:40
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    $\begingroup$ I would like to see your definition of "area" then. Defining area in terms of squares is not difficult at all; it is precisely what integration achieves. $\endgroup$ – Martin Argerami Jan 27 '14 at 13:42
  • $\begingroup$ Perhaps the one given in the Wikipedia article, Area [and well it tells axioms are indeed useful]. Integration achieves it through limit of sums of rectangle or some other shape, not square? $\endgroup$ – Sawarnik Jan 27 '14 at 13:46
  • $\begingroup$ I said squares because you did, but yes, the natural approach is to take the area of a rectangle as a definition. That's precisely what the article in Wikipedia does. And I bet that "It can be proved that such an area function actually exists" is proved using integration, or at least a limiting process. $\endgroup$ – Martin Argerami Jan 27 '14 at 13:53
  • $\begingroup$ Searching through, I found this: math.stackexchange.com/questions/350383/… Now that I think shows axioms are important. And I can take your proof, ie definition, and are link is complete, but we still need the second axiom. That said, I have the only problem with it, apart from my brain suggestion we can do things by assuming less. $\endgroup$ – Sawarnik Jan 27 '14 at 13:54

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