What is the proof for the area of a square being a^2?

[C0nfused]

Hi everybody,
I have a question about the proof that the area of a square is a^2. I have read that we use these axioms do define area:

1) Equal polygon surfaces have the same(equal) area.
2)if we divide a polygon surface in a finite number of separate surfaces then the total area is equal to the sum of these smaller areas
3)the area of a square with side length 1 is 1

Sorry for the bad English but it's not my mother-tongue.

I have also checked in the book I have about Geometry that we use the fact that the area of a square of side length a is a^2 to prove many other formulas. However, the book omits this proof. So I would like some help in this.

I have made the following thoughts:
If the side length of a square is rational number a, then a=p/q ,p,q naturals. So if S is the area of our square, we can create a square of are S' of side length c=a*q. Then we easily prove that it can be divided into q^2 smaller squares of the same area as the initial one. So from axiom 2 we can say that (q^2)*S=S'. Also c is natural as it's equal to p so the "new" square can also be divided into p^2 squares of side length 1. So from axioms 2 and 3 we get that (p^2)*1=S'.
So we get that S*(q^2)=p^2 => S=(p/q)^2=a^2.

I think the above are correct. What do u think? I can't however think of any solution for the case that the side length is irrational. Any help would be appreciated.

Thanks

 

[HallsofIvy]

The unit of area "one square meter" is DEFINED as the area of a square with sides of length 1 meter. In general, we define the unit of area to be that of a square with length 1 of whatever length unit we are using. From that it follows that a square with sides of length n (n an integer) has area n² square units because it can be divided into n*n= n² small squares one unit on a side. For the area of a square with sides of fractional length, use a "similar polygons" argument which is what you were doing. It more tedious than deep which is why it is seldom included.

 

[Hurkyl]

Very good C0nfused!

There's another important trick to learn for dealing with areas... taking advantage of the fact that if $R \subseteq S \subseteq T$ (That is, R is contained in S, and S is contained in T), then $A(R) \leq A(S) \leq A(T)$. (A(X) means the area of X)

To prove the area formula for a square S of irrational side length, what you need to do is to look at regions T that contain S, and regions R that S contains... where you already know the areas of the regions of type T and the regions of type R.

For example, a square of side length $\pi$ has an area between that of the square of side 3 and that of the square of side 4.

 

[C0nfused]

Thanks for your answers.
So if we have a square of side length a, with a irrational then we can "create" two squares, the first with side length b (b rational and b<a)and the other with side length c (c rational and c>a). If A(R) is the area of a square of side length b and A(T) the area of a square of side length c then
A(R)<=S<=A(T) so b^2<=S<=c^2 (this is true for ANY b<a and c>a). We know that infinite rational b's and c's exist as close we want to a with b<a and c>a so we can take the limits of the functions b^2 and c^2 when b-->a and c-->a. They both are a^2, so we assume that S=a^2.
It's seems ok to me but is it? Also is this the way Euclid proved it (just curious to know)?
Thanks again

 

[Hurkyl]

Yep, that sounds exactly right.

I don't know Euclid's proof -- it might have used similarity, or it might not have even been seen as a problem (I don't remember the timeline of Euclid vs. the discovery of irrationals)

This technique, though, was known as the method of exhaustion, and I believe was already known by the time of Euclid, so it certainly is plausible that they used this.

 

[mathwonk]

actually none of these arguments however tedious proves the area is a^2, only that the only possible area is a^2, assuming the area exists.

i.e. you are not proving existence. you must also prove by your axioms that no other subdivision could give a different number.

or you must have some definition of things that HAVE area.

your axioms beg the question of existence of area.

i.e. you have given some proeprties area should have but have not proved that a function exists with these properties.

so conceivably someone could find another number that is also forced by your axioms to equal the area of a square. then a square would not have area.

the ancients amy not have considred these matters.

an analogy is to prove the are function under a graph satisfies the diff eq dA/dx = f(x), and use this to compute area under the graph, but without ever defining it.


thus one can skip all the riemann sums theory and compute all the areas and volumes you want, but then you never have a theorem that these areas and volumes are well defined for any particular class of functions.

 

[C0nfused]

Thanks for your help.Here are my thoughts about what mathwonk mentioned:

We have defined that a square of side length 1 has area 1 and that all polygons have area, right? So I think that we have some definition of things that have area. We also use the axioms and find, using a specific method, a number (a^2) that can be called area as it has all the mentioned properties. So I guess that we have at least found that area exists.

Of course we haven't proved that area is unique. Area is supposed to be a number. So, as you mentioned, it must be unique for some specific geometrical object. The first axiom states that if we have for example two equal squares, then they have the same area. So, if we suppose that we can find another number c<>(a^2) that is the area of a square of side length a, then actually we "prove" that axiom 1 is wrong, or that the system of axioms we have set is inconsistent. So the problem that mathwonk mentioned is actually to prove the consistency of the above axioms? I am not at all familiar with this kind of proofs.

Any help/ideas would be appreciated