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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