Area of Square

Theorem
A square has an area of $$L^2$$ where $$L$$ is the length of a side of the square.

Integer Side Length
In the case where $$L=1$$, the statement follows from the definition of area.

If $$L \in \N, L > 1$$, then we can divide the square into smaller squares, each of side length one.

Since there will be $$L$$ squares of side length one on each side, it follows that there will be $$L\cdot L = L^2$$ squares of side length one.

Thus, the area of the square of side length $$L$$ is $$L^2 \cdot 1 = L^2$$.

Rational Side Length
If $$L$$ is a rational number, then $$\exists p,q \in \N: L=\frac{p}{q}$$. Call the area of this square $$S$$.

We can create a square of side length $$c = L \cdot q$$, and we call the area of this square $$S'$$. We then divide this square into smaller squares of side length $$L$$.

Since there will be $$q$$ squares of side length $$L$$ on each side of the larger square, it follows that there will be $$q^2$$ squares of side length $$L$$.

Thus, $$S' = q^2 \cdot S$$.

From the integer side length case, $$S' = c^2$$.

So $$L^2 \cdot q^2 = (L\cdot q)^2 = c^2 = S' = q^2\cdot S$$.

Finally, $$L^2 = S$$.

Irrational Side Length
Let $$L$$ be an irrational number.

Then $$\forall \epsilon > 0: \exists A \in \Q^+ : A < L \wedge |A-L| < \epsilon$$ and $$\forall \epsilon > 0: \exists B \in \Q^+ : B > L \wedge |B-L| < \epsilon$$.

Thus, $$\lim_{\epsilon \to 0^+} A = L$$ and $$\lim_{\epsilon \to 0^+} B = L$$.

Since a square of side length $$B$$ can contain a square of side length $$L$$, which can in turn contain a square of side length $$A$$, then $$\mbox{area}\Box B \geq \mbox{area}\Box L \geq \mbox{area}\Box A$$.

By the result for rational numbers, $$\mbox{area}\Box B = B^2$$ and $$\mbox{area}\Box A = A^2$$.

We also note that $$\lim_{B \to L} B^2= L^2 = \lim_{A \to L} A^2$$.

Thus $$\lim_{B \to L} \mbox{area}\Box B = \lim_{B \to L} B^2 = L^2$$ and $$\lim_{A \to L} \mbox{area}\Box A = \lim_{A \to L} A^2 = L^2$$.

Finally, $$L^2 \geq \mbox{area}\Box L \geq L^2$$, so $$\mbox{area}\Box L = L^2$$.

Note
Technically, this proof is circular. The use of the definite integral to represent area is based on the fact that the area of a rectangle is the product of the rectangle's width and height. That fact is in turn derived from this one.

Proof
Let a square have an arbitrary side length $$a\in\R$$.

This square is equivalent to the area under the graph of $$f(x)=a$$ from $$0$$ to $$a$$.

Thus from the geometric interpretation of the definite integral, the area of the square will be the integral $$A=\int_0^a a \, dl$$.

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