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 $f \left({x}\right) = a$ from $0$ to $a$.

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