Area of Square/Proof 1

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 $\exists p, q \in \N: L = \dfrac p q$. 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:
 * $L^2 \cdot q^2 = \left({L \cdot q}\right)^2 = c^2 = S' = q^2\cdot S$

Finally:
 * $L^2 = S$

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

Then from Rationals Dense in Reals we know that within an arbitrarily small distance $\epsilon$ from $L$, we can find a rational number less than $L$ and a rational number greater than $L$.

In formal terms, we have:
 * $\forall \epsilon > 0: \exists A, B \in \Q_+: A < L < B: \left|{A - L}\right| < \epsilon, \left|{B - L}\right| < \epsilon$

Thus:
 * $\displaystyle \lim_{\epsilon \to 0^+} A = L$
 * $\displaystyle \lim_{\epsilon \to 0^+} B = L$

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:
 * $\operatorname {area} \Box B \ge \operatorname {area} \Box L \ge \operatorname {area}\Box A$

By the result for rational numbers:
 * $\operatorname {area}\Box B = B^2$
 * $\operatorname {area}\Box A = A^2$

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

Thus:
 * $\displaystyle \lim_{B \to L} \operatorname {area} \Box B = \lim_{B \to L} B^2 = L^2$
 * $\displaystyle \lim_{A \to L} \operatorname {area} \Box A = \lim_{A \to L} A^2 = L^2$

Finally:
 * $L^2 \ge \operatorname {area}\Box L \ge L^2$

so:
 * $\operatorname {area}\Box L = L^2$