# 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.

Thus we have that the area is a function of the length of the side:

- $\forall L \in \R_{\ge 0}: \map \Area L = L^2$

where it is noted that the domain of $L$ is the set of non-negative real numbers.

## Proof

### 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 $L^2 \cdot 1 = L^2$.

$\Box$

### Rational Side Length

Let $A$ be the area of the square $S$ with side length $L$.

If $L$ is a rational number, then:

- $\exists p, q \in \N: L = \dfrac p q$

Create a square of side length $p$.

From the integer side length case, its area equals $p^2$.

Divide the sides into $q$ equal parts.

Thus the square of side length $p$ is divided into $q^2$ small squares.

As they all have side length $\dfrac p q$, each of them equals $S$.

It follows arithmetically:

- $A = \dfrac {p^2} {q^2} = \paren {\dfrac p q}^2 = L^2$

$\Box$

### Irrational Side Length

Let $L$ be an irrational number.

Then from Rationals are Everywhere Dense in Topological Space of 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:

- $\operatorname {area}\Box L = L^2$

$\blacksquare$