# Area of Square

## Contents

## Theorem

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

## Proof 1

### 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

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:

- $L^2 = S$

$\Box$

### 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:

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

$\blacksquare$

## Proof 2

Let $\Box ABCD$ be a square whose side $AB$ is of length $L$.

Let $\Box EFGH$ be a square whose side $EF$ is of length $1$.

From the Axioms of Area, the area of $\Box EFGH$ is $1$.

By definition, $AB : EF = L : 1$.

From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\Box ABCD$ to $\Box EFGH$ is the duplicate ratio of the ratio of $AB$ to $EF$.

Thus by definition of duplicate ratio:

- $\Box ABCD : \Box EFGH = \left({AB : EF}\right)^2$

That is:

- $\dfrac {\Box ABCD} {\Box EFGH} = \left({\dfrac L 1}\right)^2 = L^2$

That is, the area of $\Box ABCD$ has $L^2$ as many units as $\Box EFGH$.

Hence the result.

$\blacksquare$

## Proof 3

Let a square have a 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:

- $\displaystyle A = \int_0^a a \, \mathrm d l$

Thus:

\(\displaystyle A\) | \(=\) | \(\displaystyle \int_0^a a \, \mathrm d l\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left[{l \cdot a}\right]_0^a\) | $\quad$ Integral of Constant | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \cdot a - 0 \cdot a\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^2\) | $\quad$ | $\quad$ |

$\blacksquare$

### Warning

**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.

However, this demonstration neatly parallels the integration based proofs of the areas of other figures, for example Area of Circle.