Why the area of a square is $side^2$ I recently stumbled upon Edwin Moise's proof of the area of a square, but there is something that is bothering me in his proof - he claims that since the inequalities $(1)$ and $(5)$ are true, then $a =\sqrt{S_a}$. but why? why $(1)$ and $(5)$ imply that there is an equality between $a$ and $\sqrt{S_a}$?
I would be grateful if you could explain his proof, especially the latter part.
Thank you!


 A: Here $\alpha S$ represents the area of $S$.
Now, your question is why do $(1)$ and $(5)$ imply $a=\sqrt{\alpha S_a}$. This is a very standard philosophy in mathematics. Think about it in this way-
You have shown that $\frac p q<a \iff \frac p q<\sqrt{\alpha S_a}$. But, the choice of $\frac p q$ was also arbitrary. So, the fact that $\frac p q<\sqrt{\alpha S_a}$ whenever $\frac p q<a$ holds for all $\frac p q<a$. Just ponder upon this for a moment and you'll realise that it can only be true if $a=\sqrt{\alpha S_a}$.

Once you are done pondering, here's a mathematical proof-
If possible let $a\neq\sqrt{\alpha S_a}$.
Without loss of generality, assume $a>\sqrt{\alpha S_a}$. Now, because of the way the rationals are distributed, given any $A$ and $B$, there is a rational number between $A$ and $B$. So, let $\frac x y$ be a rational number between $a$ and $\sqrt{\alpha S_a}$. So, $\frac x y<\sqrt{\alpha S_a}$ but $a>\frac x y$ which is a contradiction to the fact that $(1)$ and $(5)$ are equivalent.
A: Here's another mathematical proof of the given fact. This is not really radically different from the last one, but it's more geometric in the sense that it makes the picture much more clear.
We will use $b=\sqrt{\alpha S_a}$ in this argument.
If possible let $a\neq b$ and $|b-a|=d$.
Now, because of the structure of the reals, for any $r\in \mathbb R$, we can find a sequence of rationals $\{q_n\}_{n=1}^\infty$ which converges to $r$. So, take a sequence $\{a_n\}_{n=1}^\infty$ which converges to $a$. Put $\epsilon=\frac d 2$. Because of convergence, for this choice of $\epsilon$, there is an $n\in \mathbb N$ such that $|a-a_n|<\epsilon=\frac d 2$. This gives us a contradiction.
Another way to see it is to construct the sequence and argue that since we have
$$\lim_{n\to \infty} |a_n-a|=0$$
that is the distance between $a$ and a rational $a_n$ can be made arbitrarily small, $a$ and $b$ must coincide.
I hope this makes the geometry clear.
