Proving rectangle's area, how does this work? At school our teacher has done something similar but I didn't understand it. Could you explain?
Let
$$ m,k \in \mathbb{N}^+ \qquad n \in \mathbb{N}^+ \qquad a,b \in \mathbb{R}^+$$
We have a rectangle $ABCD$. Here is what he have drawn, there is the explanation below:

We say that we divide the rectangle ($ABCD$) into small squares, each of side $\frac{1}{n}$. Since $a$ and $b$ are not necessarily integers, there may be squares that 'can't fit'. So we create points $A'$ and $C'$. $A'$ is where the last complete square ends on side $BA$ and $C'$ is where the last complete square ends on side $a\:(BC)$. These two define $D'$ too (you can see on the picture). We also create $A''$ and $C''$. $A''$ is where the square that 'hangs out' ends. $C''$ is the same but on the other side. These two create $D''$.
We say that we have $m$ complete squares on one side and $k$ complete squares on the other. Therefore, in the smaller rectangle, coloured orange, the area is $m \times \frac{1}{n} \times k \times \frac{1}{n}$. This has to be less (or perhaps equal) to the original rectangle's area, which is $ab$ (We want to prove this). The original rectangle's area has to be smaller than the one's which was coloured violet on the drawing: $(m+1) \times \frac{1}{n} \times (k+1) \times \frac{1}{n}$. Thus, we can write the system of inequalities:
$$ 
\begin{cases}
m \frac{1}{n} \leq b < (m+1)\frac{1}{n} \\
k \frac{1}{n} \leq a < (k+1)\frac{1}{n}
\end{cases}
$$
Multipy
$$
m\frac{1}{n}k\frac{1}{n} \leq ab < (m+1)\frac{1}{n}(k+1)\frac{1}{n}
$$
$$
mk\frac{1}{n^2} \leq ab < (m+1)(k+1)\frac{1}{n^2}
$$
Putting this aside for a moment, we can look at the areas of the three rectangles. Obviously, the orange one ($A'B'C'D'$) is smaller or equal to the original one ($ABCD$) which has to be smaller than the larger one ($A''B''C''D''$).
So:
$$
T_{A'B'C'D'} \leq T_{ABCD} < T_{A''B''C''D''}
$$
We can rewrite the rectangles with integer sides, since we have proved already for integers:
$$
mk\frac{1}{n^2} \leq T < (m+1)(k+1)\frac{1}{n^2}
$$
To be honest, at this point, I have no idea what we are doing... Why don't we say that the inner rectangle's area is $mk$, the outer one's is $(m+1)(k+1)$? Why do we even need the $\frac{1}{n}$? Why don't we use unit sided squares?
Nevermind, I'll continue with what I've been showed:
$$
\left| T-ab \right| < (m+1)(k+1)\frac{1}{n^2}-mk\frac{1}{n^2} = \\
=m\frac{1}{n^2}+k\frac{1}{n^2}+\frac{1}{n^2} < \frac{a}{n}+\frac{b}{n}+\frac{1}{n}=\frac{a+b+1}{n}
$$
As $n$ approaches $\infty$, $\frac{a+b+1}{n}$ approaches $0$.
$|T-ab| \rightarrow 0$
$T=ab$
$q.e.d.$
Could some one please explain to me this properly so that I can understand? I simply don't get what we're doing...
 A: A square with side length $\frac1n$ has a size of $\frac1{n^2}$ of a unit square. This can easily be verified, since you can fit exactly $n^2$ of such small squares into a unit square. This is why the $mk$ squares constitute an area of
$$
mk\frac1{n^2}
$$
The same story goes for $(m+1)(k+1)$.
The general idea here is that both $T$ and $ab$, the size of the rectangle in question and the numerical figure of the side lengths multiplied, are squeezed in between two areas we know how to calculate, $T_{inner}=mk\frac1{n^2}$ and $T_{outer}=(m+1)(k+1)\frac1{n^2}$, that converge to each other as the partitioning $\frac1n$ gets finer and finer. Thus $T,ab$ are forced to converge to each other. But since the do not depend on $n$ but remain constant throughout, they must always have been equal in the first place.

BTW, I would find it natural to use $\frac1{2^n}$ rather than $\frac1n$ since then $T_{inner}$ would always be increasing with $n$ and $T_{outer}$ would be decreasing.
A: Exposition: Up to minor details concerning the path you take when constructing the notion of area; be it the archaic geometry of the greek, or the more modern theory of measure - one must establish how to assign a number, representing the area of some more elementary shapes.
One way to do this, as done here: When equipped with the notion of length, we realize the size of a square (The elemntary shape which we will use) as the square of its side's length.
This choice is not a coinsidence. We (intentionally or not) understand area as a property which is translate invariant. This, along with a few other properties of our every-day notion of "area", and a few other properties of out more mathematical notion of area) surprisingly enough - boils down to a unique function, measure (See https://en.wikipedia.org/wiki/Measure_(mathematics)).
This is the more modern way to explore area. But, dropping this more advanced technologies:

How can we still assign a number to a shape that will represent its area?

This process is similar to the way we measure lengths. We begin with an imaginary ruler (a stick, a cane) treating it as our fundemental unit-length. Suppose we wish to measure the length of another cane. We work the following procedure: We place them alongside, if the sizes match, the the other cane's length is $1$ unit aswell. Suppose it is longer, we place our ruler at the begining and mark its end on the cane. We take the ruler and place its bottom part on the mark so that its new place extends its former, this way we are guaranteed to (at some stage) extend our ruler over the cane. The number of extensions gives us a numeral which indicates the least number of unit-lengths the cane's length posses. There is still some residual length we cannot examine yet, but this length is less then unit-length.
This leads us to the following difficulty, how to measure lengths smaller than unit-length?
To adress this, We take our fundemental unit-length measurement ruler device and we mark it with 9 stripes,equally spaced, indicating ten one-tenth of our ruler. How this is done - must remain a mystery and need not concern us, as we are trying to come up with a procedure of measuring lengths, or, judging this from a different prespective: comparing lengths.
This procedure allows us to now measure canes with accuracy of up to one-tenth of a unit length. Applying this procedure again and again (to the one-tenths, and to the one-tenth of the one-tenth) we understand this limiting process will eventually yield a number (The justification to this last statement comes from the more axiomatic handling of the real line and the notion of limits).
Understanding how to deal with lengths, we can now turn to measure areas (using similar techniques).
