What is the simplest way to rigorously define division (why does the standard algorithm work)? I become really nervous if I catch myself doing one process without really understanding how it works. Well, one of these processes is division. In primary school, I learned the technique of dividing (on paper), but never really understood why this "number-shuffling" of mine worked.
Recently, I got some insight into it when I learned the technique of dividing polynomials. For example, something like
$$(x^3 + 6x^2 + 3x - 8) : (x^2 + x - 2) = x + 5; \quad 2 \text{ remains.}$$
Or in the standard form for the division of polynomials $p(x) = k(x)q(x) + r$:
$$(x^3 + 6x^2 + 3x - 8) = (x + 5)(x^2 + x - 2) + 2.$$
The idea is to divide the "first part" of $p(x)$ and divide it by the "first part" of $q(x)$ ($x^3 : x^2 = x$), and acquire the "first part" of $k(x)$. Then, multiply the whole $q(x)$ with this "part", subtract that from $p(x)$ and repeat the process until you hit $r$.
Since any real number can theoretically be expressed as
$$a_na_{n-1}\cdots a_0,a_{-1}\dots = a_n \cdot 10^n + \cdots + a_0 \cdot 10^0 + a_{-1} \cdot 10^{-1} +\cdots,$$
I though to myself: "Hmmm, these polynomials simply look like longer ways to write down numbers in base $n$ ($n$ being $x$ in this case)." So I thought that I'd use the same "division method" on normal whole numbers. And I discovered something that amazed me.
My pick was $903:12 = 75.25$. Well, this is how it went:
$$(9 \cdot 10^2 + 0 \cdot 10^1 + 3 \cdot 10^0) : (1 \cdot 10^1 + 2 \cdot 10^0) = 9 \cdot 10^1 - 18 \cdot 10^0 + 39 \cdot 10^{-1} - 78 \cdot 10^{-2} \cdots.$$
What baffled me is that this infinite polynomial never actually gets to $75.25$ but approaches to it. I'm not fond of infinite series, but my guess is that after infinite iterations, the result would finally be exactly $75.25$. Well, in my case (4 iterations), the result is $75.12$, which is already not too shabby. I suspect that this kind of "limit division" could be expressed as an infinite sum. If you know how to write one down, please, do so.
I did this "personal investigation" (and discovered something amazing that I will forever cherish) to better understand how division really works, but I failed. I still don't understand.
Please rigorously define division, so I can finally be calm. Thank you! 

You might understand the division algorithm better by using simple equality:
$$\frac{x^3+6x^2+3x-8}{x^2+x-2}$$
$$= \frac{(x^3+6x^2+3x-8)-(x^2+x-2)x}{x^2+x-2}+\frac{(x^2+x-2)x}{x^2+x-2}$$
$$= \frac{5x^2+5x-8}{x^2+x-2}+x$$
We specifically chose $x$ to add and subtract from the expression, so as to cancel the $x^3$ on the left. The purpose of the algorithm is to reduce the degree of $p(x)$, eventually to a lower degree than $q(x)$.
Let $p(x) = p_mx^m+p_{m-1}x^{m-1}+\cdots+p_1x+p_0$,
and $q(x) = q_nx^n+q_{n-1}x^{n-1}+\cdots+q_1x+q_0$,
with $m \geq n$ and $p_m\neq0\neq q_n$. Then
$$\frac{p(x)}{q(x)} = \frac{p_mx^m+p_{m-1}x^{m-1}+\cdots+p_0}{q_nx^n+q_{n-1}x^{n-1}+\cdots+q_0}$$
$$= \frac{(p_mx^m+p_{m-1}x^{m-1}+\cdots+p_0)-\frac{p_mx^m}{q_nx^n}(q_nx^n+q_{n-1}x^{n-1}+\cdots+q_0)}{q_nx^n+\cdots+q_0}+\frac{p_mx^m}{q_nx^n}$$
$$= \frac{(p_mx^m+p_{m-1}x^{m-1}+\cdots+p_0)-p_mx^m-\frac{p_m}{q_n}q_{n-1}x^{m-1}-\cdots-\frac{p_m}{q_n}q_0x^{m-n}}{q_nx^n+\cdots+q_0}+\frac{p_m}{q_n}x^{m-n}$$
$$= \frac{p_{m-1}x^{m-1}+\cdots+p_0-\frac{p_m}{q_n}q_{n-1}x^{m-1}-\cdots-\frac{p_m}{q_n}q_0x^{m-n}}{q_nx^n+\cdots+q_0}+\frac{p_m}{q_n}x^{m-n}$$
The degree of the numerator has been reduced from $m$ to $m-1$. It should be clear that $\frac{p_m}{q_n}x^{m-n}$ is the only term, exactly the term that can be subtracted to cancel $p_mx^m$.
