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!

  • 4
    $\begingroup$ Division is the inverse of multiplication. $x=a/b$ if and only if $x\cdot b=a$ (and $b\neq0$). $\endgroup$
    – mr_e_man
    Commented Aug 12, 2018 at 9:14
  • $\begingroup$ @mr_e_man Maybe, I'll double-check. $\endgroup$
    – God bless
    Commented Aug 12, 2018 at 11:20
  • $\begingroup$ Whoops, I mistyped the $3$ in my first polynomial and wrote $5x$ instead of $3x$. Edit incoming. $\endgroup$
    – God bless
    Commented Aug 12, 2018 at 11:25
  • $\begingroup$ He he, it does. :) $\endgroup$
    – God bless
    Commented Aug 12, 2018 at 11:52
  • $\begingroup$ Division isn't an operation, division is a question. To divide $a$ by $b$ is to ask "what number, when multiplied by $b$, would give me $a$"? So to define division rigorously you just need to define multiplication rigorously. $\endgroup$
    – Jack M
    Commented Aug 12, 2018 at 11:58

4 Answers 4


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 $p_mx^m$.

  • 1
    $\begingroup$ Woah! This blew me away! Thank you so much! $\endgroup$
    – God bless
    Commented Aug 12, 2018 at 11:27
  • $\begingroup$ It's unclear to me how we got to the second line, where by the second line I mean this: i.imgur.com/N9a7BcB.png $\endgroup$ Commented Jan 31, 2022 at 17:17
  • 1
    $\begingroup$ @user161005 - $$\frac ab=\frac ab-c+c=\frac{a-cb}{b}+c$$ $\endgroup$
    – mr_e_man
    Commented Jan 31, 2022 at 18:03
  • $\begingroup$ @mr_e_man I think I understand it now, except one thing. What is your point here?Yes, you shown that these manipulations will decrease degree of the leading term of the dividend by 1. But these manipulations don't look like polynomial long division. $\endgroup$ Commented Feb 1, 2022 at 17:49

Division is basically partitioning an expression/value into (smaller) components. It is the number of times that the components can be contained within the expression/value.

Let's convert base $10$ to a variable $x$. So $903$ becomes $9x^2+3$ and $12$ becomes $x+2$. Then we get $$\begin{align}9x-18\\\text{____________}\\x+2\text{/}\,\,9x^2+0x+3\\\hspace{-3cm}9x^2+18x\\\text{____________}\\-18x+3\\-18x-36\\\text{________}\\39\end{align}$$

Note: Feel free to edit this attempt at writing long division with $\LaTeX$.

Now we would usually stop here, giving $$903=9\cdot10-18+\frac{39}{12}=75.25$$ But if we continue, we would arrive at $$9x-18+39x^{-1}-78x^{-2}+156x^{-3}-312x^{-4}+\cdots$$ or $$9x-18+\frac{39}x\sum_{k=0}^\infty\left(-\frac2x\right)^k$$ and since $|-2/x|=0.2<1$, we can invoke geometric series. Thus the expression becomes $$9x-18+\frac{39}{x\left(1+\frac2x\right)}=9\cdot10-18+\frac{39}{10\cdot1.2}=75.25,$$ as before.

In essence, this is just a rather interesting way to write $\frac14$ as a geometric series!

  • $\begingroup$ Thank you for your answer. I don't understand why you write $(x + 2) / (9x^2 + 0x +3)$. Shouldn't it be the other way around? And could you please define division a a very basic level (like ZFC or something) and then show me how a machine would go about executing it? This always boggled my mind the most — how to make division computable. $\endgroup$
    – God bless
    Commented Aug 12, 2018 at 9:25
  • $\begingroup$ That's the full long division sign. See here $\endgroup$
    – TheSimpliFire
    Commented Aug 12, 2018 at 14:49
  • $\begingroup$ @GregorPerčič how on earth would defining anything in terms of ZFC make life easier? Heck, how would bringing set theory in here at ALL make life easier? A machine verifiable proof could take tens to hundreds of pages to write down. No thank you. Learn the algebra - not the set theory. Also, computability has little to do with set theory. Computability theory is often considered a branch of Comp Sci, and there's a Stack Exchange site for theoretical CS if you really want to ask... That being said, look up synthetic division if you want a fast algorithm $\endgroup$ Commented Aug 12, 2018 at 23:38

TheSimpliFire has already dealt with the infinite series part of your question. Therefore, I shall restrict myself to the 'rigorous' definition of division.

One, potential approach (as you discovered) is to define division as an infinite series. Note however, that the steps to do so may implicitly involve division.

As it turns out, in rigorous math (say, abstract algebra), the concept of division (and subtraction) is usually never explicitly defined. Usually , we just define multiplication and then define multiplicative inverses (think, reciprocals) as follows: I For each, $a \in \mathbb{R^*}$, there exists $(1/a) \in \mathbb{R^*}$ such that $a \cdot (1/a) = 1$. (Here, we have defined the multiplicative inverse on the set $\mathbb{R^*}$ of non-zero real numbers.)

Of course $1$ is already defined by $1 \cdot a = a$ with the proper conditions appended.

Then you can 'define' division of $a$ by $b$ as multiplication of $a$ by $(1/b)$.

Multiplication itself, is defined recursively using addition; while addition can be defined by extending it from the natural numbers while defining them using Peano's axioms.

Note that all definitions of the elementary operations are somewhat revised everytime we proceed to a bigger set (say, from $\mathbb{Q}$ to $\mathbb{R}$).

Also, all operations are not defined for all sets. For example, division cannot be defined on the set $\mathbb{N}$, of natural numbers.

You can read up on set theory, abstract algebra and real analysis if you want to know more.

  • $\begingroup$ @mr_e_man. Ya, nice catch . My bad. Division is defined on non-zero real (or complex) numbers. That's how "division by zero is not defined". $\endgroup$
    – user0
    Commented Aug 12, 2018 at 12:15
  • $\begingroup$ @mr_e_man the error is now corrected. See edited answer. $\endgroup$
    – user0
    Commented Aug 12, 2018 at 12:17

You may be confusing two (sometimes, but not necessarily, related) concepts, namely: definition and algorithm or method.

They serve different purposes, but sometimes one may hint at the other, but there is no such constraint (hence nonconstructive definitions and axioms, or at best impracticable definitions). On the other hand, a technique, method or algorithm helps us to get at something in a particular form given a set of constraints -- usually of space and time; hence one studies such things as the acceleration of convergence of iterations, the efficiency of algorithms, etc.

Thus, whereas a definition is mainly conceptual in manner and serves us in being able to conceptually grasp some object, as it were (and is not required to do more than this), a method need not enlighten, but is usually a means to solve some problem at hand, or lay one's hand, practically as it were, on some solution, usually a real number. Thus, a definition is required to be as simple and sparse as possible in order to gain in clarity and elegance, whereas a method may be tricky and involved, so long as it helps to solve one's problem efficiently given a set of conditions -- thus the different division algorithms better than (but a little bit more sophisticated than) the long division method taught schoolchildren.

It is not clear then which one of these you want to be clear about on division, but I guess maybe a bit of both. Now, just like there are many ways to do something (algorithm), there are many ways to understand something too (definition). I can only ficus on one way of doing the latter here.

So, how shall we define division? What do we even mean? Clearly, it is some sort of operation, but what does it do? How does it behave? This is what a list of axioms is meant to elucidate. The clearest way to understand division (say of real numbers) is to see it as the contrary operation to multiplication (most of the time, in any case). If we define multiplication $×$ of real numbers $a$ and $b$ as a binary operation (i.e., it must combine two entities to produce a unique result) possessing some properties, some of which are that we want $a×b$ to always be a real number, that $a×1=a,$ for example (you may investigate the rest), then we can define division as the operation that, given that $c=a×b,$ takes $c$ and $b$ to $a,$ and one then investigates what properties it has and its relationship to multiplication. One would then have commenced on a journey towards abstract algebra.


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