Sorry if this question is too far out there, but I'm looking for a rigorous definition of the division operation. As I have seen it before, $a/b$ is the solution to the equation $a=xb$. While I am okay with this sort of defintion, it is tantamount to saying that division is the inverse operation of multiplication. (at least that's how I see it, comments on this assertion are welcome.)

Instead, what I seek is a different sort of formalization of the definition of division of real numbers. Whereas the definition above is nearly equivalent to defining division as the inverse operation of multiplication (again, maybe i'm wrong), I seek a definition that does not rely on the definition of multiplication already being asserted.

Motivation: Consider the fundamental theorem of calculus that essentially states that the indefinite integral is the inverse of the derivative. This is a highly non-trivial theorem, and while one may be able to define the indefinite integral as the inverse of the derivative (could be wrong here as well), it is much more meaningful to define the two operations separately and then show that they're actually just inverses of one another. Does anything like this exist for multiplication and division? Or are these operations just too fundamental?

Question: Does there exist a definition for $a/b$ where $a,b \in \mathbb{R}$ that does not require multiplication to already be defined?

I know, I'm grasping at straws, but maybe there is something here that really enhances my understanding of what division is in general. Or maybe there are some interesting definitions involving limits or something that I have never seen before.

  • $\begingroup$ The first question is, how did you define multiplication? (Good question BTW) $\endgroup$
    – Daniel
    Aug 5, 2015 at 15:09
  • $\begingroup$ I would define it as if it were just notation for repeated addition, but i would only know how to define it for integers, and wouldn't know how to extend it to real numbers in general. $\endgroup$ Aug 5, 2015 at 15:11
  • $\begingroup$ @PaddlingGhost: Your point about being able to use repeated addition to justify multiplication for integers but not real numbers is indeed a crucial point. One either goes through conceptual division of repetition (meaning that if I have a procedure $P$, is there a procedure $Q$ such that repeating $Q$ $2$ times is the same as performing $P$?) to get to rationals and then treat real numbers as everything that can be approximated by rationals, or one goes through geometric intuition as 'experienced' in the real world (as I chose in my answer). $\endgroup$
    – user21820
    Aug 5, 2015 at 16:04
  • $\begingroup$ Saying "I'm looking for a rigorous definition of the division operation" is misleading, because that is not what you are looking for - you are interested in a definition that does not make reference to multiplication. No, I do not believe it is natural at all to detach division from multiplication. $\endgroup$
    – anon
    Aug 6, 2015 at 23:58

2 Answers 2


One approach is to define a new notion: a Number is a pair of real numbers, $(a, b)$, with $b \ne 0$, and $(a, b)$ is "the same as" $(c, d)$ if $ad = bc$. (The idea here is that the Number $(a, b)$ represents the thing you want to call $a/b$, by the way).

We can add Numbers: $$(a, b) + (p, q) = (aq + pb, bq)$$

and multiply: $$(a, b) * (p, q) = (ap, bq)$$

and it turns out that these operations are independent of representations (i.e., if two Numbers are "the same", then when added to another Number, the results will be "the same"). You have to prove that of course.

Once you've done all this -- shown that the set of equivalence classes of Numbers is the same as the set of reals with addition and multiplication under the correspondence $r \mapsto (r, 1)$ -- you're ready to define division:

$$ (a, b) / (c, d) = (ad, bc). $$

And then you can prove that division has all the properties that you want.

This is really the idea that @Travis is suggesting in his comment, but skipping the visit to $\mathbb Q$.

BTW, Spivak's Caculus, in its last chapter, has a lovely problem about the "high school student's real numbers" in which one defines a real to be a sequence of digits, defines addition and multiplication carefully, and then proves that these form a complete ordered field. In that problem, "division" is essentially the long-division algorithm, and proving that it does the opposite of long-multiplication is a major pain in the neck. Perhaps this is, in some way, what you're seeking.

Post-comment remark

One more thing -- it's worth looking at Edwin Moise's "Elementary Geometry from an Advanced Standpoint", in which one of the later chapters describes the work of Eudoxus (I think!) on defining real numbers via (geometric) ratios. It's quite remarkable work, and hints at the idea of Dedekind cuts that were developed only after a couple of millenia.

Further post-comment remark

Here's the text of problem 29-2 from Michael Spivak's Calculus.

"This problem outlines a construction of "the high-school student's real numbers." We define a real number to be a pair $(a, \{b_n\})$ where $a$ is an integer and $\{b_n\}$ is a sequence of natural numbers from 0 to 9,. with the proviso that the sequence is not eventually $9$; intuitively, this pair represents $$a + \sum_{n = 1}^\infty b_n 10^{-n}.$$ With this definition, a real number is a very concrete object, but the difficulties involved in defining additional and multiplication are formidable (how do you add infinite decimals without worrying about carrying digits infinitely far out?). A reasonable approach is outlined below; the trick is to use least upper bounds right from the start.

(a) Defined $(a, \{b_n\}) << (c, \{d_n\})$ if $a < c$ or if $a = c$ and for some $n$ we have $b_n < d_n$, but $b_j = d_j$ for $1 \le j < n$. Using this definition, prove the least upper bound property.

(b) Given $\alpha = (a, \{b_n\})$, define $\alpha_k = a + \sum_{n = 1}^k b_n 10^{-n}$; intuitively, $\alpha_k$ is the rational number obtained by changing all decimal places after the $k$th to $0$. Conversely, given a rational number $r$ of the form $a + \sum_{n = 1}^k b_n 10^{-n}$, let $r'$ denote the real number $(a, \{b'_n\})$, where $b'_n = b_n$ for $1 \le n \le k$, and $b'_n = 0$ for $n > k.$

Now for $\alpha = (a, \{b_n\})$ and $\beta = (c, \{d_n\})$ define \begin{align} a ++ b &= \sup\{ (\alpha_k + \beta_k)': k \text{ a natural number} \} \end{align}

(the least upper bound exists by part (a)). If multiplication is defined similarly, then the verification of all conditions for a fields is a straightforward task, not highly recommended. Once more, however, existence of multiplicative inverses will be the hardest."

Note: Spivak uses a boldface "<" where I've used $<<$, and a bold plus sign where I've used $++$, but I cannot make those appear here.

So that doesn't really do what you've asked, but I think it gets to the main point: you can do all this stuff more or less algorithmically, at least if you're willing to invoke upper-bound ideas (which is somewhat the trick Eudoxus uses as well, amazingly enough). But mostly it's a pain in the neck. :)

  • $\begingroup$ Yes, that last paragraph, i think, summarizes exactly what i'm looking for. $\endgroup$ Aug 5, 2015 at 15:15
  • $\begingroup$ @PaddlingGhost: I don't agree that division should ever be thought of as long-division. Otherwise it makes no sense to ask what $\frac{1}{2+3i}$ is, or the so-called rational function $\frac{x^2+1}{x^3-4}$ where $x$ is an indeterminate. $\endgroup$
    – user21820
    Aug 5, 2015 at 16:09
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    $\begingroup$ Since PaddingGhost was explicitly asking about division for the real number system, I think that long division, properly described, is one pretty reasonable definition. Once you've gotten past that, and seen that the algorithmic and "algebraic" definitions match, you can use the algebraic definition for a while, and then choose it as the thing to generalize. $\endgroup$ Aug 5, 2015 at 17:49
  • $\begingroup$ @John Hughes if someone were to clearly define a division algorithm that will always terminate then I would certainly consider this a satisfactory answer. Even further, if one could show me how this division algorithm ends up being multiplication's inverse, I would have no choice but to mark it as the accepted answer. $\endgroup$ Aug 5, 2015 at 18:33
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    $\begingroup$ @PaddlingGhost: As I said in my comments, the division algorithm is based on inverting multiplication. At each step you choose some amount that when multiplied is less than the remainder. $\endgroup$
    – user21820
    Aug 6, 2015 at 0:22

I guess your problem is actually that you don't understand what is so good about having inverses. That is the key. We always want to know whether we can undo something. Multiplication can be thought of as a scaling (repeated addition only works for integers, but the idea of scaling works for even complex numbers), and the natural question is when it can be undone. For amounts in the real world, it seems we can sort of scale by any real value, for example by moving closer to an object it becomes larger in our visual field. We can even scale by negative real values using say a lens. We can even scale by zero, such as focusing the sun onto a single point (a point is an ideal concept; of course there will be imperfections in the lens and our setup). We intuitively see that scaling by any nonzero amount can be undone, in the sense that we can figure out the original location before scaling. However, after a scaling by zero, all points are now at the centre of scaling and we cannot tell where they came from. Division is just the name we give to the inverse of multiplication, which is the name we might give to scaling. Namely when we scale a real position $p$ by a real amount $r$, we define the resulting position to be the result of multiplying the two reals as $p \times r$. Similarly to undo a scaling so that from the resulting position of $p \times r$ we get $p$ we define the original position as the result of dividing that by $r$, namely the original position is $p = ( p \times r ) \div r$. As explained, this holds for any real values $p,r$ such that $r \ne 0$.

Same goes for subtraction as the inverse of addition. We can consider addition of a real value to correspond to translation (shifting without changing shape) by that amount in a specific direction, and subtraction of that same real value to correspond to undoing the translation, which is equivalent to translating by the same amount in the opposite direction. This time every translation can be undone, and similarly every addition can be undone. In mathematical symbols $p = ( p + a ) - a$ for any real values $p,a$. This translation concept extends automatically and naturally to complex numbers.

To be precise for complex numbers, addition of a complex number corresponds to translation of the 2d plane in some direction in the same plane. The order of adding does not matter because the order of performing translations does not. Multiplication of a complex number corresponds to a scaling and rotation about the centre $0$. (It is beyond the scope of this answer to explain the rotation part.) Again, order of multiplication does not matter because scalings and rotations can be swapped without changing the result. However, the order of additions and multiplications cannot be swapped. This is also the case for real numbers where you are basically translating or scaling a 1d line along itself.

Now the next part of the question is concerning integrals. There are different types of integrals. The one you are referring to is probably the Riemann integral, which is the natural outcome of trying to define area under a curve. Why do we want that? Partly historical curiosity and Newtonian mechanics, at least. The latter is because we intuitively know that distance is speed times time taken for constant speed. Extending it to non-constant speed results in approximation concepts that lead to the Riemann sum. Of course in the real world everything converges so we do not need to care whether velocity is integrable; of course it is! But actually in hindsight from Einstein's theory of relativity it is better to start with position and define velocity as simply the rate of change of position, similarly force as rate of change of momentum, because the conventional formulae in the reverse direction fail spectacularly. It turns out that the Riemann integral is an anti-derivative under certain conditions, but there are functions whose derivative is not integrable! (see Volterra's function)

  • $\begingroup$ @PaddlingGhost: You will also notice that the other answer is slightly circular because there is a priori no reason to consider that notion he defined! Why on earth should we treat $(a,b)$ the same as $(c,d)$ if $ad=bc$ without already having an understanding of ratio? What is a ratio? $\endgroup$
    – user21820
    Aug 5, 2015 at 16:13
  • $\begingroup$ @PaddlingGhost: The real-world geometric intuition tells us that when we scale something, shape is preserved. What does preserving shape mean? If we think carefully about it we can condense it down to ratios of lengths being preserved. What is length? Length is what is invariant under translation and rotation. We make rulers because we know that no matter where we bring our ruler, its markings remain at the same distance from each other, so that we can measure distances of other objects. Scaling preserves ratios of distances. That in a sense defines ratios. $\endgroup$
    – user21820
    Aug 5, 2015 at 16:18
  • $\begingroup$ So then you answer my question in they negative by saying no, division is simply the inverse of multiplication and here's why. So, this does answer the question, but it is unlikely to earn the green check mark as someone has already answered in the affirmative and given a suitable definition that does not involve inverses. $\endgroup$ Aug 5, 2015 at 16:31
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    $\begingroup$ Just read your comments and they were helpful. I too find problems with the other answer . $\endgroup$ Aug 5, 2015 at 16:33
  • $\begingroup$ @PaddlingGhost: You are simply wrong. The other answer used multiplication. What else does $ad = bc$ mean if not $a \times d = b \times c$?? $\endgroup$
    – user21820
    Aug 5, 2015 at 16:46

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