(This was published previously on kragen-tol.)
My friend Santi asked me why we divide by a fraction by interchanging the numerator and the denominator and multiplying; that is, why a/(b/c) = a(c/b). I wasn’t quite sure how to answer, but after thinking about it, it turns out that there are many deep and fascinating answers that involve many aspects of the universe of mathematics. Here are three different answers. Sort of.
Part of the problem is that it’s difficult to say what really counts as an explanation here, because, as Feynman explained in his famous BBC video on “Fucking magnets, how do they work?”, an explanation has to start with things that you already understand to be true. In cases like this, it’s really easy to fool yourself into thinking that you have an explanation, when all you really have is circular logic. Here, let me demonstrate.
A “group” is a set with an operation that have the four properties of closure, associativity, identity, and invertibility. Nonzero fractions, together with multiplication, are a group. It turns out that the divide-by-multiplying-upside-down thing isn’t limited to fractions at all; it’s a much more general property that applies to any group, including bizarre things like permutations under composition, three-dimensional rotations of polyhedra under composition, matrices under matrix multiplication, Gaussian integers under addition, bit strings of some fixed length under XOR, and integers under multiplication modulo a prime number!
To explain, first I will explain the meaning of the group properties. Since I’m writing this mostly in ASCII, I’m going to use “G” to mean the set, other letters to mean elements of the set, “+” to mean the operation, and two other special notations which I explain below: “0” to mean the identity element of G, and “-X” to mean the inverse of an element X of G.
Closure: if A and B are in G, then A+B is also in G. This rules out things like “numbers 1 to 10 under ordinary addition”, because 9+9 is 18, which isn’t in the set, and things like “integers under ordinary division”, since even though 6/3 is in the set, 2/3 isn’t.
Associativity: (A+B)+C = A+(B+C), always. This rules out non-associative operations like division or subtraction.
Identity: There’s an element of G which I will call 0 which has the property that A+0 = 0+A = A, for all A. We call it the “identity element”. This rules out operations like “return the left argument”.
Invertibility: Every element A in G has a corresponding inverse element -A such that A+-A = -A+A = 0. This rules out operations like multiplication on rational numbers, since zero (not the identity element, real zero) is a rational number, the identity element for multiplication on rational numbers is 1, and there’s nothing you can multiply by zero to get 1. (But multiplication on nonzero rational numbers is still fine!)
You will note that commutativity isn’t one of the group properties, even though, say, integer addition is additive; and in fact there are lots of interesting groups whose operation isn’t commutative. If you can prove a property of groups without depending on commutativity, then it applies not just to commutative groups like the nonzero rationals under multiplication, but also noncommutative groups like matrices under matrix multiplication and permutations under composition.
The question we started out with was, “Why does a/(b/c) = a(c/b), when a is a rational number and b and c are nonzero integers?” Here’s why that’s true not just for integers but actually for any nonzero rational numbers.
Let’s start by rewriting it in the notation I have above: ab becomes A+B, and a/b becomes A+-B. So we’re trying to find out if, and why, A+-(B+-C) = A+(C+-B).
Using the definition of invertibility (-), we can derive that:
B+-B = 0
From there, we can replace B with B+0, using the definition of identity (0):
(B+0)+-B = 0
Then, using the definition of invertibility, we can replace that 0 with -C+C:
(B+(-C+C))+-B = 0
The definition of associativity lets us move around the parentheses around + operations:
(B+-C)+(C+-B) = 0
Now, if (B+-C)+(C+-B) is 0, then for any element Q, by the substitutability property of equality:
Q+0 = Q+(B+-C)+(C+-B)
In particular, if we take Q to be -(B+-C), which we know exists by the properties of invertibility and closure, we have:
-(B+-C)+0 = -(B+-C)+(B+-C)+(C+-B)
The right side now begins with a thing of the pattern -R+R, which we know from invertibility is 0, so we have:
-(B+-C)+0 = 0+(C+-B)
And by the definition of identity, we now have:
-(B+-C) = (C+-B)
This is a stronger form of what we wanted to prove in the first place, since it shows that for any A:
A+-(B+-C) = A+(C+-B)
which is our original statement, but it also shows that you don’t even need the A.
So that shows that a/(b/c) = a(c/b), as long as all three are nonzero rational numbers, and in particular if b and c are nonzero integers. It also shows that the same thing is true if, for example, a, b, and c are numbers in the range of 1 to 6, with the following tables for multiplication and division in Z/7:
* 1 2 3 4 5 6 / 1 2 3 4 5 6
1 1 2 3 4 5 6 1 1 2 3 4 5 6
2 2 4 6 1 3 5 2 4 1 5 2 6 3
3 3 6 2 5 1 4 3 5 3 1 6 4 2
4 4 1 5 2 6 3 4 2 4 6 1 3 5
5 5 3 1 6 4 2 5 3 6 2 5 1 4
6 6 5 4 3 2 1 6 6 5 4 3 2 1
For example, if a=3, b=4, and c=5, then a/(b/c = 5) = 2, and a(c/b = 3) = 2 also. Try it for any three of these numbers. It will always work. (It works in Z/p for any prime p. See attached modmul.py.)
Here’s a Python example of permutations using my permutations module:
>>> from permutations import cycle
>>> a = cycle(1, 2, 4)
>>> b = cycle(2, 5)
>>> c = cycle(2, 4)
>>> a * (b * c**-1)**-1
cycle(1, 2, 5)
>>> a * (c * b**-1)
cycle(1, 2, 5)
Note that, since permutation composition is not commutative, a(1/b * c) is not the same:
>>> a * (b**-1 * c)
cycle(1, 2) * cycle(4, 5)
But wait! All this is based on my initial assertion that the four group axioms apply to nonzero rationals under multiplication. How do we know that’s true? Maybe all of the above is begging the question, when it comes to the nonzero rationals, since how do we know that nonzero rationals even have multiplicative inverses in the first place? I mean, if we assume that rationals including zero are a group under multiplication, then we can use the same argument to claim that rationals including zero can be divided in this way, but that turns out to be wrong. So first we have to show that nonzero rationals are a group!
Suppose we take as given that nonzero integers form a commutative group under multiplication, and we want to explore relations a:b of those integers, using the equality relation a:b = c:d iff an:bn = cm:dm for some nonzero integers m and n; and we define multiplication as (a:b)(c:d) = ac:bd. What can we find out?
First, we can prove pretty directly that ((a:b)(c:d))(d:c) = a:b, as long as neither c nor d is zero:
((a:b)(c:d))(d:c) =
{definition of multiplication}
(ac:bd)(d:c) =
{definition of multiplication}
acd:bdc =
{commutativity of integer multiplication}
adc:bdc =
{associativity of integer multiplication}
a(dc):b(dc) =
{definition of rational equality}
a:b
(Bloviation 0)
Which is pretty close to what we started out wanting to know. But we can also show that nonzero rational numbers, defined in this way, are a group under multiplication. We need to show closure, associativity, identity, and invertibility.
Closure: (a:b)(c:d) produces ac:bd, and since the integers are closed under multiplication, ac and bd are integers.
Associativity:
((a:b)(c:d))(e:f) = (ac:bd)(e:f) = (ac)e:(bd)f
(a:b)((c:d)(e:f)) = (a:b)(ce:df) = a(ce):b(df)
which are equivalent because integer multiplication is associative.
Identity: (1:1)(a:b) = 1a:1b = a:b since 1 is an identity for integer multiplication.
Invertibility: Bloviation 0 above shows that these relations have right multiplicative inverses (and how to compute them); you can carry out the same argument for left multiplicative inverses.
So that shows us how to compute multiplicative inverses of pairs of nonzero integers with these weird definitions of multiplication and equality. But how do we know that those pairs of integers are really “the rational numbers”?
We can translate individual integers into this pair-of-integers form as follows: t(x) = x:1. It should be straightforward to see that multiplication on these pairs corresponds to multiplication on the integers, i.e. t(x)t(y) = t(xy). And, in math, that’s really all you need: it walks like the group of fractions, quacks like the group of fractions, so it’s the group of fractions! There is no more “really” than that, in math.
I guess this is the algebraic structure you get if you assume that nonzero integers must have multiplicative inverses, and then take the set of the integers and their multiplicative inverses and extend it by transitive closure of multiplication; it’s the smallest set that includes the nonzero integers and satisfies the group axioms. I’m not immediately sure how to show that, but I’m pretty sure it’s true.
An interesting thing here is that the proof above depends on the commutativity of (nonzero) integer multiplication, from which we could directly derive the commutativity of rational multiplication. Integer multiplication is closed, associative, and has an identity, but it isn’t invertible, which makes it an algebraic structure called a “commutative monoid”. I’m not sure what happens if your numerator and denominator are drawn from some other monoid that isn’t commutative, such as quaternions with nonzero integer coefficients, or nonzero square matrices of integers of some size n.
But that’s all very algebraic and abstract. You could read all of the above and still think that there was maybe a flaw in the logic somewhere, that there might be some case where a/(b/c) isn’t really a(c/b). What about everyday intuition?
When we say n/m, we’re looking for a solution x to the equation mx = n; we want to know how many times we would have to add m to itself to get n, or looking at it another way, what number we would have to add to itself x times to get n. Spatially, we’re looking for the length x of a stick of which we would have to lay m of, end to end, to add up to n, or the number of sticks x of length m.
If m is a fraction, it’s easier to think of it as a length of a stick than as a number of sticks, so let’s go with that. It’s clear that if m is a reciprocal of an integer, like 1/3 or 1/4, then that integer is the number of sticks you need to reach a length of 1; and if you have to go twice or three times as far, you need twice or three times as many sticks, so clearly x is going to be proportional to n. Similarly, it’s clear that if m is twice or three times as long, you need half or a third as many sticks to go the same distance, so making m be 2/3 or 3/4 will make x be half or a third of what it was when m was 1/3 or 1/4.
So that kind of covers it: you can get to any fraction m by starting with the reciprocal of an integer (your denominator) and then multiplying it by another integer (your numerator), and if you watch how x changes as you do that, you can see that x gets multiplied by the denominator, divided by the numerator, and multiplied by n, your original dividend.
This Python script generated the multiplication and division table for Z/7 above.
#!/usr/bin/python
nn = 7
col = " "
def num(xx):
print "%2d" % xx,
print " *",
for ii in range(1, nn):
num(ii)
print col, " /",
for ii in range(1, nn):
num(ii)
print
for ii in range(1, nn):
num(ii)
for jj in range(1, nn):
num((ii * jj) % nn)
print col,
num(ii)
inverse = (jj for jj in range(1, nn) if (ii * jj) % nn == 1).next()
for jj in range(1, nn):
num((inverse * jj) % nn)
print