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.
An answer based on group theory
-------------------------------
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.
### Our question, restated in generic group terms ###
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.
### Applying the result to other groups ###
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][1]:
>>> 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)
[1]: http://lists.canonical.org/pipermail/kragen-hacks/2013-August/000560.html
"permutations (elements of a symmetric group) in Python"
### Why this is a little bit bogus ###
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!
Rational numbers as pairs of integers
-------------------------------------
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.
A spatially-oriented intuitive answer
-------------------------------------
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.
modmul.py
---------
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
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