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Q&A #3893

Teachers' Lounge Discussion: Dividing by zero

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From: Kevin Santosuosso <mulder_00@yahoo.com>
To: Teacher2Teacher Public Discussion
Date: 2004010713:34:51
Subject: The problem with "dividing by zero" (long, advanced)

The problem with "dividing by zero" can be very hard to explain to
younger students, because the issue itself needs some advanced thought
to understand. To answer the question fully, you need to understand
the real number system, and more specifically, the underlying
operations and choices that shape it.

First of all, let's be clear about something: there is no "division"
nor "subtraction"; at least not as independent operations. The *only*
two base operations associated with the real number system are
addition and multiplication, and these are operations defined by
certain properties. Specifically, the real number system is an example
of a *field* (see http://mathworld.wolfram.com/DivisionAlgebra.html
for the properties of a division algebra; a field is a commutative
division algebra). 

This field has two operations (addition and multiplication) that act
upon a set of elements (in this case, the real numbers) that obey
certain rules (see the link). For example, add or multiply any two
real numbers together and you get another real number. In this field,
there are two "special" elements: the "additive identity" (zero: that
is, a + 0 = 0 + a = a) and the "multiplicative identity" (one: that
is, a * 1 = 1 * a = a). Furthermore, each element in the set has an
"additive inverse" (that is, a + (-a) = (-a) + a = 0) and a
"multiplicative inverse" (that is, a * a^ = a^ * a = 1) that relate
back to their respective identity elements (and are also elements of
the set themselves). The real number system follows these rules. (Or
we could view it as if the real number system is *constructed*
according to these rules, and therefore follows them. This is probably
better.) Subtraction and division are simply the terms we give to
adding certain elements and multiplying certain elements.

It's not hard to understand that you can't add or multiply real
numbers by things that aren't real numbers and still get real numbers.
If I asked you to give me the answer to "5 plus blue" or "2 divided by
fluffy bunny", you'd look at me rather strangely. And so you should,
since "blue" and "fluffy bunny", while these things certainly exist,
aren't numbers. So I can't combine them by adding or multiplying them.
Yet we often talk about dividing by zero (that is, multiplying by the
multiplicative inverse of zero). The problem: the multiplicative
inverse of zero is not part of the field and so is *not a real
number*! You may as well be dividing by fluffy bunny.

In this context, then, zero is a unique element of the real numbers:
it is the only real number with *no* multiplicative inverse. If you
look at the definition of the division algebra, you'll notice that the
definition *explicitly forbids* the additive identity (zero) to have a
multiplicative inverse (see rule #7). The $64,000 question is: why? We
can work this out easily.

Let's try to construct our field by accepting the rules of the
commutative division algebra, but allowing 0 to have a multiplicative
inverse. First of all, we note (where a and b are any two elements of
the set):

                    a + 0 = a
              b * (a + 0) = b * a
            b * a + b * 0 = b * a
- (b * a) + b * a + b * 0 = - (b * a) + b * a
                    b * 0 = 0

We "knew" this, of course, but this is why it is so. Note that we
don't actually define a * 0 = 0. Rather, it is a direct consequence of
the basic properties of the field.

We haven't yet needed to allow 0 to have a multiplicative inverse, 0^.
Let's do so now. Then, since every product of two elements of the set
is another element of the set, for every element a there must be some
element b such that:

      a * 0^ = b
(a * 0^) * 0 = b * 0
a * (0^ * 0) = 0
       a * 1 = 0
           a = 0

That is, every element of the set is itself identically 0! So the only
way to construct the field *with* a multiplicative inverse of the
additive identity is when the set contains *only* one element: namely,
the additive identity itself, which must then be its own inverse. So
we see that, if we want the real number system, we must construct our
field by explicitly refusing to allow one element of our field (zero)
to have a multiplicative inverse at all. 

So why can't we divide by zero? Because the multiplicative inverse of
zero isn't a real number. You can't even make the statement

2 / 0 = 2 * 0^

because there is no such number 0^. So this is "indeterminate" in the
same way that "5 + blue" is indeterminate. The answer may, in fact,
exist, but whatever it is, it's *not* a real number.

I realize that this will probably not help younger students understand
the situation, because they won't have considered *why* addition or
multiplication (never mind subtraction and division) exist in the
first place. Still, maybe it will help you think about what's
happening here.


Kevin Santosuosso
Bnei Akiva Schools
Toronto, ON, Canada

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