Re: Is zero even or odd?
From: Keith Ramsay (kramsay_at_aol.com)
Date: 02/07/05
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Date: 7 Feb 2005 00:32:42 -0800
David W. Cantrell wrote:
| ...
| > I think however that it would be little enough of a problem to
| > define 0/0. There are contexts in complex analysis where z/0
| > for z<>0 is defined to be a point on the Riemann sphere called
| > "infinity". In integration theory, sometimes one incorporates
| > a value "infinity" and defines infinity*0 to be 0. Mathematicians
| > are flexible enough to take such tweaks in definitions in stride.
|
| But at least for this mathematician, "flexibility" is not needed
| here. I don't think of them as "tweaks" at all; I don't feel that
| there's anything needing to be taken "in stride".
I was assuming that one has had to deal with mathematics
using both conventions, at some time. You seem to be saying
that defining 0/0 to be 0 is the way that you've always
done it. So you were taught it that way? If so, where were
they teaching this way?
Surely you've discussed mathematics with people who were using
the convention that division by zero is always undefined? At
least on sci.math? Did the issue just never arise? Excuse me
if it seems a little hard to believe. I don't know of anybody
other than you who defines 0/0 to be 0.
If you've had to deal with the convention that division by 0
isn't defined, it goes to show how easy it was for you to
accept such changes of convention in stride.
[...]
| > It's simpler, however, not to have to refer explicitly to the
| > domain. One can say a(b/c)=ab/c wherever either side of the
| > equation is defined.
|
| Not necessarily. (Please see the paragraph below which begins with
| "Not necessarily.")
I don't know which statement of mine you're saying is not
necessarily so.
[...]
| > |As I've noted in this newsgroup on several occasions, division
| > |may reasonably be defined so that 0/0 = 0. That may be done in
| > |the context, say, of arithmetic on the one-point extension of
| > |the reals, R* = R U {oo}. [One could use C* just as well, of
| > |course. And there are other possibilities.] Multiplication is
| > |defined so that 0*oo = oo*0 = 0. Reciprocation is defined so
| > |that /0 = oo and /oo = 0. Division is defined in terms of
| > |reciprocation and multiplication: |a/b = a*(/b). Thus, we
| > |have 0/0 = 0.
| > |
| > |It happens that, in that system, the law you mentioned above can
| > |be extended nicely:
| > |
| > |a(b/c) = ab/c for _all_ a, b, c in R*.
| > |
| > |But not all laws of algebra can be extended in such a system.
| >
| > Yes, I think losing some of the properties of addition would be
| > more than compensating an inconvenience.
|
| It is not necessarily the case that defining division by 0,
| including the case 0/0, causes us to lose any properties of
| addition.
That's not what I said. I was responding to your comment about
that particular system, where one has an infinity.
| For example, consider one of the simpler systems in which
| division by 0 makes sense: the system of nonnegative extended reals,
| [0, +oo], with binary operations of addition, +, and multiplication,
| *, and the unary operation of reciprocation, /. With /0 = +oo,
| /(+oo) = 0, 0*(+oo) = +oo*0 = 0, and division defined by
| a/b = a*(/b), we then have x/0 = +oo for nonzero x and 0/0 = 0.
| Please correct me if I've overlooked something, but I cannot see
| how _any_ properties of addition are lost in that system.
I think usually in a system with an infinity, infinity+1 is
defined to be infinity, which causes cancellation a+c=b+c ->
a=b to be lost. Leaving infinity+1 undefined would mean that
addition is sometimes undefined even when the terms are
defined. This is partly why people don't always just use a
real line extended with an infinity.
| > |> One of the advantages of leaving a function undefined
| > |> at selected points is that one can express such a law by
| > |> saying that it holds when either side of the equation is
| > |> defined.
| > |
| > |That's a slip, surely. You must have meant to say "when _both_
| > |sides of the equation are defined."
| >
| > No, a(b/c)=ab/c among some other identities holds when either
| > side is defined.
|
| Not necessarily. It depends on which system we are using.
Please reread this carefully.
You were in the process of arguing for the general advantage
of defining functions for more values. I was explaining why
it is sometimes advantageous not to do so, and one of the
advantages one sometimes gets is being able to express a
law in such a way-- holding whenever either side is defined--
instead of having to state what the exceptions are.
I didn't say that if one leaves a function undefined at
selected points, one *always* gets a system in which a law
can be expressed this way. That's obviously untrue.
| Suppose we
| modify the system I just described above so that division is now
| defined by a/b = a*(/b) _iff a and b are not both zero_. Thus the
| modified system incorporates the two types of things you'd called
| "tweaks" earlier, but explicitly avoids defining 0/0. In this
| modified system, a(b/c) = ab/c does not hold when either side is
| defined. To see that it does not hold, merely consider a = 0, b = 1,
| c = 0: the left side is 0, the right side is undefined. But the
| identity does hold when _both_ sides of the equation are defined.
This is fine, but not relevant.
| There are well known examples, too, in which it makes sense to say
| that something called an identity holds only when both sides of the
| equation are defined. Let's consider an example you snipped from my
| previous response: cot(x) = 1/tan(x)is normally called an identity.
| That's true even in high-school courses in which there is no thought
| of using an extended number system. But if the high-school students
| think that the equation must hold whenever _either_ side is defined,
| they get an unpleasant surprise. Letting x = pi/2, the identity
| tells them that 0 = 1/undefined. Surely then the convention is that
| something called an identity must hold when _both_ sides are
| defined.
I'm not claiming that the term "identity" is to be defined
to exclude such identities as cot(x)=1/tan(x). But I do
claim being able to state an identity as holding whenever
either side is defined is sometimes better. For one thing,
the weaker sense of identity isn't transitive, but the
stronger one is. We get away with using the weaker sense of
identity in many cases because we're dealing with a type of
function where it happens to be an equivalence relation.
For instance, a lot of our examples have been rational
expressions. These can be regarded either as expressions
to be evaluated (using one of the conventions we've been
discussing) or as elements of C(x1,...,xn), where x1,...,xn
are the variables occurring in the expression. These two
ways of interpreting the expression don't quite match
because of cases like x/x. As an element of C(x), x/x=1.
As functions on the complex numbers, x/x and 1 usually
are taken to define functions with slightly different
domains. If we define 0/0 to be something other than 1,
then we get a function with a different value at x=0.
I think this is another reason why people tend to define
division so 0/0 is undefined, rather than some specific
value. If we take the partial functions on the complex
plane defined by rational expressions in the usual way,
the weak identity (equal where both defined) is an
equivalence relation, and the equivalence classes
correspond to the elements of C(x). They merely fail
always to be defined in for all the values of x for which
it would be natural. If we take 0/0 to be defined and
equal to some value like 0, then the functions we define
on the complex plane are defined everywhere, and the
relation of identity is an equivalence relation, but
the equivalence classes are some ring (which is not a
field, and so on).
| > There's a substantial range of manipulations
| > that one can perform without changing the domain of definition.
| > If 0/0 were some specific value this would be less true.
| >
| > |Let's consider x/x = 1, for example.
| >
| > It's also true that there are manipulations involving
| > cancellations like this, where leaving 0/0 undefined allows
| > them to be described as holding when both sides are defined.
| > If 0/0 were a specific value, one of x/x=1 and 2x/x=2 would
| > have defined and unequal left and right sides at x=0.
|
| One? If 0/0 = 0,
Please read the above again carefully. I wrote, "if 0/0 were
a specific value"-- not, "if 0/0 were 0".
| _both_ x/x=1 and 2x/x=2 would have defined and
| unequal left and right sides at x=0.
Yes. Does this not strike you as a bit of a "gotcha"?
...
| > |I suspect that you're already familiar with some specific context
| > |(such as measure theory) in which it's fairly common to take
| > |0*(some infinite element) to be 0, in order to gain such an
| > |advantage.
| >
| > Yes.
| >
| > |In the extended real system which I described above,
| > |taking 0*oo = 0, we then get 0/0 = 0.
| >
| > I don't see why this is "then".
|
| It's "then" because 0/0 = 0 follows immediately from the definition
| of division in terms of multiplication and reciprocation:
| a/b = a*(/b).
Okay, I missed your intended context. We have various
ingredients here:
0. Defining a/b to be a*(1/b).
1. Defining 1/0 to be infinity.
2. Defining 0*infinity to be 0.
3. Regarding 0/0 to be 0.
(3) follows from (2) if you already have (0-1). I didn't
realize you were taking the premises in this order.
I've seen a system with 0*infinity defined to be 0 (in
measure theory, for instance). I've also seen a system
with 1/0 defined to be infinity (in complex analysis,
where infinity is the extra point on the Riemann sphere).
I've seen a/b defined to be a*(1/b) for purposes of model
theory or something like that, to show that one can get
away with one fewer fundamental operation. (Which now
that I write it, seems very silly, but this is how I
seem to remember it.) I haven't so far as I can recall
until now seen anybody try to combine any of them, let
alone all three of them.
The motivations behind including "infinity" in measure
theory and in complex analysis seem very different. In
measure theory, one has a lot of infinite measure sets
that one wants to deal with smoothly. The integral of 0
over an set of infinite measure is still 0, and the
integral of any function, however large, on a set of
measure 0 is also 0.
In complex analysis, one has functions that tend to
infinity at certain points, but in a nice way neatly
parallel to the way that analytic functions approach zero
at points where they are zero. This seems like a different
notion of "infinity". When Caratheodory defines operations
on "infinity", he explicitly leaves 0*infinity undefined,
and I think it's probably better that he does.
[...]
| Then consider the function g(x) = x^0. With 0^0 taken to be 1, g is
| a constant function and so, of course, we should have g'(x) = 0 for
| all x. But if we differentiate g(x) = x^0 using the power rule, we
| literally get g'(x) = 0 x^(-1), or equivalently, 0/x. If this is to
| give us the correct result, by literal substitution, when x = 0,
| then we need to have 0*oo (or equivalently, 0/0) to be 0. Without
| that, the power rule needs to be modified in order for it to be
| correct, literally, when r = 0, x = 0.
I can see how this saves you a case, but it seems somewhat
like a coincidence. If we differentiate (1-cos x)/x using
the usual quotient rule, we get (x sin x - (1-cos x))/x^2
which at x=0 would be 0/0, but this time the value 0 is
incorrect. You still need to distinguish cases. (I was
about to write "differentiate between cases".)
I also note that this case of the derivative corresponds to
the exceptional case of the antiderivative, so the exceptions
are at least sort of paired up.
Keith Ramsay
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