Is zero even or odd?

[vonroach]

The Macintosh calculator returns 1. So do most Hewlett-Packard calculators
that I have tried, and at least one by Texas Instruments that I can recall.
Likewise Maple and MATLAB (but not Mathematica).

You rely on a computer? How slipshod.

 

[vonroach]

I take it you haven't been reading the thread, since I have given several
mathematical arguments and provided references to support the conclusion
that 0^0 = 1.

None of which have been accepted.

 

[David Kastrup]

None of which have been accepted.

So you think the polynomial 2 x^2 + 3 x^1 + 4 x^0 is undefined at x=0 ?

 

[Dave Seaman]

I read in sci.electronics.design that Dave Seaman <[email protected]>

But for me that is not the issue. I do not LIE on newsgroups - there is
no point - but I do make mistakes; everyone does. I suggest that your
accusation of 'lying' was, and is, unjustified, whether my post was
correct or not correct.

It is no more unjustified than your false claim that I "just want to win
the argument, come what may." I do not base my facts on my opinions, and
I strongly resent any implication to the contrary. I have endeavored to
stick to the mathematics in this discussion.

It seems that there have been interpretation issues on both sides, and
each of us has reacted by drawing unjustified conclusions. For my part,
I am sorry that I spoke in haste. However, I hope you will recognize
that you are not without blame.

 

[Dave Seaman]

No you can't go that route either.

0 factorial is an empty product, and therefore equal to 1.
0^0 is an empty product, and therefore equal to 1.
The same argument also shows that empty sums are equal to 0.

 

[Dave Seaman]

On Tue, 28 Dec 2004 09:51:12 -0800, "Alfred Z. Newmane"

(-1^1/2)^0 =?
or ( i )^0 =?

i^0 = exp(0*log(i)) = exp(0) = 1,

where log(i) is multivalued (= i*pi/2 + 2*n*pi*i), but it doesn't
particularly matter which value you choose. And (-i)^0 = 1 by similar
analysis.

 

[John Fields]

No, he's making it up as he goes along. The limit called for is a very
small number greater than 1/0 which is meaningless and undefined.
What problem called for this math abstraction? There is a practical
limit on how small `things' can get - beyond that limit,, one is just
blowing smoke through one's butthole. These are the uncertainties that
we live with.

---

1/0 is a small number?

If

1
y = lim x->0 ---
x


Then it seems to me like y gets pretty big when x gets pretty small!

 

[John Fields]

Sorry Johnny if you are going to indulge in abstractions, you have to
observe the rules.

 

[Michael Mendelsohn]

Ah, but the normal order of precedence dictates that the
multiplication be performed first, so my method first reduces k*0 to 0
by virtue of the multiplication, then the division by zero is
performed to yield the ratio of 1.

(k*0) -> 0
------ --- = 1
0 -> 0

Interestingly, your method also requires the quotient of 0/0 to be 1,
otherwise the multiplication by k wouldn't yield k as the product!

So (k*0)/0 is not equal to k*(0/0), then?
What a pity!
To defend your point, you have to abandon the associative law of
multiplication.

Cheers
Michael

 

[Matthew Russotto]

Wrong. That limit cannot exist because 0^x is undefined for all x < 0.

Doesn't matter; you can find a delta for every epsilon.

I consider it to be something more than a mere convention. In Suppes:
_Axiomatic Set Theory_, it's a *theorem* that m^0 = 1 for every cardinal
m. Since 0 is a cardinal, the corollary is that 0^0 = 1. Specifically,
it represents the cardinality of the set of mappings from the empty set
to itself.

A corollary is the very antithesis of a "convention."

If there was a proof, one which doesn't depend on a contrived meaning
for exponentiation, I'd agree with you. But then, there wouldn't have
been such a controversy if there was such a proof.

 

[Dave Seaman]

On Tue, 28 Dec 2004 18:21:18 +0000 (UTC), Dave Seaman

You rely on a computer? How slipshod.

I rely on the mathematical arguments I have previously posted with references
to recognized authoritative authors (Suppes, Lang, and the sci.math FAQ).

 

[Dave Seaman]

Doesn't matter; you can find a delta for every epsilon.

Not true. In fact, you cannot find a single delta that works for *any*
epsilon > 0. I'll let you choose any epsilon you like. You still lose.

If there was a proof, one which doesn't depend on a contrived meaning
for exponentiation, I'd agree with you. But then, there wouldn't have
been such a controversy if there was such a proof.

This is not a contrived meaning of exponentiation. It's the standard
definition for a^b where a and b are cardinal numbers, and this is a
necessary first step before extending the definition to more general
situations.

Consider 2^3, for example. It's the cardinality of the set of mappings
from the set { 0, 1, 2 } to the set { 0, 1 }, which is 8.

Besides, there is also the definition from algebra, in which x^n is
defined whenever x is a member of a monoid M and n is a natural number.
In particular, x^0 = e, the identity in M.

 

[George Cox]

1/0 is a small number?

If

1
y = lim x->0 ---
x

Then it seems to me like y gets pretty big when x gets pretty small!

You're confused. If y = 1/x then y gets pretty big when x gets pretty
small.


1
But if y = lim x->0 --- then y doesn't exist.
x

 

[John Fields]

So (k*0)/0 is not equal to k*(0/0), then?

---
Obviously, but no matter.

Consider:

If my point is that 0/0 = 1, and if k*(0/0)=k then 0/0 _must_ be
equal to 1, otherwise k would not be equal to k. That, I believe,
proves my point, my point being:

x
y = lim x->0 --- = 1
x
---

 

[John Fields]

You're confused.

 

[Michael Mendelsohn]

---
Obviously, but no matter.

Consider:

If my point is that 0/0 = 1, and if k*(0/0)=k then 0/0 _must_ be
equal to 1, otherwise k would not be equal to k.

No, that is circular reasoning:
"If 0/0 = 1 the 0/0 must be equal to 1".

If dividing by zero yields no result (i.e. it is undefined), then both
(k*0)/0 and k*(0/0) produce the same, i.e. they're both undefined.

That, I believe,
proves my point, my point being:

x
y = lim x->0 --- = 1
x

But they're not equal any more,
and (a*b)/c = a*(b/c) by the associative law.

I am trying to show to you that _assuming_ 0/0=1 (to be 100% clear,
assuming that 0/0=1 is _always_ true) leads to you having to abandon the
generality of that law.

If by associative law, you find that (k*0)/0=k*(0/0), that proves that
0/0 can't be 1,
because the equation simplifies to 1=k.

So it's either that law or 0/0=1, but not both.

Cheers
Michael

 

[George Cox]

Then what would be the proper way to write it, please?

Write what? That 1/x gets big as x gets small?

1/x -> +infinity as x -> +0

1/x -> -infinity as x -> -0.

 

[George Cox]

If my point is that 0/0 = 1, and if k*(0/0)=k then 0/0 _must_ be
equal to 1, otherwise k would not be equal to k. That, I believe,
proves my point, my point being:

x
y = lim x->0 --- = 1
x

This is true, but it is true because

x
y = lim x->0 --- = lim x->0 1 = 1.
x

The "1" on the left of the equals sign is there because the two x's
cancelled.

It is not always the case that

lim x -> A f(x) = f(A).

So the true statement

x
y = lim x->0 --- = 1
x

does not allow you to conclude that 0/0 = 1.

 

[keith]

Did those calc courses teach you that there is a +infinity and a
-infinity?

Of course. We had no issues with discontinuities either.

Correct. And I used to teach calculus around 30 years ago.

That's about when I took it. (actually 33 years).

But there is a way to say the limit exists. You have to work with the
one-point compactification of the reals instead of the two-point
compactification that is commonly taught in calculus courses. That means
you have just a single infinity, instead of a +infinity and a -infinity.

Which makes no practical sense. Engineers like the practical, even though
it's infinite. ;-)

 

[John Fields]

This is true, but it is true because

x
y = lim x->0 --- = lim x->0 1 = 1.
x

The "1" on the left of the equals sign is there because the two x's
cancelled.

It is not always the case that

lim x -> A f(x) = f(A).

So the true statement

x
y = lim x->0 --- = 1
x

does not allow you to conclude that 0/0 = 1.

But _does_ allow me to conclude that any other number, no matter how
small, and no matter what its sign will, when divided by itself, give
a quotient of 1?

OK, look at it this way:

These two number lines are behind a screen, so the numbers can't be
seen, but arranged in such a way that when a number is picked from
one, the same number will be picked from the other and those two
numbers will be divided and the quotient presented for viewing.

-1 0 1
-|-------------------------|-------------------------|-


-1 0 1
-|-------------------------|-------------------------|-

I fail to see why, at one infinitesimal spot on the line(s), the
division becomes intractable and the machine refuses to output a 1.