# Is zero even or odd?

Discussion in 'Electronic Design' started by Gactimus, Dec 20, 2004.

1. ### vonroachGuest

You rely on a computer? How slipshod.

2. ### vonroachGuest

None of which have been accepted.

3. ### David KastrupGuest

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

4. ### Dave SeamanGuest

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.

5. ### Dave SeamanGuest

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.

6. ### Dave SeamanGuest

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.

7. ### John FieldsGuest

---

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!

9. ### Michael MendelsohnGuest

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

10. ### Matthew RussottoGuest

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

11. ### Dave SeamanGuest

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

12. ### Dave SeamanGuest

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.
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.

13. ### George CoxGuest

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

14. ### John FieldsGuest

---
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
---

16. ### Michael MendelsohnGuest

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.
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

17. ### George CoxGuest

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

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

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

18. ### George CoxGuest

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.

19. ### keithGuest

Of course. We had no issues with discontinuities either.
That's about when I took it. (actually 33 years).
Which makes no practical sense. Engineers like the practical, even though
it's infinite. ;-)

20. ### John FieldsGuest

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.