Is zero even or odd?

[vonroach]

1/0 is a small number?

No it is meaningless gibberish.

 

[vonroach]

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

Confused is too mild a term. If y = 1/x. then y goes from1 to a very
small fraction as x increases from 1. If x decreases to small
fractions, then y increases. x=0 yields a meaningless result.

 

[vonroach]

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

There isn't one. Division by 0 is meaningless.

 

[vonroach]

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

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

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

Both are meaningless. Just crap piled higher and deeper as in Ph D.

 

[Michele Dondi]

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

Wow! We started from the trollish "Re: Is zero even or odd?" and now
we're arriving to Lie (news)groups...
;-)


Michele

 

[vonroach]

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

yes

 

[vonroach]

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.

x^0 = 1 except when x=0 which doesn't exist.

 

[John Fields]

The point is that you can't cancel them when x may be 0, because
division by 0 is undefined.

---
And there's the rub! My point is that it _should_ be defined, and
defined like this:


0 1
--- = 1, --- = oo,
0 0
---

However, in the lim x->0 case x approaches
0, but never really equals zero, so the cancellation may be performed.

---
I disagree.


If we say

x
y = lim --- = 1
x +oo -> -oo x

Then y will be equal to 1 for every instance of x except when x
crosses over from + to -?

In my view that's preposterous, and the mere parroting of "division by
zero is disallowed because it's undefined" a convenient dodge. No
insult intended.[/QUOTE]

It is desirable for mathematics not to allow two contradictory
statements such as 0/0=1 and 0/0=2 to be true at the same time; and your
resoning allows these statements to be true in some cases (we did this
in the Ohm's law episode, where you never said my reasoning was wrong,
it was merely not the thing you wanted to prove).[/QUOTE]

---
Actually, what I didn't want to do was to get into a long, off-topic
harangue about Ohm's law (Ohm's formula, actually. Ohm's law is an
entirely different thing and is used to determine whether conductors
are 'Ohmic'. That is, if their resistance remains constant when
current through them is made to vary), which requires that for
resistance to be _measured_ a known voltage _must_ be placed across it
and the resulting charge flowing through the resistance determined.
Your example eschewed the _practical_ requirements in order to
contrive a desired outcome and, as such, was irrelevant.

But, proceeding along that tack for a while, maybe we can use our
"exquisite" technology to advantage here by assuming that we have
managed to craft two identical zero ohm resistors (superconducting, if
you like) and that we can force equal numbers of electrons through
each of them in equal amounts of time. Then, since Q = It, the
current flowing in each resistor will be I = Q/t, and the ratio of the
currents will be I1/I2.

Since Q/t will be identical for both resistors, I1/I2 will be 1 for
any current. Now, let's say that we inject fewer and fewer pairs of
electrons per unit time into the 'rig' and that eventually we inject
none. Since we have agreed that as long as I1=I2 then I1/I2 will be
equal to 1, does 0/0 not satisfy that requirement?
---

 

[George Dishman]

Otherwise? Accept your opinion? Crazy George is really crazy!

I'll take that as a "No" then.

George

 

[George Dishman]

On Tue, 28 Dec 2004 22:13:54 -0000, "George Dishman"


So far you are assuming the result you are
trying to prove, that the ratio of numerator
and denominator is to be 1.


I like that however while the statement 0=0
is obviously true, it isn't exhaustive. We
also have

0 = k * 0

hence by your method

0/0 = limit as x->0 of (k*x)/x

hence

0/0 = k for all constant k

You still fail to prove the claim of
uniqueness for the value of 1.

[/QUOTE]

That is incorrect, multiplication and division
are of equal precedence.

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!

Not at all. I have avoided writing this as a limit
for the reason Mati gives but will follow your lead

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

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

Let

k*x
y(x) = ---
x

Clearly y(x) = k for all non-zero finite values of x and

limit y(x)
x->0 ---- = k
x

By your argument that the limit can be used to
define the value at zero, we can infer:

y(0)
---- = k
0

but of course y(0) = 0 hence

0
--- = k
0

for all finite k.

George

 

[David Kastrup]

---
And there's the rub! My point is that it _should_ be defined, and
defined like this:


0 1
--- = 1, --- = oo,
0 0

And that's nonsensical.

It gives us, for example,
1 = 0/0 = (0+0)/0 = (0/0) + (0/0) = 2

Well, so you are talking inconsistent nonsense. You must not expect
that it will impress anybody much.

If we say

x
y = lim --- = 1
x +oo -> -oo x

This notation is complete and utter meaningless hogwash. What you
_can_ say is that

lim (x->v) (x/x) = 1

for _all_ v in R (and actually also if v is +oo or -oo, which is not a
number, but a neighborhood in some sense of the word).

But you can equally well say that

lim (x->v) (2*x/x) = 2

for _all_ v in R. For v=0, both limits are of the _form_ 0/0, and
that means that any such limiting _form_ is undefined without further
qualification. But you don't need limiting arguments to show that 0/0
can't be defined: mere algebra is sufficient.

Then y will be equal to 1 for every instance of x except when x
crosses over from + to -?

Your notation above is rubbish to start with, so assigning any meaning
to it is likely to be rubbish too.

In my view that's preposterous, and the mere parroting of "division
by zero is disallowed because it's undefined" a convenient dodge.

Well, not being able to choose a consistent value is _the_ perfect
reason for leaving it undefined. You call it a dodge, other's call it
a necessity.

 

[John Woodgate]

I read in sci.electronics.design that John Fields <jfields@austininstrum

I don't have a problem with either, but it seems to be skirting that
taboo point where x = 0.

I think this thread has shown that there are two sorts of people, those
that associate taboos with 0 and those that don't.

My view is that logically-inferred values of expressions involving zero
should be accepted unless they result in contradictions. I do not
support a priori restrictions on interpretation, and I do not accept
that 'undefined' is a valid show-stopper.

 

[John Woodgate]

I read in sci.electronics.design that George Dishman

) about 'Is zero even or odd?', on Wed, 29 Dec 2004:

0
--- = k
0

for all finite k.

Yes. John Fields' value of 1 is just ONE valid solution. Not wrong, but
not the whole solution.

 

[John Woodgate]

'Undefined' is a human artefact. Anything can be defined; if the
definition results in contradictions, it's a bad definition.

And that's nonsensical.

It gives us, for example,
1 = 0/0 = (0+0)/0 = (0/0) + (0/0) = 2

0/0 can't be *defined* as 1 because, as has been repeatedly
demonstrated, that definition results in contradictions, such as the one
illustrated above.

The only definition that does not result in contradictions is that 0/0
is 'any number'. 1 is just one solution.

 

[John Fields]

You eventually lower the voltage to 0V.
That's what I achieved with the short.

The proper circuit:

+---(V)---+
| |
(-)---o---[R]---o---(A)---o---(+)

Will yield the proper results if examined using Ohm's law.

Assuming that the voltage across the resistance is 1V and the current
through it is 1A, then the resistance will be:

E 1V
R = --- = ---- = 1 ohm (1)
I 1A


Assuming that the voltage across the resistance is 2V and the current
through it is 1A, then the resistance will be: 2 ohm.

I am trying to illustrate that I can make a point that 0/0=2.
You cannot discard my point without adding extra information about your
set of conditions.

---
I thought that by interspersing your remarks among mine you were being
arbitrary. I see now that wasn't the case; thanks.
---

This extra information is not present in the 0/0 term, but it _is_
explicitly written in lim x->0 x/x and lim x->0 2x/x , respectively.



You want 1 ohm, that's what you bring into the computation. You're
setting everything up so that 1 ohm results, which means it's circular
reasoning.

---
I disagree. All I was trying to do was to set up the initial
conditions to give me an equal numerator and denominator as a starting
point for a proof that, eventually, 0/0 = 1.
---

If you hadn't set everything up that way, the 0V/0A measurement would
leave you stumped as to the value of the resistor, and 0/0=??? then.

---
0/0 would still be = 1 but yes, of course, to the rest of it. A
_measurement_ would be impossible without a known, non-zero voltage
forcing charge through the resistance.
---

Again, the set of values that are divided by itself is x/x for all x in
R\{0}, and lim x->0 x/x = 1 as well.



You are doing the limit argument nicely.
Now you arrived at x=1 not by dividing by 0 outright, but by taking
larger values and going closer to 0. I would explicitly write what you
did, so your last formula would change to
r
x = lim --- = 1
r->0 r

This is mathematical shorthand for your reasoning that any nonzero value
plugged into r/r is 1, so it makes sense to take r/r=1 for r=0 as well.

---
Yes, thanks.
---

You misunderstood my point. My point is that you have to arrive at x/x
or r/r before you plug in the zero. Conmsider: if you had x=2r/r, put in
r=0 to get x=2*0/0 and use 2*0=0 to simplify to x=0/0, then the
numerator is algebraically 0=2r and the denominator is algebraically
0=r, and then
2r
x = lim ---- = 2
r->0 r
which would lead you to conclude that 0/0=2, in this case. You set your
case up so that x=r/r, i.e. numerator and denominator are algebraically
the same *before* you plug in the zero.

---
OK, but it seems to me that if you subscribe to

2r
x = lim ---- = 2
r->0 r

when r = 0, then 0/0 _must_ be equal to 1, otherwise x could not have
been equal to 2. That is, if r/r = y and 2y = 2, then y = 1.
---

This is not true for R=E/I, which is why you can't make a measurement in
the E=0, I=0 case (unless you have extra information).


You seem to grasp that so well. You can certainly measure 0V and 0A, but
you can't determine R. For what reason do not grasp the analogy that 0/0
represents the same condition in mathematics?

---
Because it doesn't?

In the latter case we're trying to determine whether, when equal
quantities are divided into each other, the result will always be
equal to 1, while in the former we're dealing with with the interplay
between physically different entities. I.e., 1 apple/1 orange = ???

 

[Matthew Russotto]

I read in sci.electronics.design that John Fields <jfields@austininstrum


I think this thread has shown that there are two sorts of people, those
that associate taboos with 0 and those that don't.

My view is that logically-inferred values of expressions involving zero
should be accepted unless they result in contradictions. I do not
support a priori restrictions on interpretation, and I do not accept
that 'undefined' is a valid show-stopper.

So what do you do when there are multiple logically-inferred values,
as with 0^0? Or when the logically-inferred value breaks a lot of
other things, as with 0/0.

BTW, what's log 0? arctan 0?

 

[John Fields]

Sigh. Forget machines. Is the definition of division clear to you?
the result of the division a/b is such a number c that c*b = a.
That's *all* there is to it. So, the result of 0/0 is a number c such
that c*0 = 0. Can you tell me what c is?

OK, here is a little exercise for you. At x = 0, sin(x) is also zero.
And, so is any positive power of either x or sin(x). So, using the
Taylor expansion for sin(x), evaluate the following limits:

1) lim_x->0 sin(x)/x
2) lim_x->0 sin(x)/x^2
3) lim_x->0 (sin(x))^2/x

---
In the context of the problem at hand, the definition of division
isn't in question. What's being discussed is whether zero, being
equal to itself will yield a quotient of 1 when it's divided into
itself. As for the rest of it, I will follow the advice you proffer
in your dotsig and forego arguing with you.

 

[John Fields]

Otherwise? Accept your opinion? Crazy George is really crazy!

 

[John Fields]

Chuckle....0/0 is meaningless and you are wrong and bordering on
lunacy.

 

[George Cox]

<snip> hence

0
--- = k
0

for all finite k.

No 0/0 is undefined, it has _no_ value.