Is zero even or odd?

[vonroach]

On Fri, 24 Dec 2004 19:46:25 GMT, "Nicholas O. Lindan" <see@sig.com>
wrote:

0^0 is a mess in either system.

A meaningless mess. As Pauli might say `it isn't even wrong'. And
Heisenberg would add a `pile of crap'.

 

[Alfred Z. Newmane]

vonroach wrote:

On Fri, 24 Dec 2004 07:58:30 -0800, "Alfred Z. Newmane"
a.newmane.remove@eastcoastcz.com> wrote:

(Was refering to infinity (oo))

Alfy - another ambiguous inquiry. An abstract symbol useful in
several concepts.

Ambiguous? Yeah, if you take one part of what I said out of the context
it was in, leaving the rest completely out, then sure, it can be seen as
ambiguous. In fact, that could apply to just about anything, once taken
out of context.

But thanks for playing.

You aren't fooling anyone here.

 

[Alfred Z. Newmane]

vonroach wrote:

On Fri, 24 Dec 2004 07:57:21 -0800, "Alfred Z. Newmane"
a.newmane.remove@eastcoastcz.com> wrote:

j [sqrt(-1)] isn't a number but we still mix it up with numbers.

Of course it is a number, thats why we treat it as such.

Ok what it's value then?
.
square root of -1 is the number that yields -1 when squared. Another
useful abstract concept.
It is termed `imaginary' while most other numbers are just `abstract'.
They all come from the same place.

You replied to my correction post, so you know I wasn't refering to
that.

 

[George Cox]

vonroach wrote:

... most other numbers are just `abstract'.

Only "most"? I thought that all numbers were abstract.

 

[Michael Mendelsohn]

"Nicholas O. Lindan" schrieb:

"Shawn Corey" <shawn.corey@sympatico.ca> wrote
a = b
[..]
2 = 1

Now if we can just get the IRS to agree.

Posted on alt.humor.best-of-usenet:
news:ahbou=yFKxd.5373$Z47.3517@newsread2.news.atl.earthlink.net

Cheers
Michael
--
Still an attentive ear he lent Her speech hath caused this pain
But could not fathom what she meant Easier I count it to explain
She was not deep, nor eloquent. The jargon of the howling main
-- from Lewis Carroll: The Three Usenet Trolls

 

[Mike Kent]

John Fields wrote:
....

x
y = ---
x

we can see that for any value of x, as x goes to zero, y will remain
constant, and exactly equal to 1. Therefore,


0
--- = 1
0

5000 x
Likewise, if we consider y = --------
x

"we can see that for any value of x, as x goes to zero, y will
remain constant, and exactly equal to" 5000.

"Therefore,"

0
--- = 5000
0

and hence 1 = 5000.

 

[George Dishman]

"John Fields" <jfields@austininstruments.com> wrote in message
news:1pl0t01nkne3vn771q23upffooddlo5qvv@4ax.com...

On Mon, 27 Dec 2004 16:51:57 -0000, "George Dishman"
george@briar.demon.co.uk> wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:h8c0t0dvf5fq2ig0uqgaihf89ovrikmsn2@4ax.com...
On Mon, 27 Dec 2004 15:47:32 -0000, "George Dishman"
george@briar.demon.co.uk> wrote:

<snip>

... I'm starting to
think that you're not so much interested in considering 0/0 as you are
in being argumentative and confrontational. If that's the case, then
this 'discussion' is over.

OK, the point on the circuits was an aside anyway so
let's drop it.

<more snipped>

Yes, and that's why I wouldn't ordinarily use Ohm's law to try to
prove that 0/0 = 1.

Neither would I, but I'm not the OP.

---
Which is supposed to mean what, exactly?

Simply agreeing with your view.

In this case, however, to indicate to the OP that
when the current through the resistor and the voltage across it are
both numerically equal, the value of the resistance will be one ohm
and will remain one ohm as the voltage across the resistance goes to
zero.

Sure, but as has been pointed out, if you start
by assuming any other value of resistance, the
numerical value of the ratio also remains constant
at that other value. You have simply illustrated a
particular example, not proven any general rule.

---
Yes, that's what I was trying to do. (Illustrate a particular example,
that is.) By writing:

"In this case, however, to indicate to the OP that
when the current through the resistor and the voltage across it are
both numerically equal, the value of the resistance will be one ohm
and will remain one ohm as the voltage across the resistance goes to
zero."

I thought that I was making it clear that I was talking about a
particular instance by writing "In this case"... and "when the current
through the resistor and the voltage across it are both numerically
equal"... Both rather limiting.

Sorry if I confused you.
---

I think if you look at Mike Kent's reply I'm not the
only one who mistook your motive. It appeared to me
that you were trying to prove that 0/0=1. Going back
to the post to which I first replied, you said in
reply to Michael Mendelsohn:

"John Fields" <jfields@austininstruments.com> wrote in message
news:2dvvs0ppsmtl1tgcpv4rp4fd8a5cko8bak@4ax.com...

On Mon, 27 Dec 2004 02:23:11 +0100, Michael Mendelsohn
invalid@msgid.michael.mendelsohn.de> wrote:

Indeed, you got my point. You cannot determine if 0/0 should have a
finite value, and even if, you can't determine what its value should be.

---
Yes, you can. I described how below.

A little later you seemed to repeat your claim:

Now, if we go to the more general case of:

x
y = ---
x

we can see that for any value of x, as x goes to zero, y will remain
constant, and exactly equal to 1. Therefore,


0
--- = 1
0

Are you now clarifying that you only used the value
of 1 as an example and 0/0 could equally well have
other values?

Equally, you could ask what current flowing through a
superconductor produces zero volts and again the
answer is any current. Hence 0/0 can have any
value and it is therefore undefined.

---
See John Woodgate's reply.
---

I did. By undefined I mean it does not have a unique
value. I'm an engineer reading this in sci.astro (beats
me why it's here) so perhaps other people are looking at
the question with a more mathematical viewpoint. Anyway,
it seemed to me you were arguing that 0/0 has a specific
value of 1 which is not correct.

The very notion of a superconductor renders Ohm's law unfit to
characterized it,

Not at all, R is defined as V/I and for any current
(below the level that destroys the superconduction
obviously), V=0 hence R=0.

---
Yes, I should have thought through that one a little more critically.
Thanks.
---

so saying that 0/0 can have any value because Ohm's
law doesn't work for superconductors is nonsensical.

I wouldn't have chosen Ohms Law myself but since that
is the topic, you should at least get the arguments
right.

---
Are you _intentionally_ being arrogant, is it something over which you
have little control, or am I misreading your tone?

Perhaps I misread your tone. When you seemed to suggest
my comments on a superconductor as "nonsensical" and my
mention of Ohms Law as "unfit to characterized it", I
answered in kind.

If you can clarify whether or not you were suggesting
that 0/0 has the unique value of 1, perhaps we can get
away from that style.

best regards
George

 

[John Fields]

On Mon, 27 Dec 2004 20:17:28 -0000, "George Dishman"
<george@briar.demon.co.uk> wrote:

If you can clarify whether or not you were suggesting
that 0/0 has the unique value of 1, perhaps we can get
away from that style.

---
Yes. I apologize for the crankiness...

That 0/0 has the unique value of 1 is precisely what I was suggesting,
and I've written it up in

kdt0t0ddceqrf3lg4hgjcl8mf7pbp2kvar@4ax.com

(my last post to Michael Mendelsohn on sci.electronics.design)


--
John Fields

 

[Fred Bloggs]

vonroach wrote:

On Fri, 24 Dec 2004 15:34:06 GMT, Fred Bloggs <nospam@nospam.com
wrote:


was based on an off hand remark by Halmos in his General
Topology, and he was almost certainly referring to this result, the date
is about right.


Humm... off hand remark of about the right date, you say, Well it's
certainly hard to argue with that.

Not real interested in satisfying your standards of rigor. I know your
kind- shoot your mouth off with a bunch of semi-specialized terminology
and a total void when it comes to making use of it.

 

[George Cox]

Fred Bloggs wrote:

Apparently you can't pick up on "0/0" being a symbol- call it @#$%^&*,
which has been shown to be the set of all numbers.

Are you claiming that

0/0 = set of all numbers

? Because if you are you are wrong.

 

[vonroach]

On Mon, 27 Dec 2004 16:21:47 +0000, John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

You just *defined* it. 0/0 = any value.

Yeah?

 

[Matthew Russotto]

In article <t4rss0duo9eho2urcsibtq302e3s3edqkr@4ax.com>,
vonroach <hadrainc@earthlink.net> wrote:

On Fri, 24 Dec 2004 19:46:25 GMT, "Nicholas O. Lindan" <see@sig.com
wrote:

0^0 is a mess in either system.

A meaningless mess. As Pauli might say `it isn't even wrong'. And
Heisenberg would add a `pile of crap'.

Well, 0^0 is a mess. But lim x->0 0^x is well defined.

 

[Dave Seaman]

On Mon, 27 Dec 2004 20:58:35 -0600, Matthew Russotto wrote:

In article <t4rss0duo9eho2urcsibtq302e3s3edqkr@4ax.com>,
vonroach <hadrainc@earthlink.net> wrote:
On Fri, 24 Dec 2004 19:46:25 GMT, "Nicholas O. Lindan" <see@sig.com
wrote:

0^0 is a mess in either system.

A meaningless mess. As Pauli might say `it isn't even wrong'. And
Heisenberg would add a `pile of crap'.

Well, 0^0 is a mess. But lim x->0 0^x is well defined.

No, it isn't. That limit does not exist.
But 0^0 does exist and has nothing to do with limits.


--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>

 

[RP]

George Dishman wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:h8c0t0dvf5fq2ig0uqgaihf89ovrikmsn2@4ax.com...

On Mon, 27 Dec 2004 15:47:32 -0000, "George Dishman"
george@briar.demon.co.uk> wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:2dvvs0ppsmtl1tgcpv4rp4fd8a5cko8bak@4ax.com...

On Mon, 27 Dec 2004 02:23:11 +0100, Michael Mendelsohn
invalid@msgid.michael.mendelsohn.de> wrote:


snip

Your circuit:


+---------------------(V)----+
| |
(-)-----o-------[__R__]---o---(A)----o--------(+)
|____________________________|
the short

contrives to hide the resistance while purporting to use Ohms law to
determine the resistance so, quite clearly, the results obtained will
be nonsensical.

snip second circuit

In the above circuit, as in the previous, the instrumentation is
assumed to be perfect, so there will be no current required to read
the voltage.


If the instrumentation is assumed to be perfect, there
will be no voltage dropped across the ammeter and hence
the first circuit (above) is equally valid.


If that's not satisfactory then a wheatstone bridge can
be used to measure the voltage with no regard given to the impedance
of the voltmeter.


Agreed, it is easier to achieve it in reality that
way, but for the purposes of considering 0/0 that
is academic.


Neither circuit gives both readings accurately.

---
Knowing the meter resistance, the second circuit does.


The point was, knowing the meter resistance,
both circuits do.


---

snip proof that 1=1

If you want to use an Ohms Law example to
understand this, realise that when you try to
calculate 0/0, you are asking "what resistance
will allow zero current to flow when zero voltage
is applied. The answer is any resistance.

---
Yes, and that's why I wouldn't ordinarily use Ohm's law to try to
prove that 0/0 = 1.


Neither would I, but I'm not the OP.


In this case, however, to indicate to the OP that
when the current through the resistor and the voltage across it are
both numerically equal, the value of the resistance will be one ohm
and will remain one ohm as the voltage across the resistance goes to
zero.


Sure, but as has been pointed out, if you start
by assuming any other value of resistance, the
numerical value of the ratio also remains constant
at that other value. You have simply illustrated a
particular example, not proven any general rule.


Equally, you could ask what current flowing through a
superconductor produces zero volts and again the
answer is any current. Hence 0/0 can have any
value and it is therefore undefined.

---
The very notion of a superconductor renders Ohm's law unfit to
characterized it,


Not at all, R is defined as V/I and for any current
(below the level that destroys the superconduction
obviously), V=0 hence R=0.


so saying that 0/0 can have any value because Ohm's
law doesn't work for superconductors is nonsensical.


I wouldn't have chosen Ohms Law myself but since that
is the topic, you should at least get the arguments
right.

George

Though you are both essentially claiming, for the sake of argument,
that the resistor has fixed resistance, and arguing from this premise
that resistance isn't therefore undefined under any conditions, even
under the condition 0/0, this is a fallacy, since the premise is
false. OTOH your conclusion is accidentally correct.

Resistance is not determinate in reality, but varies with current,
some resistors increasing with current and some decreasing with
current. Thus under the condition E=0 and I=0, we must have R=0 orr
R=oo. So the solution to the argument about the value of 0/0 seems to
be 0/0=0 orr 0/0=oo. That is, the resistance isn't (for instance)
10ohms under these conditions, even though it may be 10ohms when 1V is
applied to it.

This is however just an idealistic solution, since in reality it is
also true that E is never precisely zero. If it were, then the
universe will have ended and there would be no discussion to be had

Richard Perry

 

[David Kastrup]

russotto@grace.speakeasy.net (Matthew Russotto) writes:

In article <t4rss0duo9eho2urcsibtq302e3s3edqkr@4ax.com>,
vonroach <hadrainc@earthlink.net> wrote:
On Fri, 24 Dec 2004 19:46:25 GMT, "Nicholas O. Lindan" <see@sig.com
wrote:

0^0 is a mess in either system.

A meaningless mess. As Pauli might say `it isn't even wrong'. And
Heisenberg would add a `pile of crap'.

Well, 0^0 is a mess. But lim x->0 0^x is well defined.

Uh, no. lim x->0+ 0^x is well-defined. But since the left-hand limit
does not even exist, lim x->0 0^x is undefined.

--
David Kastrup, Kriemhildstr. 15, 44793 Bochum

 

[vonroach]

On Mon, 27 Dec 2004 20:58:35 -0600, russotto@grace.speakeasy.net
(Matthew Russotto) wrote:

Well, 0^0 is a mess. But lim x->0 0^x is well defined.

Two entirely different expressions. N^0=1, N can't be 0.
0^n = 0, n can't be 0. limit if n->0 of 0^n is meaningless. 0 is a
`place holder between 1 and -1, it can only participate in addition,
subtraction, and multiplication; and calculus notation where it is
used to identify instantaneous rates.

 

[vonroach]

On Tue, 28 Dec 2004 04:56:23 +0000 (UTC), Dave Seaman
<dseaman@no.such.host> wrote:

No, it isn't. That limit does not exist.
But 0^0 does exist and has nothing to do with limits.

Only in your mind.

 

[vonroach]

On Tue, 28 Dec 2004 10:19:37 -0000, "George Dishman"
<george@briar.demon.co.uk> wrote:

Generally resistance varies with temperature

And a few other conditions. But true, resistance varies with
temperature (and other factors) which open the path to
superconductivity.

 

[Alfred Z. Newmane]

vonroach wrote:

On Tue, 28 Dec 2004 04:56:23 +0000 (UTC), Dave Seaman
dseaman@no.such.host> wrote:

No, it isn't. That limit does not exist.
But 0^0 does exist and has nothing to do with limits.

Only in your mind.

Here you go again.

0^0 is undefined.

 

[Alfred Z. Newmane]

vonroach wrote:

On Mon, 27 Dec 2004 20:58:35 -0600, russotto@grace.speakeasy.net
(Matthew Russotto) wrote:

Well, 0^0 is a mess. But lim x->0 0^x is well defined.

Two entirely different expressions. N^0=1, N can't be 0.
0^n = 0, n can't be 0.

Actually, for 0^n = 0, n has to be greator than 0, as it cannot be
negative too.

the valid DOMAIN for 0^n is anything greator than 0. ) and anything
below (that is, negative) are out of the domain, and therefore, invalid.