Is zero even or odd?

[John Fields]

Well, I did - do you think I'm stupid?

---
Not so far
---

The problem is, the measurement was automatic, and since there was
short-circuit somewhere (presumably in parallel to the supposed
resistor), the voltage was zero across the measurement points even
before the resistor was stolen. The current is of course measured the
proper way.

---
I'm not trying to be insulting, but would you mind explaining how the
current was measured?
---

The software computed resistance by Nick's rules and hence never noticed
anything unusual.

---
Nick's rules?

I'm in the dark about that. Clue me in?
---

Unless E=0 too, in which case the result is 1 (says Nick).

On a short circuit you can detect no voltage, but you can measure a
current.

E 0
R = --- = --- = 0
I I

This leads to a contradiction when E=I=0.

---
So it would seem, but a short across the resistor would still have
resulted in a voltage drop across the resistor equal to the parallel
resistance of the resistor and the short multiplied by the current
through that parallel resistance. Since there's no such thing as a
perfect short and the resistance of the resistor was known beforehand,
the "short" and the anomalous current (which you said you measured)
through it should have pointed to either the resistor failing shorted,
an external short developing across the resistor, or some dirty
bastard stealing the resistor and putting a short across where it was,
no?

How much was that resistor worth, anyway?

 

[Dave Seaman]

Actually, I think the physicists think its just a bit annoying, its the
mathematicians that think its the ugly bodge.

No, not at all. It's not a function in the ordinary sense, but a
generalized function. It's a linear functional defined on a certain
function space.

Or a failure of the mathematics.

Definitely not.

 

[robert j. kolker]

No, not at all. It's not a function in the ordinary sense, but a
generalized function. It's a linear functional defined on a certain
function space.

Here is an analogy. A function is like a vector. A delta "function" is
like a one form.

Bob Kolker

 

[Matthew Russotto]

I read in sci.electronics.design that Matthew Russotto


Is it? Does the limit of its differential differ as x->0+ and as x->0-?
If not, it's 'squeezed'.

I'm not sure what you mean by 'squeezed'; it's piecewise continuous.

 

[John Woodgate]

I read in sci.electronics.design that Matthew Russotto

I'm not sure what you mean by 'squeezed'; it's piecewise continuous.

Squeezed: if the limits are equal, the value of the function at the
limit point cannot differ from the limit value.

 

[Nicholas O. Lindan]

When checking it turned out that some thief had actually stolen the
resistor where 0V,0A was measured. The circuit was broken, but noone
noticed because the voltage was zero.

The circuit wasn't connected. Therefore no measurement was being
made. V = IR has no relevance. R < oo to close the circuit and
for the equation to apply.

 

[Nicholas O. Lindan]

Well, I did -

OK. Can you please produce said resistor: the one that isn't there but
that the meter is connected to an measuring the voltage of.

do you think I'm stupid?

Take a guess. Is as does.

[Words]

No more words.

 

[Kevin Aylward]

No, not at all.
Nope.

It's not a function in the ordinary sense, but a
generalized function.

Ho hum. What isn't a function? We are discussing QED.

Go back up and read what was wrote. It started with "Another is
renormalization theory in QED..." Note the lack of mention of "Dirac
function"

It's a linear functional defined on a certain
function space.

What is? QED?

<http://mathworld.wolfram.com/DeltaFunction.html>.

Oh... you mean the Dirac function. Well I know all about this, but you
are off on a tangent.

My "not at all" is regarding the infinities in QED. The Dirac function
was not part of the discussion at this point.

Definitely not.

Definitely is. The infinities in QED are unresolved.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Nicholas O. Lindan]

Referencing:

<http://mathworld.wolfram.com/ContinuumHypothesis.html>

It's a widespread belief (and one that is unfortunately perpetuated by some
popular expositions) that the cardinality of the reals is aleph_1. Not so.
The cardinality of the reals is 2^aleph_0, which is the same as
aleph_0^aleph_0. This cardinal is called c, for the cardinality of the
continuum. The proposition that c = aleph_1 is called the continuum
hypothesis, and it is known to be independent of the usual axioms of set
theory.

The latter part of the paragraph seems to support the view that
c = continuum = cardinality of the reals = aleph-0 ^ aleph-0 = aleph^1
which you claim in the first two sentences to be false.

Dazed and confused again.

 

[John Woodgate]

I read in sci.electronics.design that Kevin Aylward

Oh... you mean the Dirac function. Well I know all about this, but you
are off on a tangent.

Does it have a tangent? If so, it is very small and lives inside
capacitors. You can see in the specs, tan[delta] = 0.001, for example.

 

[denis feldmann]

Nicholas O. Lindan a écrit :

Referencing:






The latter part of the paragraph seems to support the view that
c = continuum = cardinality of the reals = aleph-0 ^ aleph-0 = aleph^1
which you claim in the first two sentences to be false.

Dazed and confused again.

Reading trouble? The last part says explcitely that the affirmation that
c =aleph_1 (i.e. the affirmation that the cardinal of IR is the
smallest strictly greater than the cardinal f IN) is not provable (or
refutable) with the usual axioms of set theory. What confuse you , there?

 

[vonroach]

This form of Euler's equation is particularly timely, because this is
the season to get pie-eyed from too many wine-gums and square eyes from
watching too much TV. (;-)

Your reference to Euler's blindness shows your usual poor taste. His
equations were more beautiful than anything produced in UK, including
the amusing wrong guesses of Newton. Wrong on nature of light. Wrong
in imaginary description of `force', `mass', and `space/time'. His
crowning clueless guess was on nature of `gravity'. Almost as
pathetic as Einstein who `borrowed' his first wife's work and spent
the rest of his life struggling to come up with something that would
top it.

 

[vonroach]

I'm lost.

Then you have come to the right door! Come in and join the others.

 

[vonroach]

Although it's called aleph-one, no-one knows whether it is
the *next* infinity after aleph-null, or whether there are other
infinities in between.

I know, but I'm not about to share that information with nitwits who
post here.

 

[vonroach]

the question is, is 2^{aleph_0} the next infinity after aleph_0? (And
generally, is 2^{aleph_{alpha}} the next infinity after aleph_{alpha}?)

No, to be or not to be is the question.

 

[vonroach]

The latter part of the paragraph seems to support the view that
c = continuum = cardinality of the reals = aleph-0 ^ aleph-0 = aleph^1
which you claim in the first two sentences to be false.

Dazed and confused again.

That is the usual outcome when one follows the abstract creations of
the mind such as mathematics too far. A sip of reality may sober you
again.

 

[Fred Bloggs]

(2 x0)/0 = 2x(0/0) . there now is that better?

You must be an idiot- we have just finished telling you that 0/0 is not
a number- it is a set. You and that other idiot are merely saying that
if it's a number then it must be a set. Why don't you try demonstrating
some intelligence by showing how the assumption of it must be a set
leads to the conclusion that it must be a number? You won't find one
with your mindless geek symbol manipulation, "nitwit".

 

[vonroach]

If in measuring a resistor, we find 0.0A at 0.0V, is the resistance 1
Ohm, then?

Er...how many resistances have you really measured? Did you read the
instructions carefully?

 

[vonroach]

Value of the resistor = 1 new ohm.

Illustration of the ability of the abstract brain to cope with
foolishness.

 

[Fred Bloggs]

Ah, the inverse , like 1/0 is inverse of 0/1? Is 0/0 the inverse of
0/0? And 1/1, the inverse of 1/1.

Inverse in the sense of function preimage, sherlock, and that is a set.
This so-called division operator is really an association of singleton
preimage sets with the number they contain. You can go ahead and make it
an operator if you want, but then you must exclude those Cartesian pairs
with 0 in the denominator- so that "undefined" literally makes sense now.