Re: Is zero even or odd?
From: Michael Mendelsohn (invalid_at_msgid.michael.mendelsohn.de)
Date: 12/27/04
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Date: Mon, 27 Dec 2004 20:58:44 +0100
John Fields schrieb:
> On Mon, 27 Dec 2004 02:23:11 +0100, Michael Mendelsohn
> >John Fields schrieb:
> >> On Sat, 25 Dec 2004 00:23:49 +0100, Michael Mendelsohn
> >> <invalid@msgid.michael.mendelsohn.de> wrote:
> >> >John Fields schrieb:
> >> >> ---
> >> >> I'm not trying to be insulting, but would you mind explaining how the
> >> >> current was measured?
> >> >
> >> >By putting an instrument into the circuit where I wanted to measure the
> >> >current.
> >>
> >> ---
> >> OK, but, unfortunately, placing the ammeter inside the short can only
> >> give ambiguous results.
> >
> >That is my point exactly. A mathematical situation where you get 0/0 is
> >ambiguous. If there is an unambiguous way to resolve it, you should have
> >thought about it before. (See below).
>
> ---
> Was that meant to be insulting?
No, sorry. The "you" should be read as "one", i.e. in the general sense.
I've no reason to insult you.
If there's an unambiguous way to resolve it, you can phrase your
computation so that the term 0/0 doesn't occur, which requires changing
things "upstream" in the computation process (canceling x/x, for
example).
> >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.
Below, you remove the short from my diagram.
However, you also remove the power supply, which achieves the same
thing.
> 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.
> If we now reduce the voltage to 0.5V and rearrange to solve for I,
> we'll now have:
>
> E 0.5V
> I = --- = ------ = 0.5A (2)
> R 1R
We'll then have I = 1V/2R = 0.5 A
> plugging that current into (1) gives us
>
> 0.5V
> R = ------ = 1 ohm
> 0.5A
R = 1V / 0.5 A = 2 ohm
> If we continue to reduce the voltage, the current and voltage will
> always be numerically equal, R will remain at 1 ohm and, clearly, will
> remain at 1 ohm even if we disconnect the voltage supply, forcing both
> the voltmeter and ammeter to read 0, in which case we'll have:
>
> 0V
> R = ---- = 1 ohm
> 0A
If we continue to reduce the voltage, the current will always be
numerically half of the voltage, R will remain at 2 ohm and, clearly,
will remain at 2 ohm even if we disconnect the voltage supply, forcing
both the voltmeter and ammeter to read 0, in which case we'll have:
R = 0V / 0A = 2 ohm
> 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
This is only true because you assumed a resistor of 1 ohm. If you assume
a resistor of 2 ohm, then 0/0 = 2.
Again, you can only state with concidence that 0/0 = 1 in this case
because you already *know* that the resistance is 1; you have not
computed it from 0/0, because the 0/0 quotient doesn't help you to know
that the resistance is 1 ohm.
> ---
> >> Consider: Since
> >>
> >> x
> >> y = --- = 1
> >> x
> >>
> >> is certainly true for x = 1, x = 0.5, x = 0.25, and doesn't seem to
> >> change as x diminishes toward, through and into the negative realm on
> >> the other side of zero, why should there be an anomaly where x = 0?.
> >
> >Because you can't see from 0/0
> >that it is the result of putting x=0 into x/x.
>
> ---
> The value of x is unimportant. What does matter is that the numerator
> and denominator be numerically equal.
No, it matters that they are algebraically equal.
When they are numerically equal, that is only useful if the number's not
zero.
You can cancel out the "x"s, and you should definitely do that if you
can, but you can't cancel out the zeroes.
> >It might have been the result of putting 0 into 2x/x,
> >in which case the result ought to be 2.
>
> ---
> It might have been, but it wasn't.
How do you know?
If I give you a resistor to measure, but no power supply, you will
measure 0A and 0V, but must the resistor always be 1 ohm, then?
> >In fact, any c = cx/x = 0/0 for any c in R with x=0,
> >so 0/0 could be any number c in R.
In other words, your above experiment could have been carried out with
any resistor of c ohm.
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
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