Re: Is zero even or odd?
From: Michael Mendelsohn (invalid_at_msgid.michael.mendelsohn.de)
Date: 12/29/04
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Date: Wed, 29 Dec 2004 13:58:54 +0100
John Fields schrieb:
> On Mon, 27 Dec 2004 20:58:44 +0100, Michael Mendelsohn
> <invalid@msgid.michael.mendelsohn.de> wrote:
>
> >Below, you remove the short from my diagram.
> >However, you also remove the power supply, which achieves the same
> >thing.
>
> ---
> I don't know what you mean, since the + and - terminals are there and
> I refer to the voltage across the resistance as being 1V.
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.
>
> ---
> Why would I want to do that? I'm specifically setting up a set of
> conditions to illustrate _my_ point, not yours.
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.
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.
> >> 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.
>
> ---
> Yes, of course. But I didn't "assume" a resistance of one ohm, I
> selected the voltage and current to force the resistance to one ohm.
> ---
>
> >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.
>
> ---
> The game being to prove that 0/0 = 1, I'm not looking so much for a
> resistance of 1 ohm as I am a set of values which when divided by
> themselves will result in a quotient of 1.
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.
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.
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.
> Let's make them each equal to 1E-40:
>
> 1E-40
> x = ------- = 1
> 1E-40
>
> Damn! That x is still equal to 1!
>
> It seems that no matter what we do, as long as the numerator and
> denominator are equal, the quotient will always be 1. So, if the
> smallest number we can come up with is 0, and if 0 = 0, then it seems
> we can say:
>
> 0
> x = --- = 1
> 0
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.
> >> 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.
>
> ---
> Yeah, good point. they have to have the same sign in order for the
> quotient to come out positive.
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.
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).
> >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?
>
> ---
> Since R = E/I, I won't be able to make a measurement, so the cat will
> be both dead and alive.
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?
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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