Re: Is zero even or odd?
From: John Fields (jfields_at_austininstruments.com)
Date: 12/27/04
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Date: Mon, 27 Dec 2004 08:21:25 -0600
On Mon, 27 Dec 2004 02:23:11 +0100, Michael Mendelsohn
<invalid@msgid.michael.mendelsohn.de> wrote:
>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?
---
>> >Presumbly that branch of the circuit had a box marked "resistor" in it,
>> >which should have contained a 100 Ohm resistor, so measured as if that
>> >had been there, but when I opened the box, I found it empty.
>>
>> ---
>> But, with the short in place, as you've shown it, it would be
>> impossible to determine whether the resistor was there or not, let
>> alone determining its value.
>
>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.
---
>> >We're talking about hypothetical boxes and hypothetical resistors here,
>> >because I am trying to model a 0/0 quotient using an electrical circuit
>> >in the attempt show to Nicholas O. Lindan ("Nick" for short) that simply
>> >assuming that this quotient is 1 is a little reckless.
>
>> >> >> you would have measured the entire supply voltage minus what was being
>> >> >> dropped across the load by the current flowing through the meter and
>> >> >> you would have concluded that by subtracting the meter current that
>> >> >> you would have had:
>> >> >>
>> >> >> E E
>> >> >> R = --- = --- = oo
>> >> >> I 0
>> >> >>
>> >> >> Which would have been right!
>> >> >
>> >> >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.
>>
>> ---
>> I don't see why.
>
>
>Because E / 0 = oo and 0 / I = 0, in the case of E=I=0, both equations
>apply, and thus both results should be valid; but oo is not equal to 0.
---
When using Ohm's law:
E
R = ---
I
The assumption is made that in order for resistance to be measured, a
voltage and a current must exist. Implicit in that assumption is that
the voltage must be applied across, and the current forced to flow
through, the resistance.
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.
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
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
plugging that current into (1) gives us
0.5V
R = ------ = 1 ohm
0.5A
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
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
---
>> 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.
---
>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.
---
>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.
---
No. No matter how you arrive at zero being both the numerator and the
denominator, once it is the quotient _has_ to be 1.
---
>Only if you know that 0/0 came from x/x, you can say that it should be
>one;
---
Again, no. Where the zeroes came from is immaterial. If they're the
only terms or the only terms left standing in the numerator and the
denominator, the quiotient _must_ be 1.
---
>but if you know as much, why not eliminate the quotient when you
>initially see x/x?
---
That's not the point of the exercise. :)
---
>If you fear the division by zero, write code that does not divide by a
>variable. ;)
---
Right... Write code that multiplies by the reciprocal of the
variable instead?^)
--
John Fields
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