Zero Ohms = Mathematically Incorrect

[John Larkin]

Sure. I don't follow rules, I make rules.

Come to think of it, I've know c

Could be. I don't program in C; it's an abomination.

How wonder how he can speak at all if he doesn't let any definitions get in
the way?

Oh, lighten up. Rigidity is no fun.

Definitions are merely things people agree to agree about. The R in
"ohm's law" is a definition of "resistance" but ohm's law isn't a
physical law because it isn't ever exactly followed, and sometimes
it's not even closely followed. So we use it when it makes sense,
namely when it produces usably close approximations and has some
predictive value. But it ain't a law, and worrying about I=E/R going
infinite is a waste of time.

I believe I noted in a couple of previous posts that some
superconductors demonstrate unmeasurable resistance, specifically no
measurable voltage drop or power loss with a finite circulating
current. I had to leave my wallet on a table this morning so I could
work right up against a superconductive magnet in a 140 GHz EPR
system. The magnet was "charged" many months ago.

John

 

[John Larkin]

No, it is wrong and the correct answer is error, not undefined.

"Error" is what a $4 calculator will tell you that 0/0 is.
"Indeterminate" is what an engineer will call it. "Undefined" is what
a mathematician might say.

The reason is that you can spell ErrOr on a 8-digit, 7-segment
display, but you can't spell UNDEFINED or INDETERMINATE.

John

 

[Bill Bowden]

"Error" is what a $4 calculator will tell you that 0/0 is.
"Indeterminate" is what an engineer will call it. "Undefined" is what
a mathematician might say.

The reason is that you can spell ErrOr on a 8-digit, 7-segment
display, but you can't spell UNDEFINED or INDETERMINATE.

John

0/0 is a "Undefined Result" according to my HP 28S calculator. Anything
divided by zero is infinite, and 0 divided by anythiing is 0, so 0/0
must be something between 0 and infinity. Make it whatever you want.

-Bill

 

[Mark Fortune]

0/0 is a "Undefined Result" according to my HP 28S calculator. Anything
divided by zero is infinite, and 0 divided by anythiing is 0, so 0/0
must be something between 0 and infinity. Make it whatever you want.

-Bill

more precisely, can be _anything_ in between -infinity and infinity
inclusive.

Mark

 

[John Fields]

0/0 is a "Undefined Result" according to my HP 28S calculator. Anything
divided by zero is infinite, and 0 divided by anythiing is 0, so 0/0
must be something between 0 and infinity. Make it whatever you want.

---
ISTM that it can only be 1.

That is, for any fraction where the numerator and the denominator
are identical, the quotient is always 1, so if:

a
--- = b = 1
a

and 'a' slides through zero, why should the value of 'b' be
different from what it was infinitessimally before 'a' became zero?

 

[Greg Hansen]

0/0 is a "Undefined Result" according to my HP 28S calculator. Anything
divided by zero is infinite, and 0 divided by anythiing is 0, so 0/0
must be something between 0 and infinity. Make it whatever you want.

Well, don't do that! sin(x)/x, sin(2x)/x, sin(x)/x^2, and
sin(x)/sqrt(x) are all 0/0 when x=0 but have different, and clearly
defined, limits. Even x^2/x is 0/0 when x=0, but has a limit when you
cancel a set of x's from the numerator and denominator. But the
calculator doesn't know how you're getting to 0/0, that depends on what
you're trying to model, so that's for the user to figure out.

 

[Homer J Simpson]

Oh, lighten up. Rigidity is no fun.

Then why do they advertise Viagra?

 

[John Fields]

Then why do they advertise Viagra?

 

[G. Schindler]

Which is pretty much what I said.... as far as you go. Using limits,
you can often, but not always, rewrite EQUATIONS that result in division
by zero to indicate what the value for that EQUATION would be when the
denominator approaches or (I think) passes through zero. Limits are
taught as part of Calculus.

While, for instance, the statement "zero divided by zero" could have an
infinite number of values it most likely has a finite number of correct
values for the equation that caused the zero divided by zero condition.

You would use limits to find the appropriate value.

Unfortunately, too many of our high school teachers left of in teaching
their students that it is carte blanche "incorrect".

BTW ... Personally, I'd have figured the other way around between an
engineer and a mathematician. But then again, I'm an engineer with a
minor in math.

Greg

 

[G. Schindler]

Exactly right!

Greg (Schindler)

 

[John Larkin]

---
ISTM that it can only be 1.

That is, for any fraction where the numerator and the denominator
are identical, the quotient is always 1, so if:

a
--- = b = 1
a

and 'a' slides through zero, why should the value of 'b' be
different from what it was infinitessimally before 'a' became zero?

but 2a/a is always 2, so if a=0, then 0/0 = 2.

In other words, 0/0 is the limit of an infine number of equations, not
all of which evaluate to 1.


John

 

[John Fields]

but 2a/a is always 2, so if a=0, then 0/0 = 2.

In other words, 0/0 is the limit of an infine number of equations, not
all of which evaluate to 1.

---
They all evaluate to 1 if the numerator and the denominator are
identical. In your example:

2a
---- = 2
a

they are not, except when a = 0, in which case:

2a 2*0 0
---- = ----- = --- = 1
a 0 0

The proper procedure, in your example, should have been to multiply
the numerator _and_ the denominator by the same quantity,:

2a
---- = 1
2a

which would have eliminated the discontinuity going through zero.

 

[John Larkin]

---
They all evaluate to 1 if the numerator and the denominator are
identical. In your example:

2a
---- = 2
a

they are not, except when a = 0, in which case:

2a 2*0 0
---- = ----- = --- = 1
a 0 0

The proper procedure, in your example, should have been to multiply
the numerator _and_ the denominator by the same quantity,:

2a
---- = 1
2a

which would have eliminated the discontinuity going through zero.

Yoy bwon't find a mathematician to agree with you. And in most
physical situations, the only sensible definition of 0/0 is
"indeterminate."

Not all circulating current in superconductive rings is precisely 1
amp.

John

 

[John Fields]

Yoy bwon't find a mathematician to agree with you.

---
That doesn't mean I'm not right.
---

And in most physical situations, the only sensible definition of 0/0 is
"indeterminate."

---
Not at all. Consider this: (View in Courier)

+V +V
| |
| +---|--[R]--+
+---|-[10R]-+ | | |
| | | 1V--[R]--+--|+\ |
NOM>----[R]--+--|+\ | | >-----+-->QUOT
| >-----+--------------|-/
DENOM>--[R]--+--|-/ |
| | |
[10R] | |
| | |
0V -V -V


While it's not a divider, it illustrates the principle that as long
as the voltages at NOM and DENOM are equal QUOT will always be equal
to 1V, even when NOM and DENOM are at 0V.

Another example would be a digital magnitude comparator, where the
A=B output would always be true as long as the word presented to
input A was equal in magnitude to the word presented at input B,
including when both words were equal to zero, or negative, as
represented in two's complement or any other scheme.

Finally, from Analog Devices' 1976 "nonlinear circuits handbook",

<QUOTE>:

Dividers

(Ratio Circuits)

Chapter 3

An analog "divider" circuit produces an output voltage or current
proportional to the ratio of two input voltages or currents. For
convenience and clarity, in this chapter, it is to be assumed that
the inputs and outputs are voltages(unless noted otherwise).

Vz Vz
Eo = K ---- = Vi ----
Vx Vx

The denominator is denoted Vx, the numerator Vz, and the
output Eo. The dimensional scale factor K (or VI), is usually 10
volts. If the ratio of the inputs is unity, the output is equal to
K.

<END QUOTE>

There's more, and some graphics, so I posted the rest of it to abse.

---

Not all circulating current in superconductive rings is precisely 1
amp.

---
I didn't say it was. What I said was that if the resistance of the
ring is zero ohms and the voltage dropped across the ring is zero
volts, then the current in the ring would be whatever was injected
into it times the ratio of E/R:

/ E \ / 0 \
Iring = Iinj * |-----| = Iinj * |-----| = Iinj * x
\ R / \ 0 /

Since we know the current propagates unhindered, without increasing
or decreasing as time goes by, and we know that there is zero
resistance in the circuit and, consequently zero voltage drop, then
x _must_ be equal to 1, and the only way that can be true is if

0
--- = 1
0

 

[John Larkin]

Yoy bwon't find a mathematician to agree with you.

---
That doesn't mean I'm not right.
---

And in most physical situations, the only sensible definition of 0/0 is
"indeterminate."

---
Not at all. Consider this: (View in Courier)

+V +V
| |
| +---|--[R]--+
+---|-[10R]-+ | | |
| | | 1V--[R]--+--|+\ |
NOM>----[R]--+--|+\ | | >-----+-->QUOT
| >-----+--------------|-/
DENOM>--[R]--+--|-/ |
| | |
[10R] | |
| | |
0V -V -V


While it's not a divider, it illustrates the principle that as long
as the voltages at NOM and DENOM are equal QUOT will always be equal
to 1V, even when NOM and DENOM are at 0V.

Except the positive feedback rails the output. So the circuit makes no
sense. It sure ain't a divider.

Another example would be a digital magnitude comparator, where the
A=B output would always be true as long as the word presented to
input A was equal in magnitude to the word presented at input B,
including when both words were equal to zero, or negative, as
represented in two's complement or any other scheme.

Not a divider either! More like a subtractor.

I agree that 0-0 = 0

Finally, from Analog Devices' 1976 "nonlinear circuits handbook",

<QUOTE>:

Dividers

(Ratio Circuits)

Chapter 3

An analog "divider" circuit produces an output voltage or current
proportional to the ratio of two input voltages or currents. For
convenience and clarity, in this chapter, it is to be assumed that
the inputs and outputs are voltages(unless noted otherwise).

Vz Vz
Eo = K ---- = Vi ----
Vx Vx

The denominator is denoted Vx, the numerator Vz, and the
output Eo. The dimensional scale factor K (or VI), is usually 10
volts. If the ratio of the inputs is unity, the output is equal to
K.

<END QUOTE>

There's more, and some graphics, so I posted the rest of it to abse.

Look closely at the text and the graphics. Both avoid a denominator of
zero.

---


---
I didn't say it was. What I said was that if the resistance of the
ring is zero ohms and the voltage dropped across the ring is zero
volts, then the current in the ring would be whatever was injected
into it times the ratio of E/R:

/ E \ / 0 \
Iring = Iinj * |-----| = Iinj * |-----| = Iinj * x
\ R / \ 0 /

Since we know the current propagates unhindered, without increasing
or decreasing as time goes by, and we know that there is zero
resistance in the circuit and, consequently zero voltage drop, then
x _must_ be equal to 1, and the only way that can be true is if

0
--- = 1
0

Jibberish.

John

 

[John Fields]

On Fri, 25 Aug 2006 10:53:32 -0500, John Fields

On Fri, 25 Aug 2006 06:44:29 -0700, John Larkin

On Wed, 23 Aug 2006 08:49:27 -0500, John Fields

On 22 Aug 2006 22:44:37 -0700, "Bill Bowden" <[email protected]>
wrote:


0/0 is a "Undefined Result" according to my HP 28S calculator. Anything
divided by zero is infinite, and 0 divided by anythiing is 0, so 0/0
must be something between 0 and infinity. Make it whatever you want.

---
ISTM that it can only be 1.

That is, for any fraction where the numerator and the denominator
are identical, the quotient is always 1, so if:

a
--- = b = 1
a

and 'a' slides through zero, why should the value of 'b' be
different from what it was infinitessimally before 'a' became zero?


but 2a/a is always 2, so if a=0, then 0/0 = 2.

In other words, 0/0 is the limit of an infine number of equations, not
all of which evaluate to 1.

---
They all evaluate to 1 if the numerator and the denominator are
identical. In your example:

2a
---- = 2
a

they are not, except when a = 0, in which case:

2a 2*0 0
---- = ----- = --- = 1
a 0 0

The proper procedure, in your example, should have been to multiply
the numerator _and_ the denominator by the same quantity,:

2a
---- = 1
2a

which would have eliminated the discontinuity going through zero.

Yoy bwon't find a mathematician to agree with you.

---
That doesn't mean I'm not right.
---

And in most physical situations, the only sensible definition of 0/0 is
"indeterminate."

---
Not at all. Consider this: (View in Courier)

+V +V
| |
| +---|--[R]--+
+---|-[10R]-+ | | |
| | | 1V--[R]--+--|+\ |
NOM>----[R]--+--|+\ | | >-----+-->QUOT
| >-----+--------------|-/
DENOM>--[R]--+--|-/ |
| | |
[10R] | |
| | |
0V -V -V


While it's not a divider, it illustrates the principle that as long
as the voltages at NOM and DENOM are equal QUOT will always be equal
to 1V, even when NOM and DENOM are at 0V.

Except the positive feedback rails the output. So the circuit makes no
sense.

---
You've never heard of rail-to-rail outputs?

Run it in LTSPICE and then come back with your complaints, OK?
---

It sure ain't a divider.

---
Geez, I thought I'd made that clear earlier when I said it wasn't.
Did you miss it? It ain't supposed to be a divider, it's just
supposed to show that identical inputs will result in a constant
output. Do you need a formal argument to prove a point when you
want to play fast and loose?

 

[John Fields]

On Fri, 25 Aug 2006 10:53:32 -0500, John Fields

On Fri, 25 Aug 2006 06:44:29 -0700, John Larkin

On Wed, 23 Aug 2006 08:49:27 -0500, John Fields

On 22 Aug 2006 22:44:37 -0700, "Bill Bowden" <[email protected]>
wrote:


0/0 is a "Undefined Result" according to my HP 28S calculator. Anything
divided by zero is infinite, and 0 divided by anythiing is 0, so 0/0
must be something between 0 and infinity. Make it whatever you want.

---
ISTM that it can only be 1.

That is, for any fraction where the numerator and the denominator
are identical, the quotient is always 1, so if:

a
--- = b = 1
a

and 'a' slides through zero, why should the value of 'b' be
different from what it was infinitessimally before 'a' became zero?


but 2a/a is always 2, so if a=0, then 0/0 = 2.

In other words, 0/0 is the limit of an infine number of equations, not
all of which evaluate to 1.

---
They all evaluate to 1 if the numerator and the denominator are
identical. In your example:

2a
---- = 2
a

they are not, except when a = 0, in which case:

2a 2*0 0
---- = ----- = --- = 1
a 0 0

The proper procedure, in your example, should have been to multiply
the numerator _and_ the denominator by the same quantity,:

2a
---- = 1
2a

which would have eliminated the discontinuity going through zero.

Yoy bwon't find a mathematician to agree with you.

---
That doesn't mean I'm not right.
---

And in most physical situations, the only sensible definition of 0/0 is
"indeterminate."

---
Not at all. Consider this: (View in Courier)

+V +V
| |
| +---|--[R]--+
+---|-[10R]-+ | | |
| | | 1V--[R]--+--|+\ |
NOM>----[R]--+--|+\ | | >-----+-->QUOT
| >-----+--------------|-/
DENOM>--[R]--+--|-/ |
| | |
[10R] | |
| | |
0V -V -V


While it's not a divider, it illustrates the principle that as long
as the voltages at NOM and DENOM are equal QUOT will always be equal
to 1V, even when NOM and DENOM are at 0V.

Except the positive feedback rails the output. So the circuit makes no
sense.

---
You've never heard of rail-to-rail outputs?

Run it in LTSPICE and then come back with your complaints, OK?
---

It sure ain't a divider.

---
Geez, I thought I'd made that clear earlier when I said it wasn't.
Did you miss it? It ain't supposed to be a divider, it's just
supposed to show that identical inputs will result in a constant
output. Do you need a formal argument to prove a point when you
want to play fast and loose?

 

[jasen]

a
--- = b = 1
a

a/a=1

but 2a/a=2

and 0/0 occurs in both equations when a=0

so which is it?

Bye.
Jasen

 

[Bill Bowden]

On Fri, 25 Aug 2006 11:20:22 -0700, John Larkin

On Fri, 25 Aug 2006 10:53:32 -0500, John Fields

On Fri, 25 Aug 2006 06:44:29 -0700, John Larkin

On Wed, 23 Aug 2006 08:49:27 -0500, John Fields

On 22 Aug 2006 22:44:37 -0700, "Bill Bowden" <[email protected]>
wrote:


0/0 is a "Undefined Result" according to my HP 28S calculator. Anything
divided by zero is infinite, and 0 divided by anythiing is 0, so 0/0
must be something between 0 and infinity. Make it whatever you want.

---
ISTM that it can only be 1.

That is, for any fraction where the numerator and the denominator
are identical, the quotient is always 1, so if:

a
--- = b = 1
a

and 'a' slides through zero, why should the value of 'b' be
different from what it was infinitessimally before 'a' became zero?


but 2a/a is always 2, so if a=0, then 0/0 = 2.

In other words, 0/0 is the limit of an infine number of equations, not
all of which evaluate to 1.

---
They all evaluate to 1 if the numerator and the denominator are
identical. In your example:

2a
---- = 2
a

they are not, except when a = 0, in which case:

2a 2*0 0
---- = ----- = --- = 1
a 0 0

The proper procedure, in your example, should have been to multiply
the numerator _and_ the denominator by the same quantity,:

2a
---- = 1
2a

which would have eliminated the discontinuity going through zero.

Yoy bwon't find a mathematician to agree with you.

---
That doesn't mean I'm not right.
---

And in most physical situations, the only sensible definition of 0/0 is
"indeterminate."

---
Not at all. Consider this: (View in Courier)

+V +V
| |
| +---|--[R]--+
+---|-[10R]-+ | | |
| | | 1V--[R]--+--|+\ |
NOM>----[R]--+--|+\ | | >-----+-->QUOT
| >-----+--------------|-/
DENOM>--[R]--+--|-/ |
| | |
[10R] | |
| | |
0V -V -V


While it's not a divider, it illustrates the principle that as long
as the voltages at NOM and DENOM are equal QUOT will always be equal
to 1V, even when NOM and DENOM are at 0V.

Except the positive feedback rails the output. So the circuit makes no
sense.

---
You've never heard of rail-to-rail outputs?

Run it in LTSPICE and then come back with your complaints, OK?
---

It sure ain't a divider.

---
Geez, I thought I'd made that clear earlier when I said it wasn't.
Did you miss it? It ain't supposed to be a divider, it's just
supposed to show that identical inputs will result in a constant
output. Do you need a formal argument to prove a point when you
want to play fast and loose?
---

Not a divider either! More like a subtractor.

I agree that 0-0 = 0

---
0+0 also equals zero, but we're not talking about addition and
subtraction here, we're talking about division, one step up (or
down) from that.
---

Look closely at the text and the graphics. Both avoid a denominator of
zero.




Jibberish.

0/0=1 is sorta like the question of where is the extra dollar?

3 guests check into a $30 hotel room and each pays $10. The hotel clerk
later realizes he overcharged the guests since the room was only $25
instead of $30. So, the clerk sends the bellhop to the room to refund
the extra $5 and each guest takes $1 and the bellhop gets a $2 tip.
Therefore, each guest has payed $9 for a total of $27 and the bellhop
got a $2 tip, bringing the total expenses to $29.

But the room was $30 paid in advance, so what happened to the extra
dollar?

-Bill

 

[Mark Fortune]

0/0=1 is sorta like the question of where is the extra dollar?

3 guests check into a $30 hotel room and each pays $10. The hotel clerk
later realizes he overcharged the guests since the room was only $25
instead of $30. So, the clerk sends the bellhop to the room to refund
the extra $5 and each guest takes $1 and the bellhop gets a $2 tip.
Therefore, each guest has payed $9 for a total of $27 and the bellhop
got a $2 tip, bringing the total expenses to $29.

But the room was $30 paid in advance, so what happened to the extra
dollar?

-Bill

That had me thinking for a while, but the way I figure it, each guest
has ended up paying £9, totalling £27 :- £25 for the room, and £2 for
the bellhop.

The £2 belhop tip was deducted from the £27 total expenses, and not
added on afterwards.

depends which way round you think of it.