FromTheRafters wrote in news:niklfm$kju$1
@news.albasani.net:

Doug Miller explained :
FromTheRafters wrote in news:nii8ff$38m$1
@news.albasani.net:

trader_4 wrote:
Say what now? Distance Traveled= Rate x Time. If the rate is zero,
then the distance traveled is zero. Again, clearly zero has meaning.

Not if you use the D=RT formula, it is 'undefined' when either R or T
is zero (I did *not* say approaching zero).

Complete nonsense. If R = 0, then D = 0T = zero. If T = 0, then D = 0R =
zero.

https://en.wikipedia.org/wiki/Division_by_zero

There's no division in D= RT. That's multiplication.

If the rate is zero by stipulating that it is so, and the distance is
zero by multiplication using Ohm's Law, the time must be infinite since
we aren't moving.

That is incorrect. In the equation D = RT, if D and R are both zero, T can -- and must -- be
any finite number (e.g. if T = 5 and R = 0, then D = RT = 0 * 5 = 0). The one thing T *cannot*
be is infinite: zero times "infinity" is what we call in mathematics an "indeterminate form" that
cannot be evaluated algebraically.

Since time is T=D/R the time is undefined by division
by zero

Correct, but not relevant to the equation D = RT which is *not* the same thing:

-- D = RT is defined for all real values of all three variables
-- T = D / R is defined for all real values of D and T, and non-zero values of R

- or - as time is infinite in the multiplication D=RT it is
undefined by multiplying by infinity.

Time is not infinite in this equation; as explained above, T must be a finite value.

You can't get around this without using calculus,

You can't get around it *with* calculus either.

and when you use
calculus you have admitted to having non-zero values where in algebra
you had stipulated zero values.

More nonsense. It does not appear that you ever actually had a course in calculus; at any
rate, you certainly didn't *pass* it...

Ohm's Law, as it relates to 'voltage
drop' will require some current 'approaching zero' rather than a
stipulated zero value.

[...]

E=IR
I=E/R
R=E/I

If you state that either I or R is zero (as is the case with open
circuit, which is not really a circuit, or zero resistance attached to
an ideal source) the formula doesn't work. There was mention of "zero"
current and in the 'superconductor' comment was about zero resistance.
My analogy with the trains was a bit of fun, but still illustrates that
dividing by zero yields the result of 'undefined' *not* 'zero'.

More utter nonsense. If I or R is zero, then so is E.

As I or R approach zero, so does E.

And when either I or R *equals* zero, so does E.

Starting with E = IR and deriving from it I = E/R is valid *if and only if* R
is unequal to zero. Likewise, R = E/I can be derived from E = IR *if and
only if* I is nonzero.

Correct, so Ohm's Law doesn't work for these zero values

That depends on how it's expressed. As V = IR, zero values are perfectly legitimate.

and calculus
must be resorted to. Then you have actual non-zero values approaching
limits.

So, we're back to my original statement that 'voltage drop' as defined
by Ohm's Law requires that there be a current flowing through the
device and energy being dissipated by that device.

I'm not getting into *that* argument. I'm just trying to correct the complete and utter nonsense
you've been spouting about algebra.