Wiring electric baseboard

Mark Lloyd wrote in news:a7m3z.9628$ snipped-for-privacy@fx02.iad:

That is incorrect. Resistance is any finite number: e.g. 5 ohms * 0 amps = 0 volts. The one thing R *cannot* be is infinite.

No it did not.

And yours is one of them.

That's because: (a) you did a mathematically undefined operation (division by zero), and (b) your calculator is constructed incorrectly, and giving you the wrong answer. The display

*should* have showed "undefined". *My* calculator does it right. Pressing the same sequence of keys gives this result: ERR:DIVIDE BY 0

My wife's calculator also does it right, with the result -E-, same as it gives for the square root of a negative number, logarithm of a nonpositive number, inverse sine of 2, etc.

Reply to
Doug Miller
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FromTheRafters wrote in news:nikm9f$mae$1 @news.albasani.net:

"Undefined"?? You have to be kidding. Do you really mean that you don't know, and can't figure out, how far it moved? that in real life if an object is not moving then no matter how much time elapses the distance it moves is ZERO?

If something is not moving, the distance it moves is NOT "undefined". The distance it moves is zero.

Absolute nonsense. If the rate R = D / T = 0, that does *not* mean that T is infinite. It means that D = 0.

How much farther do you need it "broken down"? What part of 0 * 60 = 0 do you find confusing?

It's not necessary to use calculus. 0 * 60 = 0 is 4th-grade arithmetic.

Reply to
Doug Miller

FromTheRafters wrote in news:nikq90$udm$1 @news.albasani.net:

The set of natural numbers is the infinite set {1, 2, 3, 4, ...} and its sum is likewise infinite.

And under no circumstances at all is one to the second power equal to two.

Perhaps you should wait to post again until after you've sobered up. ;-)

Reply to
Doug Miller

Doug Miller explained :

Right, so we're left with an answer of undefined.

So, if the T is as you state 'any finite number' there must be a rate of D/T and if that finite number T is zero we're back at division by zero and if that finite number T is not zero we're back at some positive or negative rate which is a contradiction of your original premise of R being zero.

Okay, you got me there. I was treating D=RT as a formula not just as an equation. My mistake. E=MC^2 so if C=0 E=0 and M doesn't matter. Got it now, sorry for wasting your time.

[snip]
Reply to
FromTheRafters

If a pointless pin still a pin? How about an infinite number of monkeys sharing 0 bananas? All right - it is a silly post. Some silliness is essential. 42. HAL is "once removed" from IBM.

Reply to
hah

FromTheRafters wrote in news:nil1vt$d8d$1 @news.albasani.net:

No, we're not, because T is *not* infinite. T can be *any* *finite* value. The one thing it

*cannot* be is infinite precisely because 0 * infinite is undefined -- but the left side of this equation, D, is NOT undefined. It is very much defined, and it is EQUAL TO ZERO.

NO WE ARE NOT! Where in the equation D = RT do you see any division anywhere? This is *multiplication*. And as I have already noted, transforming D = RT into D / T = R is valid only for NON-ZERO values of T.

Complete nonsense. D = RT. D = 0, R = 0, substitute any finite value for T and the equation is true.

No, you still don't have it. C isn't a variable in that equation. C is a constant. If M = 0, then E =

0 and C doesn't matter.

The only thing you got right here is that you're certainly wasting a lot of time.

Reply to
Doug Miller

Sure, doing it *that* way. Of course you would never get there to *all* natural numbers since it is an infinite process. It doesn't make sense in arithmetic to us seventh gradeers now just like the previous division by zero being undefined wouldn't make sense to a seventh grader. That was my point. however with some redefining of things into a different system of which seventh graders are not aware you can get different results. Results that work in that system.

In an additive group in the integers (which the naturals are a subset of) exponentiation is the repeated application of the group operator just as it is in multiplicative groups. It just so happens that repeated addition is equivalent to multiplication and we aren't used to thinking of it as exponentiation. I said exponent and right circumstances. I didn't say second power, because that would have been misleading.

Reply to
FromTheRafters

Nice! :) Watch out though, E=MC^2 can take on all sorts of funny stuff when it is just an equation where C can be set to zero and solved by seventh graders.

Reply to
FromTheRafters

Doug Miller pretended :

You got me again, what I meant was Ohm's Law is not the right tool if you are trying to use it outside of its limitations. Within its limitations you can do as I said. Using zero like you did in the example is outside of its limitations.

Within Ohm's Law you are right about that. There is no arrangemnt of the terms where V is problematic as far as I'm aware. It always seems to be in the numerator or standing alone.

Outside the limitations of Ohm's Law because the current can't be zero.

I wasn't posting it for the question, but for the answers.

Reply to
FromTheRafters

Doug Miller explained :

Which is why Ohm's Law is a formula not just an equation.

It looks just like a letter in an equation not a constant in a formula. As a formula the relationships between the terms become important just like they do in Ohm's Law.

Yeah, I guess it's time to stop now.

It's been fun.

Reply to
FromTheRafters

That's the spirit! Sorry if I upset anyone.

Reply to
FromTheRafters

For whatever it's worth...

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Reply to
1+1

FromTheRafters wrote in news:nil5s9$jon$1 @news.albasani.net:

And the difference is --?

I suppose it does -- to you. And to everyone else who doesn't understand it.

No, it hasn't.

Reply to
Doug Miller

FromTheRafters wrote in news:nil3lr$g3j$1 @news.albasani.net:

You really ought to confine your discussions to subjects you know something about.

If there are any.

Reply to
Doug Miller

FromTheRafters wrote in news:nil57k$ims$1 @news.albasani.net:

More nonsense. Nothing invalid in mathematics, or in Ohm's Law, about multiplying by zero.

You don't seem to understand the difference between multiplication by zero and division by zero.

OF COURSE current can be zero. If no current flows, I is not undefined. I is equal to zero.

And you were given correct answers repeatedly, and continued to argue with them.

Reply to
Doug Miller
1+1 wrote in news:tbWdna-Ij5YHgNPKnZ2dnUU7- snipped-for-privacy@giganews.com:

display showed 'inf'.

Apparently nothing, since it's incorrect.

The result of division by zero is undefined. It's not "equal to infininity".

Reply to
Doug Miller

On 05/31/2016 04:52 PM, Doug Miller wrote: [snip]

Your calculator throws an error because it doesn't know how to do it. I didn't need a calculator for something so simple.

I know how to find the square root of a nonpositive number (actually, both of them). Every number had TWO square roots. A little hint: when is

360 equal to 0?

BTW, every number has three cube roots too, but that may be a bit more complicated.

Reply to
Mark Lloyd

Mark Lloyd wrote in news:QXp3z.14892$ snipped-for-privacy@fx31.iad:

No, my calculator throws an error because division by zero *is* an error.

root

Yes, I do too -- but most calculators operate only on real numbers.

Not necessarily three *distinct* cube roots...

Reply to
Doug Miller
[snip]

The same number, but different numerals.

The 3 cube roots of a number have angles of 0 (positive), 120, 240 degrees. In this case, 0 at any angle is still 0.

BTW, when I mentioned 360 = 0, somehow I thought of the Xbox 360. What I actually meant there was about angles (in degrees).

Reply to
Mark Lloyd

No completion of that circuit is necessary to measure the voltage potential. The meter has an impedance in the meg ohms and for all practical purposes can be ignored for the purposes of this discussion. And there are non-contact measuring techniques and instruments as well. You can measure electric potential without a circuit. You really are in way over your head here.

Reply to
trader_4

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