In the discussion of the 100 ft of wire going to the heater load, the terms voltage drop and voltage loss relative to the wire are one and the same. Everyone here gets that, except you.

You don't need current to have a voltage drop of zero. In physics, do you need energy, acceleration, to have a velocity of zero? Ever take physics 101?

Simple algebra test for Johnny:

Distance Traveled = Rate x Time

1 - At a rate of 2m/sec, how far does the object move in one minute?

Answer: 2m/sec x 60 secs = 120 meters

2 - At a rate of 0 m/sec, how far does the object move in one minute?

Our answer: 0 m/sec X 60 secs = 0 meters.

Your answer: That is undefined! It's dividing by zero! You can't have a distance of zero!

Sorry Johnny, you fail.

That doesn't mean it is true, as it is an appeal to numbers.

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

D=RT

Undefined, because not moving (a rate R=D/T of zero) means the time it takes to go

Do I? Break it down and tell me how you arrived at your conclusion, being careful not to divide by zero or multiply by infinity.

No fair using calculus, because then you have non-zero values which approach zero.

On Tuesday, May 31, 2016 at 2:48:20 PM UTC-4, FromTheRafters wrote:

Notice again, there is no definition from you as to the alleged differences and why the voltage drop and voltage loss relative to 100 ft run of wire are not the one and the same. If you're so smart, why is that? I defined them, neither you nor your half-assed cites did.

No answer noted.

So you really are the village idiot, incapable of 7th grade math. There is no division by zero there.

I did, many times. So did Doug.

D = RT

D = 0 M/sec x 60 secs = 0

There was no division by zero, no multiplication by infinity.

You can't even understand math basics, stop already with the false BS about calculus. You're not intimidating us.

Notice again, there is no definition from you as to the alleged differences and why the voltage drop and voltage loss relative to 100 ft run of wire are not the one and the same. If you're so smart, why is that? I defined them, neither you nor your half-assed cites did.

No answer noted.

So you really are the village idiot, incapable of 7th grade math. There is no division by zero there.

I did, many times. So did Doug.

D = RT

D = 0 M/sec x 60 secs = 0

There was no division by zero, no multiplication by infinity.

You can't even understand math basics, stop already with the false BS about calculus. You're not intimidating us.

"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

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.

On Tuesday, May 31, 2016 at 6:00:17 PM UTC-4, Doug Miller wrote:

+1

+1

Rafters doing calculus that would be something. He also said that 1 squared is sometimes equal to 2. I'm sure you'd like to see that explained too.

+1

+1

Rafters doing calculus that would be something. He also said that 1 squared is sometimes equal to 2. I'm sure you'd like to see that explained too.

On 05/31/2016 11:55 PM, trader_4 wrote:

[snip]

I found an alternate universe (SC.777777777776) where 1 squared is equal to 2, but only on Tuesdays when the 2 is light pink. No one can live there, and it's dangerous to visit. There's a significant probability of your brain unexpectedly becoming 0 brains.

[snip]

I found an alternate universe (SC.777777777776) where 1 squared is equal to 2, but only on Tuesdays when the 2 is light pink. No one can live there, and it's dangerous to visit. There's a significant probability of your brain unexpectedly becoming 0 brains.

On 05/31/2016 11:55 PM, trader_4 wrote:

[snip]

I found https://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html , which MAY be of some help.

[snip]

I found https://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html , which MAY be of some help.

--

Mark Lloyd

http://notstupid.us/

Mark Lloyd

http://notstupid.us/

Click to see the full signature.

On Wednesday, June 1, 2016 at 12:07:10 PM UTC-4, Mark Lloyd wrote:

It's right up Rafter's ally. This is a legitimate case where dividing by zero is done and it produces an error. Unfortunately it doesn't prove that 1 squared can be 2. Or that V =IR is invalid when I=0.

It's right up Rafter's ally. This is a legitimate case where dividing by zero is done and it produces an error. Unfortunately it doesn't prove that 1 squared can be 2. Or that V =IR is invalid when I=0.

On Tue, 31 May 2016 06:33:36 -0400, FromTheRafters

ferget it!!! It's like a bunch of theologians arguing about how many angels can dance on the point of a pin.

Bend the pin into a circle and it becomes pointless - - - -

ferget it!!! It's like a bunch of theologians arguing about how many angels can dance on the point of a pin.

Bend the pin into a circle and it becomes pointless - - - -

On 05/31/2016 03:45 PM, snipped-for-privacy@snyder.on.ca wrote:

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.

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.

--

"Multitasking Attempted. System confused."

"Multitasking Attempted. System confused."

hah brought next idea :

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

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

On Monday, May 30, 2016 at 11:46:02 PM UTC-4, snipped-for-privacy@snyder.on.ca wrote:

BS. Put the test leads of a high impedance meter into a receptacle in your house, with no other loads on the circuit. It's an open circuit, excluding the impedance of the meter, which is in the mega ohms and totally insignificant. That small effect is less than the accuracy of the meter. And if you want to use more exotic instruments, we could measure it without even that negligible meter load.

My God, why do we need to go down even more silly rat holes?

BS. Put the test leads of a high impedance meter into a receptacle in your house, with no other loads on the circuit. It's an open circuit, excluding the impedance of the meter, which is in the mega ohms and totally insignificant. That small effect is less than the accuracy of the meter. And if you want to use more exotic instruments, we could measure it without even that negligible meter load.

My God, why do we need to go down even more silly rat holes?

On Monday, May 30, 2016 at 8:13:30 PM UTC-4, FromTheRafters wrote:

You really are the village idiot. Having no voltage drop and a voltage drop of zero are the same thing. The equation can't produce words, it produces a value of ZERO. YOU claimed that value had no meaning. THAT is what's BS. It has meaning, it means the voltage drop is zero. As I said many posts ago, it's like having 3 apples and taking 3 away. Subtraction yields a number of zero. There are zero apples. It has meaning. I could also say there are no apples. Or in your example of distance traveled a result of zero means the distance traveled is zero. I could state that as the object has not moved. Capiche? No, of course not.

You really are the village idiot. Having no voltage drop and a voltage drop of zero are the same thing. The equation can't produce words, it produces a value of ZERO. YOU claimed that value had no meaning. THAT is what's BS. It has meaning, it means the voltage drop is zero. As I said many posts ago, it's like having 3 apples and taking 3 away. Subtraction yields a number of zero. There are zero apples. It has meaning. I could also say there are no apples. Or in your example of distance traveled a result of zero means the distance traveled is zero. I could state that as the object has not moved. Capiche? No, of course not.

trader_4 laid this down on his screen :

Then what does that make you?

No, it isn't.

Undefined, not zero.

Ah, so now I can see your math skill level. You finally got one right.

How many oranges does that make, brainiac?

LOL

Then what does that make you?

No, it isn't.

Undefined, not zero.

Ah, so now I can see your math skill level. You finally got one right.

How many oranges does that make, brainiac?

LOL

On Tuesday, May 31, 2016 at 11:10:46 AM UTC-4, FromTheRafters wrote:

Explain that to us. Gfre, CL and I are on the same page. V = IR. If there is zero current, there is zero voltage There is zero voltage drop across that resistor and there is no voltage drop. It all means exactly the same thing, but then we're in the real world.

Idiot.

Again, you failed even basic math. Trivial test question:

V = IR. R is .16, I is zero.

What is V?

All the rest of us say it's ZERO. Are you sitting here, saying that the correct answer on a 7th grade math exam to that question is not zero, but "it's undefined"? Good grief.

I wouldn't be laughing, you're demonstrating to everyone that's reading this that you can't handle simple 7th grade math.

Explain that to us. Gfre, CL and I are on the same page. V = IR. If there is zero current, there is zero voltage There is zero voltage drop across that resistor and there is no voltage drop. It all means exactly the same thing, but then we're in the real world.

Idiot.

Again, you failed even basic math. Trivial test question:

V = IR. R is .16, I is zero.

What is V?

All the rest of us say it's ZERO. Are you sitting here, saying that the correct answer on a 7th grade math exam to that question is not zero, but "it's undefined"? Good grief.

I wouldn't be laughing, you're demonstrating to everyone that's reading this that you can't handle simple 7th grade math.

trader_4 expressed precisely :

The term 'voltage drop' refers to a difference in voltage potentially available to other devices in the circuit (read as 'closed circuit') because of the device dissipating some of the available energy because the current through the device causes the device to, for example, emanate heat. Zero current cannot cause this phenomenon, so there is no 'voltage drop'. If there is some other reason for there being a discrepancy between what voltage 'should' be there, it is not due to 'voltage drop' but might be a burnt open fuse or other fault.

I'm saying that Ohm's Law as it pertains to 'voltage drop' requires that the current be non-zero. Algebra is not the same as arithmetic. Algebra is about taking the entities and rearranging them according to certain rules V=IR for instance can be algebraically changed to R=V/I by dividing both sides by I or to I=V/R by dividing both sides by R. This can't be done if the numbers you choose to use are causing you to divide by zero or multiply by infinity. Ohm's Law is valid if you choose the right numbers, not so much if you don't.

Are they teaching algebra in the seventh grade now?

In arithmetic you don't really care about the relations between numbers and I suppose it is okay to overlook the problems which arise by dividing by zero and multiplying by infinity in transformed versions of the same algebraic structure.

The natural number one with an exponent of two can equal two under the right circumstances, but I wouldn't expect a seventh grader to know this or that adding***all*** natural numbers together gives -1/12.

The term 'voltage drop' refers to a difference in voltage potentially available to other devices in the circuit (read as 'closed circuit') because of the device dissipating some of the available energy because the current through the device causes the device to, for example, emanate heat. Zero current cannot cause this phenomenon, so there is no 'voltage drop'. If there is some other reason for there being a discrepancy between what voltage 'should' be there, it is not due to 'voltage drop' but might be a burnt open fuse or other fault.

I'm saying that Ohm's Law as it pertains to 'voltage drop' requires that the current be non-zero. Algebra is not the same as arithmetic. Algebra is about taking the entities and rearranging them according to certain rules V=IR for instance can be algebraically changed to R=V/I by dividing both sides by I or to I=V/R by dividing both sides by R. This can't be done if the numbers you choose to use are causing you to divide by zero or multiply by infinity. Ohm's Law is valid if you choose the right numbers, not so much if you don't.

Are they teaching algebra in the seventh grade now?

In arithmetic you don't really care about the relations between numbers and I suppose it is okay to overlook the problems which arise by dividing by zero and multiplying by infinity in transformed versions of the same algebraic structure.

The natural number one with an exponent of two can equal two under the right circumstances, but I wouldn't expect a seventh grader to know this or that adding

On Tuesday, May 31, 2016 at 3:56:21 PM UTC-4, FromTheRafters wrote:

You failed again. The assignment was to define voltage drop AND voltage loss and explain why they are not one and the same for the example we're talking about, 100 ft of wire. You defined only voltage drop and even there you have it wrong. A current of zero produces a voltage drop of zero.

A ball thrown up in the air, we can write the equation for it's velocity at any point. At the peak, the equation gives zero for it's velocity, which of course is correct. Following your logic, that wouldn't be right either. We don't say the velocity can't be defined. It is defined, it's zero, just like the voltage across a resistor with no current flowing.

But regardless of that, voltage drop and voltage loss are one and the same in our example of 100 ft of wire going to the heater load. They are both calculated the same, given by Ohm's LAw. I can't believe anyone would argue this.

You just said that with V = IR and a value of .16 for R, 0 for I, it's undefined. Sorry, you failed 7th grade math.

Did you draw that graph yet? Straight line right through the origin? What's up with that?

We are using the right numbers.

V = IR, V= I *.16

I = 0, then V = 0, there is no division by zero. You can take almost any equation, rearrange it so that a variable is in the denominator. Sure if you then try to actually do that, put a zero value there for that variable, then it's undefined or infinity. But no one is doing that in the above equation, it's just multiplication by zero.

You only need to worry about dividing by zero if it actually is happening. No one is dividing by zero, only multiplying by zero.

More diversion into the wilderness. But I'll play along. Explain to us under what circumstance 1 squared is equal to two.

You failed again. The assignment was to define voltage drop AND voltage loss and explain why they are not one and the same for the example we're talking about, 100 ft of wire. You defined only voltage drop and even there you have it wrong. A current of zero produces a voltage drop of zero.

A ball thrown up in the air, we can write the equation for it's velocity at any point. At the peak, the equation gives zero for it's velocity, which of course is correct. Following your logic, that wouldn't be right either. We don't say the velocity can't be defined. It is defined, it's zero, just like the voltage across a resistor with no current flowing.

But regardless of that, voltage drop and voltage loss are one and the same in our example of 100 ft of wire going to the heater load. They are both calculated the same, given by Ohm's LAw. I can't believe anyone would argue this.

You just said that with V = IR and a value of .16 for R, 0 for I, it's undefined. Sorry, you failed 7th grade math.

Did you draw that graph yet? Straight line right through the origin? What's up with that?

We are using the right numbers.

V = IR, V= I *.16

I = 0, then V = 0, there is no division by zero. You can take almost any equation, rearrange it so that a variable is in the denominator. Sure if you then try to actually do that, put a zero value there for that variable, then it's undefined or infinity. But no one is doing that in the above equation, it's just multiplication by zero.

You only need to worry about dividing by zero if it actually is happening. No one is dividing by zero, only multiplying by zero.

More diversion into the wilderness. But I'll play along. Explain to us under what circumstance 1 squared is equal to two.

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. ;-)

Doug Miller wrote :

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.

Sure, doing it

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.

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

If there are any.

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