Standard crank division by zero thread

[rogue]

Originally Posted by em99010pepe

I think it's time to lock the thread or leave it dead as happened in the following forums:

http://www.physicsforums.com/showthread.php?t=301139
http://www.sciencechatforum.com/view...11838&start=30
http://www.marilynvossavant.com/foru...ghlight=#14565
http://www.scienceforums.net/topic/3...equation-ever/
http://www.mymathforum.com/viewtopic.php?f=40&t=7283
http://www.sciforums.com/showthread.php?t=103447
https://nrich.maths.org/discus/messa...tml?1273932598

and so on...

I didn't expect his proof to be disemboweled in so many places. Thanks.

Although he presumes it generates interest in people trying to understand its correctness, it is only generating interest in people trying to convince him that he is wrong.

Don will continue to "appeal to a higher authority" at every opportunity. Every mathematician/brainiac in the world could tell him that he is wrong and he would tell them that he is right. There is nobody who could convince him that he is wrong.

[science_man_88]

Originally Posted by rogue

I didn't expect his proof to be disemboweled in so many places. Thanks.

Although he presumes it generates interest in people trying to understand its correctness, it is only generating interest in people trying to convince him that he is wrong.

Don will continue to "appeal to a higher authority" at every opportunity. Every mathematician/brainiac in the world could tell him that he is wrong and he would tell them that he is right. There is nobody who could convince him that he is wrong.

I kinda see what the 0/0 argument is about saying 0/0=n basic invalidates the specific proof of must be within a value because n taking on every value at once is impossible but it covers all values by it's variability after that trying to prove that another variable needs to be in a specific value is pointless because as it won't narrow the values it can cover and therefore won't prove a constraint because there's no restriction of n possible without contradiction.

[CRGreathouse]

Originally Posted by akruppa

Some of us kinda enjoy this. Admittedly it's on the same level as laughing at the handicapped kid during recess, but I'm not choosy about entertainment.

It's more like listening to a quadriplegic tell you that he actually beat the Miami Heat in last weekend's basketball tournament. The brazenness of the falsehood makes it much more amusing.

[Don Blazys]

To NBtarheel_33,

Quoting NBtarheel_33:

n/0... it is nonsensical...
n/0... It is not allowed.

I agree, 1^(n/0) is nonsensical, so it must be disallowed.

Now, you agree that 1^n = 1^(0/0) = 1 ?

Don.

[xilman]

Originally Posted by Don Blazys

To NBtarheel_33,

Quoting NBtarheel_33:

I agree, 1^(n/0) is nonsensical, so it must be disallowed.

Now, you agree that 1^n = 1^(0/0) = 1 ?

Don.

Nope. 1^(0/0) is undefined.

Paul

[Condor]

Originally Posted by NBtarheel_33

... Hence, n/0 must be undefined. It is not allowed. If we see it in the course of solving a problem, we should treat it as a STOP sign, NOT a division problem. Either something has gone wrong in our work, or we will need to resort to taking limits. You seem to (incredibly) agree with this.

No, Chapel Hill, he does not agree with that. As you can tell from the fact that he quoted the red parts out of context, completely ignoring the thrust of what you said. This is why you can't appear to agree with anything he says that is even partially true, because he will not grasp the correction you intend.

When Don says "disallowed," he means the entire expression is invalid and must be expunged from the set of true statements. When you said "not allowed," you clearly meant the division operation only. The parts Don deleted explicitly say so. This is the ultimate source of Don's error, and the one he refuses to address in any form.

Specifically, he refuses to address the fact that "indeterminate" means "depending on the limits of the indicated parts of the expression as they approach zero, the quotient could approach any value and must be determined by those limits and not division of the values themselves." For example, it is not true that "1^(0/0)=1." It is indeterminate. In fact, the expression

which is 1^(0/0) at X=0, is unbounded and goes to infinity as X goes to zero. This clearly disproves Don's latest argument.

And while an entire expression of the form n/0 is nonsensical, Don refuses to acknowledge that n/0 as one of the parts of a larger expression may not be. Since (1+X)^(1/X) approaches 1 as X approaches 0, its "continuous extension" at X=0 well defined to be 1.

[rogue]

In one of the other threads someone posed this to Don:

Now, suppose a = b = 7, say.

Is the left-hand side of the expression valid?"

to which Don replied:

Of course it is. At a=b=7, it results in the indeterminate form 0/0
which can easily be determined to be 14.

Your example has absolutely nothing to do with the identity in question
which does not result in an indeterminate form but a division by zero

Notice how he says "it results in the indeterminate form 0/0", but then in the same sentence says "which can easily be determined to be 14". What kind of twisted logic is that? How can one say that it is "indeterminate", but is then "determined" at the same time?

I think that Don believes that indeterminate means the same thing as "any number", which is clearly not true. Even when he talks about "any number" he wants to use a specific number, thus confusing the concept of "variable" with "value". This shows a lack of understanding of basic algebra.

[science_man_88]

Originally Posted by rogue

In one of the other threads someone posed this to Don:


to which Don replied:



Notice how he says "it results in the indeterminate form 0/0", but then in the same sentence says "which can easily be determined to be 14". What kind of twisted logic is that? How can one say that it is "indeterminate", but is then "determined" at the same time?

I think that Don believes that indeterminate means the same thing as "any number", which is clearly not true. Even when he talks about "any number" he wants to use a specific number, thus confusing the concept of "variable" with "value". This shows a lack of understanding of basic algebra.

I kinda see how he'd come to that except him calling it indeterminate and then not. which reduces on assuming 0/0=n is valid to and so 0/0 is both indeterminate and = 14 in his mind.

[science_man_88]

Originally Posted by science_man_88

I kinda see how he'd come to that except him calling it indeterminate and then not. which reduces on assuming 0/0=n is valid to and so 0/0 is both indeterminate and = 14 in his mind.

I forgot to mention:

Originally Posted by http://en.wikipedia.org/wiki/Division_by_zero#Fallacies_based_on_division_by_zero

every number solves the equation instead of there being a single number that can be taken as the value of 0/0

[CRGreathouse]

Originally Posted by rogue

I think that Don believes that indeterminate means the same thing as "any number", which is clearly not true. Even when he talks about "any number" he wants to use a specific number, thus confusing the concept of "variable" with "value". This shows a lack of understanding of basic algebra.

I wonder how common this misunderstanding is. Maybe it's in the math education literature?

[Condor]

Originally Posted by Condor

Specifically, he refuses to address the fact that "indeterminate" means "depending on the limits of the indicated parts of the expression as they approach zero, the quotient could approach any value and must be determined by those limits and not division of the values themselves." For example, it is not true that "1^(0/0)=1." It is indeterminate. In fact, the expression

which is 1^(0/0) at X=0, is unbounded and goes to infinity as X goes to zero. This clearly disproves Don's latest argument.

ARGH!!!!!

I mis-typed my formula, and now can't remember what it was that was unbounded. Ah, well; here's one that is bounded, but isn't 1. It's 2.71828..... The point is the same, but I'm sure Don will quote only the incorrect one: