Standard crank division by zero thread - Page 25

Christenson · 2011-04-22, 17:03

Quote:

Originally Posted by rogue

No it doesn't. n*0=0 does not imply that 0/0 = n. You can't simply divide both sides by 0.

You are basically saying that n*0 = 0 becomes (n*0)/0 = 0/0 becomes n = 0/0. In reality, you could also do this: (n*0)/0 = (n/0)*(0/0) = 0/0. If 0/0 = n (which you state to be true), then (n/0)*n = n. Divide both sides by n, now you get n*0 = 1. That isn't right because n*0 = 0.

We are trying to get you to understand that you cannot use division by 0 (even 0/0) as the basis of any proof. Show your proof (and this thread) to any real mathematician (I have a degree in math BTW) and they tell you the same that we are trying to tell you.

My world book encyclopedia, from the 1950s, has a famous "proof" in about 10 lines that 0=1. It involves dividing by 0, in the form of (a+b)/(a-b). I seem to remember a calculus class where we learned l'hopital's theorem for finding limits when we tried to divide 0 by 0.

xilman · 2011-04-22, 17:06

Quote:

Originally Posted by Don Blazys

That's because humour must have an element of truth in order to be funny.

I'm reminded at this point by the old words of wisdom:

Laugh, and the world laughs at you. Crying doesn't help much.

Paul

jyb · 2011-04-22, 17:33

Quote:

Originally Posted by CRGreathouse

I can't think of anyone on this thread (apart from Don himself) who think that he's right. CRGreathouse, Spherical Cow, rogue, xilman, condor, akruppa, flouran, axn, tichy, NBtarheel_33, rajula, R.D. Silverman, and yes science_man_88 have all made it clear that they think he's wrong.

Well, I initially thought Blazys was wrong, but then I saw this:

Quote:

Originally Posted by Don Blazys

Dividing any number other than 0, by 0
is strictly dissalowed. That's the truth.

If it's in red and underlined, then I guess it really must have some validity.

(Ordinarily I would avoid piling on, particularly when Blazys is so obviously delusional. However in this case he's invoking the silent masses who are secretly on his side to justify his continued nonsense and abuse of all who try to point out his errors, so I thought it worthwhile to try to disabuse him of at least that mistake.)

Actually, here's a serious question for you, Don. Perhaps if you genuinely think about the answer it might help you see some problems. You say that dividing any number other than 0 by 0 is disallowed. Why is that? Who disallowed it and what was the justification? Where does that "rule" spring from?

Edit: I realize that some of the things you've said above actually count as reasonable answers to the questions I just asked, without actually getting at the real point. So let me ask a more direct question: how do you define division? I.e give me a concise mathematical definition for what a/b means.

Condor · 2011-04-22, 17:38

Quote:

Originally Posted by rogue

No it doesn't. n*0=0 does not imply that 0/0 = n. You can't simply divide both sides by 0.

What Don doesn't seem to realize is that if "n*0=0 is true for any number n," it doesn't mean he gets to pick his favorite number and insert it for 0/0. It means that any number could be inserted there. That in turn means that the value of expressions involving it can't be determined (hmm... could that be what "indeterminate" means?). In particular, the only way one can "eliminate" the multiplier $\frac{\frac{{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}$ from an expression, is if you have determined that its value is, and can only be, 1. Since Don now admits this multiplier can be "any number," he is in fact admitting that it can't be eliminated.

Don should go back to his job as a night watchman at a school, and stay away from the students. We don't want to have to worry about them getting wrong ideas from him. He should also stay away from papers he sees when nobody else is around, or at least have a qualified teacher explain them to him.

Or maybe he could explain why the "proof" I added last time, that every integer z is bioth "allowed" and "disallowed" by his methods. He quite obviously added in those attacks as a means of distraction from that point.

Don Blazys · 2011-04-22, 18:10

Quoting rogue:

Quote:

n*0=0 does not imply that 0/0 = n.

Yes it does! Here's why.

If $2*3=6$ , then it must be the case that $\frac{6}{3}=2$ .

Likewise,

if $(n*0)=0$ , then it must be the case that $$\frac{0}{0}$=n$ ,

where $n$ can be any number.

That's the proper, correct and accepted interpretation of the indeterminate form $\frac{0}{0}$ .

In my proof, that "indeterminate form" is utterly trivial,
and can be easily avoided if we choose to do so.

However, if for some silly reason we choose not to avoid it,
then we must properly and correctly interpret its meaning
as being "any numeric exponent of unity".

Since 0 cannot divide any number exept itself,
we never dissallow the indeterminate form $\frac{0}{0}$
but either determine its value by some method such as l'Hôpital's rule,
or simply interpret it's meaning without involving it in any operations.

Also, since 0 cannot divide any number exept itself,
we always disallow divisions by zero such as $\frac{2}{0}$ .

Quoting rogue:

Quote:

You cannot use division by 0 (even 0/0) as the basis of any proof.

That's true. That's why, in my proof, which you can find here:

httр://donblazys.com/03.рdf

I simply avoid those trivial indeterminate forms by demonstrating that
logarithms are not even involved when $z=1$ or $z=2$ .

My proof contains no indeterminate forms, and the fact that
my opponents are making such a big deal out of indeterminate forms
so trivial that they can be easily avoided bespeaks of and is a
testament to how utterly stupid and desperate they really are!

Don.

rogue · 2011-04-22, 18:42

Quote:

Originally Posted by Don Blazys

if $(n*0)=0$ , then it must be the case that $$\frac{0}{0}$=n$ ,

where $n$ can be any number.

Wrong again. You clearly ignored my example of showing the wrongness of this statement. As I said above, the n in n*0=0 is not the same as the n in 0/0=n. For example, it is a fact that 1*0=0, but it does not follow that 0/0=1.

When you say that 0/0=n and that n can be any value, you are clearly wrong. 0/0 is indeterminate. "indeterminate" does not mean "any number". There is no correlation. This is why division by 0 (even 0/0) is not allowed in your (or any other) proof.

I don't understand why you can't reconcile this even though everyone here (including those much smarter than me and those with more advanced degrees in mathematics) have tried to convince you that you are wrong.

If you are so convinced that you are right, why don't you try to publish your results in a journal?

Don Blazys · 2011-04-22, 18:54

Quoting "Condor"

Quote:

Since Don now admits this multiplier can be "any number," he is in fact admitting that it can't be eliminated.

When did I say that the multiplier $\frac{\frac{{\ln(c)}}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}$ can be "any number"?

I never said that! Condor is not only stupid, but he is also a liar.

I said that the trivial and avoidable indeterminate form $$\frac{0}{0}$=n$
can be any number $n$ because $(n*0)=0$ is a true statement.

In fact, in my proof, I show that multiplier to be equal to $1$ and eliminate it.

I happen to be the supervisor of my department.

I have worked with the head of our math dept.
and have even been featured tn our school paper!

I would never hire someone as dumb and dishonest
as condor for my department, or even as a janitor.

A "Condor" belongs in a zoo!

Don.

Don Blazys · 2011-04-22, 19:13

Quoting rogue:

Quote:

When you say that 0/0=n and that n can be any value,
you are clearly wrong.

No, you are clearly wrong.

Here,

http://www.mathpath.org/concepts/division.by.zero.htm

read it again.

Don.

Don Blazys · 2011-04-22, 19:23

Quoting "jyb":

Quote:

You say that dividing any number other than 0 by 0 is disallowed.
Why is that? Who disallowed it and what was the justification?
Where does that "rule" spring from?

Here's a simple explanation that you might understand:

http://www.mathpath.org/concepts/division.by.zero.htm

Don.

rogue · 2011-04-22, 19:59

Quote:

Originally Posted by Don Blazys

Here,

http://www.mathpath.org/concepts/division.by.zero.htm

read it again.

The website also says "Therefore, 0/0 does not mean any particular number - or even anything until we give it some new meaning". This contradicts your statement that 0/0 = n and that n can be any value. If 0/0 has no meaning (as this website states), then 0/0 can't be equal to n because you have assigned a meaning to n. In this case, you have assigned the meaning "any number".

This website has a much better description of 0/0: http://www.math.utah.edu/~pa/math/0by0.html.

jyb · 2011-04-22, 20:08

Quote:

Originally Posted by Don Blazys

Quoting "jyb":

Here's a simple explanation that you might understand:

http://www.mathpath.org/concepts/division.by.zero.htm

Don.

Yes, but please see the edit in my previous post. I would like your definition of division. Please don't send me a link to what somebody else thinks. How do you define division?

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