Standard crank division by zero thread - Page 28

CRGreathouse · 2011-04-25, 00:14

Quote:

Originally Posted by R.D. Silverman

The issue is not 'non-mainstream'. The issue is an unwillingness to learn
from experts who know far more than he does about the subject under
discussion. He simply refuses to learn.

Right. Non-mainstream would be a like a researcher trying to prove P = NP or doing KPU set theory. The issue here is not mainstreamness but ability and willingness to admit and correct mistakes.

Condor · 2011-04-25, 11:07

Quote:

Originally Posted by rogue

... What I'm missing is how this statement "division by zero prevents the substitution" is determined.

I don't think you meant this the way I'm going to respond to, but I need to clarify a minor slip I made earlier. This is the crux of Don's argument, and the only one where there is a clear line between "right" and "wrong." Don't misunderstand me - there is nothing "right" about 95% of what Don says; but it is too meaningless to be called wrong, either. Which is how he avoids recognizing the counterarguments. It is just the fluff he surrounds this central theme with, to delude himself into thinking there is a logical progression from his assumptions to his conclusion.

But this is just plain wrong. There is nothing illegal, improper, or "disallowed" about the existence of a zero in a denominator. And Don will find no book, rule, law, or "common sense" that says it is. What is "disallowed" is evaluating such an expression by simple division. Don does this when he eliminates [ln(c/T)-1] from a denominator. Up to that point, substituting T=c is allowed, which was my slip before. It is allowed, but then the entire expression may not be defined unless you make certain assumptions.

However, once you treat it as a simple division, and "divide out" [ln(c/T)-1] from the numerator and denominator, then the substitution T=c is indeed "disallowed." It is disallowed because you have already treated a zero in a denominator as a simple division.

The reason I want to correct this, is because Don seems to think our (correct) methods, utilizing indeterminate forms, would "prevent" even z=1. Not so - that is how we evaluate these expressions without treating the zero in a denominator as a division. This doesn't "prevent" all values of z, as Don claims, it is what allows them. All.

The only thing "prevented" in any of Don's proofs is substituting c=T after dividing out [ln(c/T)-1].

akruppa · 2011-04-25, 12:05

Quote:

Originally Posted by R.D. Silverman

Don is unteachable. I suggest that he be ignored.
He is already on my ignore list.

I tend to agree on the unteachable. However, some of the postings in this thread, especially those by Charles, Rogue and Condor, make for good reading. I see Don's raison d'etre on this forum as having provoked, against all odds, some very thoughtful and informative postings.

R.D. Silverman · 2011-04-25, 15:32

Quote:

Originally Posted by akruppa

I tend to agree on the unteachable. However, some of the postings in this thread, especially those by Charles, Rogue and Condor, make for good reading. I see Don's raison d'etre on this forum as having provoked, against all odds, some very thoughtful and informative postings.

I've read the responses. The math is elementary and trivial and IMO does
not merit the amount of discussion that has been devoted to it.

CRGreathouse · 2011-04-25, 16:02

Quote:

Originally Posted by R.D. Silverman

The math is elementary and trivial and IMO does
not merit the amount of discussion that has been devoted to it.

Yes, yes, and probably. But we're just having fun; it's not like this is somehow less productive than playing NetHack.

Don Blazys · 2011-04-26, 10:15

Well, I hope that you all had as wonderful an Easter Holiday as I did!

I'm sorry if I embarrassed the lot of you when I pointed out that
the "indeterminate forms" which you all thought were a "big deal",
turned out to be utterly trivial, and a total "non-issue"!

Judging from the incredible number of posts over the last several days,
you were all indeed like busy little bees, still trying desperately to find
some "fatal flaw" in my beautiful, simple and utterly irrefutable proof
which can be found here:

httр://donblazys.com/03.рdf

The sheer amount of interest that this proof has already generated
and continues to generate is truly remarkable, and I thank you all for
helping make this thread so extraordinarily popular!

I also thank my opponents, who have all been exposed as bumpkins,
and are now absolutely obsessed with the misguided notion that
they will "somehow" be able to refute my absolutely irrefutable result!

Like I said before, if my proof was "fatally flawed", then everyone here
(including myself) would have lost interest a long, long time ago!

After all, a simple one page proof such as mine could never generate
such an incredible amount of controversy and get this much response
if it was truly wrong or in actuality "fatally flawed"!

Thus, the mere fact that after all this time, my opponents are still
taking all kinds of silly "pot shots" at my proof tells all my readers that
I must be right !

Some of the more recent remarks are most amusing, to say the least,
because it seems as if the "pot shots" are becoming increasingly silly,
almost in direct proportion to the increasing frustration of my foes!

Here are some examples...

Quoting "rogue":

Quote:

First, $\sqrt{$c^z$^2}$ has two roots.

Apparently, poor "rogue" doesn't even know that a radical without
an index number or sign designates the positive square root only!

Quoting "Condor"

Quote:

There is nothing illegal, improper, or "disallowed" about the existence of
a zero in a denominator.

And so "Condor" lays yet another egg ! Yup, according to "Condor",
there's "nothing wrong" with the expression 2/0. (And apparently,
the rest of you agree with him !)

Well, I still say that 0 cannot divide any number exept itself.

Quoting R.D. Silverman:

Quote:

I suggest that he be ignored. He is already on my ignore list.

Well, R.D. Silvermans most unusual and unorthodox style of "ignoring" me
clearly qualifies him as being an expert ignoramus, while the method
by which he "steadfastly refuses" to follow my topics and post on my
threads is definitely inspiring the rest of my opponents to do the same!

Don.

retina · 2011-04-26, 10:36

There are none so blind as those that refuse to see.

This thread is hilarious. I vote this for thread if the year. Do we even have a thread of the year?

Don Blazys · 2011-04-26, 11:04

Quoting "retina"

Quote:

There are none so blind as those that refuse to see.

Spoken like a true "retina". I agree.

Quoting "retina":

Quote:

This thread is hilarious. I vote this for thread if the year.
Do we even have a thread of the year?

Perhaps even "thread of the century".
Spread the word. Tell your friends!

Don.

bsquared · 2011-04-26, 13:28

Quote:

Originally Posted by retina

There are none so blind as those that refuse to see.

This thread is hilarious. I vote this for thread if the year. Do we even have a thread of the year?

Seconded!

Although we have
"22. 50 points for failing to respond to appropriate corrections, questions and challenges. "

We now need to add
"22b. 75 points for citing the existence of copious corrections, disproofs, and challenges as further proof of your work"

Condor · 2011-04-26, 15:39

I think I finally figured it out. Don’s mathematical knowledge has not progressed beyond arithmetic! That’s why he interprets an explanation meant for middle schoolers, who also have little background beyond arithmetic, as though it is a law that applies the same way in algebra and calculus! He doesn’t understand what a variable is, only constants. That’s why he has to interpret a/b as the number n such that b*n=a.

There is a difference between what an expression represents, and the value that expression takes on for certain values of its arguments. There is nothing inherently wrong with an expression that could have a zero in a denominator. There might be something wrong with the normal interpretation of that expression’s value at some points; but the expression itself, the meaning itself, and the offending argument values are still possible. Nothing is "disallowed."

As an example, here’s the formula for distance as a function of latitude ( $\phi$ ) and longitude ( $\theta$ ), assuming a spherical earth of radius $R$ :

$\cos{\frac{D}{R}}=\sin{\phi_1}\sin{\phi_2} + \cos{\phi_1}\cos{\phi_2}\cos{(\theta_1-\theta_2)}$

If you know all the arguments except one longitude, the expression for it is:

$\theta_1=\theta_2+\cos^{-1}${\frac{\cos{\frac{D}{R}}-\sin{\phi_1}\sin{\phi_2}}{\cos{\phi_1}\cos{\phi_2}}}$$

You can use this for any two points on the earth, unless one of them is the north or south pole. Then there is a division by zero. That doesn’t mean the north and south poles don’t exist, or that a point at a known latitude and a known distance from a pole doesn’t have a longitude. The poles do exist, and all points not at a pole have a well-defined longitude. But this expression can’t tell you what that longitude is.

Similarly, even if your expression wasn’t indeterminate, having a division by zero would say nothing about $c^Z$ except you can’t calculate it by that expression. But it is indeterminate, so you can if you assume the expression is continuous.

Don, take a night course (in your nights off as a watchman) in algebra, then calculus, and you will learn this. Until then, please don’t try to explain basic concepts that you don’t understand. Or, you could just address post #286, and try to explain why we can’t make any integer z “disallowed.” Hint: as I described above, it doesn’t mean that any z, no matter what it is, is “disallowed.” It means your expression can’t evaluate it.

Don Blazys · 2011-04-27, 11:28

Quoting akruppa:

Quote:

I see Don's raison d'etre on this forum as having provoked,
against all odds, some very thoughtful and informative postings.

Quoting Al Jolson:

Quote:

You ain't seen nothing yet.

Quoting Uncwilly:

Quote:

And a few of them have been right.

I am definitely right, and here's how my readers can be assured of that.

Way back in post #167, I issued this friendly invitation to debate.

Quoting Myself from Post #167

Quote:

All it takes is one "fatal flaw" to refute my proof. Just one !

Pick one "fatal flaw" and one "champion" from the "math community"
to debate it with me in a seperate thread, using our real names.

Everyone else just watch.

That way, not only will it be a fair debate, but it will also be
very obvious as to who wins, because we will no longer have
all of the inevitable "flaming" and confusion that occurs when
too many people post their half baked "pot shots" simultaneously.

I promise that if I lose, then I will drop my Proof of Beal's Conjecture
like a hot potato and publicly renounce it.

.

Now, this friendly invitation to debate goes out to any mathematician
from any college or university. Heck, I'm even willing to debate any
math department, the more well known and "prestigious", the better.

We know that my old friend, the great and highly esteemed Don Blasius
from U.C.L.A. (University of California, Los Angeles) is following this thread.
Maybe he can use some of his influence and connections to get Terence Tao
or even the entire U.C.L.A. math department to debate me.

Wouldn't that be fun!

Of course, that will never happen, because they know that I am right
and saw first hand how easily I cleaned the clocks of all my opponents,
and wiped the floor with them!

Don.

2011-04-26, 15:39	#307
Condor Apr 2011 31 Posts	I think I finally figured it out. Don’s mathematical knowledge has not progressed beyond arithmetic! That’s why he interprets an explanation meant for middle schoolers, who also have little background beyond arithmetic, as though it is a law that applies the same way in algebra and calculus! He doesn’t understand what a variable is, only constants. That’s why he has to interpret a/b as the number n such that bn=a. There is a difference between what an expression represents, and the value that expression takes on for certain values of its arguments. There is nothing inherently wrong with an expression* that could have a zero in a denominator. There might be something wrong with the normal interpretation of that expression’s value at some points; but the expression itself, the meaning itself, and the offending argument values are still possible. Nothing is "disallowed." As an example, here’s the formula for distance as a function of latitude ( $\phi$ ) and longitude ( $\theta$ ), assuming a spherical earth of radius $R$ : $\cos{\frac{D}{R}}=\sin{\phi_1}\sin{\phi_2} + \cos{\phi_1}\cos{\phi_2}\cos{(\theta_1-\theta_2)}$ If you know all the arguments except one longitude, the expression for it is: $\theta_1=\theta_2+\cos^{-1}\({\frac{\cos{\frac{D}{R}}-\sin{\phi_1}\sin{\phi_2}}{\cos{\phi_1}\cos{\phi_2}}}\)$ You can use this for any two points on the earth, unless one of them is the north or south pole. Then there is a division by zero. That doesn’t mean the north and south poles don’t exist, or that a point at a known latitude and a known distance from a pole doesn’t have a longitude. The poles do exist, and all points not at a pole have a well-defined longitude. But this expression can’t tell you what that longitude is. Similarly, even if your expression wasn’t indeterminate, having a division by zero would say nothing about $c^Z$ except you can’t calculate it by that expression. But it is indeterminate, so you can if you assume the expression is continuous. Don, take a night course (in your nights off as a watchman) in algebra, then calculus, and you will learn this. Until then, please don’t try to explain basic concepts that you don’t understand. Or, you could just address post #286, and try to explain why we can’t make any integer z “disallowed.” Hint: as I described above, it doesn’t mean that any z, no matter what it is, is “disallowed.” It means your expression can’t evaluate it. Last fiddled with by Condor on 2011-04-26 at 15:44 Reason: typos

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2011-04-26, 10:36	#304
retina Undefined "The unspeakable one" Jun 2006 My evil lair 3×19×109 Posts	There are none so blind as those that refuse to see. This thread is hilarious. I vote this for thread if the year. Do we even have a thread of the year?