[QUOTE=Don Blazys;260472]
And what a waaah waaaaaaaah crybaby "rogue" turns out to be!!!
He sees this thread growing and growing in popularity, and he
"just can't take it anymore!" WAAAAAAAAAH!!!! WAAAAAAAAAAH![/QUOTE]

That is funny! As this thread gains popularity you lose more credibility. I've already stated that it would be a good thing for this thread to gain popularity on google as more people could see how wrong your proof is.

Not one person here (besides yourself) is trying to defend your proof. Everyone except you recognizes its flaws.

[quote]
Quoting "rogue":
[quote]
n*0=0 does not imply that 0/0 = n.
Division by 0/0 is just as invalid in a proof as 1/0.[/quote]
Wrong![/quote]

Based upon your statement if n*0=0 implies that 0/0=n then by substituting 1 for n one could say that 1*0=0 implies that 0/0=1 (and only 1). That is so wrong it is laughable.

Honestly, have you taken (and passed) any college level math courses beyond the basics? If so, then I suggest that you show your proof to a college professor that you respect and provide him/her a link to this thread. I know that you haven't done that.

BTW, you have not refuted only thing from post 346 of this thread and you have avoided post 347 as well.

[QUOTE=rogue;260481]That is funny! As this thread gains popularity you lose more credibility. I've already stated that it would be a good thing for this thread to gain popularity on google as more people could see how wrong your proof is.

Not one person here (besides yourself) is trying to defend your proof. Everyone except you recognizes its flaws.



Based upon your statement if n*0=0 implies that 0/0=n then by substituting 1 for n one could say that 1*0=0 implies that 0/0=1 (and only 1). That is so wrong it is laughable.

Honestly, have you taken (and passed) any college level math courses beyond the basics? If so, then I suggest that you show your proof to a college professor that you respect and provide him/her a link to this thread. I know that you haven't done that.

BTW, you have not refuted only thing from post 346 of this thread and you have avoided post 347 as well.[/QUOTE]

he also avoided my asking of how he derived the "constants" that aren't constant unless n = 0/0 creates constant n.

[QUOTE=rogue;260481][Addressed at Don Blazys] Honestly, have you taken (and passed) any college level math courses beyond the basics?[/QUOTE]
No, and I don't think he has taken High School ones, either. Or in reading comprehension either, it seems, since he has difficulty with that as well. There are many examples of how he uses his first thought to rebut any flaw, completely ignoring how the thought is both incorrect and inapplicable. Some examples:

First: here's another attempt to explain division for Don, since he can't seem to get it right. I'll use smaller words and simpler concepts, separated out, so even someone with his limited attention span can see:[list][*]It is not allowed to [i]divide[/i] anything by zero.[*]Even zero.[*]That does not mean that the [i]presence of[/i] zero as a denominator is "disallowed" - just the act of reducing the term to a simpler form by division.[*]In particular, if certain limits can be defined, the term can be reduced to a simpler form [i]without dividing by zero[/i].[*]You can get [i]any[/i] value out of that. That does not mean you get to pick one you like, it means you can't pick one at all. You have to calculate it by limits.[*]In particular, it means that 0*n=0 and 0/0=n are not equivalent statements.[*]The 0/0 in his "cohesive terms" is an example where you can get a value.[*]But so is the 1^(1/0).[/list]Second, he still has not responded to posts #260, #286, and #315. If his methods were correct, any integer M could be "allowed" or "disallowed."

Third, the Beal Conjecture does not [I]require[/I] the three terms be pairwise co-prime, as he now claims. Only that the set has no common factor, which is how Don uses "coprime" (notice I don't say "requires", because he doesn't, he just mentions it in passing) in his proof "so that the only common factor possible is the 'trivial' unity." [b]This is not a requirement to be pairwise coprime, so the "proof" works the same way with four terms.[/b]

Now, it is true that you can't have a factor common to exactly two of the three terms; but that is a consequence of having three terms, not a requirement.

Still, Don thinks:[QUOTE]my proof predicts that with four terms, at least
two of them will always have a common factor if w,x,y,z > 2.[/QUOTE]
Which is the problem. Because 193^3+461^3+631^3=709^3. These terms are not merely pairwise co-prime, they are actual primes.

So Don has now admitted there is a fatal flaw in his proof, since it predicts a falsehood.

[QUOTE=Condor;260498]No, and I don't think he has taken High School ones, either. Or in reading comprehension either, it seems, since he has difficulty with that as well. There are many examples of how he uses his first thought to rebut any flaw, completely ignoring how the thought is both incorrect and inapplicable. Some examples:

First: here's another attempt to explain division for Don, since he can't seem to get it right. I'll use smaller words and simpler concepts, separated out, so even someone with his limited attention span can see:[list][*]It is not allowed to [i]divide[/i] anything by zero.[*]Even zero.[*]That does not mean that the [i]presence of[/i] zero as a denominator is "disallowed" - just the act of reducing the term to a simpler form by division.[*]In particular, if certain limits can be defined, the term can be reduced to a simpler form [i]without dividing by zero[/i].[*]You can get [i]any[/i] value out of that. That does not mean you get to pick one you like, it means you can't pick one at all. You have to calculate it by limits.[*]In particular, it means that 0*n=0 and 0/0=n are not equivalent statements.[*]The 0/0 in his "cohesive terms" is an example where you can get a value.[*]But so is the 1^(1/0).[/list]Second, he still has not responded to posts #260, #286, and #315. If his methods were correct, any integer M could be "allowed" or "disallowed."

Third, the Beal Conjecture does not [I]require[/I] the three terms be pairwise co-prime, as he now claims. Only that the set has no common factor, which is how Don uses "coprime" (notice I don't say "requires", because he doesn't, he just mentions it in passing) in his proof "so that the only common factor possible is the 'trivial' unity." [b]This is not a requirement to be pairwise coprime, so the "proof" works the same way with four terms.[/b]

Now, it is true that you can't have a factor common to exactly two of the three terms; but that is a consequence of having three terms, not a requirement.

Still, Don thinks:
Which is the problem. Because 193^3+461^3+631^3=709^3. These terms are not merely pairwise co-prime, they are actual primes.

So Don has now admitted there is a fatal flaw in his proof, since it predicts a falsehood.[/QUOTE]

if he means exponents w,x,y,z>2 then this isn't a counterexample. because gcd(3,3) = 3

[QUOTE=science_man_88;260535]if he means exponents w,x,y,z>2 then this isn't a counterexample. because gcd(3,3) = 3[/QUOTE]
Please, don't give him an opening to misinterpret what you say. Because he has always shown that he will misinterpret as much as he can to claim he is right, when it is clear to everybody he is wrong.

In a^w+b^x+c^y=d^z, he claims that either one of {w,x,y,z} must be less than 3, or two of {a,b,c,d} must share a factor. I provided a counterexample (even though his "proof" only assumes [not requires] that {a,b,c,d} do not all share a factor).

Don's "proof" has a fatal flaw that even he can't worm out of, since he claimed it predicted otherwise.

[QUOTE=Condor;260541]Please, don't give him an opening to misinterpret what you say. Because he has always shown that he will misinterpret as much as he can to claim he is right, when it is clear to everybody he is wrong.

In a^w+b^x+c^y=d^z, he claims that either one of {w,x,y,z} must be less than 3, or two of {a,b,c,d} must share a factor. I provided a counterexample (even though his "proof" only assumes [not requires] that {a,b,c,d} do not all share a factor).

Don's "proof" has a fatal flaw that even he can't worm out of, since he claimed it predicted otherwise.[/QUOTE]

then it's me who misinterprets anyways if you really don't want me to say anything take all my post from the thread and delete them.

science_man_88,

I know this is a bit harsh, but I'm going to write it anyways. Since you don't understand most of the math involved I recommend that you don't get involved with this discussion. When you state that you understand one of the steps in his logic, you are implicitly giving support to his proof.

My opinion of him has changed from crank to troll. In his last post he was deliberately trying to provoke six of us. His cherry picking and hand waving are getting rather tiresome. I suspect that he knows that he is wrong, but chooses to provoke us because we have shown his proof to be a failure (many times over).

I spent a couple of hours last night entertained by all of this and still only got about halfway through all of the posts. I have a solution:

Let's all resolve not to respond to this massive crank any further. If we don't respond, he will just go away and start a thread somewhere else. No matter what he posts, we just ignore it. This can be easily accomplished by putting him on your ignore list. I think I will do that now so that I'm not tempted.

[QUOTE=rogue;260547]My opinion of him has changed from crank to troll.[/QUOTE]

I've gone through the same transition in my thinking.

[QUOTE=CRGreathouse;260557]I've gone through the same transition in my thinking.[/QUOTE][AOL]Me too !!!!???!!!?!!!![/AOL]

Paul

Quoting "rogue":
[QUOTE]
I recommend that the thread be locked and he be banned.
[/QUOTE]
Quoting "rogue" again:
[QUOTE]
I've already stated that it would be a good thing for
this thread to gain popularity...
[/QUOTE]
Well, if it's a "good thing for this thread to gain popularity"
then why is "rogue" crying for this thread to be locked? :missingteeth:
You see how illogical he is? You see how he contradicts himself?
You see what an utterly disingenuous liar he is?

The word "rogue" is [B][I]defined[/I][/B] as "A dishonest, knavish person",
and he certainly fits that description!

Quoting "roque":
[QUOTE]
Not one person here (besides yourself) is trying to defend your proof.
[/QUOTE]
All that means is that I'm a lot smarter than the rest of you.

Quoting "rogue" (Writing to science_man_88)
[QUOTE]
I know this is a bit harsh, but I'm going to write it anyways.
Since you don't understand most of the math involved I recommend that
you don't get involved with this discussion. When you state that you
understand one of the steps in his logic, you are implicitly giving support to his proof.
[/QUOTE]
I am the OP and the host of this thread, and science man is
perfectly welcome to post on [B][I]my[/I][/B] thread any time he likes.

In my opinion, "rogue" doesn't understand the math here
any better than science_man_88. In fact, there are certain
things that science_man_88 understands [B][I]better[/I][/B] than "rogue",
such as the [B][I][U]fact[/U][/I][/B] that "[B][COLOR=red]0 cannot divide any number[/COLOR][/B] [COLOR=blue][B]exept itself[/B][/COLOR]".

[QUOTE]
Based upon your statement if n*0=0 implies that 0/0=n then by substituting 1 for n
one could say that 1*0=0 implies that 0/0=1 (and only 1). That is so wrong it is laughable.
[/QUOTE]

"(and only 1)" [B][I]doesn't apply[/I][/B] because n*0 = 0 implies that 0/0 = n
for [B][I]any[/I][/B] number n.

What's really laughable is that after all this time, you [I][B]still[/B][/I] think that
"indeterminate forms" are somehow an "issue" in my proof when in
fact they are entirely avoidable and therefore utterly trivial.

In my proof, logarithms [B][I]aren't even involved [/I][/B]when z = 1 or z = 2.
In my proof, 1^(0/0) = 1^n = 1 and 1^(2/0) = [B][COLOR=red]DISALLOWED[/COLOR][/B].
Thus, in my proof, "indeterminate forms" are a "non-issue". Period.

Quoting "condor":
[QUOTE]
In a^w+b^x+c^y=d^z, he claims that either one of {w,x,y,z}
must be less than 3, or two of {a,b,c,d} must share a factor.
I provided a counterexample (even though his "proof" only assumes
[not requires] that {a,b,c,d} do not all share a factor).
[/QUOTE]

Unlike the rest of you, I [B][I]always[/I][/B] admit it when I make a mistake.
When I wrote...

Quoting myself:
[QUOTE]
However, my proof [B][I]predicts[/I][/B] that with [B][I]four[/I][/B] terms, at least
[B][I][U]two[/U][/I][/B] of them will [B][I]always[/I][/B] have a common factor if w,x,y,z > 2.
[/QUOTE]

what I meant to write was w,x,y,z > 3.

Quoting "rogue":
[QUOTE]
...people could see how wrong your proof is
[/QUOTE]
If my proof is wrong, then why are you [B][I][U]still[/U][/I][/B] posting on this thread?

People can see that the clucking chickens in our colleges and
universities are [B][I]afraid[/I][/B] to engage me in a fair and honest debate.

Here again, is my challenge...

Quoting Myself:
[QUOTE]
[SIZE=3]Since all it would take is [B][I][U]one[/U][/I][/B] fatal flaw to refute my proof[/SIZE]
[SIZE=3]let my opponents pick one, and [B][I]only one[/I][/B] issue that they[/SIZE]
[SIZE=3]are certain constitutes a "fatal flaw", and let them also pick[/SIZE]
[SIZE=3]one and [B][I]only one[/I][/B] "champion" to debate that issue with me,[/SIZE]
[SIZE=3]in a seperate thread. Let both sides use their real names,[/SIZE]
[SIZE=3]and let the debate be moderated by someone like Garo.[/SIZE]

Hopefully, my opponent would be the math department
of some prestigious university such as Princeton or U.C.L.A,
and I urge all my opponents here in this forum who are in
college or university to approach [B][I]their[/I][/B] math dept. heads
with my challenge. (What's the matter, are you chicken?) :missingteeth:

I can virtually [B][I]guarantee[/I][/B] that they [B][I][U]will[/U][/I][/B] "chicken out" because
they will [B][I][U]never[/U][/I][/B] be able to find that [B][U]one[/U][/B] "fatal flaw".

All they will be able to come up with are a bunch of silly
"non-issues" like the ones in this thread, and using [B][I]any[/I][/B] of
those silly non-issues will cause them to [B][I][U][SIZE=3][COLOR=red]lose[/COLOR][/SIZE][/U][/I][/B] the debate! :missingteeth:

Thus, if this debate does occur, I win,
and if it doesn't occur, I also win... [B][I]by default[/I][/B]!
[/QUOTE]

Don.