[QUOTE=rogue;260133]Just because it is available via Google doesn't mean that it is factual. The only way to guarantee that your proof is factual is to have it peer reviewed in a math journal. Having various undereducated people say that "it looks right" mean nothing. There are other proofs to Beal's Conjecture that are just are wrong as yours.



Could you provide a link?

According to this [URL="http://www.wolframalpha.com/entities/famous_math_problems/beal's_conjecture/21/b5/ar/"]link[/URL], you are eligible for a $100,000 reward, or are you so rich that the reward is worth nothing to you?

BTW, in case anyone else is interested, Don tried this before over [URL="http://www.physicsforums.com/showthread.php?t=301139"]here[/URL]. The mod over there shut down the thread.

Finally, the more hits to this thread on Google, the more people will see the incorrectness of his proof.[/QUOTE]

Hey, you can also increase page rank slightly by linking from outside...maybe the mods on physicsforum will return the favor!

I keep seeing nonsense mathematics like "0/0 = n" and it is driving me bonkers! This is the crux of the problem with Blazys' entire theory, and what holds many students back from understanding division, even at a level as high as college algebra.

Think of it this way. 0/0 IS IN NO WAY a traditional division or ratio in the sense of 7/10 or 1/2 or 3456/1234. As such it cannot be even thought of in the same realm as a "normal" division or ratio having a nonzero denominator. You just can't think of 0/0 as a "division" problem, no matter how much you might want to, no matter how much it might make your homework easy, no matter how it helps you prove (or think you can prove) Beal's Conjecture.

We are all (supposedly) interested here in number theory, to one degree or another. At the very heart of number theory is the Division Algorithm, which essentially gives us a basic idea of what division means, and how it is to be performed (yes, I know that it is much deeper, and technically we should get into the field axioms, etc. etc., but none of that is really needed to debunk what has been going on in here for weeks). According to Wikipedia (which is OK for a reference here, but as my college math mentors always told me, Internet citations (and especially Wikipedia) should never be a substantive base for serious mathematical research - be a true scholar, read the journals!), the Division Algorithm says the following:

Given two integers a and b, with b *NONZERO*, there exist unique integers q and r such that a = b * q + r, and where 0 <= r < abs(b), and where abs($) is taken to be the absolute value of $.

Now, let us naively look at 0/0. Suppose 0/0 is a regular division. Let's fill out the Division Algorithm: a = 0, b = 0...WHOOPS! I can't even begin! So if we're going to be able to treat 0/0 as a regular division, we're going to first need to patch up the old Division Algorithm - maybe we could call it the Blazys Division Algorithm, right?

To look at it another way, when I was very young (say first or second grade), I thought to myself that well, I know why I can't do things like 7 divided by 0, or 34.8 divided by 0, or pi divided by 0, because that is the same thing as asking to find x such that

x * 0 = 7, x * 0 = 34.8, and x * 0 = pi,

and, while I was fifteen years away from learning the field axioms, I knew that anything times 0 was ALWAYS 0.

[QUOTE=NBtarheel_33;260191]I keep seeing nonsense mathematics like "0/0 = n" and it is driving me bonkers! This is the crux of the problem with Blazys' entire theory, and what holds many students back from understanding division, even at a level as high as college algebra.

Think of it this way. 0/0 IS IN NO WAY a traditional division or ratio in the sense of 7/10 or 1/2 or 3456/1234. As such it cannot be even thought of in the same realm as a "normal" division or ratio having a nonzero denominator. You just can't think of 0/0 as a "division" problem, no matter how much you might want to, no matter how much it might make your homework easy, no matter how it helps you prove (or think you can prove) Beal's Conjecture.

We are all (supposedly) interested here in number theory, to one degree or another. At the very heart of number theory is the Division Algorithm, which essentially gives us a basic idea of what division means, and how it is to be performed (yes, I know that it is much deeper, and technically we should get into the field axioms, etc. etc., but none of that is really needed to debunk what has been going on in here for weeks). According to Wikipedia (which is OK for a reference here, but as my college math mentors always told me, Internet citations (and especially Wikipedia) should never be a substantive base for serious mathematical research - be a true scholar, read the journals!), the Division Algorithm says the following:

Given two integers a and b, with b *NONZERO*, there exist unique integers q and r such that a = b * q + r, and where 0 <= r < abs(b), and where abs($) is taken to be the absolute value of $.

Now, let us naively look at 0/0. Suppose 0/0 is a regular division. Let's fill out the Division Algorithm: a = 0, b = 0...WHOOPS! I can't even begin! So if we're going to be able to treat 0/0 as a regular division, we're going to first need to patch up the old Division Algorithm - maybe we could call it the Blazys Division Algorithm, right?

To look at it another way, when I was very young (say first or second grade), I thought to myself that well, I know why I can't do things like 7 divided by 0, or 34.8 divided by 0, or pi divided by 0, because that is the same thing as asking to find x such that

x * 0 = 7, x * 0 = 34.8, and x * 0 = pi,

and, while I was fifteen years away from learning the field axioms, I knew that anything times 0 was ALWAYS 0.[/QUOTE]

[QUOTE]The expression

requires a value to be found for the unknown quantity in

Again, any number multiplied by 0 is 0 and so this time every number solves the equation instead of there being a single number that can be taken as the value of 0/0[/QUOTE]

from this one can say in my opinion that 0/0 is variate and hence can be denoted as a variable. it's the only thing that makes sense. oh course I'm not suggesting we use the constant but rather replace it because the rest work out well ( and yes I know someone will point out the rest depends on it).

[QUOTE=NBtarheel_33;260191]I keep seeing nonsense mathematics like "0/0 = n" and it is driving me bonkers! This is the crux of the problem with Blazys' entire theory, and what holds many students back from understanding division, even at a level as high as college algebra.[/QUOTE]

There are unmeasurable amounts of math [I]teachers[/I] who don't know that 0 is an even number (or know incorrectly that it is an odd number or that it's parity is undefined), and ...they teach!

[QUOTE="Wiki"]It is uncertain [URL="http://en.wikipedia.org/wiki/Parity_of_zero#Education"]how many teachers harbor misconceptions about zero[/URL]. The Michigan study did not publish data for individual questions. One report comes from Betty Lichtenberg, who wrote an article titled "Zero is an even number" in the journal [I]The Arithmetic Teacher[/I] in 1972. Lichtenberg, an associate professor of mathematics education at the University of South Florida, draws on her experience with a course she and her colleagues taught on methods for teaching arithmetic. She reports that several sections of prospective elementary school teachers were given a true-or-false test including the item "Zero is an even number." They found it to be a "tricky question", and [B]about two thirds answered "False"[/B].[/QUOTE]
Don't even get started about division by zero! Teachers like these go on writing letters of support that are then proudly presented on some obscure websites. If you haven't read those letters before criticizing the alleged proof, you don't know what you are missing. You won't have any further questions after you've read them.

[QUOTE=rogue;260119]Where in your proof do you derive those conditions? I'm not talking about the "pre-conditions" for the proof. Somewhere in your proof you must derive the fact that a,b,c must be co-prime and x,y,z > 2. You haven't done that. All your proof has is some faulty logic based upon some transformation using c and z. Your proof doesn't do anything with a,b,x, and y. For example, one could extend your logic to say that there are no integral solutions to:

a^w + b^x + c^y = d^z[/quote]
That is one I hadn't thought of before! Don's "proof" can be used the exact same way on four terms instead of three. Every other "parallel proof" changed something Don could point at and say "you changed that, so it isn’t the same thing." He'd be wrong, as always, but that has never stopped him from claiming to be right. But not this one, since the change uses the same fraudulent claim that repeating the "proof" for each term is what ties that number of terms together!

Well, Don, if the methods behind your proof are correct, then the same methods prove [tex]a^w+b^x+c^y[/tex] can't equal [tex]d^z[/tex] unless one of w,x,y, or z is 2 or less; or a,b,c, and d share a factor that is greater than 1. Yet [tex]3^3+4^3+5^3=6^3[/tex].

Well?

Don's rebuttals to your postings have already been reduced to pointing out that a condor is a kind of bird. If you expect anything better this time, you must be far less jaded than I am.

[QUOTE=Condor;260274]That is one I hadn't thought of before! Don's "proof" can be used the exact same way on four terms instead of three. Every other "parallel proof" changed something Don could point at and say "you changed that, so it isn’t the same thing." He'd be wrong, as always, but that has never stopped him from claiming to be right. But not this one, since the change uses the same fraudulent claim that repeating the "proof" for each term is what ties that number of terms together!

Well, Don, if the methods behind your proof are correct, then the same methods prove [tex]a^w+b^x+c^y[/tex] can't equal [tex]d^z[/tex] unless one of w,x,y, or z is 2 or less; or a,b,c, and d share a factor that is greater than 1. Yet [tex]3^3+4^3+5^3=6^3[/tex].

Well?[/QUOTE]

Note also that there ARE solutions to a^x + b^y = c^z for x,z > 2.
e.g. 243 + 100 = 343. A requirement is that 1/x + 1/y + 1/z < 1.
Don fails to use this in any way. His 'proof' proves that
243 + 100 = 343 is impossible.

Just IGNORE him. He is a classic CRANK. Trying to discuss math with him
is POINTLESS. BTW is Don Blazys his real name???

[QUOTE=R.D. Silverman;260289]Just IGNORE him. He is a classic CRANK. Trying to discuss math with him is POINTLESS. BTW is Don Blazys his real name???[/QUOTE]It may be pointless, but it sure is amusing.

Paul

[QUOTE=R.D. Silverman;260289]BTW is Don Blazys his real name???[/QUOTE]

If not, why would he have chosen it? Don like in "Don Corleone?" "Blazys" like in fire? "King of the Flames?"

Have you noticed an unbelievable combinations of ego and ignorance? Taunts of "I won" if the thread threatens to die out? Gloats over the ever expanding thread length? Unbelievable appeals to ridiculous authority (third graders, rejection letters)?

Has it occurred to anyone that the unbelievable should not be believed? That perhaps you have trolled by a master troller, the King of the Flames?

The hope is that he finally sees the errors in his proof, retracts the proof and apologizes to everyone. Blame for me expecting too much from him.

If he comes back and starts his "hand waving" arguments again, I recommend that the thread be locked and he be banned. It happened over at the other forum.

One unfortunate thing is that he will try to push this tripe on others. I can't do much about that, but many will be mislead by his proof and unable to show its faults. Hopefully some of them will discover this thread and see how we have thoroughly dismantled his proof.

He stated that he works in a high school. I hope it is as a janitor and not as a teacher. If as a teacher, it doesn't speak well for the state of education.

BTW, is there anyone here who thinks that his proof could possibly be correct?

I, for one, wouldn't ban him simply for being wrong. Not even for being appalingly wrong [I]and[/I] smug about it. However, his more recent post increasingly resorted to countering arguments with insults, so he may enjoy a health dose of banhammer soon, after all.