Quoting CRGreathouse:
[QUOTE]
If he knew basic calculus, he'd say that if c = T
then his equation (1) is indeterminate of form [TEX]0^\infty[/TEX].
[/QUOTE]
No, I would [B][I][U]never[/U][/I][/B] say anything that stupid!

Now, if CRGreathouse knew basic calculus,
then he would know that we [B][I]can't[/I][/B] let T= c in

(c/c) * c^3 = (T/T) * c^3 =
T*(c/T)^((3*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

because doing so would result in

(c/c) * c^3 = c*(1)^(2/0)

which is [B][COLOR=red]strictly disallowed[/COLOR][/B] because it involves [COLOR=red][B]division by zero[/B][/COLOR].

Also, if CRGreathouse knew basic calculus,
then he would know that we can't even let T [B][I]"approach"[/I][/B] c
because T and c are [B][I][U]defined[/U][/I][/B] as positive integers.

Moreover, if CRGreathouse knew basic calculus,
then he would know that even if we could let T approach c
the result would be c*[TEX]1^\infty[/TEX] and [B][I]not[/I][/B] [TEX]0^\infty[/TEX] as he believes!

Quoting CRGreathouse:
[QUOTE]
Ah. So...
[/QUOTE]
That's Japanese slang for "Yes, now I understand."
(But I don't think he does. Then again,
he could simply be describing himself.)

Quoting CRGreathouse:
[QUOTE]
Care to elucidate?
You could start with a definition of your "cancelled common factors".
[/QUOTE]
In an equation such as

T*a^x + T*b^y = T*c^z,

T is called the "[COLOR=black]common factor[/COLOR]" because it is
a "factor" that is "common" to all three terms.

Dividing the above equation by T results in

(T/T)*a^x + (T/T)*b^y = (T/T)*c^z

where the expression (T/T) can now be described as
the "[B][COLOR=red]cancelled common factor[/COLOR][/B]" because it represents
the original "common factor", but in a "cancelled state".

The fact that CRGreathouse and science man
didn't even know what a [B][COLOR=red]cancelled common factor[/COLOR][/B] is
and had to go "Google searching" in order to find out
speaks volumes of how ignorant they really are!

Quoting CRGreathouse:
[QUOTE]
Actually, my friends *do* think I'm crazy.
[/QUOTE]
So do I "CRNuthouse"... so do I.

Don.

[QUOTE=Don Blazys;257830]Quoting CRGreathouse:

No, I would [B][I][U]never[/U][/I][/B] say anything that stupid!

Now, if CRGreathouse knew basic calculus,
then he would know that we [B][I]can't[/I][/B] let T= c in

(c/c) * c^3 = (T/T) * c^3 =
T*(c/T)^((3*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))[/QUOTE]

That's not what being an indeterminate expression means -- you would know that if you knew high-school calculus.

I've already mentioned that the implicit domain for your expression requires c and T to be distinct and positive, while the original does not. Both my equation and your expression require c and T to be nonzero. Your expression is equal to either side of my equation on those stronger conditions, or undefined otherwise.

[QUOTE=Don Blazys;257830]Also, if CRGreathouse knew basic calculus,
then he would know that we can't even let T [B][I]"approach"[/I][/B] c
because T and c are [B][I][U]defined[/U][/I][/B] as positive integers.[/QUOTE]

Certainly I don't have to be bound by that restriction! It's like solving an integer programming problem with the simplex or ellipsoid method: you find a solution in [TEX]\mathbb{R}^n[/TEX] and then if it happens to be in [TEX]\mathbb{Z}^n[/TEX] you're done.

If I were to use college rather than high-school mathematics I would say that you could take the analytic continuation of your expression, which would be... wait for it... c^3. In fact your expression differs from c^3 only in its (implicit) domain.

[QUOTE=Don Blazys;257830]Moreover, if CRGreathouse knew basic calculus,
then he would know that even if we could let T approach c
the result would be c*[TEX]1^\infty[/TEX] and [B][I]not[/I][/B] [TEX]0^\infty[/TEX] as he believes![/QUOTE]

Oops, did I actually type that? Of course you're right, I intended [TEX]1^\infty.[/TEX]

[QUOTE=Don Blazys;257830]In an equation such as

T*a^x + T*b^y = T*c^z,

T is called the "[COLOR=black]common factor[/COLOR]" because it is
a "factor" that is "common" to all three terms.

Dividing the above equation by T results in

(T/T)*a^x + (T/T)*b^y = (T/T)*c^z

where the expression (T/T) can now be described as
the "[B][COLOR=red]cancelled common factor[/COLOR][/B]" because it represents
the original "common factor", but in a "cancelled state".[/QUOTE]

Are you leaving it in because you want to change the domain to exclude T = 0, or would you write it that way even if the domain was explicitly specified to exclude T = 0? If it's known that T is nonzero then you can simply write

a^x + b^y = (T/T)*a^x + (T/T)*b^y = (T/T)*c^z = c^z

(though a step like this would be too trivial to ever be written outside a pedagogical exercise.)

[QUOTE=Don Blazys;257830]The fact that CRGreathouse and science man
didn't even know what a [B][COLOR=red]cancelled common factor[/COLOR][/B] is
and had to go "Google searching" in order to find out
speaks volumes of how ignorant they really are![/QUOTE]

I don't need to go to Google to tell you that your term is not standard mathematical parlance. I think that science_man_88 searched it, though, and found essentially nothing on the web, mostly your drivel.

Quoting CRGreathouse:[QUOTE]
I've already mentioned that the implicit domain for
your expression requires c and T to be distinct and positive,
while the original does not.
[/QUOTE]
Before I introduced "cohesive terms", the two sides of the equation:

(c/c) * c^3 = (T/T) * c^3

were thought to have properties that are [B][I]essentially the same[/I][/B].

Now, it is known that their properties are [B][I]radically different[/I][/B],
because even though the coefficient of either term is 1,

(T/T) * c^3 can be used to derive the identity:

(T/T) * c^3 = T*(c/T)^((3*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

whereas (c/c) * c^3 cannot.

[QUOTE]
In fact your expression differs from c^3 only in its (implicit) domain.
[/QUOTE]
That's right kid.

[QUOTE]
If it's known that T is nonzero then you can simply write

a^x + b^y = (T/T)*a^x + (T/T)*b^y = (T/T)*c^z = c^z
[/QUOTE]

Well I suppose you can write a^x + b^y = c^z
if you are using math to [B][I]bake cookies[/I][/B], but since
writing (T/T)*a^x + (T/T)*b^y = (T/T)*c^z
allows us to derive the powerful and profound
logatithmic identity:

(T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)),

whose very existence [B][I]demonstrates[/I][/B] that both
Beal's conjecture and Fermat's Last Theorem are true,
[B][U]students will make up their own minds [/U][/B]as to
which terms contain more information and which terms
are more consistent with the properties of logarithms.

Quoting CRGreathouse:
[QUOTE]
Of course you're right...
[/QUOTE]
Of course.

And I must say young man,
that I am proud of you
for finally realizing that !

Don.

[QUOTE=Don Blazys;258388]Now, it is known that their properties are [B][I]radically different[/I][/B],
because even though the coefficient of either term is 1,

(T/T) * c^3 can be used to derive the identity:

(T/T) * c^3 = T*(c/T)^((3*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

whereas (c/c) * c^3 cannot.[/QUOTE]

Wrong.

I'm not surprised you'd make an elementary mistake like that, though.

[QUOTE=CRGreathouse;258401]Wrong.

I'm not surprised you'd make an elementary mistake like that, though.[/QUOTE]

[FONT=Consolas][SIZE=3]Reading through this thread (and very quickly resorting to skimming), this seems to be the crux of the matter. Don Blazys simply refuses to believe (Or is it to understand? Or to recognize what it is that others are saying?) that what he calls an identity ...[/SIZE][/FONT]
[FONT=Consolas][SIZE=3][quote](T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))[/quote][/SIZE][/FONT]
[FONT=Consolas][SIZE=3]Is really this:[/SIZE][/FONT]
[FONT=Consolas][SIZE=3][quote] (T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)) except when c=T[/quote][/SIZE][/FONT]
[FONT=Consolas][SIZE=3]The "identity" doesn't exist in isolation, it is inexorably tied to the domain he implicitly assumed when he derived it. He clearly understands that such domains are important, because he refuses to even consider using values outside of the domain he [I]explicitly[/I] assumed, even though there is no reason the identity doesn't apply to some of them as well. But he completely ignores that one step in the derivation he refers to was a division by ln(c/T), which implicitly requires a domain that excludes T=c. As far as his identity is concerned, that value does not exist, any more than T=-sqrt(3) does. By his own methods Don must admit it is his "identity" itself that is "strictly prohibited," not any antecedent of it. Not that I think he will, of course. That's why I don't dance with trolls.[/SIZE][/FONT]

[FONT=Consolas][SIZE=3]But there are trivial ways to see that this proof must be wrong. Go back to its very beginning, and make one very minor change. Instead of assuming x and y are elements of {3,4,5,6,....} as Don does, assume x=y=1. Every single step in his proof can be repeated in the same way he made them, eventually reaching the conclusion that you can't find co-prime positive integers a, b, c; and an integer z>2, such that a+b=c^z. So I guess 7+20 does not sum to 27. Truly an inspirational breakthrough! :)[/SIZE][/FONT]

[FONT=Consolas][SIZE=3]Or let Z1={1,5,7,11, ...} (i.e., all positive integers not divisible by 2 or 3), Z2={2,4,6,8} (even integers), and Z3={3,9,15,21,...} (odd integers divisible by 3). Then add another equation in the style of (1) and (2): a^x+b^y = (T*(c/T)^((Z3/3)*ln(c)/ln(T)-1)/(ln(c)/ln(T)-1))^3 to show that Z3>3 is "disallowed," but a^x+b^y=[(c/c)*c^1]^3 is "allowed." This lets you "prove" a modified Beal Conjecture - one that allows powers of 3.[/SIZE][/FONT]

[FONT=Consolas][SIZE=3](Of course, we could claim that the [B]assumption that Don's proof is valid[/B] is now disproved by reductio ad absurdum, since it leads to a contradiction of itself. But that would be too easy. :) )[/SIZE][/FONT]

Right -- actually the domain he needs is more restricted, for example it requires that c and T are positive. But if you have the appropriate domain (that is, any subset of the maximal domain c > 0, t > 0, c != T) then this follows, as any high-school student should be able to show. Blazys thinks that it's amazing, but it's really just slightly tedious.

Of course you're not the first, or the second, even just on this thread, to point out that Blazys' 'proof' would show various "obviously wrong" statements to be true. (And there have been others on different forums, no doubt.) But he can't acknowledge the flaws in his 'proof', nor can he understand the transfer arguments. I had hoped to show him the light on some of these points so that perhaps he could spend his mathematical time on projects where he isn't barking up the wrong tree, but that was naive of me to think that I could explain that -- or much of anything -- to him.

Quoting Condor:
[QUOTE]
...he completely ignores that one step in the
derivation he refers to was a division by ln(c/T),
which implicitly requires a domain that excludes T=c.
[/QUOTE]You are wrong. (And you write like a first grader!)

The domain excludes T = c [B][I][COLOR=black]if and only if[/COLOR][/I][/B] z > 2.
Otherwise, the domain does [B][I]not[/I][/B] exclude T = c,
which is why the proof works.


Quoting Condor:
[QUOTE]
Instead of assuming x and y are elements of {3,4,5,6,....} as Don does, assume x=y=1.
[/QUOTE]If you read page 2 of:

[U][COLOR=Navy]httр://donblazys.com/03.рdf[/COLOR][/U]

where it says "the Beal equation can also be represented as"...
then maybe, just maybe you will be able to figure out that
similar logarithmic identities can be derived from all three terms
and that in order to avoid redundancy and for the sake of brevity,
the "c term" was [B][I]arbitrarily[/I][/B] chosen to demonstrate the restriction
on its exponent when the other exponents are greater than 2.

Thus, if you want to assume x=y=1, then you must assume
a version of the proof where either the "a term" or the "b term"
is used to derive the logarithmic identity!

Quoting Condor:
[QUOTE]
Then add another equation in the style of (1) and (2):
a^x+b^y = (T*(c/T)^((Z3/3)*ln(c)/ln(T)-1)/(ln(c)/ln(T)-1))^3
to show that Z3>3 is "disallowed,"...
[/QUOTE]To do that, you would first have to find some [B][I]construct[/I][/B] which
demonstrates that [B][I]any [/I][/B]sum or difference of two terms is implicitly
a "cube under a third degree radical".

The logarithmic identities in the proof are [B][I]derived[/I][/B] from a [B][I]construct[/I][/B]
which demonstrates that [B][I]any[/I][/B] sum or difference of two terms is implicitly
a "square under a second degree radical" and that [B][I]precludes[/I][/B]
the possibility of finding a similar construct for cubes.

Sorry "Condor", but your arguments are both stupid and silly.
If they were valid, then I would drop my proof like a hot potato.
But at least your pseudonym makes sense. After all, a "condor" is
a bird whose reasoning power is severly limited by its "bird brain".

Don.

Quoting CRGreathouse:
[QUOTE]
But he can't acknowledge the flaws in his 'proof'...
[/QUOTE]
That's because my proof has no fatal flaw and is absolutely irrefutable.

If there was a fatal flaw, then I would drop it like a hot potato.

Only a fool would spend precious time on a result that was flawed!

Anyone who can't see that my proof is both true and correct is an idiot.

(I don't mean that as an insult. To me, that's just a fact!)

Don.

[QUOTE=Don Blazys;258577]If there was a fatal flaw, then I would drop it like a hot potato.[/QUOTE]

As is evident to everyone on this thread but you, you wouldn't. :smile:

I made a list some posts ago of those on this thread who have found fatal (vs. fixable) mistakes in your supposed proof. I could now add at least one name to that list. As I recall all but one of the disproofs were correct; one attempted disproof was itself flawed and thus not a challenge.

I'd still like to see your list of mathematicians who have "tried and failed" to find a flaw in your proof. I would hope that any college math major could find a mistake in half an hour and that any actual mathematician could find it in less than 5 minutes (depending on their reading speed!). Anyone who has put serious effort into finding a mistake and can't is demonstrating a serious lack of mathematical ability. If you've rooted out any such pseduo-mathematicians I'd really like to know.

Of course I [i]don't[/i] mean someone who looked at it and dropped it in disgust a minute later, nor someone who found a flaw that you don't acknowledge or understand -- I expect these to be fairly common.

[QUOTE=CRGreathouse;258549]Right -- actually the domain he needs is more restricted, for example it requires that c and T are positive.[/quote]
I just assumed we started with Don's domain - positive integers (which is why T=-sqrt(3) is also outside of it). I made explicit the new restriction he implied, but is still turning a blind eye to.
[quote]I had hoped to show him the light on some of these points so that perhaps he could spend his mathematical time on projects where he isn't barking up the wrong tree, but that was naive of me to think that I could explain that -- or much of anything -- to him.[/QUOTE]
It's pretty obvious, from his methods, that his concept of logic works backwards. He starts with a conclusion ("All three terms must have exponents greater than 2 at the same time") and somehow concludes whatever he can "demonstrate" about the the individual concepts in the conclusion (he "demonstrated" that z was restricted in c^z independent of a^x or b^y; and then similar things about x and y) must be related to each other and so "prove" that conclusion. You really can't think he will do it any differently when you try to educate him. Since he is already convinced his proof is right, he will always engineer some kind of twisted argument against anything that shows it isn't.

BTW - is it possible to edit posts? I don't see an icon for it. I didn't realize I had cut-and-pasted formating when I pulled that one in from the editor I prefer, and I agree it looks ugly. (Oh - is it that only the last post can be edited? This one did get an "edit" icon.)

don I can't even find a place that agrees with your definitions of Beal's conjecture:

[QUOTE]Beal’s Conjecture can be stated as follows: For positive integers: a, b, c, x, y, z,
if: a[SUP]x[/SUP]
+ b[SUP]y[/SUP]
= c[SUP]z[/SUP]
and: a, b, c are co-prime, then: x, y, z are [B]not all greater than 2[/B][/QUOTE]

[QUOTE="http://en.wikipedia.org/wiki/Beal's_conjecture"]where A, B, C, x, y, and z are positive integers with [B]x, y, z > 2[/B] then A, B, and C must have a common prime factor.[/QUOTE]

[QUOTE="http://www.math.unt.edu/~mauldin/beal.html"] If A[SUP]x[/SUP] +B[SUP]y[/SUP] = C[SUP]z[/SUP] , where A, B, C, x, y and z are positive integers and [B]x, y and z are all greater than 2[/B], then A, B and C must have a common prime factor.[/QUOTE]

that's a major flaw and we don't need to bother with looking at the rest.