Standard crank division by zero thread - Page 31

Don Blazys · 2011-04-30, 12:23

Quoting em99010pepe:

Quote:

So have you published in any of those journals?

That's a good idea. Thanks for the info!

Don.

rogue · 2011-04-30, 13:34

Quote:

Originally Posted by Don Blazys

$3^3=$\frac{3}{3}$*3^3= <br /> 3*(\frac{3}{3})^{\frac {\frac{{3}*{ln(3)}}{ln(3)}-1}{\frac{ln(3)}{ln(3)}-1}} = <br /> 3*(1)^{\frac {\frac{{3}*{ln(3)}}{ln(3)}-1}{\frac{ln(3)}{ln(3)}-1}}=3*(1)^{\frac{2}{0}}=DISALLOWED$

because 0 cannot divide any number exept itself.

You missed my finer point. All this means is that the transformation itself is "disallowed". Your transformation says nothing about the Beal Conjecture. If one were to follow your logic, then one could show that a^2 + b^2 = c^2 has no integral solution. If one were to follow your logic, then one could easily prove Fermat's Last Theorem. Obviously everyone besides you recognizes that FLT was not proven so easily.

BTW, regarding the square root, you must handle all cases in your proof in order for it to be rigorous. You can't choose one root over another.

CRGreathouse · 2011-04-30, 20:51

Quote:

Originally Posted by em99010pepe

So have you published in any of those journals?

He's already explained that the rejection from the journal he submitted to was, in fact, a commendation on the quality of his work. I'm sure further rejection letters will only puff his feathers further.

Don Blazys · 2011-05-01, 04:06

Quoting CRGreathouse:

Quote:

...the journal he submitted to was, in fact,
a commendation on the quality of his work.

"Rejection" or "acceptance" doesn't matter much to me.
What really matters to me is that my proof is both true
and correct and that the referee gave it some support.

Quoting CRGreathouse:

Quote:

...puff his feathers further.

Well, the only "feathers" I see are on that
"chicken without a head" avatar of yours!

Don.

flouran · 2011-05-01, 04:31

Quote:

Originally Posted by Don Blazys

"Rejection" or "acceptance" doesn't matter much to me.
What really matters to me is that my proof is both true
and correct and that the referee gave it some support.

If your proof was both true and correct, then most likely it would have been accepted by now by an appropriate journal.

Don Blazys · 2011-05-01, 04:34

Quoting "rogue":

Quote:

You can't choose one root over another.

No, we must choose the positive roots only.
Concepts such as co-primality are meaningless
for numbers other than positive integers.

Quoting "rogue":

Quote:

If one were to follow your logic, then one could show that
a^2 + b^2 = c^2 has no integral solution.

You couldn't "follow logic" if your life depended on it!
If z=1 or z=2, then logarithms aren't even involved!

Quoting "rogue":

Quote:

If one were to follow your logic, then one
could easily prove Fermat's Last Theorem.

"Beal's Conjecture" is the "general case" of
Fermat's Last Theorem. When I proved
Beal's Conjecture, then I automatically
proved Fermat's Last Theorem.

Don.

rogue · 2011-05-01, 04:45

Quote:

Originally Posted by Don Blazys

If z=1 or z=2, then logarithms aren't even involved!

What if z > 2? Are you saying then that there is no integral solutions to a^x + b^y = c^z for z > 2?

Quote:

Originally Posted by Don Blazys

"Beal's Conjecture" is the "general case" of
Fermat's Last Theorem. When I proved
Beal's Conjecture, then I automatically
proved Fermat's Last Theorem.

Don Blazys · 2011-05-01, 06:13

Quoting "flouran".

Quote:

If your proof was both true and correct, then most likely it
would have been accepted by now by an appropriate journal.

So far, all I sent out was one handwritten manuscript to
the Journal of the London Mathematical Society about
a dozen years ago. They replied that although the referee
gave it some support, (presumably because it is both true
and correct!) they simply didn't have enough journal space.

Quoting "rogue"

Quote:

What if z > 2? Are you saying then that there is no integral
solutions to a^x + b^y = c^z for z > 2?

If x,y, and z are > 2 and a,b,c are co-prime, then my proof
demonstrates that there can be no integral solutions.

That's why this topic is so incredibly popular! And...
That's why you are still posting on my thread.

Don.

em99010pepe · 2011-05-01, 07:40

Quote:

Originally Posted by Don Blazys

So far, all I sent out was one handwritten manuscript to
the Journal of the London Mathematical Society about
a dozen years ago. They replied that although the referee
gave it some support, (presumably because it is both true
and correct!) they simply didn't have enough journal space.

Here's the letter from 1999:

"We regret to inform you that we shall not be able to publish your paper. There is very great pressure on the publication space in the Journal at present, and although the referee gave some support to your work, we have felt obliged to reject your paper in favour of more highly recommended contributions."

I think it was a nice way to say your work is rubbish.
Why didn't you try again to submit your paper? Why?

xilman · 2011-05-01, 09:34

Quote:

Originally Posted by Don Blazys

No, we must choose the positive roots only. Concepts such as co-primality are meaningless for numbers other than positive integers.

Drivel!

-3 is co-prime to +14 because the only factors they have in common are the units -1 and +1.

3 is not co-prime to 15 because as well as both being divisible by the units, each is also divisible by +3 and -3.

Paul

Don Blazys · 2011-05-01, 10:19

Quoting "em99010pepe":

Quote:

I think it was a nice way to say your work is rubbish.

If my proof had some "fatal flaw", then
the referee would have given it no support whatsoever!

He would have simply pointed out that "fatal flaw".

Moreover, if my proof was "wrong", then
they would not have ended that letter by
suggesting that I send it to another good journal!

The fact is, my proof is simply an equation, and there is no error.

Quoting "em99010pepe":

Quote:

Why didn't you try again to submit your paper? Why?

Because I don't need to. Think about it. If you wanted to find a proof
of "Beal's Conjecture", then would you go to some university library and
search for some particular volume of some stuffy journal, or would you
simply Google search "Beal's Conjecture Proof"?

My proof is ranked number #1, #2 and #3 on Google, so any students
wishing to learn the truth about "Beal's Conjecture" and "Fermat's Last
Theorem" have easy access to it.

I also feel that threads such as this are a better vehicle for my proof
because they are "interactive", and therefore dynamic, alive and vital
with a much larger audience than any "journal" can possibly provide.

This thread is like a classroom, and my students are coming in droves.
They come to learn what mathematics is really all about, and
they come to be entertained by the spectacle of me defeating
my opponents at every turn.

Most importantly perhaps, I already did publish my proof in the online
journal "Unsolved Problems", where it can be refereed by the entire
math community!

Don.

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