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- 03-13-2017, 08:25 AM #131

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## Re: Fermat's last, and mine too.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 03-13-2017, 09:09 AM #132

- 03-13-2017, 10:33 AM #133

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## Re: Fermat's last, and mine too.

If you want to discuss mathematics with knowledgable people, then I am the person to talk to and you should listen to as I am vastely superior to you here and to my knowledge, everyone else feel free to correct me, with the highest degree in mathematics.

But in reality you do not want it.

- 03-13-2017, 10:35 AM #134

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- 03-13-2017, 11:14 AM #135

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## Re: Fermat's last, and mine too.

**Proof of Fermat’s Theorem**

(Requires two dimensions for independent integers a and b in the integer plane (a,b) so that both Fermat's expression and the Binomial Expansion can be defined as functions )

For the case for the Binomial Expansion , in the vector equation

so that even if is an integer, c is not.

cannot be an integer, since the coefficients are calculated independently for the magnitude**in this context**, where , so that c cannot be an integer for and independent integers on the integer plane , since that case is exhausted by Pythagorean triples where so that or which requires the complex expression to eliminate the multiplicative term.

This case is then easily expanded to the case for where , where only if or , in which case or but Fermat’s expression does not obtain.

This is because Fermat’s expression characterizes the metric of a Presburger arithmetic, which does not include multiplication. The set of Peano’s axioms does include multiplication, and so is complete, but complex numbers are required to eliminate the multiplicative factor in order to retrieve Fermat’s expression for , which is only possible because and are first order, so one can interchange variables. However, the factors in all include terms like where , as was possible in first order (where ) , so it is impossible to interchange variables, although via the cross product and the nullvector; i.e. and commute, satisfying that fundamental criterion for positive integers.

----------

**(in this context means with the field definition from the relativistic unit circle and its application to the plane (x,y) with the plane (a,b) a subset consisting of final states so the unit bases are established for each dimension; the comlete proof is in my pdf, not on this forum, and is available via the links in my signature).**

This means that the system of positive integers is consistent but not complete, since it is possible to express Pythagorean triples only if complex numbers are included, and in this case multiplicative products ae excluded. This shows that Gödel’s proof is valid since his characterization is only in terms of positive integers, and the system of positive integers is neither consistent or complete.

This also means that particle count in a physical system cannot be conserved (as integers) unless the system is described by Pythagorean triples, if not, the total count cannot be an integer, and so is relevant to renormalization and the metric tensor for the General Theory of Relativity.Last edited by BuleriaChk; 03-14-2017 at 11:36 AM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 03-13-2017, 02:15 PM #136
## Re: Fermat's last, and mine too.

That doesn't make sense. Example: a=2, b=3, then . Surely we all agree that 5 is an integer (whether or not you think it is rational.)

since that case is exhausted by Pythagorean triples where so that or which requires the complex expression to eliminate the multiplicative term.

This case is then easily expanded to the case for where , where only if or , in which case or but Fermat’s expression does not obtain.

This is because Fermat’s expression characterizes the metric of a Presburger arithmetic, which does not include multiplication. The set of Peano’s axioms does include multiplication, and so is complete,

but complex numbers are required to eliminate the multiplicative factor in order to retrieve Fermat’s expression for , which is only possible because and are first order, so one can interchange variables. However, the factors in all include terms like where , as was possible in first order (where ) , so it is impossible to interchange variables, although via the cross product and the nullvector; i.e. and commute, satisfying that fundamental criterion for positive integers.

This means that the system of positive integers is consistent but not complete, since it is possible to express Pythagorean triples only if complex numbers are included, and in this case multiplicative products ae excluded.

This shows that Gödel’s proof is valid since his characterization is only in terms of positive integers, and the system of positive integers is neither consistent or complete.

This also means that particle count in a physical system cannot be conserved (as integers) unless the system is described by Pythagorean triples,

if not, the total count cannot be an integer, and so is relevant to renormalization and the metric tensor for the General Theory of Relativity.

- 03-13-2017, 02:51 PM #137

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## Re: Fermat's last, and mine too.

Edit:

**No, I was right after all; I hadn't defined the Binomial expansion for n=2 as vectors (my bad, I was writing fast), which is fundamental to my proof as always (and absent, as always, from your responses).**

-------------------

I had said (without referring to vectors):

That is my bad; I still occasionally try to prove that the issue has something to do with any of the elements not being an integer. (I regress).

The proof for Fermat's theorem is for n>2, not n=2, in which case there is no first order multiple like in in the Binomial Expansion (compared with Fermat's expression where there is no multiplicative product whatever), so the dot product cannot apply. (that is the point of my vector proof).

**It has to do with rem(a,b,n) = 0 IFF a=0 or b=0 for n>2, NOT whether a or b or (a+b) to whatever power is an integer or not. So n=2 is irrelevant to my proof.**Of course, Fermat's expression can apply to real numbers where (but not integers, since otherwise would be a Pythagorean triple, which is only possible for n=2.)

I am just showing that the multiplication factor is omitted in Fermat's proof (there is no multiplication factor in the Fermat expression compared with the Binomial Expansion.

My whole focus is on n>2, not n=2 and the Binomial Expansion is ALWAYS integers, so the issue is what makes the Binomial Expansion different from Fermat's expression. Again, it is NOT whether a and b are integers or not, but that rem(a,b,n) = 0 IFF a=0 or b=0 so Fermat's expression cannot be valid (since it doesn't include multiplication that can be factored out for n>2).

That is corrected in the following post.Last edited by BuleriaChk; 03-14-2017 at 10:51 AM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 03-14-2017, 04:49 AM #138
## Re: Fermat's last, and mine too.

n=2 is definitely relevant to your "proof" because rem(a,b,n) = 0 IFF a=0 or b=0 for n=2, also. For n=2, your rem(a,b,n) equals 2ab, which can only be zero if a or b is zero. That should have been your first clue that your proof was wrong, but instead you tried to "fix" it by pretending you were taking a dot product instead of multiplying.

You have to admit, in the binomial expansion, 2ab is a product of integers, not a dot product of vectors.

Your proof does not work.

Of course, Fermat's expression can apply to real numbers where (but not integers, since otherwise would be a Pythagorean triple, which is only possible for n=2.)

I am just showing that the multiplication factor is omitted in Fermat's proof (there is no multiplication factor in the Fermat expression compared with the Binomial Expansion.

My whole focus is on n>2, not n=2 and the Binomial Expansion is ALWAYS integers, so the issue is what makes the Binomial Expansion different from Fermat's expression. Again, it is NOT whether a and b are integers or not, but that rem(a,b,n) = 0 IFF a=0 or b=0 so Fermat's expression cannot be valid (since it doesn't include multiplication that can be factored out for n>2).

Last edited by grapes; 03-14-2017 at 04:55 AM.

- 03-14-2017, 10:34 AM #139

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## Re: Fermat's last, and mine too.

**No, I was right after all; I hadn't defined the Binomial expansion for n=2 as vectors (my bad, I was writing fast), which is fundamental to my proof as always (and absent, as always, from your responses).**

(I am writing a lot of stuff in my addenda that I don't post here because of the tex difficulty; read my pdf's if you want to understand me).

**The correct paragraph should read:**

=====

For the case for the Binomial Expansion , in the vector equation

,

so that even if is an integer, is not.

cannot be an integer, since the coefficients are calculated independently for the magnitude**in this context**, where , so that c cannot be an integer for and independent integers on the integer plane , since that case is exhausted by Pythagorean triples where so that or which requires the complex expression to eliminate the multiplicative term.

----------

**(in this context means with the field definition from the relativistic unit circle and its application to the plane (x,y) with the plane (a,b) a subset consisting of final states so the unit bases are established for each dimension; the comlete proof is in my pdf, not on this forum, and is available via the links in my signature).**

**If you are going to deny the relevance of independent variables, characterized as vectors, then that is a subject for a different thread, since my proof depends on that characterization. If you continue to do that, everything you say is irrelevant to my proof, except "I don't believe in independent variables, or vectors and not even more than one dimension". That only needs to be said once, not continuously in a stream of one-dimensional irrelevant responses in my thread; any more than once, you are just spamming the thread.**(Back to my addendum for today, coming shortly)

And if you continue to deny vectors, then get the frick out of my thread and start your own, where you can ignore Descartes and spout your own theory in the echo chamber to the rest of the Village Idiots in the Peanut Gallery.

Answering your responses where you maintain this dog-headed instance on avoiding vectors as characterizing independent variables is a waste of time for everyone except the Peanut Gallery.

Last edited by BuleriaChk; 03-14-2017 at 11:36 AM.

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 03-14-2017, 01:34 PM #140
## Re: Fermat's last, and mine too.

No, you were right, you were wrong.

I hadn't defined the Binomial expansion for n=2 as vectors (my bad, I was writing fast), which is fundamental to my proof as always (and absent, as always, from your responses).

(I am writing a lot of stuff in my addenda that I don't post here because of the tex difficulty; read my pdf's if you want to understand me).

**The correct paragraph should read:**

=====

For the case for the Binomial Expansion , in the vector equation

,

One sub a? That's not a vector? Is it one, when it's squared? What's the point?

so that even if is an integer, is not.

cannot be an integer, since the coefficients are calculated independently for the magnitude**in this context**, where , so that c cannot be an integer for and independent integers on the integer plane , since that case is exhausted by Pythagorean triples where so that or which requires the complex expression to eliminate the multiplicative term.

----------

**(in this context means with the field definition from the relativistic unit circle and its application to the plane (x,y) with the plane (a,b) a subset consisting of final states so the unit bases are established for each dimension; the comlete proof is in my pdf, not on this forum, and is available via the links in my signature).**

**If you are going to deny the relevance of independent variables, characterized as vectors, then that is a subject for a different thread, since my proof depends on that characterization. If you continue to do that, everything you say is irrelevant to my proof, except "I don't believe in independent variables, or vectors and not even more than one dimension". That only needs to be said once, not continuously in a stream of one-dimensional irrelevant responses in my thread; any more than once, you are just spamming the thread.**

You, however, continue to repeat the same errors. If you think that's how we should deal with spammers--

(Back to my addendum for today, coming shortly)

And if you continue to deny vectors, then get the frick out of my thread and start your own, where you can ignore Descartes and spout your own theory in the echo chamber to the rest of the Village Idiots in the Peanut Gallery.

Answering your responses where you maintain this dog-headed instance on avoiding vectors as characterizing independent variables is a waste of time for everyone except the Peanut Gallery.

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