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

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

Originally Posted by grapes
rem can be negative
Look at rem(a,b,n) for a,b,n positive integers There are no negative integers.

Binomial Theorem

You say you have an advanced degree in Math? Fat Chance.....

(THIS is an answer to my proof? Sheesh!)

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

Originally Posted by grapes
rem can be negative
123 = 93 + 103 - 1
Originally Posted by BuleriaChk
Binomial Theorem

You say you have an advanced degree in Math? Fat Chance.....

(THIS is an answer to my proof? Sheesh!)
That didn't even take a high school education.

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

Originally Posted by grapes
123 = 93 + 103 - 1

That didn't even take a high school education.
WTF? This is just garbage. How does this relate to the Binomial Expansion?
It is true that writing mathematical symbols arbitrarily in nonsense syntax related to nothing doesn't take a high school education...

So either you're incredibly stupid or you're trolling again....

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

Originally Posted by BuleriaChk
WTF? This is just garbage. How does this relate to the Binomial Expansion?
It is true that writing mathematical symbols arbitrarily in nonsense syntax related to nothing doesn't take a high school education...

So either you're incredibly stupid or you're trolling again....
I was responding to your post.
Originally Posted by grapes
Originally Posted by BuleriaChk
FLT Document Updated (this post)

(Once more, into the breach....)

Another way of stating Fermat's theorem is:

cannot be valid for positive integers a,b,c and n for any arithmetic system that includes multiplication between integer variables a and b.

(Powers such as multiply an integer with itself in a single dimension: ,

For a=b, so that and , so c cannot be an integer.

, where the vector is not aligned with either or , and rem(a,b,n) consists of terms with only multiplicative products. Therefore, Fermat's expression cannot be valid unless these multiplicative products vanish; i.e. a=0 or b=0.
rem can be negative
If you can't do the math, get a calculator.

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

Originally Posted by grapes
I was responding to your post.

If you can't do the math, get a calculator.
rem(a,b,n) is never negative in the Binomial Expansion for a,b,n positive integers. WTF are you talking about?

And again, you completely ignored the vector characterization and used childish arithmetic.

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

Originally Posted by BuleriaChk
rem(a,b,n) is never negative in the Binomial Expansion for a,b,n positive integers. WTF are you talking about?
Fermat's Last Theorem. The binomial theorem has nothing to do with the proof of Fermat's Last Theorem.
And again, you completely ignored the vector characterization and used childish arithmetic.
Still haven't got out your calculator?

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

Originally Posted by grapes
Fermat's Last Theorem. The binomial theorem has nothing to do with the proof of Fermat's Last Theorem.
There is no point in going on here. Grapes' responses have degenerated into childish nonsense and inane propositions empty of any technical or mathematical content. (I suspect it is because he knows I am right and doesn't have the courage or ethics to admit it. But that's ok; my analysis is on my website for the world to see... time will tell...)

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

Originally Posted by BuleriaChk
There is no point in going on here.
OK, fine by me, you weren't participating in the conversation anyway. Thread closed.
Grapes' responses have degenerated into childish nonsense and inane propositions empty of any technical or mathematical content. (I suspect it is because he knows I am right and doesn't have the courage or ethics to admit it. But that's ok; my analysis is on my website for the world to see... time will tell...)

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

Originally Posted by BuleriaChk
There is no point in going on here. Grapes' responses have degenerated into childish nonsense and inane propositions empty of any technical or mathematical content. (I suspect it is because he knows I am right and doesn't have the courage or ethics to admit it. But that's ok; my analysis is on my website for the world to see... time will tell...)
He is correct though, binomial theorem has nothing to do with Fermats, they are distinct and unrelated.

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

Originally Posted by emperorzelos
He is correct though, binomial theorem has nothing to do with Fermats, they are distinct and unrelated.
This comment is just stupid.

c1=g1(a,b) = (Fermat's Expression)

c2=g2(a,b) = (The Binomial Expansion).

Fermat's expression is the integer metric for a Presburger arighmetic (without multiplicative products like

The Binomial Expansion includes Fermat's expression but also those multiplicative products in Rem(a,b,n), and therefore is consistent as a metric for the two dimensional (real number) field (a,b)) with Peano's axioms. However, Fermat's expression requires that a=0 or b=0 for n>2 if c is to be an integer (for fields, division is always possible so is always possible, so the field of positive real numbers is consistent and but not complete (Godel's proof is based on positive integers), since it requires complex numbers for subraction (). The key issue is that one needs two fields (Cartesian coordinates) to define any function y=f(x) at all, not to mention z=g(x,y) as in the Binomial Expansion for z a single valued resultant variable.

Cartesian coordinates can express real numbers without the necessity of Dedekind cuts in a single dimension number line.

STR provides a way of generating continuous fields (without using Dedekind cuts) and for positive fields, also provides trigonometric functions (and negative fields if Dirac is included).

One liner responses from the Peanut Gallery with no intellectual content whatever in responses to posts like this (where I can provide Wiki links as necessary) are merely spam and bloating my thread's intellectual content with nothing other than complaints indicating that the Peanut Gallery have no idea WTF I am talking about.

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