# Thread: The Relativistic Unit Circle

1. ## Re: The Relativistic Unit Circle

Originally Posted by grapes
Which means c<a+b

That's true for all triangles

Except it totally invalidates your entire proof
Not those for which a < (c + b). Or b < (a + c) or for which c > (a + b).

You don't understand vectors at all. Or Cartesian coordinate systems, for that matter.
Or even variables..

And nothing you've said invalidates anything, it just shows how stupid, ignorant, and arrogant you are. I'm just beginning to realize how little you know about the fundamentals.

(I don't even need triangles for my proof, since n > 2 for Fermat's Theorem. For triangles in my proof, n = 1 and a,b,c are scalar coefficients to vectors, and not necessarily unit vectors. That is why you have no fricken idea what my proof is about. You know nothing about vectors, and my proof depends on them, by the independence of (a,b).

The vectors on the rhs don't even need to be orthogonal. In fact they aren't in most cases. In fact, they can point in any direction whatever. You're trying to reference magnitudes of vectors for the triangle inquality, but you have to specify how they are connected in relation to a basis for my proof, which requires the specification of an origin at (0,0) and coordinate axes with unit vectors .

Where did you go to high school, anyway?

(I was just concerned with Pythagorean triples for the exception when n = 2, which is irrelevant to the proof of Fermat's Theorem)

For circles and areas of rectangles, n = 2.

Village idiot. I can't believe you're a moderator.

History will judge; there are other readers...

2. ## Re: The Relativistic Unit Circle

Originally Posted by BuleriaChk
Not those for which a < (c + b). Or b < (a + c) or for which c > (a + b).
There are no triangles that have c > (a+b)

None

And all triangles have
a < (c + b) AND b < (a + c) AND c < (b + c)
You don't understand vectors at all. Or Cartesian coordinate systems, for that matter.
Or even variables..

And nothing you've said invalidates anything, it just shows how stupid, ignorant, and arrogant you are. I'm just beginning to realize how little you know about the fundamentals.
Wow
(I don't even need triangles for my proof, since n > 2 for Fermat's Theorem. For triangles, n = 1 and a,b,c are scalar coefficients to vectors, and not necessarily unit vectors. That is why you have no fricken idea what my proof is about. You know nothing about vectors, and my proof depends on them, by the independence of (a,b).

The vectors on the rhs don't even need to be orthogonal. In fact they aren't in most cases. In fact, they can point in any direction whatever. You're trying to reference magnitudes of vectors for the triangle inquality, but you have to specify how they are connected in relation to a basis for my proof, which requires the origin of a coordinate system.

(I was just concerned with Pythagorean triples for the exception when n = 2, which is irrelevant to the proof of Fermat's Theorem)

For circles and areas of rectangles, n = 2.

Village idiot. I can't believe you're a moderator.

History will judge; there are other readers...

3. ## Re: The Relativistic Unit Circle

It is for that reason that I introduced the vector relationship early on:

At the time I was concerned about the signs of the cross products in rem(a,b,n), but now realize the change in signs was due to the vectors changing sign when commuting, not the scalars, which commute arithmetically, so that vectors can be summed to null vectors while their coefficients remain positive.

But any equation with is totally irrelevant to Fermat's theorem. And specifying a and b as integers is irrelevant as well, since the Binomial theorem works for positive real numbers in general, and so does my proof, as does my proof by the relativistic unit circle.

4. ## Re: The Relativistic Unit Circle

Originally Posted by BuleriaChk
It is for that reason that I introduced the vector relationship early on:
What reason?

Because you don't understand simple trigonometry?

At the time I was concerned about the signs of the cross products in rem(a,b,n), but now realize the change in signs was due to the vectors changing sign when commuting, not the scalars, which commute arithmetically, so that vectors can be summed to null vectors while their coefficients remain positive.

But any equation with is totally irrelevant to Fermat's theorem.
Don't be foolish.
And specifying a and b as integers is irrelevant as well, since the Binomial theorem works for positive real numbers in general, and so does my proof, as does my proof by the relativistic unit circle.

5. ## Re: The Relativistic Unit Circle

Originally Posted by grapes
What reason?

Because you don't understand simple trigonometry?

Don't be foolish.
The reason was to introduce vectors necessary to precisely define the interaction coefficients in the Binomial Expansion.... if that were an issue. It wasn't to me, and I doubt it was an issue to Fermat. It is only an issue with idiots who to prove Fermat's theorem on a single number line, and can't figure out where they go wrong.

Uh, you mean geometry?
(trigonometry requires, uh, sin and cos)...

If you can't speak vectors, you have no idea what I am talking about....

You may think you know what YOU are talking about, but its my proof, you're just nattering on about the fantasies in your head. You are not speaking to my proof.

To the reader (if there are any left by this point)... I'm going to stop responding to Grapes at this point, since he has his head so far up his ass.

But I will respond to anyone not in the Peanut Gallery......

6. ## Re: The Relativistic Unit Circle

Originally Posted by BuleriaChk
The reason was to introduce vectors necessary to precisely define the interaction coefficients in the Binomial Expansion.... if that were an issue. It wasn't to me, and I doubt it was an issue to Fermat. It is only an issue with idiots who to prove Fermat's theorem on a single number line, and can't figure out where they go wrong.

Uh, you mean geometry?
(trigonometry requires, uh, sin and cos)...
OK, you don't understand simple geometry
If you can't speak vectors, you have no idea what I am talking about....

You may think you know what YOU are talking about, but its my proof, you're just nattering on about the fantasies in your head. You are not speaking to my proof.

To the reader (if there are any left by this point)... I'm going to stop responding to Grapes at this point, since he has his head so far up his ass.

But I will respond to anyone not in the Peanut Gallery......

7. ## Re: The Relativistic Unit Circle

Note on vectors

(From my pdf paper)

Scalar operations on a single dimension as coefficients of a vector mean that the elements in the coefficient refer to the same unique number means that and refer to the same unique number for single valued functions.

That is, the coefficient is embedded in a single number line for real numbers.. . In differential geometry and the General Theory of Relativity, that number is a position on a geodesic (number line, possibly curved). If two number lines are not connected (do not "touch"), there is no common origin, and the vectors are called "affine".

If the lines never touch and are in the same plane (i.e, the space is "flat"), then the concept of "parallel transport" applies in GTR.

For STR, the unit vectors are not affine; ie they "touch" in a flat space (For Fermat's Theorem, at the origin of and , and in general at and ( or in a space-time diagram).

It is at the origin that interactions occur.

Anyone who has studied differential geometry, GTR, or tensor analysis will understand this.
--------------------------------------
Not from my paper (but relevant):

Those coefficients represent the Peanut Gallery and me, and for sure, our world lines are far from touching ....
It is at the origin that meaningful interactions occur.

(I'm just waiting now..... I think I may have gotten thru to some other resources.... We'll see...)

(I am reminded of Wittgenstein's comments in his "Remarks on the Foundations of Mathematics" vis a vis infinite series.... )

(In my paper on "The Creation of the Universe (almost everything you need to know)" on my website I show why must be interpreted as a density... in relation to the covariant and contravariant relation of mass and energy.

8. ## Re: The Relativistic Unit Circle

Originally Posted by BuleriaChk
Note on vectors

(From my pdf paper)

Scalar operations on a single dimension as coefficients of a vector mean that the elements in the coefficient refer to the same unique number means that and refer to the same unique number for single valued functions.

So, when you say "s = f(x,y,z,....)" you mean that s equals f(x,y,z,....)?

You might be onto something
That is, the coefficient is embedded in a single number line for real numbers.. . In differential geometry and the General Theory of Relativity, that number is a position on a geodesic (number line, possibly curved). If two number lines are not connected (do not "touch"), there is no common origin, and the vectors are called "affine".

If the lines never touch and are in the same plane (i.e, the space is "flat"), then the concept of "parallel transport" applies in GTR.

For STR, the unit vectors are not affine; ie they "touch" in a flat space (For Fermat's Theorem, at the origin of and , and in general at and ( or in a space-time diagram).

It is at the origin that interactions occur.

Anyone who has studied differential geometry, GTR, or tensor analysis will understand this.
--------------------------------------
Not from my paper (but relevant):

Those coefficients represent the Peanut Gallery and me, and for sure, our world lines are far from touching ....
It is at the origin that meaningful interactions occur.

(I'm just waiting now..... I think I may have gotten thru to some other resources.... We'll see...)

(I am reminded of Wittgenstein's comments in his "Remarks on the Foundations of Mathematics" vis a vis infinite series.... )

(In my paper on "The Creation of the Universe (almost everything you need to know)" on my website I show why must be interpreted as a density... in relation to the covariant and contravariant relation of mass and energy.
What is the meaning of a word?
Let us attack this question by asking, first, what is an explanation of the meaning of a word; what does the explanation of a word look like?
-- Wittgenstein, The Blue Book

9. ## Re: The Relativistic Unit Circle

Originally Posted by grapes
So, when you say "s = f(x,y,z,....)" you mean that s equals f(x,y,z,....)?

You might be onto something

What is the meaning of a word?
Let us attack this question by asking, first, what is an explanation of the meaning of a word; what does the explanation of a word look like?
-- Wittgenstein, The Blue Book
No shit, Dick Tracy...

It is the "Remarks on the Foundations of Mathematics" that are most interesting in this context.
(Some passages in the Investigations, as well)

(I just updated my pdf with more content)...

10. ## Re: The Relativistic Unit Circle

Counting

Two philosophers, A and B are counting positive widgets. At some point they stop counting and claim they have a final result.

A claims “Aha! I have counted all the positive widgets in the Universe, and the number is .

B claims “No! That cannot be. I have counted all the positive widgets in the Universe, and the number is .

How can we decide which one is right?

Again, we can assign a weight to each widget count . This gives us a picture of the final condition at:

(so it doesn't matter if the widgets are positive or negative).

The conclusions from the Relativistic Unit circle, Quantum Field Theory, Fermat’s Theorem, GTR, and Godel’s theorem follow immediately.

If , there are no widgets. ( is a figment of the imagination ) .

If , then only A is counting, so A must be right ( )

If the product exist,s then there is an argument over who is right or wrong

If , the result of the count cannot consists of unit widgets.

If , not all the widgets have been counted.... (the final condition must be re-initiated as an initial condition.)

To introduce negative widgets, one must use the imagination:

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