# Thread: Dimensions, sets, and Fermat's Theorem

1. ## Dimensions, sets, and Fermat's Theorem

There is a more complete discussion of the substance of this thread in my .pdf at:

The Relativistic Unit Circle

Consider the following syntactical expression:

c = a + b

What this expression states in its simplest form is that the result is predetermined. For 6 = 3 + 3 = 4 + 2, the expression is merely a way of subdividing the elements in a set in which there is a metric ( 6 > 3). This can be expressed in a number line related to the unit element where

(The Distributive Law)

One can think of the unity element as the basis for relating all other elements in the set. This means that the variables "a", "b", and "c" are syntactical elements that show an operational relationship; in the above case, one of subdividing the set in such a way that the function "+" relates the elements a and b to the final element c, but the left and right hand sides of the equality refer to the same number: c = c and c = (a+b)

The single number line can be expressed as a matrix of one dimension: |c| =|a+b| = |a| + |b|

Now consider the case where we begin with a single number line {set} designated by a and add a second number line represented by b.

{a} -> {a} +{b}

Since each set {a} and {b} each contain all possible elements, a distinction must be made between the sets. This is done by Cartesian coordinates (Cartesian products), where the third element {c} now refers to the resultant set created by the relationship between the two distinct sets {a} and {b}.

The disjoint sets (e.g., positive integers) {a} and {b} are often expressed as an ordered pair (a,b). For real numbers, a and b are replaced by x and y in the ordered pair (x,y).

That is, the element {c} does not belong in either {a} or {b} if a and b are disjoint (independent) sets.

For set theory, this is expressed by the Cartesian Product:

For vectors, this is expressed by the vector notation: where i and j are unit vectors in each distinct set, and r is the resultant that belongs in neither set; a, b, and c are called scalars.

In matrix notation, the distinct sets {a} and {b} have the representation: , and the resultant of the operation "+" (now vector addition) is represented by
. The arithmetic operation for the "length" (vector magnitude) of the resultant satisfies the relation:

, the triangle equality, which is complete since there are no other sets or elements thereof.

If a and b are positive integers, this means that the Binomial Theorem is a vector relation in two dimensions, not an arithmetic relationship in a single dimention, since rem(a,b,n) are elements that are not members of either set or so Fermat's Theorem is proven. That is, rem(a,b,n) cannot reside in either {a} or {b} since it consists of interactive elements of both sets.

It is impossible to prove Fermat's theorem in one dimension, since the sum (a + b), and c can refer to the same unique positive integer, and likewise the expression:

2. ## Re: Dimensions, sets, and Fermat's Theorem

Has it occured to you that the fact you must constantly refer to your irrelevant, insigifnicant pathetic "ignore list" that you basically admit defeat instantly because it's completley irrelevant and contains primarily of people who are significantly more educated than you?

3. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by BuleriaChk
Consider the following syntactical expression:

c = a + b

What this expression states in its simplest form is that the result is predetermined. For 6 = 3 + 3 = 4 + 2, the expression is merely a way of subdividing the elements in a set in which there is a metric ( 6 > 3). This can be expressed in a number line related to the unit element where

(The Distributive Law)

One can think of the unity element as the basis for relating all other elements in the set. This means that the variables "a", "b", and "c" are syntactical elements that show an operational relationship; in the above case, one of subdividing the set in such a way that the function "+" relates the elements a and b to the final element c, but the left and right hand sides of the equality refer to the same number: c = c and c = (a+b)

The single number line can be expressed as a matrix of one dimension: |c| =|a+b| = |a| + |b|

Now consider the case where we begin with a single number line {set} designated by a and add a second number line represented by b.

{a} -> {a} +{b}

Since each set {a} and {b} each contain all possible elements, a distinction must be made between the sets. This is done by Cartesian coordinates (Cartesian products), where the third element {c} now refers to the resultant set created by the relationship between the two distinct sets {a} and {b}.

The disjoint sets (e.g., positive integers) {a} and {b} are often expressed as an ordered pair (a,b). For real numbers, a and b are replaced by x and y in the ordered pair (x,y).

That is, the element {c} does not belong in either {a} or {b} if a and b are disjoint (independent) sets.
So {c} does not belong in either
For set theory, this is expressed by the Cartesian Product:

For vectors, this is expressed by the vector notation: where i and j are unit vectors in each distinct set, and r is the resultant that belongs in neither set; a, b, and c are called scalars.
So, r belongs in neither
In matrix notation, the distinct sets {a} and {b} have the representation: , and the resultant of the operation "+" (now vector addition) is represented by
. The arithmetic operation for the "length" (vector magnitude) of the resultant satisfies the relation:

, the triangle equality, which is complete since there are no other sets or elements thereof.

If a and b are positive integers, this means that the Binomial Theorem is a vector relation in two dimensions, not an arithmetic relationship in a single dimention, since rem(a,b,n) are elements that are not members of either set or
So, rem(a,b,n) is neither, just like {c} or r. That's because it is a component of c

Remember?
so Fermat's Theorem is proven.
Give it up. Your proof is ridiculous
That is, rem(a,b,n) cannot reside in either {a} or {b} since it consists of interactive elements of both sets.

It is impossible to prove Fermat's theorem in one dimension, since the sum (a + b), and c can refer to the same unique positive integer, and likewise the expression:

4. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by grapes
+
Give it up. Your proof is ridiculous
You were doing ok until this part.

Stop jumping so eagerly to false conclusions - but keep working on it (i.e., throwing shit against the wall to see if something will stick...

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

That actually proves the theorem.

The resultant rem(a,b,n) cannot be a part of the plane with the sets {a} and {b} (represented by the ordered pair (a,b) except in the case of the Binomial Theorem for n = 2 (where 2ab is an area, and thus within the plane).

For n > 2 rem(a,b,n) IS part of the resultant set {c} (It is a scalar in an extra dimension identified by the k vector (making a 3D volume).

(This volume cannot be a part of ({a},{b}). That is, the plane of positive integer pairs defined by (a,b))

Which PROVES the theorem.

(sigh)

5. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by BuleriaChk
You were doing ok until this part.

Stop jumping so eagerly to false conclusions - but keep working on it (i.e., throwing shit against the wall to see if something will stick...

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

That actually proves the theorem.

The resultant rem(a,b,n) cannot be a part of the plane with the sets {a} and {b} (represented by the ordered pair (a,b) except in the case of the Binomial Theorem for n = 2 (where 2ab is an area, and thus within the plane).

For n > 2 rem(a,b,n) IS part of the resultant set {c} (It is a scalar in an extra dimension identified by the k vector (making a 3D volume).

(This volume cannot be a part of ({a},{b}). That is, the plane of positive integer pairs defined by (a,b))

Which PROVES the theorem.

(sigh)
EEEEXCEEEEPT!!....you have not established there must be a link between them. That is the logical step you're missing.

6. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by BuleriaChk
You were doing ok until this part.

Stop jumping so eagerly to false conclusions - but keep working on it (i.e., throwing shit against the wall to see if something will stick...
I know you mean this ironically (smiley face there), so I won't have to point out the next level of irony. Your "proof" involves no mathematics, just casual introduction of mathematical concepts. When you need a quantity to be zero, you declare it to be zero--how many times have you claimed 2ab had to be zero for positive integers? Just because, in your words, a and b are represented by independent vectors and multiplication is represented by dot product and therefore 2ab vanishes

That's just ridiculous

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

That actually proves the theorem.

The resultant rem(a,b,n) cannot be a part of the plane with the sets {a} and {b} (represented by the ordered pair (a,b) except in the case of the Binomial Theorem for n = 2 (where 2ab is an area, and thus within the plane).

For n > 2 rem(a,b,n) IS part of the resultant set {c} (It is a scalar in an extra dimension identified by the k vector (making a 3D volume).

(This volume cannot be a part of ({a},{b}). That is, the plane of positive integer pairs defined by (a,b))

Which PROVES the theorem.

(sigh)

7. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by grapes
I know you mean this ironically (smiley face there), so I won't have to point out the next level of irony. Your "proof" involves no mathematics, just casual introduction of mathematical concepts. When you need a quantity to be zero, you declare it to be zero--how many times have you claimed 2ab had to be zero for positive integers? Just because, in your words, a and b are represented by independent vectors and multiplication is represented by dot product and therefore 2ab vanishes

That's just ridiculous
(for grapes): It is your inability to understand simple vectors w.r.t. sets that is ridiculous..

When one makes the statement "for all positive integers a and b", then to make a distinction between them, they must belong to disjoint (distinct) sets. That is, when the sets are disjoint, they do not "interact"; there is no relation between the elements. This is the case for Pythagorean triples, where the legs of the triangle are perpendicular, so the dot product vanishes.

(for grapes): For the case n=2 (Binomial Theorem) 2ab does not vanish because the legs of the triangle are not perpendicular. However, 2ab is an area, and is in the same plane as the ordered pair (a,b). So Rem(a,b,2)=2ab 0.

(for grapes): The case of 2 is not relevant to the proof of Fermat's Theorem (n > 2)

For n > 2, the terms in rem(a,b,n) has the general form of the Binomial theorem (e.g. ).

I use "absolute value" of cross products merely to indicate positive integer multiplication, but also that these products can never be in either {a} or {b}, so therefore could never be in the plane (a,b). Therefore they can be represented as a "non-interacting" vector w.r.t. (a,b) by a third vector

The sets in the Binomial Theorem are then related by vector addition, with the resultant vector not orthogonal to either i, j, or k, but a mixture (e.g., the hypotenuse of the right triangle for n = 2 (Pythagorean) - or the longest leg of the triangle for Binomial case n = 2 .

(for grapes): Once again, the case of 2 is not relevant to the proof of Fermat's Theorem (n > 2)

(Once again, for grapes) (sigh):

For n > 2, then the full expression for the Binomial Theorem is represented by:

,

which holds for all members of the sets (I couldn't use squiggly brackets for the powers because tex wouldn't let me).

Since no element in {rem(a,b,n)} = 0, then is not in the plane , much less in the sets or so Fermat's Theorem is proved.

(sheesh)

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

8. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by BuleriaChk
(for grapes): It is your inability to understand simple vectors w.r.t. sets that is ridiculous..

When one makes the statement "for all positive integers a and b", then to make a distinction between them, they must belong to disjoint (distinct) sets. That is, when the sets are disjoint, they do not "interact"; there is no relation between the elements. This is the case for Pythagorean triples, where the legs of the triangle are perpendicular, so the dot product vanishes.

(for grapes): For the case n=2 (Binomial Theorem) 2ab does not vanish because the legs of the triangle are not perpendicular. However, 2ab is an area, and is in the same plane as the ordered pair (a,b). So Rem(a,b,2)=2ab 0.

(for grapes): The case of 2 is not relevant to the proof of Fermat's Theorem (n > 2)

For n > 2, the terms in rem(a,b,n) has the general form of the Binomial theorem (e.g. ).

I use "absolute value" of cross products merely to indicate positive integer multiplication, but also that these products can never be in either {a} or {b}, so therefore could never be in the plane (a,b). Therefore they can be represented as a "non-interacting" vector w.r.t. (a,b) by a third vector

The sets in the Binomial Theorem are then related by vector addition, with the resultant vector not orthogonal to either i, j, or k, but a mixture (e.g., the hypotenuse of the right triangle for n = 2 (Pythagorean) - or the longest leg of the triangle for Binomial case n = 2 .

(for grapes): Once again, the case of 2 is not relevant to the proof of Fermat's Theorem (n > 2)

(Once again, for grapes) (sigh):

For n > 2, then the full expression for the Binomial Theorem is represented by:

,

which holds for all members of the sets (I couldn't use squiggly brackets for the powers because tex wouldn't let me).
No, it does not hold for all members. Even in your ridiculous notation, it only holds for c=a+b. That's a long way from "all members"

Since no element in {rem(a,b,n)} = 0, then is not in the plane , much less in the sets or so Fermat's Theorem is proved.

(sheesh)

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

9. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by grapes
No, it does not hold for all members. Even in your ridiculous notation, it only holds for c=a+b. That's a long way from "all members"
It holds for

c={c}, a = {a}, b = {b}, notationally.

For the conceptually challenged (grapes and others):

,,

, ,

(I just figured how to do sets in tex....

,

What other members do you suggest... ?
(aside from your ridiculous value judgements).

(all members, taken individually -
that's the meaning of variables -
perhaps you need a refresher course in 7th grade algebra?)

Here ya go:

Variable

Well, ok, that may be too advanced for you. Here's another link:

Variable

(Make sure you check out the "More" link on the page...

10. ## Re: Dimensions, sets, and Fermat's Theorem

Originally Posted by BuleriaChk
Originally Posted by grapes
No, it does not hold for all members. Even in your ridiculous notation, it only holds for c=a+b. That's a long way from "all members"
It holds for

c={c}, a = {a}, b = {b}, notationally.

For the conceptually challenged:

c{c}, a{a}, b{b}

rem(a,b,n), rem(a,b,n)

("tex" has limitations on squiggly brackets)

(all members, taken individually -
that's the meaning of variables -
perhaps you need a refresher course in 7th grade algebra?)

Here ya go:

Variable

What other members do you suggest... ?
(aside from your ridiculous value judgements).
Values of c that do not equal a+b

Values of a,b,c that satisfy the triangle inequality

In other words, pretty much almost all the possibilities, all the ones that you have ignored in your "attempt" at a proof of Fermat's Last Theorem

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