Interactive Online Tutoring Services

With the advent of concept sharing of a point we may have a definition of division by zero.

It seems we can start differently and have a simpler classification of surfaces both 2 and 3 dimensional.

The explanation:

1. Crossings in R^3 or S^3 (positive space)
2. Label 1+,1-,ect. The locations of D-joinings/crossings (mixed space)
3. Place a, b at each D.
4. Move to negative space: move crossing now a sharing as we are capable of having 2 parts or more at a vertex with a and b always moving along.

1+,1-,ect. Are the specific locations in R^3 where we put the other concept sharing diagram back together.

But in the moving diagram they have some freedom. They can travel along as they are labels, reachable through moves of the diagram (that is moves are reversible) or they can stop at a specific pair of locations and the a, b pairs move on forward. This is still reversible as I can get back to this D, as I can reverse from forward motion. At the end, after I go all the way back I come back to the same location.

So 1+ and 1- label the crossing 1, where the original locations can be traced back from. I am placing the labels on specific locations. I mean that this is where the crossing is so that I can trace back to the original location which matches moves in this space then are fully reversible and take full advantage of the new freedoms. Then all these diagrams are equivalent and complete as long as we don’t cut the diagram or change the order of locations.

Then R1,R2,R3, isotopy have their equivalents as well. I have rotation of the locations, creation of labels, joinings, movements of joinings and labels through sharings. And that’s all (complete).

Let there be another diagram D(2) in R^3 and we wish to compare this to the original diagram. Move it to mixed space and then to negative space. Concept share it and one part of it moves off. If we can make a congruency between this and the other diagram in negative space then the two diagrams in positive space are also congruent.

So we need to look for a match of the labels, joinings when we simplify the diagrams.

If they are the same, then the information should be contained in one diagram. That is, movements of one diagram should be able to produce the second one. So we need to look at one diagram to see if we can produce another(using all moves available)

Let’s look at T. To cut down a lot of the moves, let’s apply a simplifying method.

Now I only need to apply S1,S2,S3 and isotopy(2) and I can obtain any diagram of T. Then reform the diagram to go back to R^3.

But the only new diagrams I can obtain will be through the application of the new moves to the T-invariant. That is all possible moves of the invariant. None yield the unknot or any achirality.

The new moves are rotation (which isn’t usually considered) and movement of labels and joinings through sharings. 

So starting with the standard diagram for the Trefoil knot. I place this in the new space and I have the sharings D1,D2,D3. These have the associated moving parts a1,b1,a2,b2,a3,b3. Then I move the outer loops inward (green and red) forming the sharings Q1,Q2,Q3,Q4,Q5,Q6. These have the associated moving labels r1,s1,r2,s2,r3,s3. Then I form the joinings (a1(r1), (b2(s2), (a3(r3), (b1(s1), (a3(r2), (b3(s3) (all red). Now this can be further decomposed to eventually form the “circle”.

Showing the third move. After reduction, only Q-types are involved.

Here I show why we must keep to the same orientation when we reform a joining after we undo it in negative space.

Before reading this it might be useful to read up on Math Concept Sharing seen here: https://calctutor.ca/category/concept-sharing/math-concept-sharing/

The New Plane of Numbers:

We can build another set of numbers and combine this with the new plane which was previously created. In this way create a number plane rather than a number line.

Start with place(2) space, free of any places, and an arbitrary number of points(2)(these are the points which can move) placed at some origin. We can have two number axes perpendicular to each other at this origin.

Now we can build a new number line by moving two points(2) out to some unit length of 1 unit. Do this for both axes. So that in each direction we have a distance of 1 unit marked, now also I have 1 point(2) at 1 and mark it with 1’. This is a location (0,0) moved from the origin, so represents some negative distance. 1’ is identical to 1, except that if I want to I can move these numbers in the place(2)a axis as they are associated with points(2) I can have operations like addition and multiplication on the 1’, 2’, 3’…etc. as well mirroring the usual operations on ordinary numbers. 

In the first quadrant we can make a number (1’(1’) by moving a part at 1’ on one axis and the other part at 1’ on the other axis to combine at the place(2)  (1,1). But actually move 1’(1) and1’(2) and give it  the number(2), (1′(1′). To do this we have to move two parts out to each axis, but this is okay because we can have an arbitrary number of parts there.

Now we can make (2′(2′) by the same process. Move two parts from the origin out 2 units and move one part up 2. units from the x axis and one part from the y axis to come together at the number (2′(2′).

We can go on making a midline of new numbers (1′(1′), (2′(2′), (3′(3′),….

Right now we are only interested in the numbers in this number plane that makes sense to us. That is, the numbers on the midline. 1’ and 3’ could also share but there would be no higher number here at the first level.

Similar to negative numbers, these could have some meaning, but I won’t explore this here.

We can have operations on these numbers such as addition (1′(1′) +(1′(1′)= (2′(2′) = 2’(1′(1′)- if we combine the new number system with the system of numbers point(2) numbers-and multiplication (1′(1′)* (1′(1′)= (1′(1′) or (1′(1′) *(2′(2′) = (2′(2′)= 2’*(1′(1′) as well.

But here is something interesting, since we are combining the two point(2) number systems (1′(1′) =sqrt(2’), if I move the lower axis to be congruent to the midline. Now we have to make this space different from that of which we are already aware. Let any distance be negative as has been described. Here we do not have a positive distance.

Then since (1′(1′)^n=(1′(1′) then sqrt(2’) =sqrt(2’)*sqrt(2’) =sqrt(2’)*sqrt(2’)*sqrt(2’)… (this is for the moving sqrt(2’) not the fixed sqrt(2)-not seen in this geometry). And we have an equivalence of these numbers. This is also seen intuitively as I am moving parts out to the different place(2)’s and these parts are equivalent for a certain set of numbers in this space.

Fermat’s Last Theorem:

Now I want to take a look at the general statement of Fermat’s Last Theorem, seeing if I can find a way to show its truth. It is said that it is unlikely that Fermat had a proof but my ideas could have been known at his time( since in a sense they are basic), so maybe with this new understanding he worked out what I did.

Many people have tried to prove the theorem in the past with elementary methods that would have been available at Fermat’s time. But here I am introducing something new and powerful which didn’t exist before, but might have been known and then subsequently lost.

Now consider that I might have three numbers a, b,c with c related to a and b. I am thinking specifically of a-b=c or a^2-b^2=c^2 or a^3-b^3=c^3…..

Let these be extended to the second form of numbers so that I have a’^n(1′(1′), b’^n(1′(1′), c’^n(1′(1′)=a’^n(1’(1’)-b’^n(1′(1′), n is some natural number (first level natural) all on the midline.

Now look at a’(1′(1′), b’(1′(1′) and c’(1′(1′). C’ might or might not be a natural number so we call it (c’^n)^(1/n) in order that I might operate with c’^n a natural number, being a’^n-b’^n the difference of two natural numbers.

(c’^3)(1/3)(1’(1’) is between b’(1’(1’) and a’(1’(1’). A’ is the largest number. If we focus for a while on the cubic case.

We can ask if a and b are lattice points(2). Then is (c’^3)(1/3) a lattice point(2)? I can move the midline so it is congruent with the horizontal axis.

Since (1′(1′)^n=(1′(1′) and that means sqrt(2’)=sqrt(2’)*sqrt(2’)…(this is for the mobile sqrt(2’) and not the still one-not seen in this geometry)

Does (c’^3)(1/3)*sqrt(2’)^n=m’*sqrt(2’) for m’ a natural number ? (n is an ordinary natural number) Since these are the only equivalent parts in the space.

Then geometrically we can move and expand the line b’*sqrt(2’)——–(c’^3)^(1/3)*sqrt(2’)———a’*sqrt(2’) by factors of sqrt(2’) and look to see if the point(2) (c’^3)^(1/3)*sqrt(2’)^n falls on any lattice point(2), m’*sqrt(2’). A’ and b’ and n can vary.

So an equation we can come up with is m’=(c’^3)^(1/3)*sqrt(2’)^(n-1).

But m’ (c’^3) and 2’ are not points but points(2) which can move. Let (c’^3) and 2’ all vary on the places(2) axis. These are fixed at their values in place(1) so the equation still holds. Places(2) numbers taking on the exponents from the points(2), in this way taking over from the points(2).

We can ask if there exists place(2)’s, p, on the axis which I can take all points(2) to and the equation still holds? Move all points(2) to a single place(2) which could be any place(2) thus creating the generality of the equation holding in place(2).  For then the equation will hold in place(2) as well as place(1).

The equation has to hold here for both points(2) and places(2) like it does originally for both points(2) and places(2) otherwise I can’t form a map back and forth in the extended space. This shows the allowable values of n, a natural number through the exponent of p, a general place(2).

Then let c’^3 and 2’ and m’ move together on the horizontal place(2) axis to m so that at some place(2),p: p’ = (p’^)(⅓)*(p’)^(½)(n-1).  I can reverse the motion of the points(2), 2, c^3 and m’ and the equation would work for these values of n. At the beginning where I have the points(2) similar to being fixed, the equation works for both points(2) and places(2). Where I bring all the points(2) together at the same place(2) this must also be so since this is a one-to-one correspondence, it will only work for these values of n.

Looking at the exponents then, we can ask is there any n for which 1=(1/n)+(½)(n-1)? In the cubic case, i.e. let n=3. For which the answer is no. For the squared case 1=(½)+(½)*1 and the linear case 1=1+0. Any case past cubic is also not possible. So for the cases which work, I can work backwards to the equation I started with.

So for these numbers (1′(1′), (2′(2′), (3′(3′),… Fermat’s Last theorem seems to hold. If we project these to the number axis on either the left or below we will get the same result for regular numbers as all numbers(2) have their counterparts in places(2).

Here is the achirality of the knot 4(1) shown in negative space.

Here is a better post on simplifications.