### Plane of new numbers/Fermat’s last theorem

**As a basic introduction to a new geometry, consider two points existing together but not forming one point. This must be possible since two points with no extent would have still have no extent, but there could be two points here (we have to consider the number 2, also having this property). Then we need to remove the possibility of one point being there. This necessitates another level to the space. I call this placement space or place of place space or more simply place(2) space. Points(2) (which can move) are defined in the place(2). This is an uncovering of space which already exists and not an abstraction. I think this is 17th century accessible.**

**For more details it would be good** **to read up on Math Concept Sharing seen here: **

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 consider the midline as continuous and the axis to be discreet allow any natural number or any multiple of sqrt(2′) on the axis. Now we have to make this space different from that of which we are already aware. Let any distance be negative as we are moving away from zero, but not positively. 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 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’(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** w**e can ask if we make 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’)^q=m’*sqrt(2’) for m’ a natural number ? Since these are the only equivalent parts in the space.**

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

**But m’, (c’^n) and 2’ are not points but points(2) which can move(at the other level). Let (c’^n) and 2’ and m all vary on the places(2) axis. These are fixed at their values in place(1) so the equation still holds.**

**2′ also can continue to vary since it is equivalent to multiples of sqrt(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).**

**Then let c’^3 and 2’ and m’ move together on the horizontal place(2) axis to m so that we have them all together at some place of places.** **These places of places (place(2)’s)can be numbered. Let them have numbers associated with them which mirror the number system given to the first level.**

** Then p’ = (p’^)(1/n)*(p’)^(½)(q-1). P is not equal to zero or one.** **Looking at the exponents then, we can ask is there any n, q for which 1=(1/n)+(½)(q-1)? Q is some natural number. In the cubic case, i.e. let n=3. For which the answer is no. n=4, 8, 16… will work also but this translates down to n=2 in the usual space. For the squared case 1=(½)+(½)*1 (n=2,q=2) and the linear case 1=1+0 (n=1,q=1). So these are the cases which will work. **

**So Fermat’s Last theorem holds. **