Mathematics

Mathematical Thesis

This is my attempt to show that, in regards to zero, our current calculations are not correct. Firstly I shall give three definitions to which I can refer back to.

1. “Group”- A group is a given number of variable positions that have not been assigned values.

2. “Variable Position” - A Variable Position is a single spot, a given distance from two others, on a number line.

3. “A Number”- A single Variable Position that has been assigned a value, relative to the sum of all Variable Positions from zero, in it’s own number line.


Every number we have can be understood both as a number and as a group. It follows that multiplication and division are operations in relations to these very ideas.
That is to say, that when we have the expression (2 x 3), the “number” three does not actually exist in this expression. It is quite literally the group of 3, which we are seeing. The above expression can be rewritten as this.
( 2 + 2 + 2 ) . Or it can be written as this, ( x + x + x), with an assigned value of 2. This is to say that, ( x,x,x ) is the group of 3, with out any value’s assigned. It follows for division that we have the expression ( 6 / 2 ), is actually the number 6 subtracted to equal parts and assigned to the group of 2, and then take the value of one x in the entire group as your answer. It can therefore be rewritten as this.
( 3, 3 ) and the value of one x, in the group of 2 is 3, therefore ( 6 \ 2 ) equal’s 3. This operation can be simply understood for all Whole numbers. Integers and rational, and irrationals have different fundamental axioms, and I shall address these latter.
With this understanding of multiplication and division, I shall then move to zero, which is understood to be a whole number. More importantly I shall begin by showing that zero has a group.
If the group of 3, is understood as ( x,x,x ). Then I must relate exactly how this is so, in order to show exactly what the group of zero is. It is at this point that I shall refer back to my first definition. By my definition we can understand how this is so. But there is another way to explain myself.
That is to say, that if we have a number line such as this.
(--------0---------1------------2-----------3------) it is essentially the same number line as this.
(--------x---------x------------x-----------x------). The difference here being only, that the first number line has all X’s assigned with values, and the second has no value’s assigned. It is with this second number line that I show that by definition each X is a variable position.
The number 2 in the first number line occupies one variable position, or a single “spot” on the number line. But the value assigned is reached by a sum of all variable position from zero on its own number line. So 2 is (x,x) as a group. 1 has no variable positions before it, FROM zero, but it does occupy a single variable position. This can show then that zero, while it has no values assigned, occupies by itself, a single variable position. That is to say then that variable positions that actually come before zero, that they are then assigned a negative value, instead of positive.
The operation ( 1 + (-1)= 0). Zero as a variable position must be counted in the number line, relative to this equation, or the answer would in fact be -1.
This then is my evidence for zero actually being represented as a group. (x) is the group of 1, as well as the group of zero. If we assigned a value of 1 then the variable position becomes the number 1. If we assign it the value of none, then the variable positions becomes zero.
Accepting that zero, as a group is therefore (x), then we must reapply this idea to multiplication, and division.
The expression (0 x 2), if zero is the number, or value, and it is assigned to the group of 2, or (x,x). Then we have ( 0 + 0 ). The sum is zero. However if we have 2 as a value and assign it to the group of zero or, (x), then we simply have ( 2 ). Therefore (2 x 0 = 2), and (0 x 2= 0).
This would then follow for division. ( 0 \ 2 ). The value of zero subtracted equally and assigned to (x,x), or the group of 2, is therefore (0 , 0). And the value of one variable position is zero. However if we have ( 2 \ 0 ), and 2 is the value subtracted equally and assigned to (x), then we have (2). And the value of one variable position in the group is 2. Therefore (2 \ 0 = 2), and (0 \ 2 = 0).
My Axioms are then as follows.
1. A (as a number) x 0 (as a group) = A
2. 0 (as a number) x A (as a group) = 0
3. A (as a number) \ 0 (as a group) = A
4. 0 (as a number) \ A (as a group) = 0