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In the previous part, we used "ζ" and "ξ" to represent to the new number we have invented, turns out they can be combined or unified. In this chapter, we will try to unite the 2 to make the whole system more concrete.
The New Number: Z， the Zerota Number
let's define that
n × 0^k = Z(uk,n) = Z(+k,n)
n ÷ 0^k = Z(dk,n) = Z(-k,n)
where "u" and "d" are just characters to denote "up" and "down", or we can use positive and negative.
The Crab move
According to the table below, then this means:
when we multiply by 0, we will go right,
when we divide by 0, we will go left.
... | Z(-2,n) | Z(-1,n) | Z(0,n) | Z(+1,n) | Z(+2,n) | ... |
or in other words :
Z(m,n) × 0^k = Z(m+k,n) , k > 0
Z(m,n) ÷ 0^k = Z(m-k,n), k > 0
also, it is important to note that:
Z(0,n) = n
They are not the same.
in this case, when something multiplies with 0, on the "real number line", it is zero, but on another dimension, it is something else.
if A and B are different, then A × 0 = B × 0 on the "real number line", But in Zerota Theory, it is important to remember that :
A × 0^k ≠ B × 0^k, A ≠ B
and also, if the power of Zero is different, then they are different, in other words
A × 0^m ≠ A × 0^n
Z(m,A) ≠ Z(n,A)
Factor Scale
and let's assume that B = fA or "B is f times bigger than A", we will have :
fA × 0^k = B × 0^k
f × A × 0^k = B × 0^k
f × Z(k,A) = Z(k,B)
from here we can see the Zerota Number of B is f times bigger than the Zerota Number of A.
So we can conclude that :
Z(k,B) = f × Z(k,A) ↔ B = fA
eg: Z(13,8) = 2 × Z(13,4) because 8 = 2 × 4
another thing worth mentioning would be:
We can pull a factor out or push a factor into the Zerota Number, or
Z(k,fA) = fA × 0^k = f × A × 0^k = f × Z(k,A)
Z(k,fA) = f × Z(k,A)
Addition, Subtraction, Multiplication, and Division
Turns out we can also add 2 Zerota Numbers together, or we can subtract one from another if they have the same power.
Z(k,A) ± Z(k,B) = A × 0^k ± B × 0^k = (A±B) × 0^k = Z(k,A±B)
Z(k,A) ± Z(k,B) = Z(k,A±B)
Then let's try having a Zerota number multiplied by another Zerota Number :
Z(p,A) × Z(q,B) = A × 0^p × B × 0^q = (AB) × 0^(p+q) = Z(p+q,AB)
Z(p,A) × Z(q,B) = Z(p+q,AB)
And let's also try division as well
Z(p,A) ÷ Z(q,B) = (A × 0^p) ÷ (B × 0^q) = A × 0^p ÷ B ÷ 0^q = (A÷B) × (0^p÷0^q) = (A÷B) × 0^(p-q) = Z(p-q,A÷B)
Z(p,A) ÷ Z(q,B) = Z(p-q,A÷B)
Summary
let's take a break here, these are what we have at the moment, keep an eye out for Part 3.
"becoming Zerota"
n × 0^k = Z(uk,n) = Z(+k,n)
n ÷ 0^k = Z(dk,n) = Z(-k,n)
"moving crab"
Z(m,n) × 0^k = Z(m+k,n) , k > 0
Z(m,n) ÷ 0^k = Z(m-k,n), k > 0
Z(0,n) = n
"They are not the same"
A × 0^k ≠ B × 0^k, A ≠ B
Z(m,A) ≠ Z(n,A)
"factor Scale"
Z(k,B) = f × Z(k,A) ↔ B = fA
Z(k,fA) = f × Z(k,A)
"Add,Sub,Mul,Div"
Z(k,A) ± Z(k,B) = Z(k,A±B)
Z(p,A) × Z(q,B) = Z(p+q,A×B)
Z(p,A) ÷ Z(q,B) = Z(p-q,A÷B)
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