00.--à
premise
Considerations of Too mythics as:
"2(I^2)" :~ "1" :~= 2(1/2) in right-angled isosceles triangles…
VALUE, "2(I^2)" being translations of +/-(THE SUM of TWO waves):
thesis
The value of FOUR waves: (4 : "5") = "1" + 2(I^2) :~ "2" :~ "0":
+/- ( sin^2 (a + b)+ cos^2 (a + b) ) :~ 2("1") :~ 2( 2(I^2 :~ "1/2"))
or, 2(+/- (sin a + cos b)) = 2("1")…
hypothesis
"Too" Myth
s:A = F (x) : F (y) ~ "Too" = F (I^2}(I^4):F (Y)/"1"- (I^2)(I^4) (I^2):(I^4) x (I^2)x
DISCOURSIVE PROOFS…
After Pythagoras, in any isosceles triangle:
Let k = cos^2 a + sin^2 a :~ 1
do 1
If I^2 :~ ½
root Y = 2(-Y)
k = 2(I^2)
do 2
dk/da = 4i
dk = 4i. Da = 1
do 3
4I = 1/da = 2(2. Root(I^2)) = 2(2. 1/root 2) = 4 (1/root 2)
and, 4i. Da :~ 1 :~ (da)^2 :~ (4I)^2 :~ 8….
act 1
cos^2 a + sin^2 a :~ 2. (1/root 2)(1/root 2) = 2. I. I
therefore, I :~ 1/root 2
play 1
From Pythagoras: cos^2 a = sin^2 a = 1/root 2
and, if m = a constant such that m. 1/m = k :~ 1
2 :~ I^2 :~ m :~ 1/m = ½
act 2
2 :~ m^0 = m^-1 and the root of "1" = I^2 = ½
act 3
every finite number thus has infinite translatable signifieds…
Now 2(I^2) :~ "1" :~ "2" :~ 1. 1/1 :~ 1."1" + 1."1" which transposes into, 1. "1" :~ root 2
play 2
now, also 1 + 1 : root 2 = (root 2)^2. ½
which, from our translation = 1/(I^2). (I^2)
"2" is really a set = "(1 + 1)"/(1.1) = (1^2) + (1^2) :~ "(I^2)" + "(I^2)"…
play 3
Thus, "1" & "1" = 2….
play 4
i.--à
premise
|1/(I^2) | : 2
thesis
8 = 1/(k^2)
do 1
k = 0.35355
= (0.7071)/2
= i/2 :~ I^3
do 2
since
4(k^2) = I^2 = ½
and
4k = 2I = ½
k = ½(root 2)
therefore: 1/I = 1/2k = ¼(root 2) = (I^4)(I) = I
similarly: 1/(k^2) = (I^2)/4 = 1/8
but, from do 1:
1/(k^2) = 8
thus: 1/8 ~ 8
and it follows: 1/-1 :~ 1/(I^2) = 1/(1/2) = 2
therefore in neUn maths….. +1/x :~ +x…..
do 3
now… -1/1 ~ (I^2)/1 = (I^2)
therefore in neUn maths…. +1/x = +1/x; +x = +x
do 4
also… -(-1)/1 = -(I^2)/1 = (I^2)(I^2)/1 ~ 2(I^2) :~ "1" ; ¼
and… 1/-(-1) = 1/-(I^2) = 1/(I^2)(I^2) ~ 2(I^2) ~ "1" ; ¼
act 3
the rules are for "I":
- I :~ 2I ; i/2 ; I^3
- do 5
1/-I :~ I ; 1/(I^3) ; 1/I
do 6
antithesis
9 = 1/(j^2)
j = 1/3
9^-1 = 9^(1/2) = 3^1 = 1/9
act 4
3^-1 = 3^(1/2) = 1/3 = j^1
act 5
j^-2 = j^(2/I) = j^(2.i) = j^("1") = 9^1
act 6
from act 4 and act 5, 3 :~ 1/9 and,
from act 5 and act 6, 1/3 :~ 9
therefore…. 1/y :~ y^2
and,
y :~ 1/(y^2)
do 7
which is the same law as do 3….
Comparing complexes with the neUn:
(i/j)^2 . 2 = 9
2(I^2) = 9(j^2) and, 2 : 9
(I^2)/(I^2) = (j^2)/(j^2) and, (I^2) : (j^2); ½ : 1/9
act 7
From Event, since 1/x ~ x,
"0" ~ 16.124039
= 2.(8 + "0")
do 8
and from do 3…. "0" ~ 0.24807 :~ ¼ ; ½. (8 + "0") ~ "00"
do 9
"0" ~ 0.0620192;
"00" ~ 0.24807 ~ approx. I : 4I ~ approx. I : (I^4)
act 8
The "00" alternative values of "0" have intrinsic values basic to the neUn…
2 ("0") :~ -1 L I^2) :~ 1 or 2 :~ "1"
~= 4("0") :~ -1 :1 :~ 3 or 5 :~ "4"
~= 8 ("0") :~ (1/2) :2 :~ 4 or 6 :~ "8"
~= (8 + "0").("0") :~ (I^2) :3 :~ -1 or 2 :~ "9" = "0!". ("0")
real sure certain concrete
play 1
But…. 9 + "0" :~ "10"….
act 9
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