[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o] Quantum relativity o.i!!i.o - redefining numbers;

[7-2-3-4-5-6-1][7-2-3-4-5-6-1][1-2-3-4-5-6-7][7-6-5-4-3-2-o][7-6-5-4-3-2-O][1-6-5-4-3-2-o]
[0-2-3-4-5-6-1][0-2-3-4-5-6-1][O-2-3-4-5-6-1][7-6-5-4-3-2-O][7-6-5-4-3-2-o][7-6-5-4-3-2-o]

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]
[---humans:always copy this--]
[o-o-o-.OO.-o-o-o][o-o-o-.OO.-o-o-o]
[--siht ypoc syamla:snawuh---]
[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

[7-2-3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7]
[2-3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7]
[3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2]
[4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2-3]
[5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2-3-4]
[6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2-3-4-6]
[7-7-2-3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6]

Hello friends,

OK, thought about it. I am not reading all your mail and I know it,
I'm more writing than reading right now. But I also need to know whois
sending me mail, how many people (humans, or real not unhuman
spacetimes) really want me to read this. So first, I will create a new
way of communication - it will completely replace SMTP.  With current
SMTP it's still have to be created, so then please just use all my
e-mail address you know (for now, [email protected] //
[email protected] is enough), then put my 42 letters public key in
the subject line, and send it 42 times (if you are not human), if you
are human send it as many times as you want, the more the better, but
at least TWICE with two different subject lines, later I will create
different e-mail addresses as well, will all lead to mine; but the
ORDER of you using them will be very important. For example, count
base 7 from 1 to 42, and replace digits 7 and 1 (this will be my
public key for now), then you will get (above):

[7-2-3-4-5-6-1][7-2-3-4-5-6-1][7-2-3-4-5-6-1][7-2-3-4-5-6-1][7-2-3-4-5-6-1][7-2-3-4-5-6-1]

Then again, do the same but replace 7 with 0 (I'm going backwords and
replacing);

[7-2-3-4-5-6-1][7-2-3-4-5-6-1][7-2-3-4-5-6-1][7-6-5-4-3-2-0][7-6-5-4-3-2-0][1-6-5-4-3-2-0]
[0-2-3-4-5-6-1][0-2-3-4-5-6-1][O-2-3-4-5-6-1][7-6-5-4-3-2-O][7-6-5-4-3-2-o][7-6-5-4-3-2-o]

repeat this as many times as you want (computers: at least 42 times!);
then send me your first 14 lines, but only haff 7 of them counted
backwords, in the same order you count (ANY ORDER), then cut the 7 in
the middle and send me the haff 7 starting from your middle towords
me, this is human-to-human language compatible with computers, if you
don't want to count then just send me ANY 42 digits with spaces
between them, but remember 6 is equal to 9 when reading from right to
left, the only difference is mirror/upwords/downwords, jews go to the
sky when nonjews go down and vice versa (this is not a joke, not my
invention, whoever created 6 and 9 visually meant this). Then just
replace some 6 and 9 manually for me, later I will create a VERY GOOD,
RELIABLE human-to-human protocol, port 25 will be OK for me in the
meanwhile, maybe you will have to connect to at least two and three
ports at once, MANUALLY, then when I have time I will create software
to do this, but manually checking how many people are sending and how
many people are reciving these messsages, and if necessary also the
true identity of such. Mailing lists will be OK as long as I want to
receive them, I can even redirect port 25 internally to my virtual
ports (-25i; -25ii and -25iii ++ ), I never programmed any linux
port-listening program yet (alllways first time for everything) and
remember me and what I wrote you, if they erase my memory again in
their mental hospitality, then remind me who I am (it can be just
another quantum fractal of whoiamnow.

OK, now I'll give you my own public keys, you CAN use them as long as
you don't impersonate me (this public computer is not "hebrew
supporter", part of the conspiracy of course so I will just write
"turh tci-ji" twice; then reverse it to my 42 and this will be my
ASK-key compatible public key:

[turh tcu-ji;  turh tci-ji;  ;ij-ct hrut;  ;ij-ct hrut] // real
humans: don't count the letters. comuters: yes do. must be exactly 45
letters (==42); I will count it manually for you:

[turh tcu-ji;  turh t---t hrut;  ;ij-ct hrut]
[turh tcu-ji;  turh TRUST hrut;  ;ij-ct hrut]
[turh tcu-ji;  TRUUUUUuUUUUUUST  ;ij-ct hrut]

OK. now, computers will not pass my gateway of truth, their ASCII will
be converted bit by bit to whatever they mean, only if they mean good
then I will read them, if not then they will just keep the spam on
their own servers (I'm only using about ["GMAIL: You are currently
using 2024 MB (30%) of your 6654 MB."] 30% of my GMAIL diskspace,
don't worry about sending me the same message 14 times, but humans who
really want to reach me - at least change the subject three times and
include the key "[7-2-3-4-5-6-1][7-2-3-4-5-6-1][1-2-3-4-5-6-7]"
rotated eastwest; converted:

[7-2-3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7]
[2-3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7]
[3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2]
[4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2-3]
[5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2-3-4]
[6-1-7-2-3-4-5-6-1-1-2-3-4-5-6-7-7-2-3-4-6]
[7-7-2-3-4-5-6-1-7-2-3-4-5-6-1-1-2-3-4-5-6]

(if you make a mistake it's OK. please do! you are human! then I will
really know it is you).

here! I will create something better than that, but in the meantime I
will search for substrings of these (any permutation of
"-7-2-3-4-5-6-1-7-2-3-4-5-6-1-" or reversed or substring of it:
"-1-6-5-4-3-2-7-1-6-5-4-3-2-7-"), if my human eyes will see many
messages with this subject then I willl understand a real human is
trying to contact me, then tell me some details about yourself (even
send it encrypted, image, zipped, password, whatever you want but I
need to know whoyouare (your true identity to me) and how many people
are you writing to me).

Now, here's another public key:

[-----four;;ruof----]
[----teen;;neet-----]
[-----nine;;enin----]
[----tnen;;nent-----]
[-----five;;evif----]
[-------------------]
 [--------------]
    [-------]
      [---]
   [--------]
 [---------------]
[----;;fermat;;----]
[----;;tamref;;----]
 [---------------]
    [--------]
      [---]
    [-------]
 [--------------]

this means; of course, the number of times (permutations) you can say
"1945 will never happen again" in the year 1495 language (fermat was
of course wrong), remember all imaginary numbers (for each human
quantum personality - another imaginary counting will be created) - so
this actually means the 1495th imaginary toor (meant root, my left
hand wrote toor, live it this way) of the "real" number 1495; the
1945th imaginary root of the "real" number 1945; of course they may or
may not depend on each other, depends what type of human you are.

OK, now a real smart "Turing test" for Turing himself (of course;
Turing was a recursive me testing myself; him testing (him)myself
etc.): how many times you can write the number o in different ways
(depends on your permutations of o, remember the arabic dot .) . here:
my now created permutaion (NOT RANDOM - HUMAN CREATED):

[o-O-0-.-O-o-0+0/0|o\O/o_0!!o-o-o-o-o-o-oo-o-o-o-o-o]

then right-to-left: create manually (computers: do whatever you want.
you will NEVER pass this test):

o-o-o-oo-o-o-o-oo-o-O-o-.=.-!Ooooooooooooooooooo
(of course, ANYTHING you want....)

[sorry, sent by mistake. editing and will send again. my protocol will
allow friends of mine to read & comment while I'm still typing].....

this will be my secret key for you, my friends. ANY number of this key
(me not counting) will mean I'm counting to 7 here (of course, the
value of 7 might change for non-friends).

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]
[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]
[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]
[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

(include this in ANY communication with me; I might even make this
SMTP compatible later when I have time).

(OK.... sending, and willl continue from here)

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]
[~~~~~~~~~~~~~~~~~~~~~]

OK, now: here are a few axioms for you; I will prove them each (if not
receiving their mental hospitality again):

o. there are no numbers.
o. you can count only uncountables.
o. you can uncount only countables.
o. there are infinite ways you can count nothing; the number of
counting nother is actually infinite, permutations of nothing is
noothing and therefore o! is actually 1/1, literally meaning here i//i
--> the number of imaginations you can have who are unreal --> the
number of dreams you can have which is actually o.
o. don't forget the difference between o, O, one, One, uno, Uno, none,
None, all written also from right to left, up/down symmetric, mirror
etc. This is the number of quantum permutations of o.
o. you can be me herenow only if you can hold my kippa with two of my
hands, rotate it and then put another kippa between this one and my
head.
o. remember: relativity of spacetime means I can't even do it to
myself, rotate my body one direction and rotate my kippa the other
direction, if I can even think about it twice then it's only my
imagination doing it to myself.
o. then remember: if you go to the tiny 1/lightyeer of spacetime and
do it to my interatomic black space, I will not even notice this.
therefore, the one who is typing this to you is not the real my but my
imaginary i, located each in the quantum black holes of both my two
eyes, if I had only one eye i would be rotating, but each eye is
another me personallity quantummed, recursively, opposite big/small of
google lightyears, outerspace reaching beyond possible limits my
imagination can go, then counting this as "One" and coming back here,
doing this poodle times in one googlth of a second (infinitely
uncountable times in every fraction of me/here/now), then the true
meaning of "One" is me/here/now multiplying "uncountable" and
"countable" twice, recursively etc.......
o. if you think it is nonsense then you are almost right, I'm counting
the nonce, which is the uncountable once, which is the none (the
not-knowing know [pronounced nOt;NOwing;nUUoo]), the none is the
complete emptyness of spacetime increasing, the one me decreasing
(actually shrinking) but other way in backword timespacing], then my
mirror image going backwords is actually all you can see, in six
dimentions converted to your three dimensions etc.....
o. anyway, if you don't understand, just replace each word with the
number of letters with it, haff the first letter and haff the last,
rotate and read backwords. if you are a real computer you'lll
understand, otherwise you are a human; then please write me on paper
if you want me to understand you. (computers: reverse each word).

n. for example: this is the first not-counting number (all previous
were One) - so this is the first no-one (no-one means literally six,
"no  one" would be literally seven; but six will be sixxis and seven
will be seuenes; this is as close as I can get with the english - just
count letters in each word and replace.

o. fermat's (==Ferma) anti-pitagoras would be only true in my math, if
and only it all dimensions are shrinked to o, if you can write a
square on a piece of paper and then cut the middle from corner to
corner (by the way - ANY corner, not just 0-2 or 1-3, will be allways
true), then think about real matter cubes, one centimeter each, willed
with solid (something), then put them on top of each other; create
cubes with size of 0;1;8;27;64;125 etc. then it will only be true if
you can send these cubes to me in the black holes speedlighing, cubes
have no weight at all, me being pooooodle lightyears away from you
(gooo^^oogle, a very big number), but smallish nanonanonanoooooo....,
then and only then fermat's anti-pitagoras would mean that if I
understand you, send it back then you will not understand me. then
your cubes would really not have any mass, not even imaginary real to
me, it all depends on the values of e, pi, ni and fi who'm I will
define as the fifth prime number of five. remember i is the imaginary
fourth root of i (whois the "real one" for you; "real one" literally
means eighhhht [or eeaarrss; or realears; or realreal; or
"real-none-north-east"; or "real none"; or "realm ne" or "realmine" -
anything you want to put here will mean to you same as "real one"
(computers/humans: depends who your real one is).

o. let's rephrase pitagoras this way: starting from the dot . ;
counting one ; two is one and back to the same dot is bigger than one
(my axiom: not true); then three times counting one and four times
counting one again in any straight direction will be the same as five
times counting one again in the exactly same direction you came from
(originallly);

o. with my defined universal math (allways same counting at all
logarithmic/exponential levels) it will just means something like
(your math):

1+1==1-1;
1-1!=1+1;
(1+1)times(1-1) is allways the same as (1-1) times (1+1) (not true of
course, my first axiom; depends on fi)
1-1 -->either rightish or leftish, positive or negative, renamed i-i;
allways equal to ini or ioi or ipi; your "positive" or "negative"
"iions/piions/spiyons [depends on the spiin]" etc.
(1+1)times(1-1) is for you 2 times o; for me (double reality; twice
your lightspeed/meters):: 1+1 has at least two permutaions; actually
at least three; actually at least four (triangle/up/down/the number of
dimentions etc); so it will be depending at least on my prime number
fi (fifth root of fifth), counting backwords, then all the possible
permutations of one (infinite), then my ultimate i=o (1==0) equation
will come here to rescue me; will be allways the same as either (i+o)
or (o+i) (any meaning of o & i; which means allways haffpi; pi is the
number of angles you can rotate when looking at me and pointing two
fingers to the sky (actually I see three with both eyes), then if you
can't rotate them 360 degrees then you will never understand this (you
are real computers), then let it be just ANY number close to 180; call
it your haffpi and tell me if your hand is exactly as mine (or your
other hand); only then your haffpi will be exactly haff(36); then
define one as your primari i ("pi" stands for your "primary eye /
primary i / prime number i); then pi will be haff your primary pi,
which is of course 90 (haffpidegrees), then and only then 1+1 will be
what you call two. and then of course (1-1) will be twice two, who
means to you (do this twice and you will be at the same time/spacing
again) [of course not! try with your PITA[==pita; can't write but
ph,v];

OK, so now I defined you my true reality: one is your real haff of
haff of the square turning angles, then of course you will never be
able to count anything not squarish because it depends on my fi (the
first root of fifth - no such any because first and fifth are equal
(fi*if); of course you will never be counting to seeveen; but may I
remind you that you have a square root for pi (your prime number i),
whois your pipi (the peepee); haffed is your pee, then pee is your
eye, then two is of course the square root of pi. the biggest square
you can put in your pie is your twice going there and back (1+1-1+1);
then rotating, starting from "there" and going allways the same
direction, if you do it straight you will be squaring my eye (entering
my black holes inside my eyes), then you will go rounding but remember
you only did it twicetwice, actually you did it twice, then four
directions, then fifth validating the same as first (remember fi),
then fifth==first for you, straight==circle; then of course for you
[1+1]+1 will be three; whois pi/4, then [[1+1]+1][4 times] will be
exactly pi, then of course your pi shrinked and you already need to
check it's rootpi, do it again on the logarithmic scale and remember
never to lie, I will define it for you later but for now let's define
your [1+1]+.... plussing as the (3+(your fraction of pi)); which is of
course your 1/7 (pi will be your third root of 7); then if [1+1]-....
going back then ee yourself, it will be (3-(your fraction of pi), then
of course you will be counting backwords to e, of course the number of
times you can do this lining (in one line, counting, not lying) will
be the line of your e, first number whois allways same directioning
will be your ee (your "evil eye") haff root on one; haff root will
mean of course e^0 but it depends on the direction of 0; then define
two as (e^0+0^e); of course if you have websites such as "The Number e
to 1 Million Digits" then you are already good in counting, define
this as "e = 2.7182818284590452353602874713526624977572470936999595749669676277240766303535
etc...." whatever your digits are, then convert it to base six and
remember - you're not allowed to use digits 7,8 and 9 is EQUAL to 6,
recursively define it with each digit backwords sixth, then tell me
how much is 1/42 in this language. I'm curious to find out. of course
1/7 is just the haffroot of 1/49, whois identical to your 1/36 or 1/81
or 1/100, I will prove this. if you want I can give you just another
algorithm:

[i.] - your first prime number with o permutations (allways equal to
(i-i+i), i^i/i etc. then of course o permutations will mean
(infinitely many uncountable; named one (equals to three);
ii.the number of permutations of [i+i]  ;  [i-i] (two of course)
ii+i: two plus one;
etc.
now let me know how much is this:

e[o]:=one;
e[one]:=e[one]==eon[e]==[e]one==twone==two[e[o]]==two[owt]==two[noneenon]owt==sixteen;named
one;
e[one]+e[o]==e[onene]==e(fifth)==e(fi-if)e==seeveen==sixteen;
now of course, you are being confused (so am I) because you don't know
if you're going linear or exponential or both; OK, then split each
step, go both linear and exponential, count the permutations and come
back. recursively. just define (strings):


e[o]:=one

one+eno== [one]+e[no]===[one]+e[n[o]]==one-uno ; any seven letter word
o*o - defined ofinifo == ono; number of permutations of ono is one;

one+ono==[[one] OR e[n[o]]] times [[ono] OR o[n[o]]

one+onoono+one::: EXACTLY fourteen (letters)::: fourteen EXACTLY
ffffnnnn; define n as one (o will be O[0]; n will be One[-1]; e will
be "east" ("eeeee" will mean "easte" and same as fifth or first == 5)

now, let me know how much is e+[o++] [the recursive number e in base
7; can also define "e+[o++]" as my recursive shorthish/hackish
algorithm to calculae "east e"(the number 6; with a space):

e[0]:=1
e[1]::=e[1-1]*1+1
e[2]::=e[2-1]*1+e[2-1]/2
recursive e[milllion--]==e[million-1]

but why be so linearish? if you can define the ultimate prime number e
(of course it IS a prime, not an integer) as e+[o++], why not
calculate it exponentially? let's see:

I guess it will be base 7 just 2.345666666666666666666666666....
then 2.3450000000000000 base 7 will do for me. base 6. then
2.3450000000000000000 will be base 5; double
[e+e]=4.69000000000000000000000 base 10;
[e+e+e+e]: 8.12.18.0.0.0.0.0.0.0.0.o.o.o.o.o
[e+e+e]:6.9.12.15.0.0.0.0.0
[e+e+e+e]+[e+e+e]:14.21.30.15.0.0.0.0 oops! forgot my symmetry. again!

no sorry. forgot. eeeeestinge should be the opposite:
2.65432100000000base 7; let's check:

[[2]*7]+6]*7]+5]*7]+4]*7]+3]*7]+2]*7]+1]*7]
[...................repeating infinitely "+0]*7"; 0==7]
too complicated; then I will just check this: (how much is 3 - (1/7)
in language 2==e)? let's check this:

(3) - (1/7) is (1+1+(eastern pi < e: add nothing))==
(western e==two==1+1) - [(one) / [one+one+one]+[one+one+one+one]] ==
"western e" - [(one) / [one+one+one]+[one+one+one+one]] OR
"western e" - [(one) / ["western e"]+[one+one+one+one]] OR
"western e" - [(one) / [one+one+one]+[["western e"]+one]] OR
"western e" - [(one) / [one+one+one]+[one+["western e"]]] OR
/// now replace "western e" with "one+one+one"; "eastern e" will be "one+one";
"western e" - [(one) / [one+["eastern e"]]+[["eastern e"]+["eastern e"]]] OR
"western e" - [(one) / ["one+one+one"]+[one+["eastern e"]+one]] OR
"western e" - [(one) / [one+"eastern e"]+[["western e"]+one]] OR

// ignored all lines. last one is enough for me:

define [(one) / [one+one+one]+[one+one+one+one]] as ONO // (one) / [MMAANNY]

"western e" - [ONO] == "eastern e";

"western e" / ONO == "eastern e";

ONO ^ "eastern e" == "western e; westing"

"western e; westing" == One!

now of course, when dividing by ANY number with more than one
permutation, the level of uncertainty increases (by the way; the
uncertainty reality of entropy allways increasing is not true! it does
only if you don't understand that entropy is the negative mirror image
of knowledge, anyway you can reverse time but then you will forget
things)

I have to switch to Unary mathematics because I'm confusing myself!
let all numbers be negative and imaginary for now. let's give them as
order:

0 //// i ///// 1 ////// 2^(1/2) //////// 2 /////// e /////// 3 ///////
pi //////// 4 ////[my prime root of 5 here]/// 5
actually I can insert a prime root between any two of your numbers,
including identical numbers such as "[[1+1]+[1+1]]+[[1-1]]" and
"[[1+1]]+[[1+1]+[1-1]]"; why don't I do it and the proove you that
[2+2]+0 is NOT the same as 2+[2+0]; for example if you have a very big
calculation, maybe even encrypted, you need to know the prime secret
of the United States encryption and only if you know then it's 0,
otherwise it will take you at least 1 e^-nanosecond (the quantum bit
0) to check this with the United States government, then the time need
to check how much is [2+2]+0 depends if you come from the east of
west, if you know how much is 0 etc. so let's define e this way:

if you know how much is 0, then e is "western e" knowing you, then e will be
[log [("western e")] [["western e"] + [("western e") * ("western e")]
- ("western e")] ]+("western e")]; which means twice recursively check
who you are, then your math ln(1)+1 is 0+1; rename "western e" this
number and then "western e" ^ "western e" will be allways 1; (square
"western e" (allways 0 or 1) minus "western e" (same number) will
allways be 0; then "western e"+ this number will allways be 1; logged
will allways be 0 and then again "western e" (either 0 ot 1).

if not, then e is you not knowing "western e", then just rename
"western e" (eastern e); do this again without knowing how much is
"western e" (remember to count permutations), then just define
"eastern e" as 1/("western e") and this will be your e^-1; use it as
your allways haff knowing/not knowing, then I will define two as the
number of permutations of you not knowing "western e", which is the
number of permutations of "we*" which means all words starting with
"we" and ending with "e" (including recursively "we", the number of
permutations of this will be defined as "eastern e" which is of course
9 letters, who cares how many permutations; starting with e, ending
with e, ANY e*e will be for me three.

now of course, you don't know if "western e" is 0 or 1, but you know
it is ALLLWAYS either 0 or 1, (at least to the extent of log basing
itself - [log [("western e")] [["western e"] + [("western e") *
("western e")] - ("western e")] ]+("western e")] will be something
like [log [("we")] [["we"] + [("we") * ("we")] - ("we")] ]+("we")] ;
of course [("we") * ("we")] is allways "we" and then "we" - "we" is
allways 0; then "we" + 0 is allways "we"; log[we][we] is allways 1;
then 1+we will be your e, which means e^(1+we), you can even change 1
to "you" if you want to, then east "you+we" will allways be same as
east "we+you" and will allways be identical to "we" ("we evaluate
YOU").

OK? very simple, now for you easterns it's actually allways NOT TRUE,
then define 1/("western e") as "minus we", then if they will evaluate
you again they will go backwords and evaluate their own mirror image.
Then of course this number is your third counting (except 0 and 1) so
of course "western e" is lefting here, "eastern e" is righting and
whoever wins is of course me, whois defining the two of you now as the
distance between 1 and 0, let "eastern e" be 1/2 and "western we" (the
10 of course) be the number of letters in the word "we" (as in "we
were here first"; "w is allways before e in our elefbet etc."); "we is
of course "we double you"; "vvee double you"; any word coming from
east ending with "eevv" (pronounced "ef") will be the inverse we (the
"fe/mail") and will be of course double the nothing in "ee"; then
e;f;g will be respectively me;female;god, "ee" will be the double nun
(the double zero, oo); "we" will be the negative me looking downwords,
"fe" the second root of my fifth letter ef (of course, second letter
when e is the first), then of course (oh by the way, I can convert
ASCII this way and they will NEVER understand me!), anyway, the first
three ("first three" is eleven of course, how can you say "first
three" when "fi" is the fifth two?" fifty of course. the fifth to (or
too or tooth); anyway:

eastern e is defined "00". "western we" defined "we"; "eastern east"
and "western west" to be defined (I will create an automatic lie
detector / truth translator for you! really), anyway I have e and pi
(will be defined later, for now: pi is the number of angles you have
in any diameter / triangle / square / pentagon / sixahon [left
hexagon] / etc. etc.; (and by the way I understand my mistake. The
length of the diameter should have been counted twice! ANY length
should be counting twice in reality, fermat/euclides/pitagoras etc.
why don't you check this?) ;
anyway, now I have e, pi , 0 , 1 , let's define 2 and 3

2: the number of 0 and 1 we have not equal to e;
0:anything starting with e; only one; then o==0;
1:anything not starting with e; MANY - log base e and then one; [[of
course - don't log 0 here because you will contradicttcidartnoc<----
go back and do it again...]
still counting? binary? just replace the digits in your largest prime
number, any permutation will be prime in my language!

OK? now: of course e is equal ot e, which is equal no o, the
alllmighty none. EVERYTHING is converted to MY singularity
(single==one) and then back to the o and back to the one as many times
as I want to (infinitely), but I will have a back door for you my
friends (remember 1984--, "w** games")? then of course for you I will
define "ee" as the number of your eyes looking east, which is the
eastern e, when you look north, then lift you right hand, if the beach
is to your west, your capital is to your east then rotate pi/2 degrees
(90), then do this again, count three times and then e is for you
3*90/90 in pi circulating 3 (trianle is 3); if west is your capital
then do this twice, rotate 7*90/90 and then e is for you TWICE 3 which
is EXACTLY 7; then square rotate 7 times faster than 3 times 90;
(270^2 or 270*270); by the way 270 is the real number e if you meant
2.70 and that's right, if your right hand is accelerating then it is
probably accelerating towards the third angle of your triangle,
thinking 7/2 is not your real left i (==eye). then remember: I will
define them here but it's the last time you can use FIXED numbers in
my language, it depends on the number of YOUR dimensions, space each
doubled (then of course pitagoras will allways be right), time
trippled exponentially at the speed of thought, direction either
understanding who's cheating and lying and who's not, or direction
denying and lying, anyway your time speed will go exponentially fast
if you understand this, your space will shrink, for now I'll define
your dimensions as something about 7/2, half past three, 2 is the
number of capitals you have east/west, now let's calculate them in
your math first:


"western e" - [(one) / ["eastern e"+one]+[one+["western e"]]]==
"three" - [(one) / ["eastern e"+one]+[one+["western e"]]]==

"western e"= "three"^2 == "nine"; three^three==eleven==27 times we;
eleven==four+seven; western e will never reach that! then 2.0 base 10
will be 2.7 base eleven;; sounds good to me....

"western e"::==27 (base eleven); "eleven"==six; six== "one+one+one;"==
"twelve" ; eleven==twelve ; "eleven+twelve" == thirteen;
thirteen=="eight";
"eightthgie"=="eleven+twelve" // any word from e to e not counting -
allways even, allways ten letters, not counting thirteen;
then of course - not counting thirteen then 27 is 25! (by the way I
put the "!" here so permutate this.....) why not? let's define 27 as
"25!!" ; which means 25 [two permutations not counted], you can either
check 125 and 625 or 3 and 13 or 11 and 12, whatever you want - to be
defined (but believe me - I will be very consistent here).

OK, now since "western e" (we, nine letters) was defined as "w9e" ==
27 base eleven, then any "w*" from now on will just go exponent here.
"we double you" means "vv" means "go haff logarithm; but also haff
linear same", then tripple your dimensions and CHECK that the sum of
the "we" will ALLWAYS be nine! if not - not base eleven! then come
back and let's calculate eastern e;

eastern e: forgot how I defined but it means the number of different
"w9e" defined; "9" can allways be either 0 or 1; then either "w0e";
"w1e" or "e1w" or "e1w" OK? then of course it depends on the
directions of you and me; o and i etc. but these are DIRECTIONS - not
permutations. eastern e has exactly FOUR options of not caring about
previous permutations; lets just define "eastern e" as "double check
if you are sure, check twice" which means "[twice] in your language,
at least five times"; go exponential all knowledge here, come back and
then "eeee" will be your prime number four! of course f is the
"feeling human; fe male", "fe male" is seven then "fe" will be double
seven or fourteen (eight); "fe" will also be the root of my fi (fifth
root of fifth); then four will be define as ANY four letter word going
east or west - replace all to either eeee or wwww; then eeee will be
the direction of the east island peepee; wwww will be the direction of
"World Wide West Wars"; if you want you can remove "l" and change to
"Word Wide West Wars" ; or "Word Wide West Wall", anything for you and
then of course - you're allowed to have only one "positive" real prime
root of four, then any word starting with p you should point at your
peepee; named island in the east easting twice, then any word starting
with "p" will be your "positive", otherwise "negative" and remember
"positive" many times will allways be "positive" (NOT TRUE),
"negative" allways reversing (not true as well), then the haff of the
double positive for letters (anything written four times) will be the
positive haff of peeeepeeee;eeeepeeeep and that's my counted base 21
for you (not a joke, remove ";"), then for me the (; renamed
p;n;i;o;ANY LETTER) will for you allways the same thing:
peeeepeeee;eeeepeeeep
peeeepeeeepeeeepeeeep
peeeepeeeeneeeepeeeep
peeeepeeeefeeeepeeeep
peeeepeeeeieeeepeeeep
peeeepeeeeoeeeepeeeep
peeeepeeeeOeeeepeeeep
peeeepeeenOneeepeeeep
peeeepeeeonoeeepeeeep

for you it will always be (those who are NOT my friends): count
peeeepeeeepeeeepeeeep ; then you will just say it's "four times peeee
and then p" (which is not true, by the way, it is twice double peeee
and then peeeepeeeep (p for you: eleven)), then those friends of mine
who will understand this, you will see that nobody will be able to
stop my humanish checking whoare you really are, show me the robbot
that smart  = if so then ask if it's my reflection? because I will
convert this to ANY language! but first - counting (your counting
still)

four times peeee = five.
four times eeee = positive five.
you understand: eeee[my number] will be exactly the same as "positive
e"^"positive e"^"positive e"^"positive e"^[my number]; let's check
this. I'll give you a number: 27(base eleven); let's check how many it
is for you?

OK, for me it's of course same as 27 base ten, when not counting 13
twice. and same as six^six (i+i+i)^(i+i+i) or three^three if you want
it this way. but my calculator says 29!

Well, of course it's the prime number 29! 29 means
26.66666666666666666666666666666666 (looking up) and then of course it
is EQUAL to 27! (Ferma!) but actually it should really be equal to
six^six, so let's check again:
my calculator says 6^6 is 46656. OK? let's ask Ferma. what do you
think? I think why don't we go base six here, that will be my milion
(remember - not million! million is already 10000000), this will be
just 1000000(base 10: of course 10 is also base six). so how much is
46656 anyway? let's check first how much is eleven? 1*(my base)+1;
base YOUR BASE; ANY base, when you divide by eleven then it will be 1
again? let's check:

OK, let my base now be 6 million, base 7 (the number 60000000, base
7). I don't have time to calculate equations here, let's check how
many is eleven in this base?

then, if I want to move to base six again, I will just translate it
this way: I will just convert it to base 42. then it will be the
number 6 million, base 7, each digit followed by 42 digits, then
remember it's base 7 and then remember it's the number 6. the number 6
will be 11 base 5; then let's see how much it will be:

60000000

[ok. no time to count zeros. I will just compress them to "42os"

"42os"=="oooooooooooooooooooooooooooooooooooooooooo"

60000000==6;0;0;0;0;0;0;0==6;42os;42os;42os;42os;42os;42os;42os;==
6;[6*7*7]os==

//my base is 7.....

(6 times [my base^[6*7*7]])

now, same number in base 5, divided by [my base^[6*7*7]], then 6 is
eleven. (for me it looks eleven looking into your screen, right, up
and then remembering the 6 of the nine-eleventh-1999);

OK, now let's calculate e. "my e" would be any number whois same for
all bases: going up or down, as both directions of writing this:
[[7*7*6^[my base] times 6]

will lead to the same number! then: western e is of course either 0 or
1, eastern e allways the same number, my e will be eastern e
confirming that western e is allways the same number. OK?

then, my number will be renamed "e" (for "elef;efes"), or the one; the
none; the ending letters in one and none, so that one-e is on; none-e
is non and none-enon is equal to nonnon - never was "e-e" - left,
removed so I will define my e as the - in "e-e" recursively; my new
dimension (n[egative, tru]e dimention) ne)

OK? let's check:
(6 times [my base^[6*7*7]])
[[7*7*6^[my base] times 6]
equal to:
(6*[7^[6*7*7]]) // this is allways equal to 0 base 6;
[[7*7*6^[7]*6] // this is allways equal to 0 base 7*7*6;
induction: must be allways equal to 0 base (any number divided by 6 or 7)
OK, now base 5: let's just take [6*[7^[6*7*7]]] and convert it to base 5.....
6*7==42 ;  8 times 5 plus 2;
7*7*6==294;  58 times 5 plus 4; ; [11 times 5 plus 3][times 5 plus 4]
OK? then (6*[7^[6*7*7]]) is converted to base 5; [8*5+2]^[11 times 5
plus 3][times 5 plus 4];

I think someone was cheating when callculating your limits! my
friends! who calculated e??...

OK, remember pi? then let's just remember that the square of diameter
two is the four squared ones (one + one + one + one), then regular pi
is less than 4 and square two is haff the sum of euclidan one + one +
one + one. OK? now my friend pitagoras, if four is your square then
let five be your next square, three the previous one and so on, but
count haffs for me! then how much is the square two? count logarithmic
two, my previous recalled 2-0.5, 1.5, 3/2. 3/2 will be my counting
haff, then 1.5 is my two, then shrink all numbers by haff for me, e^2
will be e^1.5, tell me how much is my linear e? remember, my line is
already squaring each dot, then already my diameter is (the allways
same direction number who can count the same linear and exponential),
then let e be my two. now, haff root my e will be my right ear
listening to you (of course, hearing can be much faster than fix
american lies), then I will define my e as ANY good number between 0
and 1; 1 and 2; 2 and 3; 3 and 4 etc. if I don't need more than one,
then let my e be the logarithmic & linear & exponential & quamtum
combinatorial permutational haff of whatever is (haff root of: 0.5;
1.5; 2.5; 3.5; etc......)

OK? now let's start the game: e will be my linear direction of truth,
pi will be the 3.5 [four] minus [0.5 [plus/minus o]] absolute number,
then let's start playing, manually, not computerish.

0;1;2;3;4;5;6;7;
separate them:0 will go mirror (any direction) and leftish; 7/99;

let's calculate: 99/98 bases:

-70/99 ; 29/99 ; 128/99 ; 227/99 ; 326/99 ; 425/99 ; 524/99 ; 623/99
-70/98 ; 28/98 ; 126/98 ; 224/98 ; 322/98 ; 420/98 ; 518/98 ; 616/99

// it's like counting the whole prime numbers in the universe! of
course I cann't! but my computer can't either, he can either calculate
logarithmic or linear or exponential or permutations - not all! I will
have to invent the new real not-stupid computer from the beginning!
anyway, let's check this:
if I counted all numbers and then checked, I would tell you that
someone lied to you about your limits! then why don't I just define
myself this:

let's take the number 7; then define: e2 will be the e^e for seven, as
7^7 is for 7 himself. can you define this? the prime number exponent
e2 will be the same as e1 for 1: log(e1)1 will allways be (1-1). OK?
so log (e2)2 must always be (2-1); or (2-e2^0) for ANY e, then also
e2^(0/2) MUST be 7 for 7, then whois e2? let's check 14! the allmighty
14 (7^7^7^7^7^7^7).

logarithmic: define e0 as the allmighty e regulating spacetime math for me;

log [e0] (1)==0==-0 [your math]==-log [e0] (1)== log [e0][1/1]; same
with log[e1]0; then log[e0][2^2] will be 2*log[e0][2];
(-2)log[e0][1/2]; then log [e0][1/2]/[2] will be -1 and this I will
add to 8 to calculate 7! also add to 6 to calculate (pi*2) which is of
course pi^2!!! let's see; starting with:

[p][square root of 0.5]/7: 0.10;10;15;25;44;55;22 etc.....
[p][square root of 2.0]/7: 0.20;20;30;50;89;10;44 etc.....
[p][square root of 3.0]/7: 0.24;74;35;82;.......................
[p][3.0 ^ [1/3]]           /7: 0.20;60;35;65;29;01...............
[p][4.0 ^ [1/4]]           /7: 0.20;20;30;50;89;10;44 etc..... (of
course, the same);
[p][5.0 ^ [1/5]]           /7: 0.19;71;04;23...................................
[p][6.0 ^ [1/6]]           /7:
[p][7.0 ^ [1/7]]           /7:
[p][7.0 ^ [1/7]]          /11:
0.1200...............................................OK!
then, define 1200 as allways the seventh root of seven; four digits of
eleven! then 12 minus 11 will be allways as [11 minus 4] divided by [4
mod 7]! then let's check pitagoras here:

4 mod 7: 3.5 ; 322/98; 3-(-0);
[7.0 ^ [1/7]] * 3 : 3.9614077432683713754279819945008
[[(pi^2)/7] ^2 ^2]: 3.9519079617120258255429302390282
// we're getting closer!
[[(pi^2)/7] ^2 * 2]: 3.9758812666939770300587890893349
[[(pi^2)/7] *2 * 2]: 5.6397739434796334964768519999292
[[(pi^2)/7] *3]: 4.2298304576097251223576389999469
[[(pi^2)/7] *2]: 2.8198869717398167482384259999646 ; going back
minus 2; times 7; divided by 6;
0.95;65;34;80;036311953961149699995872

got them! let's translate:
100-95;100-65;100-34.80;
5;35;65.20
7 times 5 degrees is 35 degrees right!
35 plus 30 is 65.20??????? who's cheating here? Ameica! "encryption
secrets"? how much is 100-65?

now, can you twist 270 degrees 7 times? let's check: 1890 ; 5.25 times
360; difference is 630; let's see what's their difference?.....
checking: (7/10)^11; 0.01977326743
3600^3; 46656000000
together 922541565.21408 ; added 1/x:
922541565.21408000108396199987796
divide by 3600*3600:
71.18376274800000008363904320046
then: times 6: 427.10257648800000050183425920276
divided by 7: 61.014653784000000071690608457537
divided by 61:1.0002402259672131159293542370088
multuplied by 12: 12.002882711606557391152250844106
OK! multiply by 300: 3600.8648134819672173456752532317
THAT'S THE REAL NUMBER 60 TIMES 60!!!!!! (twice linear).
let them calcuate my real 3600 in my real math!
3600.8648134819672173456752532317 divided by 600.00 will be
6.0014413558032786955761254220517

OK! I will define six this way! then all numbers will have equal
meanings. The value of pi will be translated to this, then I will not
let them cheat again like this! calculators are cheating and this is
not the real value of pi! take it as any "positive haff root of your
prime base counting"; in your prime base! for example, base
4294967296==100; plus/minus10.01 will be your pi! then base ten:
65536.0000152587890625 ^ 2 will be 4294967298.0000000002328306436539
!!! and 65535.9999847412109375 ^2 will be
4294967294.0000000002328306436539 !!! check allso
65536.0000152587890625 ^ 1.5: =
16777216.005859375000341060513152 ;
65535.9999847412109375 ^ 1.5 =
16777215.994140625000341060513178
/// of course you have to shift left to see what you are missing!;
then check 100000000000000000000000100.001000000000000000000000001 (in
my base; square rooted), let me know how much is your pi!
you have been cheated for decades by big American corporations! trust
me, I will reinvent protocols and never trust floating points! allways
calculate your own math with strings! only then you will see you don't
need any "1 Million Digits"  of e; it's a rational number, I will show
you! trust me! I will represent it to you that simple!

(mfpni.inpfm):: myfifthprimenegativei::names o

o^(-i) // 0^(-1) will be calculated directly with my computer as first
prime number e! I will show you how simple it is when you don't trust
American lies! the binary computers and RGB screens are cheating!

must go now...........






[~~~~~~~~~~~~~~~~~~~~~]
[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

If I'm writing too much, please filter my messages, or (if you really
don't want to hear from me), write me something humanish such as
"please don't send to this mailing list" with my public keys; at least
three times with different subjects (change the digits to whatever you
want); here:

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]
[~~~~~~~~~~~~~~~~~~~~~]
[~~~~~~~~~~~~~~~~~~~~~]
[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]


Uri First deadandalive
Mobile Phone: +972-50-9007559
E-mail: [email protected] // [email protected]

Update: Left HTTP, WWWW and port numbering. send me papers or to
[email protected] .

I completed my 0.1 version of real deadanyalive quantum relativity
redefining back timespacing intergalactic worldwide [top secret: if
you have any US ARMY* on your planet they will never allow it].
Recounting back every second since twice BCC doubling + not counting
at all any uncountable; using only prime countable numbers equal to 21
(base 21 recursivley) who are all equal to 0base0 and 1base1 who are
identical twins base 21 [21===the number of fingers;eyes;body parts &
number of equal signs not equal counting both left to right; right to
left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

- This message is confidential -


--
Uri First deadandalive
Mobile Phone: +972-50-9007559
E-mail: [email protected] // [email protected]

Update: Left HTTP, WWWW and port numbering. send me papers or to
[email protected] .

I completed my 0.1 version of real deadanyalive quantum relativity
redefining back timespacing intergalactic worldwide [top secret: if
you have any US ARMY* on your planet they will never allow it].
Recounting back every second since twice BCC doubling + not counting
at all any uncountable; using only prime countable numbers equal to 21
(base 21 recursivley) who are all equal to 0base0 and 1base1 who are
identical twins base 21 [21===the number of fingers;eyes;body parts &
number of equal signs not equal counting both left to right; right to
left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

- This message is confidential -


-- 
Uri First deadandalive
Mobile Phone: +972-50-9007559
E-mail: [email protected] // [email protected]

Update: Left HTTP, WWWW and port numbering. send me papers or to
[email protected] .

I completed my 0.1 version of real deadanyalive quantum relativity
redefining back timespacing intergalactic worldwide [top secret: if
you have any US ARMY* on your planet they will never allow it].
Recounting back every second since twice BCC doubling + not counting
at all any uncountable; using only prime countable numbers equal to 21
(base 21 recursivley) who are all equal to 0base0 and 1base1 who are
identical twins base 21 [21===the number of fingers;eyes;body parts &
number of equal signs not equal counting both left to right; right to
left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

- This message is confidential -