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

From: Uri <[email-masked]>
Date: Thu, 24 Apr 2008 19:17:58 +0200

[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-masked] //
[email-masked] 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-masked] // [email-masked]

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

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-masked] // [email-masked]

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

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-masked] // [email-masked]

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

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 -

---

From: Uri <[email-masked]>
Date: Thu, 24 Apr 2008 19:45:30 +0200

[+o+]

Oh by the way,

my human friends,

here for you real humans, who are not spammers, here's another
definition of e (the regular one): "+o+" (plus;efes;plus), can also be
"+O+" or "+0+" (+o+ looks nicer); which means: "just calculate the
symmetric sum of all numbers counting from here both directions, all
positive, then this will be your real e. if you write it my base, then
each base you have to define with two rational numbers, like this:

0.0
1.1
2.2
3.3
4.4

//these are your integers, regulated symmetric, then for you it will
be [(0.0)+(10.1)(base 1)+(100.01)(base 2) etc.; but remember you are
just adding "0.0" and then "10.01" and "100.001" symetrically here!
then it will just be "11111111111111110111111111111111111" for you,
then remember the bases are also exponential! for example 100 is 10^10
(binary:2^2) but base ten it will be already 100=ten^ten! not ten^two
and let computers understand this! then the first turing test for
spammers will be just "the dot", define your own language first. let
them define their language with the number of dots. then +o+ ---> let
them calculate in their own language, if I ask for the first 10000000
digits they can give me the first 10 million digits of e (if they
understand me!) (of course I can't read it) but then I will understand
them! I will "ask them" to "correct my mistakes" and then understand
their language! and for you humans, just remind me what is e, how you
calculated it, the sum of all permutations written backwords, then of
course e^7 will be the sum of all permutations up to 7! backwords
(like minus (o!1!2!3!4!5!6!7!) but written in a way such as
[1/0[1+1/1[1+1/2[1+1/3[1+1/4[1+1/5[1+1/6[1+1/7[........... (forever);
for me I can just define it as sum(i/o++) and shortish just +o+ is OK.
my protocol will be humanish: connect directly to my computer, or
write a computer program whom I can trust, can answer my humanish
questions for you and I will make them very simple for friends, very
impossible for not-wanted-mail and very not-relying on filters of
American + Israeli start-ups of Nazzdaqqqs lyiers etc.

Only those of you who are not working in USA corporations will be able
to receive trust from me. Sorry! I don't even trust my own mirror
image! only humanish trust and not binary (VVVVee evaluate
you!).......

by the way, going exponential means "sum all linear"; then "io++oi"
will just mean the same as +o+ on the linear level; then this number
"io++oi" will also be the computer programming representation (no need
for recursion), then of course just sum all linear numbers....

[~~~~~~~~~~~~~~~~~~~~~]

Uri First deadandalive

---

From: Uri <[email-masked]>
Date: Fri, 25 Apr 2008 01:45:50 +0100

[+o+]

Hi friends,

this issue is very urgent so I'm writing you again. I closed my
website, left HTTP not working. I will define it later. When I can I
will make a much better everything - speedy communication, mail,
everything - not depending on me and will keep working without me -
completely open for changes - like Linux. If you want to help please
do; but I need more people whom I can trust. If you want to be my
friends, you or your friends then please use my public keys, create
your own (human readable) public keys and introduce yourselves. When I
have time I will maybe create a platform for doing this - in the
meantime please do it manually. I also need money but this is urgent
and I can't start a new job right now. If you can't encrypt and don't
trust computers, then please send me a paper mail - I need to know who
you are personally.

I thought about it, pitagoran and fermat were both right - they
defined numbers and math the way they wanted it to be, so in THEIR
math of course they are both right - they will both me wrong in my
quantum math - I will prove this. I will also show you why a binary
computer will never be able to understand math. Turing knew this,
actually he worked for the USA Army and everything related to Turing,
his recursively testing himself, my splitting my personality (more
than twice - not such a number) - this is all related to the 1905-1945
conspiracy / 2005-1945 backword conspiracy / 1984-1948 BBIAWY etc. I
will explain this. But first, a new matrix:

abcd
efgh
ijkl
mnst
opuv
wxyz
// wrote this ommitting some letters, left this, have to check why.

abcd / -[0123] / אבגד / alpha betta gamma delta : allways five letters!
efgh / 0123 / הוזח
ijkl / 0123 // sounds like "i!!! jeuu kiii llll youuu" / יייי
mnop / 0123 // "klmn"-->"כלום" ; hebrew: כלמנ
qrst / 0123 // קרשת or קרסט
uvwx // 0123 // sounds "you vve double[vvee] you!! exit!
yz // yez!zevel! // -2; -1;

the letter "T" (capital t) in the two is of course number 15 which is
not surprising at all! (counting from e0; otherwise 20 counting from
a1); and 1111 binary; also in matrix:

[e;f;g;h] [0;1;2;3] [0000;0001;0010;0011]
[i;j;k;l] [0;1;2;3] [0100;0101;0110;0111]
[m;n;o;p] [0;1;2;3] [1000;1001;1010;1011]
[q;r;s;t] [0;1;2;3] [1100;1101;1110;1111]

the letter "T" (as in "True") is the largest number, right/down for me
(jew; 5 fingers); then next matrix:

[u;v;w;x;y;z] [0;1;2;3;4;5] [1;2;3;4;5;6]

"w" means twice "v" (pronouced "double "u", spelled "double v"); "u"
is the first letter in this matrix, 0 or 1 (depends how you count),
then v is "straighing" u and the "double v / double u";

then o; of course "the o"; but not ASCII 0, of course. I will explain
it later. In english when you say "o" (have been in south africa) or
write "o" you always mean "0" and not "one";

n: nine;
e: Aleph; elef; efes; oooo [==0000]; אמת ("emet")

i: 0100(four) in english; ten in hebrew י; actually 10base10 (binary /
decimal) and 0100 means the same thing, I will explain this. for now,
just think you can't have positive 10 and negative 10 represented both
as the 10 can you? how do you count twenty? or even 10+10? you
cann't!!!

OK, enough for the matrix for now. let's just say "two" is a three
letter word and I will replace it with "ttt" for computers who are not
friendly with me. they will have to replace it again. BUT I want to
tell you - it's an imaginary number! there is no such a "real" number
two! I can prove this! ok, lets see:

how did goodold pitagoras count this? he didn't have a watch, didn't
count time, but he KNEW that he can count space indefinitely, then of
cousre one centimeter or on lightyear or one nanonanonanometer or
1/one-light-year would all be the same for him! let's say he counted
intches.

then his paper was flat, no kippa, no Dom, then he first counted the
dot [.] - the arab o; no dimensions and ALL dimensions; then a
"straight line" (no such thing, defined as a x-ray of light) to
another dot, then one (pitagoras intch - let's call it a pinch). OK?
so he already had two dots when he counted his first line! But, he
REMEMBERED the first dot because his memory was not quantum, he didn't
remember how much time but he remembered it the same way I remember
the first "r" of the word "remember" [==eighhhht] when I type it. if
he diedandalived / replaced by another fraction of himself then he
wouldn't notice - he remembered the first dot, then one straight line
was his third memory. then looked back, then (at least his eyes, or
one of them) went back to the first dot, then it was already 3 dots -
the first, the second and the third==first (mod two); actually it was
(mors!!) ._._. which means dot;line;dot;line;dot (symmetric: :-:-:),
then he reasoned: "odd" things are allways dots (o + d , the ultimate
"end"; see matrix above: d==david; do; did; -o;-i;-0;-1;ד;the number
4;.... whatever.........
....and "even" things are allways "lines" (actually that's just his
memory of "many dots going same direction";;;; "direction" will just
be ddddddddd (d times nine);)

OK? do he had 3 dots; 2 "lines"; one understanding: this is my first
line, the dot! counting backwords from 3 to 1 (actually from 5 to 0,
counting 5 "things", 4 "dimensions", 3 "dots", one "line", . )
then he did it again, but from the middle (or did it again to the
other opposite side, 180 degrees of his world wide or haff counting
and then 90 degrees, then divided, counted and of course found out
that:

1. haff + one is three;
2. one + one is two;
4. two (diameter) square is big;
3. haff + one is allways straight 180/2 degrees;
of course, then why doesn't it add? because he had to count
logarithms! thenlogarithms he did, counted and concluded something
like this:

0. this is allways the first dot.
1. this is allways a line.
2. this is allways the second dot.
3. this is allways the second line.
4. this is either the first dot, or a new one.
5. if this is the first dot, then 4 times 90 degrees will allways be
back to the first dot (of course, 4 is already recursive logarithm #5)
// by the way I misspelled lgarithm which means algorithm is the same
word to me // "L-garithm" // and by the way a good algorithm has to be
at least logarithmic to the size of the information (binary log of the
number)
6. if this is not the first dot, then it a triangle - there is more
than one line etc.
then of course, rules for "straight" triangles, then "not straight",
sinus/cosinus/ etc. etc. etc.
BUT, how many times did he count before he created the "pi", his
circle? at least 5 times 4, which is 6 times 5 (not logarithm), or 5
times 5 - MANY times. and then he put pi in the middle, between 3 and
4. OK! I will calculate how much is his logarithmic pie (hint: "i"^5
allways equal to first "i", complex, then "i" will be the fifth
complex root of the logarithmic pi (didn't even define logarithm base
yet!)
e: but of course, logarithmic base will be allways POSITIVE, then must
be at least four times pi (the positive i), "imaginary positive i"),
whois exactly on the same paper of pitagoras (or was it maybe euclides
["east euclides"; "eeee eeeeeeee"]) to the north when the first line
was to the east, as I am sitting right now (tel-aviv beach to me
left).
OK? so e should be at least 4 times the pi; pi was defined by
pitagoras to be "one circle divided by one diameter", but he forgot
somethings:
1. he diametered twice (d-metered) and circled only one.
2. going from here to the moon and back will take much less time in a
circle than straight line.
3. of course, you can NEVER go straight line from here to the moon
because THERE ARE NO STRAIGHT LINES!
4. same with the nanos.... nuclear; atoms etc.
and then, the square - if diameter was "1+1", then square diameter
(haff both directions) was much bigger this his pie. OK? so let's see
how much is his diameter;

[1-1right]+[1-1top]+[1-1left]+[1-1bottom]+[1-1right;again] =
{"pitagoras"} --> "my diameter" -- named twice / two "1+1" (one plus
one he didn't define, there is no such thing on the logarithmic level
as "1+1")
anyway, this number,
"[1-1right]+[1-1top]+[1-1left]+[1-1bottom]+[1-1right;again]" was his
first "two" - a true straight line (he could have cutted by haff,
maybe different number, will check later...)

then what he did - he actually counted dimensions! he counted every
dimension twice! (on the logarithmic level), which means five times
(with the dots), then counted the number of dimensions as three, time
was nothing to him (dimension o), OK! let's count this way.

if we understand this, then already we understand there is no such a
number as two, or bi, "be i", or "biinaari" (eeeeight), at least not a
"prime" number, a "prime counting" number etc. then he defined "two"
as the opposite of "haff" (logarithmic, remember), then of course
twice-logarithmic-haff will be his two, or one. (the linear one). OK?
now, the number two was defined as the fifth root of pi, or 31.4
(something), or (3.14 times Ten), or the fourth root of sixteen (1;6
== one plus, one minus) == "opom"; =="o;+1;o;-1" which is
"oooo;ooo1;oooo;1111", this number translates binary of course to 15,
so "two" is either 1.4, or 15, or 31.4 ^ (1/5), or "3.14"^1/5 , or
["3.14 - 3"] times ten , whatever you like to do first, to two is
actuallly a DIMENSION (0;1;2----> dimension number 2) where rotating
the paper is not changing its size (of course, depends on the size of
the paper, try lightyears or nanoone/lightyears).

just went to the toilet (pee;pee) and then of course - something
written in TWO languages - one right to left, one left to right, (left
to write), one on the top, one on the bottom, of course: to understand
this you need to understand at least two languages (to++: at least
three), with numbers written always leftish, otherwise we would never
know where our directions are: left,right,west,east,up,down,capital
east (ירושלים), capital west (Wash your D.C), dots going right, .com
going left, o rotating, m going up (Mcdonnnnalllds) etc. you must know
all this to understand this! so;

pi was defined by pitagoras as the number of flat papers dimensions he
can rotate and count only one ("radius"), whois twice the original
radius, then of course "radiation" can come from there, "radio waves"
etc, then he defined 3.14* to be the number of dimensions of the flat
paper (logarithmic, relative to haff renumbered two, one was his fifth
understanding, then of course he defined two to be a "prime number"
(fifth root of 32 - of course logarithmic counting number oiiiii),
then i was his dimension of "allways use the same direction of truth",
then he wouldn't need anything more then ONE linear dimension, ONE
logarithmic dimension, name this as Two. so for me, Two will be the
number who's haff logarithmic truth is equal to the haffway between
odd and even, 0 and 1 (renamed o and i), then of course ALL PRIME
NUMBERS must exist on both levels of truth. Then,

pitagoras counted and defined his first axiom: 3*3 + 4*4 == 5*5; which
means literally that haffpi(90 degrees)plusone is Twice as much as
four times haffpi and then haffone;let's translate this (you know my
friends - people do lie ; math never does. but don't trust your binary
computers! I will define the ultimate human turing test (not
CAAAPPTCCHA!)

here:
haffpi::==90; divisors: 9;10 (-1;+1 respectively); 3;3;twicefive; ==
3;3;ttttteeee or 3;3;2;5 or 3;3;5;2 or 3;3;TTTTTeeee or 3;3;5;4;
binary 10;01;10;10; up;do;up;up (do==dowwwnnn) - ANY divisors you want
except of course 7!

four times haffpi: four times "ninety" or four * six (sixxis) or
twentyfour (==ten) or 360...

OK, translate this: "haffpi plusone is twice as much as four times
haffpi and then haffone";
counting letters:computers!
"six seven two five two four two four five six three four eight"
"hhhhhh ppppppp ii ttttt aa mmmm aa ffff ttttt hhhhhh aaa tttt
hhhhhhh" (Google spelling checker: read this!)
translate again:
"three five three four three four three four four three five four eight"
again:
"five four five four five four five four four five four four five"
it converges! my first two numbers are four and five (actually five
and four), then:
"five four five four five four five four four five four four five"
will translate to:
"four four four four four four four four four four four four four"
(thirteen times four), "eight five four", "five four four", "four four
four";13 times 3 if of course 39, 13 times 4 is 52, together is 91, so
here's your truth coming: haffpi plusone is actually ninety one. NOT
60 (90 degrees, two thirds) but 91!!! like the number of days in
Appril, May June; ; September, October, November (months 09,10,11 etc.
/ months 04,05,06 ....) OK? of course I can also convert it the math
way, but I will let smart spamming computers understand each word by
the number of letttteeerrrs in it! you'll see.

Translate my logarithmic: entropy level:

"haffpi plusone is twice as much as four times haffpi and then haffone"

// (2\\pi + One //2 == 4 * 2\\pi ; One // 2)
translate RTL:
(2 \\ One ; pi//2 * 4 == 2\\ One + pi // 2)
left to right allways - (no + in my language! log scale!)
2 --- 1 --- pi --- 2 --- 4 --- 2 --- 1 --- pi --- 2
OK, that's the first noble true - I don't know how much it is, but it
is NINE numbers (including pi) defining 4 to be at the same direction
of nonninnon "i". OK? then twice "2-i-pi-2" or "2-pi-i-2" from both
sides, mean 4 is the strongest permutation of the ninth logarithm of
two, when we already know 1;pi and 4. OK? that's true. Nobody defined
three here!!! nobody divided by three (you can't devide by 3 on the
log level (can you, Fermaaaaa???)) so of course, the log level 1 (One)
will be defined as the paper flat level of EVERYTHING! One will be the
dotted o (arabic .) and the circle o will be arround the dot and not
vice veras (which translates to you as "alllways i>>0" or "1>0" (NOT
TRUE!))

OK, now you have my fifth noble truth - then pi must be something
between 2 and 4 right? Of course you can't 360 twice; you can't even
seventh! (actually seventh is the correct word and not seven; seventh
or Shabbat) OK? then since you can't of coure 360 twice or seventh,
let's check how much is 360 in base 7;

[calculator says: 51.428571428571428571428571428571 ; lying! but
remember 13 times 4 is 52 (my calculator), also 13 times 3 is 39, then
of course not counting 13th would be something like
45.428571428571428571428571428571 (didn't count 13 twice), about haff
the 91 (actually 90.857142857142857142857142857143 now (my calculator
converted 2.8 to 3)! actually maybe I should check 364 / 7?
calculator says EXACTLY 52; OK? don't count 13 EXACTLY 7 times (twice
3.5) then you will get EXACTLY 45; but why use calculators? let's
convert manually:]

360/base 7 is 3*7*7 + 6*7 + 7 right? but 6 is (-1) in base 7; then
3*7*7 -7 + 7 or 147;
but wait! I can only divide and not multiply; then let's just say
360/1000 in base seventh is 0;3;6;0 which means binaari
0000;0111;1111;0000 (o;haff;minus 1;o) which leads to my conclusion
that maybe it was originally MEANT to be this number pi! just 3 and a
haff, or 3 hours and 60 seconds, (by the way, I heard degrees also
have minutes and seconds), then maybe it's just 3 times 60 hours,
translates for me to 180 hours, probably that's the real original
pie....

so then, how much is 180 base seven? one, then (seven plus one), then
none. one seventh, then seven sevenths plus one seventh, then stop.
Or, 180/1000 will be one;minus two!, zero! of course! the sin/cos
thing! go right, then left two, then stop! right one; left two, count
three! stop! then pie must be 180 if counting exactly one pinch
(pitagoras inch); then haffpie leftish while turning either 180 or 270
degrees or I don't know how many....

anyway, 360 and 180 and 270 -- all sum to nine, then it will be "-1"
in binary; then pie will circulte "-1" e when counting the sinus, but
counting to 3 (not 4!) since counting even numbers is not allowed
here, then counting 3 will be equal to four and then five - piitagoras
counting back four and then counting five!

of course, what he did was counting haff (actually, metering haff
pinch), then one pinch, then defining the 60 - the "golden 60" and
then (of couse, haffed 60 will be the golden 30) then of course, 60
and then minus 120 will be the same as 0; let me translate this:

haff+twothirds(90)==60; then rotate twice 60 down (in no time!!!),
then same again - will be defined exactly at twice not haffing! same
dot! then of course! then you have the 3 dimensional tryangle
60/60/60; then 6 times the 60 will be 360 (now I understand how he got
this), then of course two thirds of 90 is 60, let's translate it to
logs:

log(1/2)+log(2)+log(1/3)+log(90)==
log(1/2)+log(60)-log(60)-log(60)==
log(1/2)-log(60)+log(1/2)==log(1)!!!

log(1) is of course allways 0; then log(1/2)+log (1/2) must be the
same as log(1/2)-log (1/2) (allways 0); log (60)-log(60)-log(60) also
o; OK! see what I'll do:

first log will be allways 1/2; second log allways 1/2/2; etc.
then: log ((1/2)/2) + log (2/2/2) + log ((1/3)/2/2/2) + log (90/2/2/2/2) ==
log ((1/2)/2) + log (60/2/2) - log (60/2/2/2) - log (60/2/2/2/2) ==
log ((1/2)/2) - log (60/2/2) + log ((1/2)/2/2/2) == log (1/2/2/2/2)!!!!

OK, now we're already on the logscale, I don't need to write log..log...log:
((1/2)/2) + (2/2/2) + ((1/3)/2/2/2) + (90/2/2/2/2) ==
((1/2)/2) + (60/2/2) - (60/2/2/2) - (60/2/2/2/2) ==
((1/2)/2) - (60/2/2) + ((1/2)/2/2/2) == (1/2/2/2/2)!!!!!

you see? first line is [o+o+o+o]; next line is [o+o-o-o]; last line is
[o-o+o], and then all equal to (1/2/2/2/2) which is of course
o/10/10/10/10 (binary!) then of course, base pi:binary:[o+o+o+o]
translated to o/10/10/10/10 which means: first o divided by 10
(binary), second by 100, third by 1000 and fourth by 10000; which
means ALL BINARY NUMBERS ARE BETWEEN O AND 1!!! the largest binary
number (base pi:two) is the alllways repeating itself (1/2) on the
level (1/2/2); (1/2/2/2); (1/2/2/2/2) etc. so this will be actually
the binary prime numbers:
100000000000000000000000000000000000......... ///(1/2)/2)
0100000000000000000000000000000000000........///(1/2)/2/2)
001000000000000000000000000000000000000........///(1/2)/2/2/2) etc.
then of course! my mistake! first bit defined (pitagoras!!) will be
the "dot", which is exactly log (1/2); then the "least significant
bit" (is that how they call it?) will be the "plus/minus sign" (1/0)
of log (1/2); then 10.00 will be exactly 2/2/(1/2)/2)/2/2/2/2/2/2/2/2
(from right to left) which means "go up to one, then two steps back",
then if log(1) is 0, then go back twice and define yourself and
log(negative-haff-haff-of one!) then of course,
negative-haff-haff-of-one will be the imaginary 2i whois [i] the
negative 0.25 (-1/4) whos log is -0.25; then twice this negative log
will be two! then of course nobody knows how to haff the minus, so it
was just renamed 0.5 but that's not true! it's the "minushaff" three
times, (-i)^3; twice, logged, this defined as your prime number two.

now of course, since we did all this then we already know that

(1/2/2/2/2) allways leads to the same location, define this as the
ultimate truth of the fifth root of pie (pie ending - allways devided
by the fourth prime number - four); log equal to 0; then of course -

((1/2)/2) - (60/2/2) + ((1/2)/2/2/2) == 0;
((1/2)/2) + (60/2/2) - (60/2/2/2) - (60/2/2/2/2) == 0;
((1/2)/2) + (2/2/2) + ((1/3)/2/2/2) + (90/2/2/2/2) == 0;
then of course (now pitagoras language:
1/4 + 2/4 + 1/24 + 45/8 == 0;

6/24 + 12/24 + 1/24 + 135/24 ====0;

not true! so why don't we invent the square level? (some truth, but
should have been log level):

anyway, pitagoras meant something like this:
[haff + one (adding different dimensions)] is exactly haff of [the
other side, not counted, but looks equal to one (actually it must be
converted to linear, one means "haff + haff" which is of course
(1/2)+(1/2/2)], here:

[haff + haff + haff] is exactly haff of [haff + haff] + [haff + haff]
which means literally, three in exactly haff of four (which is true,
0111 is exactly haff of 01111 (remember, negative: the haffs continue
forever, think ASCII 127/128 counting),
or to put it this way - "two" is defined as "haff + haff" when "one"
is converted to "haff + haff"; but the previous one (logarithmic)
means that haff + haff is three quarters [or (1/2)+(1/2/2)] ; then the
first [haff + haff + haff] actually means "haff + three quarters";
1.75; and the second [haff + haff] + [haff + haff] actually means "one
+ one"; then how come "haff + three quarters" be equal to "one + one"?

well of course, in the logarithmic level who cares because they are
all equal to zero? but pitagoras is smart. thinks and then of course!
double the speed of light and then; 1/4 + 3/8 will be twice (1/4 + 1/8
+ 1/16 + 1/32)?
then, - 1/4 + 3/8 will be - 1/4 + 1/8 - 1/16 + 1/32 right?
is (-1;+3) equal to (-1;+1;-1;+1)? no, forgot something...
(3/8) was actually "haff + haff"; which means actually;
the number "- 1/4 + 1/8 - 1/16" will be equal to "- 1/4 + 1/8 - 1/16 +
1/32" right?
well, of course it is! since the pi is the magic wheel of truth, then
of course: let pi be whatever makes this equal! then;

"- 1/2/pi + 1/2/pi/pi - 1/2/pi/pi/pi" = "- 1/2/pi + 1/2/pi/pi -
1/2/pi/pi/pi + 1/2/pi/pi/pi/pi"

// of course, nobody will rotate the wheel more than 6 times! let's
check: [mod - 1/2/pi ]

[- 1/2/pi ] == [+ 1/2/pi/pi] == [ - 1/2/pi/pi/pi] ==  [+ 1/2/pi/pi/pi/pi]

then of course, "/pi" rightish must be identical to "- 1/2/" rightish;
pi\ leftish will be pi\2\1 - ; then

\pi\pi\pi\pi\2\1\ + == - 1/2/pi;

(logscale, lineary) pi + pi + pi + pi + 2 + 1 + == - 1 - 2 - pi;
circulate; (add as many pies as you want, up to 60)

(any number of) pi + 2 + 1 + o === o - 1 - 2 - pi (fo rebmun yna)

then of course - pi is your prime number three logarithmed! your
number of dimensions when you can circulate either 3 or 4 times with
the same level of minus revolving (of course -2 times -3 will allways
be +6....LOL), maybe even 5 or 6 times (7 not tested
-->pitagoras:"sorrrry"!); then of course pi will be defined in such a
way that also (5-4) will be equal to (4-3) and (3-2) and (2-1) and
(1-0) etc. (at least on the square level), then prime number 3 will
just be the number of times you can fit sixty in haff a pie... or the
number of times you can fit haffsixty in haff-the-real-pie (but never
say 91); then of course! pie is the allmighty everything, then why
don't we define it as the root of allways (o - 1 - 2 - ...... )
(repeating at least 7 times), then of course pi*pi*pi*pi will allso be
defined as (-3)*(-4)*(-5)*(-6) ; of course then (-3) and (-5) will be
reversed (+3)*(-4)*(+5)*(-6), then all of them will be reversed
[either (-3)*(+4)*(-5)*(+6) or (-3)*(+4)*(-5)]; then (-3) and (-5)
will be reversed again; and then: of course (+3)*(+4)*(+5)*(+6) will
be the correct order; forever counting our spheres of knowledge, black
holes, speed of light (I will recalculate it for you), all because of
the paper pitagoras had [and ferma, I'm not even with you yet];;;

ooops, we forgot to define +2. how much is +2? let's see:
how many times 60 fits in 180? -->3;
how many times 180 fits in 360? -->2;
then why not define 2 as the limit of (-2) rotating leftwords? 3 will
be (-2), then 180/3 will be -360; same number as 60; then (-360)/60
will be of course (-360)/(-10); then binary 1.8, rotate righting,
1.8000 + 0.18000 + 0.018000 etc.... (allmost 2); actually I can define
it exactly your math but I told you - just take the fifth root of 32,
but remember that the only base pi is counting is base (-1)!!!!!! pi
is actually EQUAL to logarithmic base -1!!!! then of course, ("any
number")^-1 will be the same as ("any number" -1); or (1/"any
number"), this is by definition the magic imaginary pie! logarithmic,
binary, ANY base counting - 32 is actually 3210;0123 which means allso
6543210;;0123456 base 7 - there is no base ten, this is just another
American lie. so if you count from the 3 to the 2, remember it's -32
and not +32. There is no such number! (real). you can only count
imaginary numbers!

then how much will (pi * (-pi)) will be?

let's check logarithmic scale: (- pi/2 + pi/2/2 - pi/2/2/2 +
pi/2/2/2/2 - pi/2/2/2/2/2 .....) will allways be the same thing right?
then, - pi/2 will allways be equal to both 0 and 1 right?

then, (we;ne;pi) will allways be totally 0; pi will just be the
opposite of we right?

then it turns out that --2^(1/2) must be allways the same as
---3^(1/3) in base pi.
calculator says
--2^(1/2)::1.4142135623730950488016887242097
---3^(1/3)::microsoft:: "invalid input or function";
3^(1/3)::1.4422495703074083823216383107801
pi:3.1415926535897932384626433832795
1/7:0.14285714285714285714285714285714
pi^(1/pi)::1.4396194958475906883364908049738
pi^^0:::1 // works!
pi^^1:::3.1415926535897932384626433832795
pi^^2:::9.8696044010893586188344909998762 /// 7*7*2/2/5 +++
pi^^3:::31.006276680299820175476315067101 // of course! pi defined as
the fifth root of (1-32)!!!
////sorry meant third! should be the fifth though! third root of
(1-32) defined as pi! ;
allso means the sixth root of [-1/2-0-0-0-0+1/2/2/2/2/2/2] ; divided
by [-1/2] then: [-0.484375] divided by -0.5] [decimal] will be 0.96875
; this number minus one will be (1/(-32)), then 9;6;8;7;5 is the magic
order of counting negative pi; convert: -1;-4;-2;-3;-5 !!! you see!
(-5)^5 in your base! try 7! will be: // doesn't work for seventh! I
know why! it will work only for 10;000;000 base seven! have to be both
logarithmic and linear. Anyway, pi is dividing between "odd" and
"even"; "odd" will be defined righting growing and "even" will allways
be exponents of 2. I understand! pi mixed two dimensions into ONE
dimension of "THE REALITY!" which is not true! I will let people know.
pie is a lie!!!!!
pi^^4:::97.409091034002437236440332688705
pi^^5:::306.01968478528145326274131004344
pi^^6:::961.38919357530443703021944365242
pi^^7:::3020.293227776792067514206493072 ///
50.338220462946534458570108217867 times 60.00000000000000000000
pi^^8::: 9488.5310160705740071285755039068
pi^^9:::29809.09933344621166650940240124
pi^^10: 93648.047476083020973716690184919
pi^^11: 294204.01797389059710569564200718
pi^^12: 924269.18152337418622257917035848
log[pi][baseten[1000000]:::12.068795205528365632588332283732
1/[log[ten][pi]]:: 2.0114658675880609387647220472887 :: here's the
baastaard! defined by pi!!!
1/[ln[pi]-1]::6.9094229856614265446876058088678
ln[2]::0.69314718055994530941723212145818

[1/[ln[pi]-1]] / [ln[2]] :: 9.9681902768179553955697130035972
do you need more evidence pi is a conspiracy? again:
ppi::=square root of [9.9681902768179553955697130035972] ::=
3.157244095222597038465624806571

[1/[ln[ppi]-1]] // 6.6800479019552125518104124334207
it converges to 6+2/3!!!!! pi is a baseten conspiracy. For pitagoras
it was probably just 3........

anyway, prime numbers are DEFINED this way: if they suit the pitagoras
3/4/5 - then they are prime. If not - they are not. pitagoras and
fermat were both right - and both wrong. They can't count quantum
numbers like this, and they can't count quantum meters of kilos. I
will check and let you know how long a pie radian is. And also, the
meters, the kilos - I will let you know how they are defined. Of
course they don't work for lightyears, nanos, nuclears - it's the TOP
SECRET american conspiracy! They have their reasons for you not to
know: encryption / nuclear secrets / monopolising ARPA NET, IP
numbers, spying, "not supporting Hebrew" software, money money
money.... believe me - there is somebody (maybe a robbot) at the DDCC
capital doing this. it's not a mistake and not a coincidence.

nagasaki == 8 (nnnniiii)
hiroshima == 9 (hhhshiaaa)
1;9;4;5 - how can you count that? base 10: +1;-1; then base 5 -1;0?
or maybe: base 5 1;-1;-1;0?
counting the seconds backwords and then - now?!?!?! Turing, who are you?? human?

I will tell you - this counting, 3,4 and 5 will allways be prime. 2
was converted to (1+1), I can define the whole number system without a
"+" (using - only, rotating, converting etc). permutations don't like.
There is no such thing as "exactly equal two permutations of the same
thing" (except 1/0 maybe). two permutations is already prime number
three.

now, if you want more primes or to divide existing numbers - it's very
easy with my math. More primes - just continue with the 3/4/5 example:
4 will allways work! then 16 + (4-i)(4-i) will allways be 32+(32-ii);
64-ii your math, then multiply (64-ii) with (64+ii) and you will get
(4096 - iiii); (from here they took the definition of pi maybe).
then: fourth root of  (4096 - iiii) :: 8 + 1 ; 9; this number minus 4 will be 5;
OK? now, instead of (5-4) choose ANY odd number;
instead of (5-3) choose ANY even number;
instead of (4-3) choose ANY odd number, but not the same as haff (5-3),
then you can get ALL dimensions of odd prime numbers, even prime
numbers (2; all others are multiplications of two), etc. Actually if
you do it with more dimensions you will get more numbers, but not
integers. simensions mean both more polynomials, more exponents but
the real dimensions are the exponents counting! don't forget this and
don't forget fermat designed this as to not to work his way! Actually
I also checked sinus/cosinus, they can have values of +-0; +-1/2; +-1
only if and only if the angle is a divisor of 360. (times 30). I
checked this and it's true! it's designed this way! you can never
rotate a quantum wheel 7 times! not in this math! you can of course,
but I will show you later. have to go now....

Uri.

[~~~~~~~~~~~~~~~~~~~~~]

Uri First deadandalive

---

From: Shlomi Fish <[email-masked]>
Date: Fri, 25 Apr 2008 05:09:22 +0300

On Friday 25 April 2008, Uri wrote:
> website, left HTTP not working. 

Hi Uri!

I'm sorry, but I'm not interested in seeing these extremely long (over 100KB 
now!) messages from you in my incoming email, especially considering the fact 
that they're completely non-sensical.

You obviously need to be in touch with a mental health professional who will 
help you, because you're no longer in control of yourself. I suggest you to 
seek an expert of http://en.wikipedia.org/wiki/Cognitive_behavioral_therapy . 

If you don't want to get yourself banned or moderated from another forum or 
two, you should stop posting this nonsense and get yourself some professional 
help.

Regards,

	Shlomi Fish (who probably has http://en.wikipedia.org/wiki/Bipolar_disorder 
and has found the book 
http://www.shlomifish.org/philosophy/books-recommends/#feeling_good to be 
very helpful[1].) 

[1] - Feeling Good is intended primarily for people with depressions, although 
it can be applied to those with anxieties and hypomanias as well. Uri, in 
your case, I'm not sure you have Mania, so please consult a Cog.-Behav. 
Psychologist and see how he can help you.

-----------------------------------------------------------------
Shlomi Fish       http://www.shlomifish.org/
http://www.shlomifish.org/humour/ways_to_do_it.html

The bad thing about hardware is that it sometimes work and sometimes doesn't.
The good thing about software is that it's consistent: it always does not
work, and it always does not work in exactly the same way.

---

From: "Arik Baratz" <[email-masked]>
Date: Thu, 24 Apr 2008 20:43:50 -0700

On Thu, Apr 24, 2008 at 7:09 PM, Shlomi Fish <[email-masked]> wrote:

[snip]

> two, you should stop posting this nonsense and get yourself some

I don't know about that. I kinda enjoy these messages. It shows me how
different other people's realities can be from my own.

I do however think we need some reality overlap with other people to get by
and not be locked up in a mental hospital, and I really hope that Uri has
that.

-- Arik

---

From: Uri <[email-masked]>
Date: Fri, 25 Apr 2008 19:01:10 +0200

.
+o+
-o-
i.i
ni.in
eni.ine
nine.enon
none.enin
mom.wow
wow.mon

[+o+]

[R.W*]

Thanks for your support my friends, and I will explain in more details
later. It's very urgent and I'm not able to reply to each question
individually, but I promise I will create a much better way of
communication, not relying on ANY american STAANDAARDS, not ASCII
(actually NASCII - Not American Standard etc...), NOT COMPATIBLE WITH
AMERICAN ASCII, SMTP, HTTP, WWW. .COM etc. But! will be able to use
this network (ARPA.NET) as one of the possible communication between
us. I will show you and prove everything - I will create a computer
who can understand ANY human language, or ANY other computer language,
but first I will need to define my interpretation of the universal
language, the Truth.

But fisrt, before [made a mistake fith fisrt, left it this way -
that's one of the ways to identify humans]; first, before starting:
remember - never identify as other people, only your True identity
(can be nicknames, even anonymously, but not ME). If you are identify
as me, my computer will recognize this as define you as an ENEMY,
which means literally "any me", "ending me", "en; die; me" etc - you
will either become REALLY me or go back to my past. You are also not
allowed to do anything bad to me or my body, you or your body or
ANYBODY. ok? so let's start:

First, if you are real humans, you must do this first on a paper. Take
a paper and define who you are, what you want, what you know, what are
your colours (in all your languages), number (count them), define
everything to yourself. You can actually visualise this imaginary,
that's OK, but I'm not sure it will work for you. Also, never ask a
computer to select random numbers for you, if you want to select
anything - you must select them yourself. It will be VERY easy for me
to find whois cheating with random generated computer numbers - they
are allways quantum equal to 0 or 1. So let's define 2 as the number
of different quantum equal numbers computers can randomize (that's
two), the number of letters of "two" is three, now I will define for
you numbers (of course, you will convert from your counting to my
universal counting and vice versa).

But first - the dot [.] . this will be used to redirect SMTP programs
to "nothing", [remember, sendmail understands "the dot" as the last
nothing in your mail (if only a dot)]. then "the dot" will be the only
thing whois really nothing, the n.o.thing, which means you can put the
dot anywhere without changing the meaning of anything. Don't remember
the ASCII number for "the dot", but I am inventing NASCII - the Never
will be compatible with Not American Stand***** ; which means NASCII
will NOT even be compatible with itself - that's the secret. you will
see how it works.

Now, "the space", good old %20 / Hex20 / 32dec etc. This will just be
equal to TWO dots, two literally means three, or four, or five, or ANY
number of dots. Then, the plus [+] and minus [-] will respectively
mean the same, they will be EQUAL (but not identical), then plus will
be just pppp and minus mmmm - replace each word with the number of
letters / the first letter. I will explain this.

ok? now, ANY number below 3 I will define to myself, including my
language, the dot, o, 1 2, O, i two, -i, 0 -1, etc. by the way, they
have a VERY GOOD REASON to define different ASCII numbers for 0, o and
O, 0 (if I remember correctly) is ASCII 48, or Hex%30, so literally
speaking you can start counting from Hex33. (or any other way you
define 3). [remember: they can change ASCII, but they will NEVER be
able to change NASCIII]. They WILL change your ascii when they want to
make us not understand each other - the whole "Visual Hebrew" / not
compatible with Hebrew fonts is a conspiracy (of course - use our
"Microsoft Word Editor" etc.....)

by the way, those of you who think my vision of conspiracy is
unrealistic - add a big "I HATE YOU / YOU ARE MY ENEMY" (in capital
letters) to the subject of your e-mails so I will know where to
redirect you..... (actually my computer will do it for you..... I will
also test you in עברית / or whatever your natural language would be)

OK. now, the pi from yesterday fits well into the word "conspiracy" -
so this will be my first public key for you. "conspiracy" (spelled:
ten letters) will be ANY word with more than nine letters, converted
to my definition of the word "conspiracy" (I can allways change this,
this is human-to-human language):

"conspiracy" == "cons+pi+racy" [4+2+4] == "CONSTANT PIRACY" (8+1+6) ==
"American constant conspiracy" == "American constant pi piracy" ==
"American constant pi racing" == "American consbiracy" etc. - written
from left to right, right to left, then translated many times, and
only when it translate to my same work - it will pass the gateway
(actually, maybe EVERY word will do the same thing). So,

right now let's rename it "conspipiracy", which means literally that
10 is EQUAL to 12, EXACTLY, and vice versa, of course this is also
true if you remember that binary digits 1,0,2 are all symmetric, then
"two" is defined at the number of symmetric equal binaary digits, then
of course 12 is equal to 10, 01 is equal to both and you can already
have six permutations of writing the same number (01; 02; 10; 20; 12;
21). Of course, you can switch to 4/2 mode, then all these numbers are
EQUAL (quantumly speaking) to 41; 42; 40; 24; 41; 14 etc. (infinite
possibilites).

OK, so first, define your own digits (also NUMBERS) 3, 4 and 5
respectively, who are EQUAL to ASCII/hex 33,44 and 55 (not a
coincidence - you can also spell "cons33racy" / "cons44racy" /
"cons55racy" and then count the letters in ASCII (of course, then
10==12 will become more equal, especially when I'm also writing to you
"cons333racy" / "cons444racy" / "cons555racy" / "cons3333racy" /
"cons4444racy" / "cons55555racy" etc.)
OK? so define the numbers / digits 3, 4 and 5; now here's the language:

.
345
(nothing - you have nothing to say - this is allways assumed before the dot)
(. [dot] - you want to say something (actually this can already mean
EVERYTHING, so if I understand you you can stop here)
345 - your first 3 letter word who is not twice symmetric. For example
"one", "two", "uno", but not "ono" - "ono" means to me o;n;o who are
identical to "o;o;n" or "n;n;o" - any permutation will do here, then I
will just understand that your 3 is equal to 5 (actually it's true -
they are all equal).
But, if you already have equal 3 and 5, why don't you double your
speed of light, then just send me the same thing twice? for example:
if you want to say "ono" then you can only say one "o" and haff "n";
then just send me "on" or "no" and I will understand. Actuallly it's
better you send me ALL the permutations of "o" and "no" you know
(pronounced - "you no"), which means either all the dimensions of 3
and hafffour, or 5 and hafffour - if they are equal to you. If I'm not
sure they are equal or not, I will ask you again. In the meantime -
let's assume they are not (3,4 and 5) may or may not be equal to you).

OK? so now - ANY three-word-letter will be converted to my word "one"
(in my language - everything can change) , ANY four-word-letter will
be converted to my word "one+one" and ANY five-letter-word will be
converted to my word "one+one+one". confused? don't be. "one" is
"3=3"; "one+one" is "3=3;4=4" and "one+one+one" is "3=3;4=4;5=5".
actually, if you want you can start with YOUR one; I don't care. you
define three,four and five the way YOU want! but remember: when you
write "one+one+one"; I'm reading it twice - RTL AND LTR, which means
"eno" and "one" are ientical for you, then n will be my four;
converted: "o4e+o4e+o4e", then recursively reading it again and
compressing. eventually I will compress EVERYTHING you write me into
one quantum, I will show you - but one is at least three.

OK? then first define your word "one", in the order of 3;4;5. but
remember - 3;4;5 are actually mod 3 (or mod 33; or mod 3-3) which
means they can rotate and mean the same thing to you as they can to
me, then 0;1;2 will be created by me, all numbers above 5 will be
created by me as well, but also remember this:

3 is the negative imaginary prime number who is allways equal to the
LARGEST (negative, closest to o) prime number who is the permutation
of two (literally any three letters), or one (same thing), then if you
have the word "one" and want to add ONE dot - then it's already four
[.one / o.ne / on.e / one.] OK? so three is the LARGEST permutation
number of "01"/ or "o/i", which means [.01 or 0.1 or 01.], also
revolving backwords, same as ".10 or 1.0 or 10." (told you - I am not
counting dots - I will convert google.com to google[google--["."]]com
(literally means: "google" + "the number google counting to o - the
number of dots [.]" and then "com" (then of course, "com" will be
converted to "345"; or ASCII 33;44;55).

(by the way the number google is not very big logarithmically, but it
is combinatory big for you (not for me), then Google.com of course
will be converted to my representation, but when you write to me I
will receive the same word in my language - you will see).

OK? so now respectively - ANY number is the negative permutations of
the previous number, actually divided by the previous number but
negative; actually it's the "set power" of the different numbers - a
new dimensional counting. because "two" is already a three dimensional
number - can't write "two" in two letters without defining them first
- I showed you the 01; 02; 10; 20; 12; 21; are all equal, then 10 and
10 are equally two, then it can either be i-i or i+i or -i-i or any
combinations of the number i; or the number o (but not both), the i
and o are my own prime numbers, adding two, all three are symmetric,
then start from your three.

OK? now, first write me the digits 3;4;5 (in ANY order - you must do
this at least 3 times - if not using paper - then let them computer do
it for you; rotate and switch directions and do it again) OK?

then each e-mail you will send to me at least 3 times, but if you send
it once I will just redirect it for you to three different locations -
independent, and start counting from your 0/1/2 letters respectively.
Then I will also communicate your message both forward and backwords,
here's what I will do (my computer ; if you want you can also write me
on paper - I can understand this even if not physically there):

[this is like saying "hello" but in a friendly way, NON-STAANDAARD way
- not SMTP, not ASCII - remember you can say "hello" to many people
and not allways mean the same thing, sometimes also "hi" or "Hello" or
"Helllo!" etc, you can program your computers and help me with this
because I don't have time to program everything, if not then I will do
it myself but it will just take more time]

me: [waiting... that's my hello]
you: [waiting].... (or not: you don't have to)
me: . (that's my hello if I don't recognize you)
you: . (the first letter will be interpreted as your dot) / OR:
you: 10 (means nothing to me, two dots) / OR:
you: 345 (any three letters will mean you are counting - especially if
you start with "345", then I will either wait for you to count or
count you:

now: I will communicate you to yourself as what I am understanding.
for example, I will have a few processes, each of them communicating
"345" to each other in different orders. But if you want to be
friendly, you will first define this:

[3-3-3-33-3-3-3-3----3-3-3-3-3]
[4-4-4-4--4--4--44-4-4-4-4---4-]
[-5-5-5-5-5-5-5-5--5--5--5-5---5]
(me not counting - any number of 3;4 and 5 (all others - just dots - I
don't care if you write "-3" or "--3" or "---3" or
"------------------------------------------------------3" - it always
means the same to me)
then - rotate them a little:
[3[4[5[3[4[5[34-3-4-4-3[5-33[[4-3 (now - I'm ignoring anything except
3/4/5 - will maybe check later what you meant).

now - YOU don't have to do this each of you personally. You can also
communicate with your friends, then if I remember you then I will
understand you - but I WILL NOT HAVE ANY MEMORY OF MY OWN! I'm a black
hole remember! then just remember each other, then communicate with me
as if I am new, then define how many colors you have (in your own
language, translated to 3/4/5):

[red-green-blue-yellow-black-orange-shit-white-air-water] (actually
they don't have to be just colours, they can mean anything). If I
don't understand you, I might listen (remember - we are speaking at
MANY times the speed of white light), or I might respond in your
language. For example, if you write me "one-one" or "one-minus-one";
then I might respone "one" which means, literally, ANY WORD COUNTING
LETTERS, STARTING WITH o AND ENDING WITH e; n will be my negative
number who is to be added before your three. Got it? if you think I am
stupid then translate it to your own language and say it to your image
in the mirror, then think about me being your image and then translate
"how many times can I lift my right hand pointing eastwords, my left
hand pointing westwords, then rotate both directions simultaneously
and accelerate , stop, do it again and count angles", then do it
TWICE, define the number of angles as twice 360, and don't forget to
divide by the number of angles I did it (your mirror image) and then
trsnalate it to your language "stupid", whois of course six, then
three will already be your sixxed; then reversed and inserted into my
system of understanding you. (sixxed;;dexxis or dexxis;sixxed -
remember to count letters).

I'm writing you all this in public, in case I forget you then just
remind me "you said this" and copypaste me, then I will remember you
were speaking my language, but first we have to define our
understanding;

so - you define to me your numbers 3;4 and 5. I rotate them, remember
and define to you my same numbers 3, 4 and 5 (who are ANY permutation
of the numbers 0, 1 and 2) plus ANY permutation of the numbers 3, 4
and 5 - actually ANY permutation of ANY numbers or ANY letters, for
example in language 42: I will convert all your numbers to twice as
many as they are, then in base 42 (whois identical to base 2.1) I will
define you your own numbers. For example:

"fourtytwo" = defined: "nine" (count the letters);
"twentyfour" = defined: "ten"
"o minus ten is equal to nine" // which means : one (or none) minus
ten is equal to nine, each number is actually the negative of it's
(plusone) number; when symmetric also negative of both (plus one) and
(minus one), then of course (-4) is EQUAL to (-5) or (-3) (all numbers
are negative, binary, base 3, imaginary and not equal to themselves),
which means "4-4" is equal to "5-5" is equal to "3-3" is equal to
"4-4" (all equal to -0; all equal to -1 (divided ligarithmic), all not
equal to +0 (think about the amplitude changing, press your screen and
let it move like a מטוטלת - same same);

OK? then: when you get at least 3 numbers who are not equal to each
other, go back to the fourth and then start playing. Remember: the "-"
sign is the equal sign, the "=" is the double equal (--) which means
this number is NOT equal (think about my imaginary i; -i and --i in
base 3 - when --i is 1). then - I will define you my numbers (defining
now) as the order of magnitude of NOT being equal, from the LARGEST
number to the smallest (the smallest - to be defined. for now just
call it "one" (means 3):

"-0-1-2-3-4-5-6-7-8-9 etc........"
this is the regular counting: BUT "-9" is also identical to -5 (-1
base -3); then -6 is identical to -3 and ALSO to -5; actually I can
convert this to binary digits:

"-0-1-1-1-1-1-1-1-1-1-1-1-1-1...............(alllways -1)"

which means - either it's your first negative zero or not (zero is of
course z-eno - z-negative one - the zone);

if it'n not your first negative zero, then I will be searching for
your last negative zero - then send me you message backwords (either
simultaneously, or rotate: one letter from the beginning, one from the
end, etc. YOUR counting!

since we understand what equality is, then of course we can remove
"-0-1" here, then start again:

"-2-3-4-5-6-7-8-9 etc........"
translated:

"-2-1-1-1-1-1-1-1-1-1-1-1................."

then of course, starting from ANY number and continuing from ANY
number will ALLWAYS mean the same thing. So now - my first axiom:

I will never count your same number twice!
If you want to say something like

"repeat this again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again and again and again and again and again
and again and again and again"; then I will NOT count how many times
you said "and again"! then you can just replace this with any
meaningful word, like:

"repeat this again [..... and again......]" - you see - human readable
(I'm NOT counting your dots) - then now's the secret - if you
understand this, then you will be able to send me ANY amount of
information - even as much as 42 terawords / bytes (not joking) - in
as small as one little sentence, when we already understand the same
language. But, you will have to quantum compress it this way - at
least three times, in YOUR language, if you have friends who speak the
same language then share your knowledge with them too - then I will
convert you to my language - and remember I and My are not just me
personally, but ANY black hole, and entity, and quantity, any Math - I
will prove to you this.

But first - if you're still using ASCII - maybe I will try to be
compatible with this too. Then why don't you take my keyword
"conspiracy" (define your own keywords if want to), then translate it
to "cons.pi.racy", then translate it to "cons.3-3.4-4.5-5.racy", then
translate it to "3-3.4-4.5-5", then translate it to "at least 3
permutations of "3-3.4-4.5-5": for example:

"3-3.4-4.5-5"
"5-5/3-3.4-4"
"5.5+3-3=4-4"
"4-4!5-53-33"
"3_3=5-5-4-4"
"4+4-3-3_505"

(from here I will take your own language, then remember: taking the
number 3, 4 and 5 (in any language) - then defining ANY $o, $n, $e
such as:

$o$o ; $n$n ; $e$e

will mean the same to you as:

- 3^3 - 4^4 - 5^5;

(first number (negative) will allways be largest, then second, then third)

or:

3*3 + 4*4 == 5*5;

(pitagoras: growing from memory counting, equal)

then only ONE of these number will allways be 3, ONE will allways be
4, ONE will allways be five. For example:

"one"; "one-plus-one"; "one-plus-one-plus-one"

translate: 3*3 + 4*4 == 5*5; or - 3^3 - 4^4 - 5^5; (same thing!):

[using PHP : .==+]

PHP: define "o"; "n"; "i", "p", "e", "-" ; translate:

"one"."one" ;
"one-plus-one"."one-plus-one" ;
"one-plus-one-plus-one"."one-plus-one-plus-one" ;

(doing this manually: by hand):

replace "o"; "n"; "i", "p", "e", "-" - "xyzXYZ"; // all others are
just dots (ignoring): + reversing;
"one"."one" == "xyYxyY" ; "YyxYyx"
"one-plus-one"."one-plus-one" ; = "xyYZXZxyYxyYZXZxyY" ; "YyxZXZYyxYyxZXZYyx"
"one-plus-one-plus-one"."one-plus-one-plus-one" ;
"xyYZXZxyYxyYZXZxyYZXZxyY" ; "YyxZXZYyxZXZYyxYyxZXZYyx"

suppose you want to cheat me, define "owt", "two" and "tvvo" instead
of 3;4 and 5. and even write "minus" instead of plus, or even your own
new math - ASCII "?" OK? then give me three numbers - *, ^ and %. I
will know who is your first! how?

you define them your order, then define ANY order that is allways the
same as the order you're counting them. For example - (oh wait - why
don't I just take from your real "largest prime numbers" list:

(of course, I will find more primes for you here now!) LOL!!!

"2325826571" // means "2^[32,582,657]-1"
"5822326571" // means "5^[82,232,657]-1"
"6572325821" // means "6^[57,232,582]-1"
"2876553221" // means "2^[87,655,322]-1" /// of course, they will all
be equallly primes as 3;4;5 are, except the 4 whois 4 times a prime
(square of two primes) together;

now, you want to cheat me, then just replace all your digits the way
you want, I don't care, as long as you remember them (it's your
language!)
then:
rename "2876553221" as : "e.f.g.h.i.j.k.l.m.n"
rename "6572325821" as : "e.f.g.h.i.i.i.i.i.j.k.l.m.n"
rename "5822326571" as : "e.f.g.h.i.i.o.o.o.o.i.i.i.j.k.l.m.n"
rename "2325826571" as : "e.z.x.w.e.c.q.a.s.w.e.d.n"
(just chose ANY letters for you!)
OK? now, don't tell me this, even don't tell me anything, just rename
all of them as "e.n", let's start the game:

me: "conspiracy!"
you: [your language]
me: "cons...pi...racy!"
you:                         (you say nothing)....
me: "you said "cons.........racy.........""
you: "your first word"
me: you said "your first word" + + "drow tsirf rouy" // now rotating
you in my mirror....
"yy/dd/our first wor//row tsirf rou//"
replace all with dots, except "yy" and "dd"
me: you said "dydydydyydyd" (which means actually
"....................................." or "dyd...ydy")
OK? now: translating you (remember: all you said is
"y......................d" or even "yo......................." or even
"y............................." ; I will take y as your 3, then ANY
letter as your 4, then ANY dot as your 5 (or vice versa: any order):

now converting you: "y" ----> "Z" // 3.3 (can choose any for you)
"d"--------->"Y" // 4.4
"."--------->"X" // 5.5

now you are saying "e.n" - I don't have an idea how many numbers or
what do they mean....
so I will translate this to "..." --> 5.5 or 4.4 or 3.3;
but, "e.n" is NOT identical to "n.e" - OK, then "n.e" will be something else:

"e.nn.e"---->"5544" OR "4433" OR "3355" OR "5533" OR "3344" OR "4455"
- ignoring the dots...
"n.ee.n"----->cut and reverse. will take 4 options (can take as many
as I want to) and add my own dots (anything not 3, 4 or 5 is a dot):

"n.ee.n"-----> "5;5;4;4" OR "3;3;4;4" OR "5;5;3;3" OR "4!43!3"

now - trying to understand you?...... "n.e..e.n" can be any (at least
three) numbers, let's check the one you represent with as FEW letters
as you can - this will be my BIGGEST prime number for you (for now),
renamed 3 (already counting all prime numbers below 3 - 0;1;2
respectively - whether or not you consider them primes - also counting
all special numbers pi, e etc.)

now, I will rename my "one" "cons.one.racy" to not to confuse with
your (actually I'm not confusing, it's the same), then "con" will be
my "one" (my constant "one").

OK? then:"n.e..e.n" will be ANY number, such that
"n.e..e.n"*"n.e..e.n"+"n.e..e.n+con"*"n.e..e.n+con"=="n.e..e.n+con+con"*"n.e..e.n+con+con";

of course - whatever my "+con" is, it's bigger (as string) than your
"n.e..e.n", then I will just add more dots here:...... (define
anything you want - I don't care):

"n.e..e.n"*"n.e..e.n"+"n.e.......e.n"*"n.e.......e.n"=="n.e............e.n"*"n.e............e.n"

OK? removed my constant. now - remembering - e..e is your three,
e.......e is your four, e............e is your five.

but you said "e.n" - what did you mean? (could just have been three
colors - yellow, purple and orange - your YPO screen).

OK? then let's just rename them again (remember - I don't have any
idea what you said): yellow, purple and orange turn out to be all six
- let's just rename them all six, choose my own colour black - this
will be five, back - this will be four , awk - my three for h...ing in
time. OK? remember: b*k, b*k, a*k - black, back and awk - 5, 4 and 3 -
all renamed kkkkk;kkkk;kkk for you.

here: let's translate: (you said ): none; // .
(you said k): one; // i
(you said kk): on; // n
(you said kkk): 3 // == aaa
(you said kkkk): 4 // == bbbb
(you said kkkkk): 5 // == bbbbb

OK? already (in my language) I have different meanings for your same
word (ANY) - any word with 3 letters - even three dots "..." - renamed
"aaa.bbbb.bbbbb"

now, let's take kkkkk options: (as many as you want):

"aaa.bbbb.bbbbb";
"kkk.bbbb.bbbbb";
"aaa.kkkk.bbbbb";
"aaa.bbbb.kkkkk";
"kkk.kkkk.kkkkk";

now, you didn't even tell me your numbers. but I remember "your first
word", and I rememver the colours yellow, purple and orange - you
didn't tell me (this from my own knowledge): then yyyyyy, pppppp,
oooooo will be your colours, then your first word would be either yyyy
or dddd, let's translate it to your colours: then either of these
options:

yyyyyypppppp ; or yyyyyyoooooo ; or ppppppyyyyyy or oooooopppppp (any
order, any number of permutations), let's just say oooooopppppp is my
choice here, because it reminds my my first number one - whois BIGGER
than your three (negative, imaginary i), then I will just rename my
one as your ooooooppppppyyyyyy (which happens to be my pppppp (six
dots, ...... or the "pi" in my "cons..pi..racy"), the "y" being last
letter in this word, then y will be allways you spelling conspiracy to
me backwords....... OK?

then now, ooooooppppppyyyyyy will one "one" which means literally
"0;1;2; start counting from here", and remember, your first 3 numbers
are your direction - mod 3 or divided by or ANY direction.... then now
your direction of "one" will be "opy", (or .py or ..y), which means y
will never be your 3. now check: let's take your numbers, the first
one "2^[32,582,657]-1" (it says "44th Known Mersenne Prime Found!!");
also says "9,808,358 digits" [http://www.mersenne.org/], also says
[September 4, 2006], let's just translate this to my language:
counting: 0/1/2/3/4/5/6/7/8/9 .... from left to right... let's see if
I can understand this....

"2^[32,582,657]-1" - same as "2325826571" // removed all dots...LOL!!!

convert to base kkk [remember: H...ing]:  "232;582;657;1.."

first, rename (not to confuse myself):::0/1/2/3/4/5/6/7/8/9 will be:
".אבגדהוזחט"

counting/replacing:
"232;582;657;1.."
"בגב|החב|והז|א.."
OK? now: replace each digit with my evaluating you..........if below 3
then dots, above 5 then dots too (can do this any way I want to)
...................................
"..3....5........5......."
OK, then you are first counting 3, and 5, then 5. The order doesn't
matter. it just means that your 4 is equal to 5... then. your first 5
will be 4, your second 5 will be 5, my order. renaming you:

"232;482;657;1.." see? now your 5 is either 4 or 5 to me, first 5 is
my 4, which means actually (5-3)(3-5)OK? renamed you.

now, let's just check. rename you back to "232;(5-3)(3-5)82;657;1.." ;
then cut; then rotate... you
are.......1.....2....3....2.....(5-3)(3-5)........8......2.......6.......5.....7......
etc. (if you forgot 9 then here will be my 9), my dot will allways be
o (0). OK?

now: counting you....one......2one......3one......2one.....5one.....3one.......
.....3one.............5one.............8one............2one................6one...............5one...............7one.............1one..............9one..........nine-one-one...........nine-one-one-one..............
etc.

OK? replacing you..............only 3/4/5 counting............

[..........................]3[...........................]5[.........................]3[.........................]3[......................]5[.........................................................................................................................]5[....................]
etc. OK?

removing all dots, remains:: 3/5/3/3/5/5 OK?
now, cutting: 3/5 ; 3/3 ; 5/5 - here are your numbers 3.3, 4.4 and 5.5
by order (from 3 to 4). who's who? I don't know. let's check:

kkk.kkkkk ; kkk.kkk ; kkkkk.kkkkk

well, of course that's their order: kkkkkk / kkkkkkkk / kkkkkkkkkk but
how would I know this? let's check pitagoras ("pitagoras" ==
"pitabread" - allways 9. then just rotate your pita bread when you
remember [-4[-5]]/[-4[-5]] is allways equal to [-4[-3]]^[-4[-3]] and
[-3[-5]]/[-3[-5]] (translating: minus one will allways be equal to
minus nine), or in your language: 1^1 and 1^1 is the same, 2^2 is NOT
and therefore the number 4 is NOT a prime (in my language: 4;5;3 are
allways equal), here let me put it this way: kkk.kkk will allways be
shorter than kkk.kkkkk and kkkkk.kkkkk in your language, not allways
in mine (it depends who's asking and what's kkk; for example if
hkkkkking will ask me if we can go bkkkk when we lkkkk bkkkk into our
past inside bkkk holes, then of course not! but if hkkking will ask me
if we can go bkkk when we lkkk bkkk into our past inside bkkkk holes,
then of course YES! now show me to google speller who will understand
this! kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.......................

OK anyway, let's show you just an example: I don't have time for
calculating new primes (allthough it's very easy), so let's just try
to understand your language first: I will quantum leap back to my past
(some 10/20 minutes ago)... take this number:
"2^[32,582,657]-1"

and replace it with the number of kkkk's I think you were meant
probably (the computer could do this as well, by defining your
language each time again, compressing and checking):

"kk;kkk;kk,kkkkk;kkkkkkkk;kk,kkkkkk;kkkkk;kkkkkkk;k"

OK, now the most important thing is that you are GOING BACK (the "-1)
which I'm not allowing you to, then let's double your speed and I will
go backbackfor you:

// when I double, I will convert all numbers below 3 to "yellow" ---
not to confuse me!

// all numbers will be doubled mod 3. (27 doubled exponentially)

// numbers more than 7 will be renamed "cons.pi.racy" mod 3

"kk;kkk;kk,kkkkk;kkkkkkkk;kk,kkkkkk;kkkkk;kkkkkkk;k" -->doubled and converted:

"yy;cons....kkkkkkkk....racy;cons..kkkk..racy;cons...kkkkkk...racy;yyyy;cons.....kkkkkkkkkk.....racy;cons..kkkk..racy;yyyy;kkkkkk;yyyy"

OK? now let's convert to your language:

"2^[8;4;6;4;pi;4;1;6]-1"; converting:
"2^[8;4;6;4;0;4;1;6]-1"// OK! now let's define this number as the
biggest ten-digit-number who is dividing your prime [original:
"Mersenne 44"]. Now, let's divide:

(from left to right, ANY-BASE-CHANGING):

"2^[32,582,657]-1"
--------------------------
"2^[8;4;6;4;0;4;1;6]-1"

OK, define them again this way:

your number: "2^[3^[2^[5^[8^[2^[6^[5^[7]]]]]]]]-1"
my number: "2^[8^[4^[6^[4^[0^[4^[1^[6]]]]]]]]-1"

divide again:
"2^[3^[2^[5^[8^[2^[6^[5^[7]]]]]]]]-1"
----------------------------------------------
"2^[8^[4^[6^[4^[0^[4^[1^[6]]]]]]]]-1"

OK now, too lazy for your math, then I will just convert your digits
to my digits and define your number and my number as both two and one
(of course, also none - the o)!!!

;-2-2;-3-8;-2-4;-5-6;-8-4;-2-0;-6-4;-5-1;-7-6;-1-1;
(all pairs are equal);

==
;o-o;5-5;o-o;o-o;4-4;o-o;o-o;4-4;o-o;o-o

how not surprizing - you don't have 3-3! OK, let's define either -8-4
or -5-1 as your 3-3. now check! whoever is your 3-3 will be my 4-4 and
vice versa. split! define my 5-5 as 3;3 both 4-4 as 5-5 OK?

;o-o;5-5;o-o;o-o;4-4;o-o;o-o;4-4;o-o;o-o
;o-o;3-3;o-o;o-o;4-4;o-o;o-o;4-4;o-o;o-o
;o-o;3-3;o-o;o-o;5-5;o-o;o-o;5-5;o-o;o-o

now, remove all dots, except the last one whois "-1"; convert to "-one";

5-5;4-4;4-4;-one;
3-3;4-4;4-4;-one;
3-3;5-5;5-5;-one;

convert backwords:

-one................3-3;4-4;5-5.............this will be your order right!

OK, rotate your order backwords and convert your number:

"2^[32,582,657]-1"
here:
"2^[756,285,23]-1"

now, one of these number has to be bigger, right? OK, define mine as bigger
//"2^[756,285,23]-1";
now, rotate to the 8th;

// "2^[85,23,75,62]-1"

OK. now, let's define this decimal, otherwise you will not understand
me? (maybe?), then remove all dots,
"2^[32582657]-1"
"2^[75628523]-1"
"2^[85237562]-1"

your number is the smallest one of these numbers, right? (positive).
then let's define the number "[2^[32582657]-1] * [2^[32582657 *
75628523 * 75628523 * 75628523 * 85237562 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562]-1]" as my ANY number who is
dividing your number. then; the division would be [2^[32582657 *
75628523 * 75628523 * 75628523 * 85237562 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562]-1], then since your number
[2^[32582657]-1] is supposed to be prime, and the other number is
bigger - then it is either a prime or not a prime. then let's define
the one not a prime:

[2^[32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562]-1]+1: not a
prime right?

divided by two - let's count how many?

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562

divided by four - how many?

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562 / 2

not interested. then let's mod all the primes 3/5/7 and count again:

convert to base

mod 3: 32582657==2 ; 75628523==2 ; 85237562 ==2;
mod 5: 32582657==2 ; 75628523==3 ; 85237562 ==2;
mod 7: 32582657==2 ; 75628523==5 ; 85237562 ==4;

OK? now, replace mods:

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562 mod 3 == 2 ** 2
** 2 ** 2 ** 2 ** 2 ** 2 ** 2 ** 2 ** 2 == (-1) ** (-1) ** (-1) **
(-1) ** (-1) ** (-1) ** (-1) ** (-1) ** (-1) ** (-1) == (----------1)
(your math).

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562 mod 5 == 2 ** 3
** 3 ** 3 ** 2 ** 2 ** 2 ** 2 ** 2 ** 2 == (-3) ** (--3) ** (--3) **
(--3) ** (-3) ** (-3) ** (-3) ** (-3) ** (-3) ** (-3) ** (-3) ==
(---------------3) (your math).

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562 mod 7 == 2 ** 5
** 5 ** 5 ** 4 ** 4 ** 4 ** 4 ** 4 ** 4 == (-5) ** (-2) ** (-2) **
(-2) ** (-4) ** (-4) ** (-4) ** (-4) ** (-4) ** (-4) ** (-4) ==
(-----------1) (your math).

OK, so in base 3 and 7, the smallest prime number will be your -3, in
base 5 the smallest prime number will be my (-3/2), now let's convert:

the biggest prime number who is dividing [2^[32582657 * 75628523 *
75628523 * 75628523 * 85237562 * 85237562 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562]-1] will be defined as my 4!!!!

the number 2^[32582657]-1 [Mersenne #44] will be defined as my 2!!!!
the number whois 2^[32582657 * 75628523 * 75628523 * 75628523 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562 * 85237562 *
85237562]-1] will be defined as my 32!!!!!
then, 2^[32582657 * 75628523 * 75628523 * 75628523 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562 * 85237562]-1]
divided by my 4, divided by my 4 again will be defined as [Mersenne
#44] (counting all bases!!!);

now, define them back to their decimal values. remove 32582657 bits,
then 2^[32582657]-1 [Mersenne #44]  will be defined as the smallest
prime number who is dividing whatever this number is (not defined -
remember!)

2^[32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *
85237562 * 85237562 * 85237562 * 85237562 * 85237562]-1]

OK! now, switch to base Mersenne #44 (positive prime); then define this number:

2^[88888888888888888;7777777777777;6666666666666;3]-1 --> this will be
my number (1/Mersenne #44) in base 32!

convert to decimal (2 defined as [[Mersenne #44] ^ [Mersenne #44]]:
divide by ([[Mersenne #44] ^ [Mersenne #44]] /2) [of course - not an
integer!)

then, compress; 2^[8..8;7..7;6..6;3]-1
convert to decimal again: 2^[8;888;888;777;776;663]-1

BUT - I need to check if it's a decimal prime right? otherwise you
wouldn't believe me!
OK - let's check how much is this number in decimal? 8888888777776663?

8888888*1000000000 +
77777*10000 +
666*10 +
3;

OK, now lets rename all digits and negative them - check whois a prime!

;a;b;c;d:

[- aaaaaaa*1000000000 - bbbbb*10000  - ccc*10 - d][times the digit "i"]
is not divided by (all the numbers from [Mersenne #44] to 2) //
[Mersenne #44]!!!!!!!
(that's at least 7 times permutated!)
checksum:
triangle:
ooooi;
oooii;
ooiii;
etc.

[- aaaaaaa*1000000000 - bbbbb*10000  - ccc*10 - d][times the digit "i"]

sum(d);sum(ccc*10); sum(bbbbb*10000) ; sum (aaaaaaa*1000000000);
go base 21 / 42: add o's

aaaaaaaoooooooooooooooooooooooooooo;
bbbbboooooooooooooooooooo;
cccooooooooooo;
dooooo;
count: a7;b5;c3;d1 +
ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo.......................................

OK? now, 105^[a7;b5;c3;d1]-1 must be a prime in base Mersenne #44;
(105 will be (Mersenne #44)*(Mersenne #44)+5)

then of course, Mersenne #44 will be number 5; I just need to find
numbers 4 and 3!!!
define them this way: 105/104/103 ok?

now equation: XYZ; Z==Uri#105; Y=Uri#104; X= Uri#103; M==Mersenne #44;

Z=[(M*M)--55ii][M*M*M*M*M*M*M];
Y=[(M*M)--44pipiii][M*M*M*M*M*M*M];
X=[(M*M)--33ii][M*M*M*M*M*M*M];

Z*Z==Y*Y+X*X;

[M*M*M*M*M*M*M][(M*M)--55ii]*[(M*M)--55ii][M*M*M*M*M*M*M]
==[M*M*M*M*M*M*M][M*M*M*M*M*M*M][(M*M)+5i][(M*M)-5i] ==
[M*M*M*M*M*M*M][M*M*M*M*M*M*M][M*M*M*M-5*5i];

[M*M*M*M*M*M*M][(M*M)--44pipiii]*[(M*M)--44ii][M*M*M*M*M*M*M]
==[M*M*M*M*M*M*M][M*M*M*M*M*M*M][(M*M)+4pii][(M*M)-4pii] ==
[M*M*M*M*M*M*M][M*M*M*M*M*M*M][M*M*M*M-4*4pipii];

[M*M*M*M*M*M*M][(M*M)--33ii]*[(M*M)--33ii][M*M*M*M*M*M*M]
==[M*M*M*M*M*M*M][M*M*M*M*M*M*M][(M*M)+3i][(M*M)-3i] ==
[M*M*M*M*M*M*M][M*M*M*M*M*M*M][M*M*M*M-3*3i];

OK? now - let's define the number of binary digits of each (in
decimal) - let Mersenne #44 be the biggest?....

of course - logging here!

Z: [7+][7+][-5 - 5] // 0 binary digits!!!
X:

[7+][7+][-5 - -4]; // 1 binary digit!!!!

OK, then let X (binary) be the next binary prime Mersenne #44 is now,
BUT ALLWAYS true for ANY prime number!!!

define: Mersenne #44:

0/1/2/3/2/5/8/2/6/5/7/1

// ok let's make that decimal.............. binary numbers are not real anyway!

counting, replacing, starting with e:

0/1/2/3/2/5/8/2/6/5/7/1
e/f/g/h/gg/j/m/ggg/k/jj/l/ff/
OK. number equals to "efghggjmgggkjjlff" // same number repeating
itself must be marked!
now, checking smaller primes! replace digits with 0/1/2/3/4/5/6/7/8/9.......

"e/f/g/h/gg/j/m/ggg/k/jj/l/ff/" - replace with

/0/1/2/3/44/5/6/777/8/99/s/tt/

replace rightwords:

/00/1/22/3/444/5/6/77/8/9/s/tt/

the difference:

/o.o/1/2.2/3/44.4/5/6/77.7/8/9.9/s/tt/

right to left:

/o.o/1/2.2/3/4.44/5/6/7.77/8/9.9/s/t.t/

OK looks symmetric to me. 44.44 will be next prime gg; switch back:

/o.o/1.1/2.2/3.3/g.g/5.5/6.6/g.g/8.8/9.9/s.s/t.t/

switch back:

now equation: XYZ; Z==Uri#105; Y=Uri#104; X= Uri#103; M==Mersenne #44;

define XYZ the same:
/o.o/1/2.2/X.X.YY.ZZ/6/7.77/8/9.9/s/t.t/
/o.o/1/2.2/YXZ.XZY/6/7.77/8/9.9/s/t.t/
/o.o/1/2.2/Z.Z.YY.XX/6/7.77/8/9.9/s/t.t/

see? I'm starting to understand their language!

OK, let's just replace their "2" with "212", lets see if it's prime for them;

[2^321;258;212,657-1] - would probably be a prime too. let's check!

0/1/2/3/2/1/2/5/8/2/1/2/6/5/7/1

e/f/g/h/g.g/f.f/g.g/j/m/ggg/f.f/g.g//k/jj/l/ff/

OK?

/0.0/1.1/2.2/3.3/2.2/1.1/2.2/5.5/2.2/2.2/1.1/2.2/1.1/2.2/6.6/5.5/7.7/1.1/0.0/

reverse:

/0.0/1.1/7.7/5.5/6.6/2.2/1.1/2.2/1.1/2.2/2.2/5.5/2.2/1.1/2.2/3.3/2.2/1.1/0.0/

OK symmetric; convert:

0;1;7;5;6;2;1;2;1;2;2;5;2;1;2;3;2;1;0
reverse:
0;1;2;3;2;1;2;5;2;2;1;2;1;2;6;5;7;1;0

convert to their decimal:

[2^[3;2;1;2;5;2;2;1;2;1;2;6;5;7]-1]
[7^[5;6;2;1;2;1;2;2;5;2;1;2;3;2]-1]

let's see. can [7^[5;6;2;1;2;1;2;2;5;2;1;2;3;2]-1] be divided by 7? I
don't think so....

OK, now rename: 7==2; define 2==7; that's my prime! [base 70000000000007!]

[2^[5;6;4;1;4;1;4;4;5;4;1;4;3;4]-1]

now, let's compress it; 0/1/2/5/6/4/1/4/1/4/4/5/4/1/4/3/4/1/0
reverse, switch 4/2:

0/1/4/2/5/6/4/1/4/1/4/4/5/4/1/4/3/4/1/0

rename: e/f/g/h/i/j/g/f/g/f/g/g/i/g/f/g/k/g/f/e

reverse: e/f/g/k/g/f/g/i/g/g/f/g/f/g/j/i/h/g/f/e

symmetric [e/f/g][/k/g/f/g/i/g/g/f/g/f/g/j/h/] |
[h/j/g/f/g/f/g/g/i/g/f/g/k/][g/f/e]

cut middle: replace: g "." --> [e/f]..[k/f/i/f/f/j/h]

e==0;
f==1;

// tired. will define all later!!!!! (or computer will do it for me)...........

Uri First deadandalive

[~~~~~~~~~~~~~~~~~~~~~]

--

Uri First deadandalive

[Robert Wallner, read your mail]; create a special human key for me if
you want to. Using R.W;