By the way, I wrote very long; if you want it short then let it be this way. I'll define you my new AXIOM [there is no such thing as a prime number, but it is a true axiom]:
base 10 (binary 10==01+01): // actually that's base prime 10*10; base 4; either 10 or 01 can be primes; but not both; base 10 ^ (ANY NUMBER n): either n is a prime (number of digits) or 10n is a prime, not both; (induction.......) for EVERY n; there must be at least One prime number in base binary 2 digits (binary can be any base prime); if n is not a prime then this will be the first (smallest) prime number with n digits. for example: the number 11 (base ten): 1011 is the "smallest" prime base 100 (four binary digits); then also 1101(13dec) is but not 1111 (because 100 is not a prime). BUT you can of course switch positive/negative, then 100 WILL BE a prime. But if you want you can count: ANY number of digits (counting); the FIRST (smallest) prime this binary representation will be this digit's prime number. for example: 1:1 10:10 11:101 100:1011 101:10001 110:1000111 111:10000011 1000:10000001 1001:100001001 1010:1000000111 1011:10000000101 etc. etc. now notice - it is very regular and not very difficult to define this. ANY prime number of n digits is either n or the complement of n in some way; if n is even then it must be some sort of a complement since you don't allow even primes except 10; then the complement will just be dividing digits to (n+1)(n-1); for example if n is 4; the 15 will be the complement; 11 will be the first prime bigger than 8 (2^4)/2. And counting backwords, then (15-4) will allways be prime. so first: 1. for ANY number 2*2*m*m whois even and square prime (but m is any not even number, including one; for example four), [2^[[[2*2*m*m]]-1] will be equal to [([2*2*m*m]+1)([2*2*m*m]-1)];; also ([2*2*m*m]-1) will be divided by (2*m+1)(2*m-1) and then [2^[[[2*2*m*m]]-1] - [(2*m-1)(2*m+1-1)] will always be inverse of the prime number ((2*2*m*m)-1) (have to generalize, but this is the idea. Should work also for more than ^2 and ferma was not true (but leave the flat euclidan numbers behind). 2. 11 will always be a prime number, therefore binary 1011 will alwayys too be the inverse of the first prime number, then of course if you have a prime such as (n) digits of 1 (for example 11;111;11111 etc.), then (n digits times 1)+(n digits times 0)+(n digits times 1)+(n digits times 1) will allways be a prime! (by the way you can inflate the (n digits times 0) to even 2^n; 3^n; ANY exponent here! then actually ANY prime number times 11 times (the same prime number times 11; square rooted) will always be a prime! (any multiplication of primes to the exponent of 1.5 will be also a prime). Then if you have a prime, just multiply it by 11; inverse; multyply again and you will have a prime number too! for example, the number "Mersenne 44" [2^32582657-1] [remember; Americans know this already. It's top secret "encryption technology"...... (or ANY prime number; you can just put the digits here), or any multiplication of primes - you can divide them. take [2^32582657-1] (your ANY PRIME) and write it like this: ;;1101;; [[2^32582657-1]^3] + [2^32582657-1]^2 + 1] and you have your next prime! of course instead of ^3 you can also write 4;5;6;7 etc...... remember to check and it should just look like this: [mmmmmmmmmmmmmmmmmmmmmmmannny ones;] [twwicccccee]; [mmmmmmmmmmmmmmmmmmmmmmmannny zeros;] [mmmmmmmmmmmmmmmmmmmmmmmannny ones;] or vice versa: ;;1011;; or actually: 100000000000001000000000000000000000000000000001 then you can just contact them with each other; each time add an inverse of a prime number (including 1 or 0), then you will just receive something like this: [any strings of prime numbers][inverse prime numbers][.....] will allways be prime. For example: [2003663613·2^195000+1] ; [2003663613·2^195000-1] whoever they are; [as strings in binary representation starting with 1 or 0; ending with 1 or 0]: concat them as many times as you want, each time you concat a number - reverse it (the complement); then you have your next primes! I can even do this by induction: m,n = any given primes, binary; digits defined as digits(m) and digits (n) [[2^digits(m)]-1-m]: also a prime (if m not divided by 4; if first digit is 0 then complement first) [[2^digits(n)]-1-n]: also a prime (if n not divided by 4; if first digit is 0 then complement first) // of course you can generalize it - doesn't have to be base 2; then: digits(m) will be defined as the first log(base2)m downwords + the first log(base2)m upwords; if difference is exactly one then take the upwords (actually when converted to negative numbers you can take negavite numbers of digits as well; commmpress! // same with n; now of course, by induction 0;1;2 we define whether primes or not; 3;4; you decide; defining 4 not a prime here: then ANY number you want! ==m // is it 4? not a prime by definition (your definition, not mine). 0;1::same// otherwise: [ //this you don't have to do actually ] calculate square root [can be extended], if it's an integer then for you - not a prime! 2: multiply m by m as many times as you want; then by 2; then calculate square root (will not work for two or any even number); received result n; then either n is a prime if and only if the square root of 2 is a prime. then - need more primes? just use this number as your base and create as many primes as you want! you can use multiple numbers m;n; etc. but don't contradict yourself, otherwise you will not receive any prime. but allways: 0. [start with 0] 1. // ignore 4. 4 ^ the previous number; 256. then this number plus one; and [(this number minus one) divided by five] will allways be primes! [4^256] (this one too: same same) [4^[4^256]]same; then generally speaking: any number who can be divided to divisors, it's exponent can be divided the same. then of course exponent and multiplication means the same thing in my math... but do you have any prime bigger than this? for example take Google; [2*5]^[[2*5]*[2*5]] let's define their 1/Google: 0.1^[[2*5]*[2*5]]; then, number of binary digits of each: about 332,333? OK define Moogle[0]: 0 define Moogle[1]: 4 ^ [Moogle [0]]; // 1 define Moogle[2]: 4 ^ [Moogle [1]]; // 4 define Moogle[3]: 4 ^ [Moogle [2]]; // 256 .... etc; then define: function Moogle[x]::always return 4^x; then define: function PrimeMoogle[x]::always return Moogle[x]+1; ... then define: RecursiveMoogle[x]: always return Moogle[RecursiveMoogle[-1]] of course I defined Poodle as Googleplex; why don't I use it? Poodle=[2*5]^[[2*5]^[[2*5]*[2*5]]]; also same as [0.1^[-[0.1^-[[2*5]*[2*5]]]]; now define RecursiveMoogle[Google] and RecursiveMoogle[Poodle] // I don't mind how much time it takes, I define it theoreticallly; and then the whole series of PrimeMoogle[x] and PrimeMoogleLess[x] will always return primes, at least for numbers from 16 and above! (2.5 and above; start for 3 if you want integers); then calculate PrimeMoogle[RecursiveMoogle[Google] + RecursiveMoogle[Poodle]] now; if they are not prime numbers base 2; then base 5 they will be! I will base them YOUR BASE NUMBER who is a divisor for them! you can only have ONE first prime divisor for them, if it's not 2 then 3 or 4 or 5. Then here are your biggest known prime numbers! The first divisor based this! Actually I can base one base two, the other one base five and then create more - they will ALLWAYS be primes! start with 257, then go up and now replace my 4 with your next prime number between 3 and 5. this is uncountable and therefore you can just start with 0; then count up and then instead of 4 write down your next (increasing prime number); then for example take this list; http://primes.utm.edu/lists/small/1000.txt start with 2 3 5 7 11 13 17 19 23 29 ; and then each number you will define with the previous prime number base; then define my function: ProoMle[x]: ProoMle[x-1]^ProoMle[x-1]+1; then if Moogle was not always a prime (your counting), then ProoMle[x] by definition will be! let's check: ProoMle[0]: defined 1; ProoMle[1]: defined 0; ProoMle[2]: 1^1+1==3; // defined 3. ProoMle[3]: 2^2+1==5 // of course; I can define it to be 3 too! // defined 5. //count only primes. ProoMle[4] // defined 4; ProoMle[ProoMle[3]]; // 5^5+1 [3^3+1=28; // 7 is my secret number! already defined here! (notice; this is not related to anything I did!) ProoMle[5]: // ProoMle[ProoMle[4]] is ProoMle[ProoMle[ProoMle[3]]] is ProoMle[ProoMle[ProoMle[ProoMle[2]]]] and then of course 3 is defined; Then you can never count the primes. the prime numbers are themselves uncountable! there are as many "prime" numbers as there are "real" numbers but there are no "imaginary" numbers! Only one entity whois my 1.5 or 0.5 or -0.5 and they are all equal! When you understand this then ANY number can be your prime. anyway, for you my friends I say you can have as many prime numbers as you want, but it you think Americans don't know how to divide your numbers - think again. It's very easy! And Ferma - you are right. You can't have straight three-dimentional triangles on one flat paper who will never fold with the speed of truth, my friend. But if you can think about my Kippa - there are no limits! In my new algebra you can do what you want. I will prove to you that you can! Here I am already prooving it now! 10^[YOUR ANY NUMBER] plus i[MY ANY NUMBER] will allways be equal to my ANY OTHER NUMBERS I WANT MINIMIZED TO o. Check this! if you can imagine yourself rotating inside a black hole then so do I. then why are your mirrors green when pointing to each other? Why not bluish or reddish? It's your lying RGB Americans. Think! There are no limits to the number of colours you can see - just depends on the RGB blood test you force people from other planets to have. Think about it! They are really not you! from other planets! you will not be able to notice this if you are not their friends! Think about it, open your horizoning (h/o reasoning) and understand that two is an irrational number too. Uri -- Uri First deadandalive Mobile Phone: +972-50-9007559 E-mail: [EMAIL PROTECTED] // [EMAIL PROTECTED] Update: Left HTTP, WWWW and port numbering. send me papers or to [EMAIL PROTECTED] . I completed my 0.1 version of real deadanyalive quantum relativity redefining back timespacing intergalactic worldwide [top secret: if you have any US ARMY* on your planet they will never allow it]. Recounting back every second since twice BCC doubling + not counting at all any uncountable; using only prime countable numbers equal to 21 (base 21 recursivley) who are all equal to 0base0 and 1base1 who are identical twins base 21 [21===the number of fingers;eyes;body parts & number of equal signs not equal counting both left to right; right to left and all 21 dimensions of nothing]. Read my autoreply for more information [my HTTP/SMTP not working]. - This message is confidential -
