3) Give another proof of the infinitude of primes by assuming that there are only finitely many primes, say p_
357003
if you take the number N and add one it must be prime... \n\ntake any number it can be writen as a product of primes, so in this case let us just examin N+1/p_j\n\nclearly N/p_j is an integer, and 1/p_j is clearly not an integer. \n\nSo if N were not prime then there would exist p_j such that N+1/p_j was an integer.   But that can not be true, therefore the assumption that there are only finitly many primes is incorrect, and the opposite is true, there are infinitly many primes QED
27160
If you are referring to twin prime numbers (primes separated by only 2), then no, it has not yet been proven that they are infinite in number. It has been conjectured, but the proof has been elusive.
787081
If n is contains an odd prime factor, this is part of an unsolved problem called Beal's conjecture. This says that if any\nsuch solution exists, then a and b cannot be coprime.\nIf n is a power of 2, it is a multiple of 4, since n> 2.\nA glance at Wikipedia also shows that this\ncase is also part of Beal's conjecture.\nAt the present time no counterexamples to\nBeal's conjecture are known.\nIncidentally, there is a $100000 prize for the\nfirst proof or counterexample to Beal's conjecture.
774052
LetP(n) denote the statement\n1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)/6\nwhen n=1,P(1)=1(2)(3)/6=1 checks\nlet P(k) be true\ni.e.1^2+2^2+3^2+.......+k^2=k(k+1)(2k+1)/6\nto prove\nP(k+1)=(k+1)(k+2)(2k+3)/6 is true\nP(k+1)=1^2+2^2+3^2+.......+((k+1)^2\n=>[1^2+2^2+3^2+......+k^2]+(k+1)^2\nnow 1^2+2^2+3^2+......+k^2=k(k+1)(2k+1)/6\nsubsttituting\n=k(k+1)(2k+1)/6 +(k+1)^2\n=(k+1)[k(2k+1)+6(k+1)]/6\n=(k+1)(2k^2+7k+6)/6\n=(k+1)(k+2)(2k+3)/6\nhence proved\n\n2.try by your self
246941
5, you are listing the prime numbers
1004221
For the moment we don´t even know if there is an infinite number of Fibonacci primes....\n\nHere are the first fifteen:\n\n2,3,5,13,89,233,1597,28657,\n514229,433494437, 2971215073, 99194853094755497,\n1066340417491710595814572169,\n19134702400093278081449423917,\n475420437734698220747368027166749382927701417016557193662268716376935476241\n\nLooks considerably less dense than the fraction prime numbers / natural numbers
113690
No.  Deductive proofs start from a "given", that is, a fact accepted as true or a theorem previously proven to be true.  But other kinds of proofs start with an assumption that the theorem to be proved is not true, and from that assumption derive some contradiction, thus proving that the theorem itself is true.\n\nAn example would be the proof that there is no highest prime number, which goes like this. Assume that there is a highest prime number.  Call it x.  Now consider the number x! + 1.  (X! is X times X-1 times X-2 all the way down to 1.)  Since x! is divisible by 2, x! + 1, being just 1 more than x!, is not divisible by 2.  Similarly, x! + 1 is not divisible by 3, 4, or any other integer up to and including x.  Since x! + 1 is not divisible by any number up to x, either it is prime, or it's prime factors are larger than x.  But that contradicts our assumption that x is the largest prime.  Thus, our assumption is false.  QED.
252368
Hi there,\nWe will always be guaranteed a period of consecutive days when exactly 129 books are sold. The reason the number 129 is important is clear below. The proof uses what is known as the pigeonhole principle. [If n>m pigeons are put into m pigeonholes, then there exists a hole with more than one pigeon.] As usual, the trick is in constructing the "pigeons" and the "pigeonholes".\n\nLet B_i (for i=1,2,...,365) be the number of books sold during the period from the first day to the ith day inclusive (for example, B_3 represents the number of books sold in the first three days). Then we have:\n1 <= B_1 < B_2 < ... < B_365 = 600\nWe have strict inequalities inbetween as at least one book is sold each day. Now for the trick. Add 129 to this sequence of numbers to get:\n130 <= B_1 + 129 < B_2 + 129 < ... < B_365 + 129 = 729.\n\nWe choose B_1, B_2, ..., B_365, B_1 + 129, B_2 + 129, ..., B_365 + 129 as our pigeons. There are 730 of them, each being a positive integer between 1 and 729 inclusive. The numbers from 1 to 729 will be our pigeonholes. Thus by the pigeonhole principle, two of the numbers in our sequence of 730 numbers must be the same. As B_1, B_2, ..., B_365 are distinct, and B_1 + 129, B_2 + 129, ..., B_365 + 129 are distinct, one of the B_i's must be equal to one of the B_i + 129.\n\nSuppose that B_n = B_k + 129, where 1 <= k < n <= 365. Then we have B_n - B_k = 129. So between the (k+1)th day and the nth day inclusive, exactly 129 books were sold.\n\nCheerios,\nMike
571480
A mathematical statement can be proven in several ways. Your method of proof will depend on the statement. For the statement above (assuming you mean integers by numbers) the proof would be somthing like this\n\nlet P = {x: x is a prime number}\n\nlet M1 = min(P)\nlet M2 = min(P - M1)\nlet M2 = min(P - M1 - M2)\n\nthis means that M1, M2, M3 are the 3 smallest prime numbers. We know that \nM1 = 2, M2 = 3, M3 = 5\nthis implies M1 + M2 + M3 = 2 + 3 + 5 = 10 > 6\ntherefore for every integer N > 6,\nN = P1 + P2 + P3, P1,P2,P3 are prime numbers
25903
Given an integer N, N is prime if 2^N is congruent to 2 (mod N).
900983
Necessary and sufficient condition is a is not the neutral element  and m is a prime superior to 2
654672
Bingo!  I've got it!\n\nConvert from base-3 to base-10:\n3, 5, 8, 11, 16...\n\nSubtract n from each term:\n2, 3, 5, 7, 11...\n\nThis is the sequence of prime numbers!  So the generating sequence would be the nth prime plus n:\n2+1, 3+2, 5+3, 7+4, 11+5, 13+6, 17+7, ...\n\nContinuing the sequence:\n3, 5, 8, 11, 16, 19, 24, 27, 32, 39, 42, 49, 54, 57, 62, 69, 76, 79, 86, 91, 94, 101, 106, 113, 122, 127, 130, 135, 138, 143, 158, 163, 170, 173, 184, 187, 194, 201, 206, 213, 220, 223, 234, 237, 242, 245, 258, 271, 276, 279, 284, 291, 294, 305, 312, 319, 326...\n\nIn base-3:\n10, 12, 22, 102, 121, 201, 220, 1000, 1012, 1210, 2000, 2010, 2022, 2120, 2211, 2221, 10012, ...\n\nAm I right?
7873
well in the demostrantion of the Real numbers...\n\nall the numbers and relations depend of the existence of 1... suck that there are some rules..\n\nso we even demostrate that 1 exists... and that 1 > 0... after that we can prove the existence of any other Real number..
537907
This site has a table of primes p where 1 < p < 1,000,000,000,000\n\nIf you want to know N, you'll have to count them.
28467
I just had a math packet dealing with this topic.  There are many equations out there that deal with finding primes, but all so far fail at a certain point in the sequence.  The method you refer to, called the Sieve of Eratosthenes, was developed by its namesake in Greece sometime between the 500s and 350 B.C. and is so far the only proven method to find primes.  Sorry, but unfortunately, there is no other accurate choice.
7923
There isn't one. Mathematicians are working on it, but they have yet to find a pattern that covers all prime numbers (except for the fact they are divisible only by themselves and one.) There are patterns that cover some prime numbers, such as paired primes (where the difference between the two prime numbers is 2, such as 5 and 7), but there is no currently known way to quickly produce or check primes, which has important implications for the RSA algorithm (used in PGP for example), which relies on the complexity of primes for its security.
832736
No other two numbers besides ONE and itself can equal to ONE. for example\n\n7 is prime because only\n\n1*7 can equal to 7\n\n1 is prime because only\n\n1*1 can equal to one