   The lowest factor of any number is prime
  So I just had a realization, probably elementary math, but I want to make sure I'm correct.The lowest factor (other than 1) of any number is prime:---------------------------------------------------18: (1), 2, 3, 6, 9, (18)2 is the lowest factor other than 1.15: (1), 3, 5, (15)3 is the lowest factor other than 1.---------------------------------------------------It's kind of obvious. Is there a formal proof?   
  Due to the fundamental theorem of arithmetic, we see that every integer number has a unique prime factorization. Noting this, we can also state that every factor of a number is either prime or decomposable into smaller primes. Therefore the smallest factor of any integer greater than 1 is a prime.   ;  If the smallest factor is not prime, then you can further decompose it into prime factors, which will be:- smaller- factors of the initial number.And depending on who you ask, 1 is prime.   ;Quote:Original post by bobofjoeDue to the fundamental theorem of arithmetic, we see that every integer number has a unique prime factorization. Noting this, we can also state that every factor of a number is either prime or decomposable into smaller primes. Therefore the smallest factor of any integer greater than 1 is a prime.Just what I wanted.   ;  To form that into a proof, you need to use proof by contradiction.Suppose that there is an integer such that its smallest positive factor greater than 1, say k, is composite. Then there are primes p_1, ..., p_n and positive integers e_1, ..., e_n such that k = p_1^e_1 * p_2^e_2 * ... * p_n^e_n. Let p > 1 denote min(p_1,p_2,...,p_n). Then p is a prime factor of the original integer smaller than k, a contradiction.Quote:Original post by FrunyAnd depending on who you ask, 1 is prime.1 is almost never considered prime in mathematics today. In particular, the fundamental theorem of arithmetic would be false if 1 was prime.   