Hi all, I am spending some time learning discrete mathematics on my own using the MIT OpenCourseWare materials. On page the second to last page of the Chapter 2 notes from here: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2010/readings/ [Broken] it states CHAPTER 2. THE WELL ORDERING PRINCIPLE Theorem 2.4.1. Every natural number can be factored as a product of primes. Proof. The proof is by Well Ordering. Let C be the set of all integers greater than one that cannot be factored as a product of primes. We assume C is not empty and derive a contradiction. If C is not empty, there is a least element, n ∈ C, by Well Ordering. The n can’t be prime, because a prime by itself is considered a (length one) product of primes and no such products are in C. So n must be a product of two integers a and b where 1 < a,b < n. Since a and b are smaller than the smallest element in C, we know that a,b /∈ C. In other words, a can be written as a product of primes p1p2 ··· pk and b as a product of primes q1 ··· ql. Therefore, n = p1 ··· pkq1 ··· ql can be written as a product of primes, contradicting the claim that n ∈ C. Our assumption that C ≠ ∅ must therefore be false. (the notation may be unclear, so it may be easier to read from the link) Could someone explain this concept a bit further? I couldn't figure out this proof because I would have never made this assertion. How is a prime considered a length one product of primes? Say I have a prime, 3, what primes is this a product of? Thanks in advance. All help is appreciated.