Recent content by Sharps

I got it. Any n with unique prime factorization containing only one prime factor (not counting its multiplicities) will be totally ordered. Nice.

Yah, I was thinking about that. I know that all natural numbers greater than one have a unique prime factorization. So, I suppose that if D(n) contains 2 primes, they will not divide one another. Can I go from there?

That's my mistake...typo. I meant to say that all prime numbers can be written as a 2^s + 3^t. Also, I've found a counterexample to that criterion, namely 125. So it's back to the drawing board. What I know: 1.) For all n such that n is prime, D(n) is totally ordered. 2.) Not much else...

Homework Statement Let n be a positive integer and let D(n) be the set of all positive integers d such that d divides n. For example, D(10)={1,2,5,10}. It's easily shown that the divides relation (xRy if x divides y) is a partial ordering on D(n), but for what n is the D(n) totally...