Suppose that is no $ n\in \mathbb{N}$, n>1 which divides every element of S. Then there exist $ a,b\in S$ with (a,b)=1. It is trivial to prove that for every $ k,l\in \mathbb{N}$, $ ka+lb\in S$. Let q>ab. Then if (k,l) solution of the equation ka+lb=q (1) there is either k>b or l>a.(2) Also we notice that if (k,l) solution of (1) then (k+b,la) is also solution of (1) This means that (1) has infinite number of solutions. Suppose that there is no solution in positive integers. Let $ (k_{1},l_{1})$ the solution where $ k_{1}$ takes the maximum negative value possible. Then $ l_{1}>0$ and also $ k_{1}+b>0$ and $ l_{1}a<0\Rightarrow l_{1}<a\Rightarrow k_{1}>b$ (from (2)) contradiction because we defined $ k_{1}$ as negative. So for every q>ab (1) has a solution in positive integers so $ \mathbb{N}/S\subseteq \left \{ 1,2,...ab\left. \right \} \right. $ contradiction. So the proof has finished.
