Skip to main content
4 events
when toggle format what by license comment
May 9, 2017 at 14:56 comment added Emil Jeřábek Expanding Liviu Nicolaescu’s argument, we have $a\le na\le a$ for all positive integers $n$.
May 9, 2017 at 14:32 comment added Liviu Nicolaescu Note that if $\leq$ is such an order, then $a=\min\{a,a\}\leq a+a=2a\leq \max\{a,a\}=a$, $\forall a\in\mathbb{N}^k\setminus \{0\}$.
May 9, 2017 at 14:28 review First posts
May 9, 2017 at 14:59
May 9, 2017 at 14:27 history asked gm01 CC BY-SA 3.0