# positivity in ordered ring

###### Theorem.

If  $(R,\,\leq)$  is an ordered ring, then it contains a subset $R_{+}$ having the following :

• $R_{+}$ is under ring addition and, supposing that the ring contains no zero divisors, also under ring multiplication.

• Every element $r$ of $R$ satisfies exactly one of the conditions   $(1)\,\,r=0$,    $(2)\,\,r\in R_{+}$,    $(3)\,\,-r\in R_{+}$.

Proof.  We take  $R_{+}=\{r\in R:\,\,0.  Let  $a,\,b\in R_{+}$.  Then   $0,  $0, and therefore we have  $0,  i.e.  $a\!+\!b\in R_{+}$.  If $R$ has no zero-divisors, then also  $ab\neq 0$  and  $0=a0,  i.e.  $ab\in R_{+}$.  Let $r$ be an arbitrary non-zero element of $R$.  Then we must have either  $0  or  $r<0$  (not both) because $R$ is totally ordered  .  The latter alternative gives that  $0=-r\!+\!r<-r\!+\!0=-r$.  The both cases that either  $r\in R_{+}$  or  $-r\in R_{+}$.

Title positivity in ordered ring PositivityInOrderedRing 2013-03-22 14:46:40 2013-03-22 14:46:40 pahio (2872) pahio (2872) 12 pahio (2872) Theorem msc 06F25 msc 12J15 msc 13J25 PositiveCone TopicEntryOnRealNumbers