positivity in ordered ring


If  (R,)  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)rR+,    (3)-rR+.

Proof.  We take  R+={rR:  0<r}={rR:  0r 0r}.  Let  a,bR+.  Then   0<a,  0<b, and therefore we have  0<a+0<a+b,  i.e.  a+bR+.  If R has no zero-divisors, then also  ab0  and  0=a0<ab,  i.e.  abR+.  Let r be an arbitrary non-zero element of R.  Then we must have either  0<r  or  r<0  (not both) because R is totally orderedPlanetmathPlanetmath.  The latter alternative gives that  0=-r+r<-r+0=-r.  The both cases that either  rR+  or  -rR+.

Title positivity in ordered ring
Canonical name PositivityInOrderedRing
Date of creation 2013-03-22 14:46:40
Last modified on 2013-03-22 14:46:40
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 12
Author pahio (2872)
Entry type Theorem
Classification msc 06F25
Classification msc 12J15
Classification msc 13J25
Related topic PositiveCone
Related topic TopicEntryOnRealNumbers