positivity in ordered ring
Theorem.
If (R,≤) is an ordered ring, then it contains a subset R+ having the following :
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R+ is under ring addition and, supposing that the ring contains no zero divisors, also under ring multiplication.
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Proof. We take R+={r∈R: 0<r}={r∈R: 0≤r∧ 0≠r}. Let a,b∈R+. Then 0<a, 0<b, and therefore we have 0<a+0<a+b, i.e. a+b∈R+. If R has no zero-divisors, then also ab≠0 and 0=a0<ab, i.e. ab∈R+. 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 ordered. The latter alternative gives that 0=-r+r<-r+0=-r. The both cases that either r∈R+ or -r∈R+.
Title | positivity in ordered ring |
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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 |