- Boxes
- definitions
- Ellipses
- theorems and lemmas
- Blue border
- the statement of this result is ready to be formalized; all prerequisites are done
- Orange border
- the statement of this result is not ready to be formalized; the blueprint needs more work
- Blue background
- the proof of this result is ready to be formalized; all prerequisites are done
- Green border
- the statement of this result is formalized
- Green background
- the proof of this result is formalized
- Dark green background
- the proof of this result and all its ancestors are formalized
Let \(a\) and \(b\) be elements of a left ordered semigroup \(S\) that are not one.
Then \(a\) is said to be Archimedean with respect to \(b\) if there exists an \(N\in \mathbb {N}^+\) such that for \(n {\gt} N\), the inequality \(b {\lt} a^n\) holds if \(b\) is positive, and the inequality \(b {\gt} a^n\) holds if \(b\) is negative.
A left ordered semigroup \(S\) has large differences if for all \(a,b\in S\), the two following implications hold
if \(a\) is positive and \(a{\lt}b\), then there exists \(c\in S\) and \(n\in \mathbb {N}^+\) such that \(c\) is Archimedean with respect to \(a\) and \(a^n*c \le b^n\)
if \(a\) is negative and \(b {\lt} a\), then there exists \(c\in S\) and \(n\in \mathbb {N}^+\) such that \(c\) is Archimedean with respect to \(a\) and \(a^n*c \ge b^n\)
Let \(M\) be a linear ordered cancel commutative monoid without anomalous pairs and let \(G\) be its Grothendieck group. If \(M\) has a positive (negative) element, then for any element \(\frac{a}{b} \in G\) where \(a,b \in M\), there exist \(a', b' \in M\) such that \(a'\) and \(b'\) are positive (negative) and \(\frac{a}{b} = \frac{a'}{b'}\).
Let \(S\) be a linear ordered cancel semigroup without anomalous pairs such that there exists a positive element of \(S\). Then for all \(x \in S\), there exists a \(y \in S\) such that \(y\) is positive and \(x*y\) is positive.
Similarly, if there exists a negative element of \(S\) then for all \(x \in S\) there exists a \(y \in S\) such that \(y\) is negative and \(x*y\) is negative.