corollaries of basic theorem on ordered groups
Corollary 1 Let be an ordered group. For all , either or .
Corollary 2 Let be an ordered group and a strictly positive integer. Then, for all , we have if and only if .
Proof: We shall first prove that implies by induction. If , this is a simple tautology. Assume the conclusion is true for a certain value of . Then, conclusion 4 allows us to multiply the inequalities and to obtain .
As for the proof that implies , we shall prove the contrapositive statement. Assume that is false. By conclusion 1, it follows that either or . If , then so, by conclusion 1 is false. If then, by what we have already shown, so is also false in this case for the same reason.
Corollary 3 Let be an ordered group and a strictly positive integer. Then, for all , we have if and only if .
Proof: It is trivial that, if , then . Assume that . By conclusion 1 of the main theorem, it is the case that either or or . If then, by the preceding corollary, , which is not possible. Likewise, if , then we would have , which is also impossible. The only remaining possibility is .
Corollary 4 An ordered group cannot contain any elements of finite order.
Let be an element of an ordered group distinct from the identity. By definition, if were of finite order, there would exist an integer such that . Since , we would have but, by Corollary 3, this would imply , which contradicts our hypothesis.
|Title||corollaries of basic theorem on ordered groups|
|Date of creation||2013-03-22 14:55:12|
|Last modified on||2013-03-22 14:55:12|
|Last modified by||rspuzio (6075)|