Talk:Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element

Doesn't the proof admit a free strengthening to strict inequality? --Lord_Farin (talk) 19:17, 3 January 2013 (UTC)

...that is, for $n > 1$. Might take a course on "thorough reading"; I've heard good things about it. --Lord_Farin (talk) 19:19, 3 January 2013 (UTC)


 * Yes, we want that one too. Even better if the proofs don't duplicate terribly much. --Dfeuer (talk) 19:22, 3 January 2013 (UTC)


 * In fact, I'd be okay with having that one instead, if you can come up with a sane name for it. --Dfeuer (talk) 22:28, 3 January 2013 (UTC)


 * Maybe simply "Strictly Positive Power is Inflationary Mapping", which although not specifying the domain for which this holds, is a nice bit shorter than the current version. Result seems to apply for positive elements in an ordered monoid. --Lord_Farin (talk) 22:34, 3 January 2013 (UTC)


 * Strictly increasing isn't strong enough for this purpose. I proved something related to that elsewhere recently, and you're welcome to generalize. But for this it needs to exceed each element at some point. The purpose is to apply the intermediate value theorem to prove Existence and Uniqueness of Positive Root of Positive Real Number. Dfeuer (talk) 22:46, 3 January 2013 (UTC)


 * Ah yes, it was Product of Positive Strictly Increasing Mappings is Strictly Increasing --Dfeuer (talk) 22:52, 3 January 2013 (UTC)


 * Are you sure we're discussing the same thing? --Lord_Farin (talk) 22:54, 3 January 2013 (UTC)


 * Yes? The identity function on an ordered set is always strictly increasing. Restricted to positive numbers, it's also positive. Since the product of positive, strictly increasing functions is strictly increasing, x*x*x*...*x is strictly increasing. The point of *this* theorem is to show that every element of the ring has a successor in the range of the power function. --Dfeuer (talk) 23:00, 3 January 2013 (UTC)

I misunderstood your earlier comment, then. But that $\circ^n x \succeq x$ is precisely that $\circ^n$ is increasing... --Lord_Farin (talk) 23:05, 3 January 2013 (UTC)


 * I finally understand you. You mean increasing as $n$ increases! Yes, and in fact we can put that in strict terms, which is better! Then that proves that $\circ^n x > 1$, which is effectively what I need to prove. Dfeuer (talk) 23:09, 3 January 2013 (UTC)


 * I actually meant that the mapping $x \mapsto \circ^n x$ is increasing on $\uparrow(1)$ (upper closure of $1$, be it weak or strict). Your desired inequality follows from transitivity. --Lord_Farin (talk) 23:14, 3 January 2013 (UTC)


 * Ummmm what? I still don't see how anything about it increasing as $x$ increases has to do with it. Can you be more explicit? --Dfeuer (talk) 23:24, 3 January 2013 (UTC)


 * Apologies. I meant Definition:Inflationary Mapping throughout. --Lord_Farin (talk) 23:26, 3 January 2013 (UTC)


 * Yes, that clarifies matters greatly! I wasn't familiar with that term. You say you think it holds for any ordered monoid, but I don't think we have a definition of ordered monoid. --Dfeuer (talk) 23:34, 3 January 2013 (UTC)

Definition:Ordered Semigroup with an identity element :). Just tedious, not hard. --Lord_Farin (talk) 23:35, 3 January 2013 (UTC)

The fact that it strictly increases with $n$ seems stronger, and potentially useful. I don't know if that is usually proved on its own or as a corollary of something more important. --Dfeuer (talk) 23:42, 3 January 2013 (UTC)