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 (Strictly) Increasing Function", 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 of Root. Dfeuer (talk) 22:46, 3 January 2013 (UTC)


 * Ah yes, it was Product of Positive Strictly Increasing Functions 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)