Talk:Reciprocal Function is Strictly Decreasing

Okay, so maybe I didn't word this too well when I raised the issue ... but as it stands the statement is fallacious. It's s.d. on either $\R_{>0}$ or $\R_{<0}$ but not on $\R_{\ne 0} = \R \setminus \left\{{0}\right\}$ because e.g. $\dfrac 1 {-1} < \dfrac 1 1$. Sorry, I kinda assumed that would be taken care of. --prime mover 16:49, 3 August 2012 (UTC)


 * ... and yes I know, Ordering of Reciprocals is wrong too. Needs to be looked at. --prime mover 16:52, 3 August 2012 (UTC)


 * Oh, I was only thinking $x, y > 0$ or $x, y <0$, not one of each. My b. To fix the proof I'd have to prove it both for positive and negative but I wouldn't know how to prove that the union both "sides" of the decreasing function is decreasing. Probably better to just scrap the second proof. --GFauxPas 22:23, 4 August 2012 (UTC)


 * Hint: the point is that $\dfrac 1 x$ is not continuous. It's discontinuous at $0$. See how you get on. --prime mover 05:33, 5 August 2012 (UTC)