Talk:Reciprocal Function is Strictly Decreasing/Proof 1

This is mis-stated.

The reciprocal is strictly decreasing on $(-\infty..0)$ and on $(0..\infty)$ but it is not strictly decreasing on $(-\infty..0) \cup (0..\infty)$.

Both $-1$ and $1$ are in $(-\infty..0) \cup (0..\infty)$ but $\frac 1 {-1} < \frac 1 1$. --prime mover (talk) 22:18, 24 December 2012 (UTC)


 * Oh, that's why I had it the way it was before. I didn't remember why I did that... --GFauxPas (talk) 22:21, 24 December 2012 (UTC)


 * In the exposition it is also a very good idea to specify in as much detail as required exactly what domains it is decreasing - because at the moment the very statement of the theorem is wrong. --prime mover (talk) 22:25, 24 December 2012 (UTC)


 * Do you mean even on the main page, or just the part that's transcluded? --GFauxPas (talk) 22:26, 24 December 2012 (UTC)


 * Oh it's not a transclusion, nm --GFauxPas (talk) 22:26, 24 December 2012 (UTC)


 * I still think this does not go far enough. I think that in order to make sure that the same mistake isn't made again (it was made back in August first time) some words are actually put in place to explain why it does not hold on $(-\infty..0) \cup (0..\infty)$. Sorry but this is important. Basic, yes, but important. I'll leave it to you. --prime mover (talk) 22:32, 24 December 2012 (UTC)


 * Under a "warning" header, or on its own page as a counterexample, or...? --GFauxPas (talk) 22:48, 24 December 2012 (UTC)


 * Don't know. Put it up as a "warning" for now, it can be refactored later if we want. Important thing is it's in there. --prime mover (talk) 23:05, 24 December 2012 (UTC)