Talk:Group/Examples/inv x = 1 - x

The set $S$
In that book of algebra and many other, the example most similar to that one described in this page is the one in which $S=\{x\in\R : 0\le x <1\}$. I think this group it's usually called "the reals modulo 1".


 * Be that as it may, but that is not the group which is being defined in this page, or indeed in : $\S 26 \mu$. Maybe there's a mistake in the book, but it distinctly says:
 * $S = \left\{{x \in \R: 0 < x < 1}\right\}$ and so $0 \notin S$
 * The inverse of $x \in S$ is $1 - x$, NOT $1-x-\operatorname{floor}(1-x)$.
 * If this is a mistake in Clark, then fair enough, it needs to be announced as such. (This will have the positive effect of providing a source for confused students - like me - who can not make this example work.)
 * What we ought to do is start a new page called Real Numbers under Addition Modulo 1 form Group in which this concept is developed formally as a group (and I think some of the work has already been done there, somewhere in Analysis, possibly), perhaps linking from / to this result, with some sort of comment to the effect that it is suspected that Clark may have made a mistake and may have meant this group.--prime mover 01:59, 20 August 2011 (CDT)

I found the group you were talking about. Check the book of Alan Clark, the exercise just above this one it's the same group but with a tranlation $x\to x+1$ and a dilatation of $x\to 2x$--Dan232 16:18, 5 December 2011 (CST)


 * ...or a dilation of $x \to x/2$. That would put the numbers in the correct range, I'd still want to check it was actually a group still. I'd need to think about it, but not tonight. --prime mover 16:38, 5 December 2011 (CST)


 * The suggested shift does not constitute a group homomorphism and so it can't be the operation sought after. --Lord_Farin 22:40, 23 July 2012 (UTC)


 * One would say an error was made but to prove that no group exists with these properties is probably very hard. --Lord_Farin 22:59, 23 July 2012 (UTC)

The proof is up; some tidying and linking is still to be done. I think that it is also valuable to determine directly that the operation $\circ$ determines a group (the identity element and inverses are done, only associativity essentially remains). --Lord_Farin (talk) 19:38, 9 October 2012 (UTC)


 * Good job. --prime mover (talk) 22:04, 9 October 2012 (UTC)