# Question on which set to use in showing something is a group

• Apr 17th 2011, 08:03 AM
usagi_killer
Question on which set to use in showing something is a group
Hi guys, I just have some questions on this problem regarding some details, any clarification would be greatly appreciated!

http://img714.imageshack.us/img714/299/problemty.jpg

Cheers!
• Apr 17th 2011, 08:27 AM
abhishekkgp
Quote:

Originally Posted by usagi_killer
Hi guys, I just have some questions on this problem regarding some details, any clarification would be greatly appreciated!

http://img714.imageshack.us/img714/299/problemty.jpg

Cheers!

if we are set out to see whether <R,*> is a group or not then a*0 remains undefined since '*' was defines only on R\{0} . so one can't be sure that <R,*> is a binary structure to begin with. So i would say that one can't conclude that <R,*> is a group if have * as defined in the question.
• Apr 17th 2011, 09:12 AM
Deveno
the author of the book should be publicly flogged for posing such an ill-formed question.

as posed, <R,*> is not a group, because * has not been defined on all of R. but this is probably not the question the author meant to pose.

he probably meant to ask if <R-{0},*> was a group.
• Apr 17th 2011, 10:06 AM
usagi_killer
thanks for that!

I thought it might've been a typo as well... i will email my lecturer who set this question (Itwasntme)
• Apr 18th 2011, 12:17 AM
Swlabr
Quote:

Originally Posted by usagi_killer
Hi guys, I just have some questions on this problem regarding some details, any clarification would be greatly appreciated!

http://img714.imageshack.us/img714/299/problemty.jpg

Cheers!

You say e/|e| is the identity element since x cannot be 0 and thus e is defined...however, it is not defined uniquely. An identity element is always unique (if, say, x and y are identity elements then xy=x and xy=y, so x=y), so something is amiss.

Remember that e is a real number, so try and place it on the number line. e=...? This will help you to see what is amiss...