SOLUTION: define a binary * on R*=R without 0 by
a*b =ab if > 0 and a/b if a<0.
Determine whether R* is a group.
attempt:
it is closed
it has identity 1
but inverse i got two differen
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-> SOLUTION: define a binary * on R*=R without 0 by
a*b =ab if > 0 and a/b if a<0.
Determine whether R* is a group.
attempt:
it is closed
it has identity 1
but inverse i got two differen
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Question 845451: define a binary * on R*=R without 0 by
a*b =ab if > 0 and a/b if a<0.
Determine whether R* is a group.
attempt:
it is closed
it has identity 1
but inverse i got two different inverses and i cannot do the associative. Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! If , then since (1/a)*a = a*(1/a) = 1. If , then . In each of these cases, the inverse is unique.
To show for all , we have four cases:
Associativity holds in all four cases, so the group operation is associative. Hence (R, *) is a group.
