document.write( "Question 845451: define a binary * on R*=R without 0 by
\n" ); document.write( "a*b =ab if > 0 and a/b if a<0.
\n" ); document.write( "Determine whether R* is a group.
\n" ); document.write( " attempt:
\n" ); document.write( "it is closed
\n" ); document.write( "it has identity 1
\n" ); document.write( "but inverse i got two different inverses and i cannot do the associative. \n" ); document.write( "

Algebra.Com's Answer #509453 by richard1234(7193) $\"\"$ $\"About$

You can put this solution on YOUR website!
If

0\">, then

since (1/a)*a = a*(1/a) = 1. If

, then

. In each of these cases, the inverse is unique.\r
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\n" ); document.write( "\n" ); document.write( "To show

for all

, we have four cases:\r
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0 \Rightarrow a*(b*c) = (a*b)*c = abc\">
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0, b < 0 \Rightarrow a*(b*c) = (a*b)*c = \frac{ab}{c}\">
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0 \Rightarrow a*(b*c) = (a*b)*c = \frac{a}{bc}\">
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\r
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\n" ); document.write( "\n" ); document.write( "Associativity holds in all four cases, so the group operation is associative. Hence (R, *) is a group. \n" ); document.write( "

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