My Math Forum What is \mathbb{R}-\{-1\}?

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 January 23rd, 2013, 08:52 PM #1 Senior Member   Joined: Sep 2009 Posts: 251 Thanks: 0 What is \mathbb{R}-\{-1\}? I have been asked to help a friend with his homework. He has questions, but no textbook, so I can't look up what symbols mean. The question is Show that $\mathbb{R}-\{-1\}$ is a group under the operation $x*y=x+y+xy$ What is $\mathbb{R}-\{-1\}$? Is it all reals except for negative one? If yes, how on earth do I prove that it's a group? Thanks.
 January 23rd, 2013, 09:25 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: What is \mathbb{R}-\{-1\}? That is a standard interpretation of the symbols. You prove something is a group by... Wait for it... The DEFINITION! Closed under operation "*" Has identity Has inverse Associative
January 23rd, 2013, 10:11 PM   #3
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Re: What is \mathbb{R}-\{-1\}?

Quote:
 Originally Posted by The Chaz ...You prove something is a group by... Wait for it... The DEFINITION! ...
I knew THAT. I wasn't sure how to do those steps. But it was easy once you confirmed the definition.

 January 23rd, 2013, 10:13 PM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: What is \mathbb{R}-\{-1\}? So what's the inverse of element "c"?
January 23rd, 2013, 10:29 PM   #5
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Re: What is \mathbb{R}-\{-1\}?

Quote:
 Originally Posted by The Chaz So what's the inverse of element "c"?
$c^{-1}=\frac{-c}{1+c}$

 January 23rd, 2013, 11:53 PM #6 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: What is \mathbb{R}-\{-1\}? in fact, ?:R-{0}--->R-{1} given by ?(x) = x-1 is a group isomorphism, since: ?(xy) = xy - 1 = xy + x - x + y - y - 1 = xy - x - y + 1 + (x-1) + (y-1) = (x-1)(y-1) + (x-1) + (y-1) = (x-1)*(y-1) = ?(x)*?(y). this gives us a way to find c^-1: it is: ?(1/(?^-1(c))) = ?(1/(c+1)) = 1/(c+1) - 1

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