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This question already has an answer here:

I need to show that the following is an abelian group:

$$x*y = \frac{x+y}{xy+1}$$ on the set $\{x \in \Bbb R \,|\, -1 < x < 1\}$.

I have been working on this problem, trying to show closure. I know that we need to show that $|x+y|<|xy+1|$ for all $x, y \in (-1,1)$. Can I assume that the max value that the expression yields is $1$ if we take $x=1$ and $y=1$? And the lowest value that is possible is when $x=-1$ and $y=1$? Or am I going about this the wrong way?

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marked as duplicate by André 3000, 6005, Mark Bennet, user147263, Adam Hughes Jun 22 '15 at 19:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I haven't found the thread I was thinking of, but here's a relevant link. $\endgroup$ – André 3000 Jun 22 '15 at 4:08
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Hint One can show directly that the operation $\ast$ is conjugate to the usual addition operation on $\Bbb R$ via the hyperbolic tangent function, which satisfies the suggestive identity $$\tanh (s + t) = \frac{\tanh s + \tanh t}{\tanh s \tanh t + 1}.$$

Alternate hint One can rearrange the desired inequality $x + y < xy + 1$ (note we've dropped the absolute value signs) as $$0 < xy - x - y + 1 = (1 - x)(1 - y).$$

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  • $\begingroup$ Could you expand on the alternative hint? I'd like to catch this but I couldn't follow. $\endgroup$ – YoTengoUnLCD Jun 22 '15 at 5:40
  • $\begingroup$ Sure: Since $|x|, |y| < 1$, we have $|xy| < 1$ and so $|xy + 1| = xy + 1$. Hence, when $x + y \geq 0$ we have $|x + y| < |xy + 1|$ as desired. To handle the case $x + y < 0$, we can replace $x$ and $y$ respectively with $-x$ and $-y$. $\endgroup$ – Travis Jun 22 '15 at 8:38

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