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## Homework Statement

Show that [tex] G = \left\{x | -1 < x < 1 \right\} [/tex] with the relation

[tex] a\ast b = \frac {a+b}{ab +1}[/tex]

## Homework Equations

1) Definition of a group.

A group is a set G equipped with an operation * and a special element e in G, callled the identity, such that

i) for every a,b,c in G

[tex]a\ast( b \ast c) = (a \ast b})\ast c [/tex];

ii) [tex]e\ast a =a [/tex] for all a in G

iii) for all a in G, there is a' in G with [tex] a' \ast a =e[/tex]

2) Definition of a group number 2

A group is a nonempty set G equipped with an operation that satisfies the following axioms:

1. if a,b are in , then [tex] a \ast b[/tex] is in G

2.for every a,b,c in G

[tex]a\ast( b \ast c) = (a \ast b})\ast c [/tex];

3. [tex]e\ast a =a [/tex] for all a in G

4. for all a in G, there is a' in G with [tex] a' \ast a =e[/tex]

## The Attempt at a Solution

I already showed that G is a group using the first definition of a group but the first axiom of the second definition is difficult to show.

The second definition is a stronger definition which leads me to think that the first definition is not correct.

Which of these definitions is the correct one ?

Also does anyone know any tricks for showing that

[tex] -1 < \frac {a+b}{ab +1} < 1 [/tex] for a,b in G ?

Basically, how to satisfy axiom 1 of the second definition.

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