Show the corespondence (x,y)->xoy is composition in set G = (-1,1)

xoy=(x+y)/(1+xy)

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You should start by considering the following inequalities, such that:

`x in G => x in (-1,1) => -1 < x < 1`

`y in G => y in (-1,1) => -1 < y < 1`

Adding `x + y` yields:

`-1 - 1 < x + y < 1 + 1 => -2 < x + y < 2`

You should perform the following multiplication, such that:

`-1< x*y < 1 => x*y in (-1,1) => x*y in G`

You need to add 1 to all inequality, such that:

`1 - 1 < 1 + x*y < 1+1 => 0 < 1 + x*y < 2 => 0 < (1 + x*y)/2 < 1`

You need to divide the inequality `-2 < x + y < 2` by the positive amount `(1 + x*y)/2 in (0,1)` such that:

`-2/2 < (x + y)/(1 + xy) < 2/2 => -1 < (x + y)/(1 + xy) < 1 => (x + y)/(1 + xy) in G => xoy in G`

**Hence, evaluating if the law of composition `xoy = (x + y)/(1 + xy)` is valid for the set `G = (-1,1)` yields that `xoy in (-1,1)` holds for `x,y in (-1,1)` .**

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