# Given the set G=(0,1) and the law of composition x*y=xy/(2xy-x-y+1), Verify if e=1/2 is the neutral element of the group G.

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### 2 Answers

The neutral element for the law of composition given by x*y = xy/ (2xy - x - y +1) is one for which any element x gives x*(1/2) = x

x *(1/2) = (1/2)x/(2(1/2)x - (1/2) - x + 1

=> (x/2)/(x - (1/2) - x + 1)

=> (x/2)(1/2)

=> 2*x / 2

=> x

Also the law of composition is symmetric with respect to x and y.

**As x*(1/2) = x, we have (1/2) as the neutral element for the law of composition x*y = xy/ (2xy - x - y +1).**

The property of a neutral element of a group is:

e*g = g*e = g, where g is an element of the set G.

We'll apply the composition law and we'll replace e by 1/2:

x*(1/2) = x

We'll replace x*(1/2) by x/2/(2x/2 - x - 1/2 + 1)

We'll compute the expression to check if it yields x.

x/2/(2x/2 - x - 1/2 + 1) = x/2(x - x - 1/2 + 1)

x/2(1 - 1/2) = x/(2/2) = x/1 = x

**We notice that the expression x*(1/2) = x, therefore e = 1/2 represents the neutral element of the given group, whose law of composition is x*y=xy/(2xy-x-y+1).**