x=3
3*y=1/2(3y-3-y+3)=y
1/2*2y=y adevarat
y=3
x*3=1/2*(3x-x-3+3)=x
1/2*2x=x adevarat
3*x=x*3=x 3 element neutru
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x=2 Se cere xo a.i. 2*xo=3
2*xo=1/2*(2xo-2-xo+3)=3
1/2*(xo+1)=3 xo+1=6 xo=5
xo*2=1/2(2x0-xo-2+3)=3 1/2(xo+1)=3 xo=5
elementul opus lui 2 este 2
c) fie x=2k+1 si y=2n+1
x*y=1/2[(2k+1)*(2n+1)-2k-1-2n-1+3]=1/2(4kn+2n+2k+1-2k-1-2n-1+3)=
1/2(4kn+2)=2kn+1=2K+1∈H unde K=kn
Deci h este parte stabila
