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The reals mod the integers are an abelian group


Let G={xR | 0x<1} and for x,yG let xy be the fractional part of x+y, i.e., xy=x+yx+y. Prove that is a well-defined binary operation on G and that G is an abelian group under .


Solution: Before we begin, note that x+n=x+n for all xR and nZ as follows. For all integers k, we have kx+n iff kx+n iff knx iff kx+n.

We first show that is well defined. Suppose x,yG. There are two cases for x+y. If x+y=0, then we have 0x+y<1 and so xyG. If x+y=1, we have x+y1. But since x,y<1, x+y<2. Thus xyG. So is indeed a binary operator on G.

We now show that is associative. To that end, let x,y,zG. Then we have the following. (xy)z= (xy)+z(xy)+z= x+yx+y+zx+yx+y+z= x+yx+y+zx+y+z+x+y= x+y+zx+y+z= x+yy+z+zx+y+z+y+z= x+y+zy+zx+y+zy+z= x+(yz)x+(yz)= x(yz). We now show that 0 is the identity element of G under . If xG, then x=0. Thus we have x0=x+0x+0=x and 0x=0+x0+x=x. We now show that every element has an inverse. If x0, then 1xG. Then x(1x)=x+(1x)x+(1x)=11=0 and similarly (1x)x=(1x)+x(1x)+x=11=0. Certainly, if x=0 then 00=0. Thus G is a group under . It is also clear that xy=yx. Hence G is an abelian group under .

Linearity

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