1. ## [SOLVED] Group Theory Help Please!!!

Hi,

Just a quick group theory question . . .

Q Let G := { x "element of" R : 0 <= x <1} and for x,y "element of" G.
Let x*y be the fractional part of x+y ie ( x*y = x+y - [x+y] where [n] is the greatest integer less than or equal to n. Show that * is a binary operation on G and G is abelian group under *.

Im just not sure what y can be , x has to be a real between 0 and 1, but it doesnt say about y only that is a part of G. also i assume the identity is 0???

and the inverse of x ????

chave

2. Originally Posted by chave
Hi,

Just a quick group theory question . . .

Q Let G := { x "element of" R : 0 <= x <1} and for x,y "element of" G.
Let x*y be the fractional part of x+y ie ( x*y = x+y - [x+y] where [n] is the greatest integer less than or equal to n. Show that * is a binary operation on G and G is abelian group under *.

Im just not sure what y can be , x has to be a real between 0 and 1, but it doesnt say about y only that is a part of G. also i assume the identity is 0???

and the inverse of x ????

chave
If y is an element in G, then $0 \leq y \leq 1$. That is what it takes to be an element of G.

EDIT: Should have been $0 \leq y < 1$. My bad.

3. thanks and the inverse of x would be??? -x seems the only thing that would work with the operation.

any other suggestions for the inverse of x????

4. Originally Posted by chave
thanks and the inverse of x would be??? -x seems the only thing that would work with the operation.

any other suggestions for the inverse of x????
What is the identity element $e$ you found?

$x^{-1}$ has to satisfy the property of $x*x^{-1}=e$

5. i assume it was zero because for

x*e = x = e*x

0 satisfies this but i cant get an inverse of x that satisfies

x*inv(x) = e

which makes me think i dont have the correct e or havent found the inverse yet???

thats where im stuck. what do you think? -x would work but it doesnt fall within 0 <= x < 1.

6. If x and y are both less than 1, then x+ y must be less than 2 and [x+y] is either 0 or 1. Basically x*y is "sum modulo 1": if x+ y itself is less than 1 then [x+y]= 0 and x*y= x+y while if x+ y is larger than 1 [x+y]= 1 and x*y= x+y- 1, the "fractional part" of x+y.

If e is the identity for this operation, then x*e= x+ e- [x+ e]= x which means we must have e= [x+e]. Yes, e= 0 works! Now, if x*y= x+y- [x+y]=0 then x+y= [x+y] which means either x+y= 0 or x+y= 1. Since both x and y are positive, the first is impossible and we must have y= x-1= 1- x.

7. thanks a lot HallsofIvy

totally get it now