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Math Help - [SOLVED] Group Theory Help Please!!!

  1. #1
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    [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 ????

    thanks a million in advance

    chave
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  2. #2
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    Quote Originally Posted by chave View Post
    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 ????

    thanks a million in advance

    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.
    Last edited by chabmgph; February 8th 2009 at 10:18 AM. Reason: Typo
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  3. #3
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    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????
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  4. #4
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    Quote Originally Posted by chave View Post
    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
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  5. #5
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    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.
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  6. #6
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    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.
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  7. #7
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    thanks a lot HallsofIvy

    totally get it now
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