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Math Help - Groups - discrete math

  1. #1
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    Groups - discrete math

    In G={x∈R | x>1} is defined operation * with x*y = xy-x-y+2. Show that {G,*} is Albelian group
    thanks
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  2. #2
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    Re: Groups - discrete math

    Quote Originally Posted by kljoki View Post
    In G={x∈R | x>1} is defined operation * with x*y = xy-x-y+2. Show that {G,*} is Albelian group thanks
    What is the identity for this operation?
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  3. #3
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    Re: Groups - discrete math

    we need to show 4 things:

    1. * is an associative binary operation on G. this means two things:

    a) x*y has to be in G
    b) for any 3 elements x,y,z of G, (x*y)*z = x*(y*z)

    2. there is some element e in G, with x*e = e*x, for ALL x in G.

    3. for any x in G, there is some element y in G with x*y = y*x = e.

    4. for any x,y in G, x*y = y*x.

    so, let's look at #1 first. suppose x,y are in G. this means x,y are both real numbers greater than 1. we need to see if x*y is also a real number greater than 1.

    well, x*y = xy - x - y + 2, and xy - x - y + 2 is certainly a real number. but is it greater than 1?

    xy - x - y + 2 = (x - 1)(y - 1) + 1.

    now x > 1, so x - 1 > 0.

    similarly, y > 1, so y - 1 > 0. therefore (x - 1)(y - 1) > 0, so (x - 1)(y - 1) + 1 > 0 + 1 = 1. so, yes, x*y > 1. this show closure, so * is indeed a binary operation on G.

    so now we can turn our attention to whether or not * is associative:

    (x*y)*z = (xy - x - y + 2)*z = (xy - x - y + 2)(z) - (xy - x - y + 2) - z + 2

    = xyz - xz - yz + 2z - xy + x + y - 2 - z + 2

    = xyz - xz - yz - xy + x + y + z (2z - z = z, and -2 + 2 = 0)

    = x(yz - y - z) - (yz - y - z) + x

    = x(yz - y - z) - (yz - y - z) + 2x - x - 2 + 2

    = x(yz - y - z + 2) - x - (yz - y - z + 2) + 2

    = x*(yz - y - z + 2) = x*(y*z).

    so * is indeed associative. what do you suppose e and x-inverse might be?
    Thanks from kljoki
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    Re: Groups - discrete math

    Oh how I wish we could allow the student to show some effort.
    Giving a premature solution in my opinion is counterproductive.
    Thanks from emakarov
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    Re: Groups - discrete math

    x*e=(xe - x - e + 2) (xe = x)
    =x - x - e + 2 = -e + 2 + x - x = -e + 2 + ex - x = ex - e - x + 2 = e*x ??????
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    Re: Groups - discrete math

    Quote Originally Posted by kljoki View Post
    x*e=(xe - x - e + 2) (xe = x)
    =x - x - e + 2 = -e + 2 + x - x = -e + 2 + ex - x = ex - e - x + 2 = e*x ??????
    You are making my points!
    Do you even know what the identity for this operation is?
    If your answer is no, then you have no idea what the question is about.
    Am I wrong?
    If not, please explain!
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    Re: Groups - discrete math

    Quote Originally Posted by kljoki View Post
    x*e=(xe - x - e + 2) (xe = x)
    =x - x - e + 2 = -e + 2 + x - x = -e + 2 + ex - x = ex - e - x + 2 = e*x ??????
    The important part of the identity group axiom is not that x * e = e * x, but that both of those expressions equal x. Also, note that x * e should equal x, but xe (with regular multiplication) is not necessarily x.

    We have that x * e is a polynomial of x. Use the fact that two polynomials are equal for all values of x iff their corresponding coefficients are equal. From there you can find what e is.
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    Re: Groups - discrete math

    Quote Originally Posted by Plato View Post
    You are making my points!
    Do you even know what the identity for this operation is?
    If your answer is no, then you have no idea what the question is about.
    Am I wrong?
    If not, please explain!
    I don't have idea, only this was given to the exam :/
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  9. #9
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    Re: Groups - discrete math

    we are looking for some real number e > 1, with the property that:

    x*e = e*x = x for any real number x > 1. you should prove that x*y = y*x yourself, it's not hard, and you might learn something. but if we take that as given, we only need to find some e with x*e = x (since then e*x = x*e anyway, so will have to equal x).

    to prove this, we need to use the definition of *.

    x*e = xe - x - e + 2.

    we want this to equal x, so we set it equal to x, and try to solve for e:

    xe - x - e + 2 = x

    now, solve for e, using ordinary algebra. if, for some strange reason, you need to divide by x - 1 at some point, rest assured that this is ok, since x > 1, x - 1 > 0, so in particular it never equals 0 (so we can divide by it).
    Thanks from kljoki
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    Re: Groups - discrete math

    So x*e = e*x = x
    xe - x - e + 2 = x
    xe - e = 2x -2
    e(x - 1) = 2x - 2 => e = (2x - 2)/(x - 1)

    and x*y = y*x = e
    x*y = e
    xy - x - y + 2 = e
    xy - y = e + x - 2
    y(x - 1) = e + x - 2 => y = (e + x - 2)/(x - 1)

    and what about x*y=y*x ???
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  11. #11
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    Re: Groups - discrete math

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

    you're fine up to here.

    but.....


    2x-2 = 2(x-1)

    do you think you might be able to simplify a little?
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  12. #12
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    Re: Groups - discrete math

    ups ))
    so e=2
    and should i leave y in next proof like this y = (e + x - 2)/(x - 1) or change for e=2 so y = x/(x - 1)
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  13. #13
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    Re: Groups - discrete math

    now start with:

    x*y = 2 that is:

    xy - x - y + 2 = 2

    and solve for y in terms of x.

    you should arrive at y = x-1 = x/(x-1) easily. note the x-1 in the denominator is ok, since x > 1. there *is* one thing you need to be sure of, however.

    you need to check that x/(x-1) is *in* G, that is: that x/(x-1) > 1, when x > 1. first check that top and bottom are positive, then "cross-multiply".

    then check your answer by computing:

    x*(x/(x-1)). if you've done everything right up to here, you should get 2.
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