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Math Help - Vector space and 10 axioms related question

  1. #1
    Senior Member x3bnm's Avatar
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    Vector space and 10 axioms related question

    I'm stuck at a problem related to vector space detection. I know how to solve it but don't understand a particular information.


    We know that if a set of objects follow 10 axioms of vector space than that defines a vector space.

    Say we have a problem:

    The set of all positive real numbers with operations
    <br />
x+y = xy\,\,\,and\,\,\, k.x = x^k <br />
    And asked to verify if the set is vector space or not.



    Now one of the axioms to be satisfied is:
    For each u in V, there is an object -u in V, called a negative of u, such that
    <br />
u + (-u) = (-u) + u = 0<br />

    For the above problem an object u = x then -u = \frac{1}{x}
    So, u + (-u) = x.\frac{1}{x} = 1 = \mathbf0 = 1 = \frac{1}{x}.x = -u + u
    Now my question: why's 1 = 0 How's that possible? In what way 1 = 0 ?

    Shouldn't we say that this set is not a vector space because it violates one of the axioms?

    I can't seem to find the answer. Can anyone help me on this?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by x3bnm View Post
    I'm stuck at a problem related to vector space detection. I know how to solve it but don't understand a particular information.


    We know that if a set of objects follow 10 axioms of vector space than that defines a vector space.

    Say we have a problem:

    The set of all positive real numbers with operations
    <br />
x+y = xy\,\,\,and\,\,\, k.x = x^k <br />
    And asked to verify if the set is vector space or not.



    Now one of the axioms to be satisfied is:
    For each u in V, there is an object -u in V, called a negative of u, such that
    <br />
u + (-u) = (-u) + u = 0<br />

    For the above problem an object u = x then -u = \frac{1}{x}
    So, u + (-u) = x.\frac{1}{x} = 1 = \mathbf0 = 1 = \frac{1}{x}.x = -u + u
    Now my question: why's 1 = 0 How's that possible? In what way 1 = 0 ?

    Shouldn't we say that this set is not a vector space because it violates one of the axioms?

    I can't seem to find the answer. Can anyone help me on this?
    x+0=x(0)=0=(0)x=0+x

    -x=0

    I think you can do this.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by x3bnm View Post
    I'm stuck at a problem related to vector space detection. I know how to solve it but don't understand a particular information.


    We know that if a set of objects follow 10 axioms of vector space than that defines a vector space.

    Say we have a problem:

    The set of all positive real numbers with operations
    <br />
x+y = xy\,\,\,and\,\,\, k.x = x^k <br />
    And asked to verify if the set is vector space or not.



    Now one of the axioms to be satisfied is:
    For each u in V, there is an object -u in V, called a negative of u, such that
    <br />
u + (-u) = (-u) + u = 0<br />

    For the above problem an object u = x then -u = \frac{1}{x}
    So, u + (-u) = x.\frac{1}{x} = 1 = \mathbf0 = 1 = \frac{1}{x}.x = -u + u
    Now my question: why's 1 = 0 How's that possible? In what way 1 = 0 ?

    Shouldn't we say that this set is not a vector space because it violates one of the axioms?

    I can't seem to find the answer. Can anyone help me on this?
    x + y = xy
    so we know that u + 1 = u(1) = u.

    Thus 1 is the zero element for addition. Not to be confused by 0 which is the usual zero element for addition in the real numbers. We can verify this with another problem:
    k*x = x^k. So -u = (-1)u = 1/u. Thus we have u + (-u) = u*1/u = 1 as required.

    -Dan
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  4. #4
    Senior Member x3bnm's Avatar
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    Thank you dwsmith and topsquark for taking the time explaining this.

    I understand both of your solutions. Again thanks a lot.
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