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Math Help - (\mathbb{Q}, +) by |\frac{a}{b}| < 1 a,b\in\mathbb{Z} and b\neq 0 is this a group

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    (\mathbb{Q}, +) by |\frac{a}{b}| < 1 a,b\in\mathbb{Z} and b\neq 0 is this a group

    (\mathbb{Q}, +) by \left |\frac{a}{b}\right | < 1, \ a,b\in\mathbb{Z} and b\neq 0 is this a group.

    Identity is 0.
    Associative is true.

    Inverse (I am not sure but I don't think this works).
    If we let  -\frac{a}{b} be the inverse, we have a problem since
    \displaystyle\left |\frac{a}{b}\right |=\begin{cases}\frac{a}{b}, \ \frac{a}{b}\geq 0\\-\frac{a}{b}, \ \frac{a}{b}< 0\end{cases}

    So is this not a group or is this approach to the inverse wrong?
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    MHF Contributor Drexel28's Avatar
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    Re: (\mathbb{Q}, +) by |\frac{a}{b}| < 1 a,b\in\mathbb{Z} and b\neq 0 is this a group

    Quote Originally Posted by dwsmith View Post
    (\mathbb{Q}, +) by \left |\frac{a}{b}\right | < 1, \ a,b\in\mathbb{Z} and b\neq 0 is this a group.

    Identity is 0.
    Associative is true.

    Inverse (I am not sure but I don't think this works).
    If we let  -\frac{a}{b} be the inverse, we have a problem since
    \displaystyle\left |\frac{a}{b}\right |=\begin{cases}\frac{a}{b}, \ \frac{a}{b}\geq 0\\-\frac{a}{b}, \ \frac{a}{b}< 0\end{cases}

    So is this not a group or is this approach to the inverse wrong?
    Are you asking whether the set A=\mathbb{Q}\cap[-1,1] is a subgroup of \mathbb{Q}? Obviously not since .5+.5=1.
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    Re: (\mathbb{Q}, +) by |\frac{a}{b}| < 1 a,b\in\mathbb{Z} and b\neq 0 is this a group

    Quote Originally Posted by Drexel28 View Post
    Are you asking whether the set A=\mathbb{Q}\cap[-1,1] is a subgroup of \mathbb{Q}? Obviously not since .5+.5=1.
    I am trying to prove if it is a group or not.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: (\mathbb{Q}, +) by |\frac{a}{b}| < 1 a,b\in\mathbb{Z} and b\neq 0 is this a group

    Quote Originally Posted by dwsmith View Post
    I am trying to prove if it is a group or not.
    And, it's not since it isn't closed under addition.
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