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

$\displaystyle (\mathbb{Q}, +)$ by $\displaystyle \left |\frac{a}{b}\right | < 1, \ a,b\in\mathbb{Z}$ and $\displaystyle b\neq 0$ 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 $\displaystyle -\frac{a}{b}$ be the inverse, we have a problem since

$\displaystyle \displaystyle\left |\frac{a}{b}\right |=\begin{cases}\frac{a}{b}, \ \frac{a}{b}\geq 0\\-\frac{a}{b}, \ \frac{a}{b}< 0\end{cases}$

So is this not a group or is this approach to the inverse wrong?

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

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Originally Posted by

**dwsmith** $\displaystyle (\mathbb{Q}, +)$ by $\displaystyle \left |\frac{a}{b}\right | < 1, \ a,b\in\mathbb{Z}$ and $\displaystyle b\neq 0$ 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 $\displaystyle -\frac{a}{b}$ be the inverse, we have a problem since

$\displaystyle \displaystyle\left |\frac{a}{b}\right |=\begin{cases}\frac{a}{b}, \ \frac{a}{b}\geq 0\\-\frac{a}{b}, \ \frac{a}{b}< 0\end{cases}$

So is this not a group or is this approach to the inverse wrong?

Are you asking whether the set $\displaystyle A=\mathbb{Q}\cap[-1,1]$ is a subgroup of $\displaystyle \mathbb{Q}$? Obviously not since $\displaystyle .5+.5=1$.

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** Are you asking whether the set $\displaystyle A=\mathbb{Q}\cap[-1,1]$ is a subgroup of $\displaystyle \mathbb{Q}$? Obviously not since $\displaystyle .5+.5=1$.

I am trying to prove if it is a group or not.

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** I am trying to prove if it is a group or not.

And, it's not since it isn't closed under addition.