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

1. (\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?

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

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 $\displaystyle .5+.5=1$.

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

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.

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

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.