# *aa3.2 Let Q be the group of rational numbers under addition

• MHB
• karush
In $Q$,$\langle\frac{1}{2}\rangle+\langle\frac{1}{4}\rangle=\langle\frac{1}{2+\frac 12}{4}\rangle$In $Q$,$\langle\frac{1}{2}\rangle+\langle\frac{1}{4}\rangle=\langle\frac{1}{2+\frac 12}{4}\rangle$2f

#### karush

Gold Member
MHB
aa3.2

Let Q be the group of rational numbers under addition
and let $Q^∗$ be the group of
nonzero rational numbers under multiplication.
In $Q$, list the elements in $\langle\frac{1}{2} \rangle$,
In ${Q^∗}$ list elements in $\langle\frac{1}{2}\rangle$

ok just had time to post and clueless

Last edited:
Hi karush,

Can you provide some attempt, or at least some initial thoughts please?

Hi karush,

Can you provide some attempt, or at least some initial thoughts please?

In Q,
$\langle\frac{1}{2}\rangle =\left\{\cdots,-2,-\frac{3}{2},-1,-\frac{1}{2}, 0,\frac{1}{2},1,\frac{3}{2},2\cdots\right\} =\left\{\frac{n}{2}\vert n \in \Bbb{Z}\right\}$

In Q,
$\langle\frac{1}{2}\rangle =\left\{\cdots,-2,-\frac{3}{2},-1,-\frac{1}{2}, 0,\frac{1}{2},1,\frac{3}{2},2\cdots\right\} =\left\{\frac{n}{2}\vert n \in \Bbb{Z}\right\}$

Yep. That is correct. (Nod)

How did you find it?
Can you also find $\langle\frac{1}{2}\rangle$ in $\mathbb Q^*$?

Yep. That is correct. (Nod)

How did you find it?
Can you also find $\langle\frac{1}{2}\rangle$ in $\mathbb Q^*$?
In $Q*$
$\langle\frac{1}{2}\rangle =\left\{\cdots, 4,2,1,\frac{1}{2},\frac{1}{4}, \cdots\right\} =\left\{2^n\vert n \in \Bbb{Z}\right\}$
sorta!

$Q$ and $Q^\ast$ as above.
Find the order of each element in $Q$ and $Q^\ast$

ok is this like $4=2^2$

Last edited:
In $Q*$
$\langle\frac{1}{2}\rangle =\left\{\cdots, 4,2,1,\frac{1}{2},\frac{1}{4}, \cdots\right\} =\left\{2^n\vert n \in \Bbb{Z}\right\}$
sorta!

$Q$ and $Q^\ast$ as above.
Find the order of each element in $Q$ and $Q^\ast$

ok is this like $4=2^2$

Nope.
The order of an element is the number of times we have to add (respectively multiply) it before we get the identity element.
So in $\mathbb Q$, how many times do we have to add $\frac 12$ before we get $0$?
And how many times do we have to add $0$ before we get $0$?