**Definition 1:**
Let $(G,\leq)$ be a nonzero partially ordered Abelian group with order unit $u$. (Recall that $u\in G$ is a order unit if, for every $g\in G$, there exists $N\in\mathbb N$ such that $-Nu\leq g\leq Nu$.) For any $g\in G$, define the quantities (as it is done in proposition 4.7 of the book [Partially Ordered Abelian Groups (by K.R. Goodearl)](http://books.google.ca/books?id=dXUsGysPqRUC&dq=Partially+Ordered+Abelian+Groups+with+Interpolation&source=gbs_navlinks_s))
\begin{align\*}
p(g)=\sup\{k/m:k\in\mathbb Z;m\in\mathbb N;ku\leq mg\}
\end{align\*}
and
\begin{align\*}
r(g)=\inf\{l/n:l\in\mathbb Z,n\in\mathbb N,ng\leq lu\}.
\end{align\*}
**Question:**
It is clear that $-\infty<p(g)\leq r(g)<\infty$ for any $g\in G$. What would constitute an example of a group with an order unit $(G\_0,u\_0)$ and an element $g\_0\in G\_0$ such that $p(g\_0)<r(g\_0)$.


**Definition 2:**
We call a function $s:G\to\mathbb R$ a *state* on $G$ if $s$ is a group order-homomorphism such that $s(u)=1$. 


**Remark 1:**
It can be shown that for any $p(g)\leq q\leq r(g)$,
there exists a state $s$ on $G$ such that $s(g)=q$. Therefore, there cannot exist a unique state on $(G\_0,s\_0)$ (as is the case for totally ordered Abelian groups, so any example involving $\mathbb R$ with its usual order is out of the question).


**Remark 2:**
It can be shown that, for any state $s$ on $G$ and element $g\in G$,
we have that $p(g)\leq s(g)\leq r(g)$. Therefore, it is enough to find a group with an order unit $(G\_0,u\_0)$, an element $g\_0\in G\_0$ and two states $s\_0$ and $s\_0'$ on $G\_0$ such that $s\_0(g)\neq s\_0'(g)$.



Example involving partially ordered Abelian groups
Let $G=\mathbb{R}^2$ with order defined by 


$(a\_1,a\_2)\leq (b\_1,b\_2) \iff a\_1\leq b\_1$ and $a\_2\leq b\_2$.


Let $u=(1,1)$ and let $g\_0=(1,2)$. 


Then $p(g\_0)=1$ and $r(g\_0)=2$. 


