How to show nonassociativity of the positive rationals under a binary operation defined in terms of max and min? Consider $\mathbb{Q}^+$ with the usual $\leq$ relation and the binary operation $\circ$ defined as:
$$p \circ q = max(p,q) + \frac{1}{2} min(p,q)$$
A book that I'm reading states that the operation $\circ$ is not associative with respect to $\mathbb{Q}^+$, with verification left to the reader. I have tried to verify this by picking random triples $(p,q,r)$ of positive rational numbers as examples, but in each case I end up with the result that $p \circ (q \circ r) = (p \circ q) \circ r$.
For example, take $p=4.5, q=18.1, r=7$:
\begin{align}
4.5 \circ (18.1 \circ 7) &= 4.5 \circ (max(18.1,7) + \frac{1}{2} min(18.1,7))\\
&= 4.5 \circ (18.1 + 3.5)\\
&= 4.5 \circ 21.6\\
&= max(4.5,21.6) + \frac{1}{2} min(4.5,21.6)\\
&= 21.6 + \frac{1}{2} 4.5\\
&= 23.85
\end{align}
and then:
\begin{align}
(4.5 \circ 18.1) \circ 7 &= (max(4.5,18.1) + \frac{1}{2} min(4.5,18.1)) \circ 7\\
&= (18.1 + \frac{1}{2} 4.5) \circ 7\\
&= 20.35 \circ 7\\
&= max(20.35,7) + \frac{1}{2} min(20.35,7)\\
&= 20.35 + \frac{1}{2}7\\
&= 23.85
\end{align}
Am I not picking the right triples of positive rationals for a counterexample, or am I misunderstanding somehow?

Select $q<p<p+q/2<r$, for example, $q=2,p=3,r=5$.
We get $(p\circ q)\circ r=r+p/2+q/4$ and $p\circ(q\circ r)=r+p/2+q/2$.
