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 $p\circ(q\circ r)=r+p/2+q/2$.