Incorrect Solution for Problem 7 of Pinter's Book of Abstract Algebra, Chapter 2? I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = \frac{xy}{x+y+1}$$
The book solution:
$$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$
$$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) = \frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$
However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:
$$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} $$
$$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right)  =\frac{x\left(\frac{yz}{y+z+1}\right)}{x+\frac{yz}{y+z+1} + 1} $$
So the denominators cancel out, the two groupings are equivalent, the operator is associative.  This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something.  Sorry for bothering people with such a basic question.

As far as I can tell it's associative.  You've proved it twice yourself already, but my nitty-gritty arithmetic is:
\begin{align*}
(x * y) * z &= \left(\frac{xy}{x+y+1}\right) * z \\
 &= \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} \\
 &= \frac{\left(\frac{xyz}{x+y+1}\right)}{\frac{xy+(z+1)(x+y+1)}{x+y+1}} \\
 &= \frac{xyz}{xy+(z+1)(x+y+1)} \\
 &= \frac{xyz}{xy+xz+yz+x+y+z+1} \\
 &= \frac{xyz}{yz+(x+1)(y+z+1)} \\
 &= \frac{\left(\frac{xyz}{y+z+1}\right)}{\frac{yz+(x+1)(y+z+1)}{y+z+1}} \\
 &= \frac{x\left(\frac{yz}{y+z+1}\right)}{x + \frac{yz}{y+z+1} + 1} \\
 &= x*\left(\frac{yz}{y+z+1}\right) \\
 &= x*(y*z).
\end{align*}
