In Charles Pinter's A Book of Abstract Algebra, Chapter 2: Operations, Section B. Properties of Operations, Question 7 part ii, an operation * is defined on the set of real numbers: $x * y = \frac{xy}{x+y+1} $ I need to find if it is associative, so: $$ \begin{align*} x*(y*z) &amp;= x* \frac{yz}{y+z+1} \\\\ &amp;= \frac{\frac{xyz}{y+z+1}}{x+\frac{yz}{y+z+1}+1} \\\\ &amp;= \frac{\frac{xyz}{y+z+1}}{\frac{xy+xz+x+yz+y+z+1}{y+z+1}} \\\\ &amp;= \frac{xyz}{xy+xz+yz+x+y+z+1} \\\\ \end{align*} $$  Then the same answer for $(x*y)*z$. This is not the same as the answer given at the end:   What have I overlooked?
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.
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