Hw2 Problem 5

### Problem 5

Determine whether the binary operation $*$ defined on $\mathbb{Q}$ by letting $a * b = ab/2$ is commutative. Is $*$ associative?

Solution

Since $a, b \in \mathbb{Q}$, we can write $a = \frac{p}{q}$ and $b = \frac{m}{n}$ where $p, q, m, n \in \mathbb{Z}$ and $q, n \neq 0$

Now, for an operation to be commutative $a * b = b * a$. $a = \frac{p}{q}$ and $b = \frac{m}{n}$ so we can substitute for $a$ and $b$, which yields,

(1)
\begin{align} a * b = \frac{\frac{p}{q} \cdot \frac{m}{n}}{2} = \frac{pm}{2qn} \end{align}
(2)
\begin{align} b * a = \frac{\frac{m}{n} \cdot \frac{p}{q}}{2} = \frac{mp}{2nq} = \frac{pm}{2qn} \end{align}

By properties of integer multiplication $a * b = b * a$. Hence, the operation is commutative.

To show the operation is associative, we must introduce another rational number, $c = \frac{x}{y}$ where $x, y \in \mathbb{Z}$ and $y \neq 0$.

For an operation to be associative $(a * b) * c = a * (b * c)$. Now,

(3)
\begin{align} (a * b) * c = \frac{\frac{pm}{2qn} \cdot \frac{x}{y}}{2} = \frac{pmx}{4qny} \end{align}
(4)
\begin{align} b * c = \frac{\frac{m}{n} \cdot \frac{x}{y}}{2} = \frac{mx}{2ny} \end{align}
(5)
\begin{align} a * (b * c) = \frac{\frac{p}{q} \cdot \frac{mx}{2ny}}{2} = \frac{pmx}{4qny} \end{align}

Hence $(a * b) * c = a * (b * c)$ and the operation is associative.