**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.2**

Determine which of the following binary operations are commutative.

- The operation $\star$ on $\mathbb{Z}$ defined by $a \star b = a-b$.
- The operation $\star$ on $\mathbb{R}$ defined by $a \star b = a+b+ab$.
- The operation $\star$ on $\mathbb{Q}$ defined by $a \star b = \frac{a+b}{5}$.
- The operation $\star$ on $\mathbb{Z} \times \mathbb{Z}$ defined by $(a_1,b_1) \star (a_2,b_2) = (a_1 b_2 + b_1 a_2, b_1 b_2)$.
- The operation $\star$ on $\mathbb{Q} \setminus \{0\}$ defined by $a \star b = \frac{a}{b}$.

Solution:

**(1)** Not commutative since $$1 \star (-1) = 1 – (-1) = 2$$ but $$(-1) \star 1 = -1 – 1 = -2.$$

**(2)** Commutative since $$a \star b = a + b + ab = b + a + ba = b \star a.$$

**(3)** Commutative since $$a \star b = \frac{a+b}{5} = \frac{b+a}{5} = b \star a.$$

**(4)** Commutative since \begin{align*} (a_1,b_1) \star (a_2,b_2) = &\ (a_1 b_2 + b_1 a_2, b_1 b_2)\\ = &\ (a_2 b_1 + b_2 a_1, b_2 b_1)\\ = &\ (a_2,b_2) \star (a_1,b_1).\end{align*}

**(5)** Not commutative since $1 \star 2 = \frac{1}{2}$ but $2 \star 1 = 2$.