# Decide whether or not a given binary operator is associative

**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.1**

Determine which of the following binary operations are associative.

- 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 associative since $$(1 \star 1) \star 1 = (1-1)-1 = 0-1 = -1 $$ but $$1 \star (1 \star 1) = 1-(1-1) = 1-0 = 1.$$

**(2)** Associative since for all $a,b,c$, we have \begin{align*} (a \star b) \star c =&\ (a + b + ab) \star c\\ = &\ (a + b + ab) + c + ac + bc + abc\\ = &\ a + (b + c + bc) + ab + ac + abc\\ = &\ a \star (b + c+ bc)\\ = &\ a \star (b \star c) \end{align*}

**(3)** Not associative since $$(1 \star 0) \star 2 = \frac{1}{5} \star 2 = \frac{11}{25}$$ but $$1 \star (0 \star 2) = 1 \star \frac{2}{5} = \frac{7}{25}.$$

**(4)** Associative since for all $(a_1,b_1)$, $(a_2,b_2)$, $(a_3,b_3)$, we have \begin{align*}&\ (a_1,b_1) \star \left[(a_2, b_2) \star (a_3,b_3) \right]\\ = &\ (a_1,b_1 \star (a_2 b_3 + b_2 a_3, b_2 b_3)\\ = &\ (a_1 b_2 b_3 + b_1 a_2 b_3 + b_1 b_2 a_3, b_1 b_2 b_3)\\ = &\ (a_1 b_2 + b_1 a_2, b_1 b_2) \star (a_3,b_3)\\ = &\ \left[ (a_1,b_1) \star (a_2,b_2) \right] \star (a_3,b_3)\end{align*}

**(5)** Not associative since $$(2 \star 1) \star 2 = 2 \star 2 = 1$$ but $$2 \star (1 \star 2) = 2 \star \frac{1}{2} = 4.$$