Exercise 1.1.2 (Commutative laws)

Proof.

(a)

The operation $\star$ on $\mathbb{Z}$ is defined by $a\star b=a-b$.

Then

$$\begin{array}{llll}\hfill & 0\star 1=-1,& \hfill & \\ \hfill & 1\star 0=1\ne -1.& \hfill & \end{array}$$

Therefore $\star$ is not commutative.

(b)

The operation $\star$ on $ℝ$ is defined by $a\star b=a+b+\mathit{𝑎𝑏}$.

Then for all $a,b\in ℝ$,

$$\begin{array}{llll}\hfill a\star b& =a+b+\mathit{𝑎𝑏}& \hfill & \\ \hfill b\star a& =b+a+\mathit{𝑏𝑎}=a+b+\mathit{𝑎𝑏}.& \hfill & \end{array}$$

Therefore $\star$ is commutative.

(c)

The operation $\star$ on $\mathbb{Q}$ is defined by $a\star b=\frac{a+b}{5}$.

Then for all $a,b\in ℝ$,

$$\begin{array}{lll}\hfill a\star b=\frac{a+b}{5}=\frac{b+a}{5}=b\star a.& & \hfill \end{array}$$

Therefore $\star$ is commutative.

(d)

The operation $\star$ on $\mathbb{Z}\times \mathbb{Z}$ is defined by $(a,b)\star (c,d)=(\mathit{𝑎𝑑}+\mathit{𝑏𝑐},\mathit{𝑏𝑑})$.

Then, for all $(a,b),(c,d)$ in ${\mathbb{Z}}^2$,

$$\begin{array}{llll}\hfill (a,b)\star (c,d)& =(\mathit{𝑎𝑑}+\mathit{𝑏𝑐},\mathit{𝑏𝑑})& \hfill & \\ \hfill (c,d)\star (a,b)& =(\mathit{𝑐𝑏}+\mathit{𝑑𝑎},\mathit{𝑑𝑏})=(\mathit{𝑎𝑑}+\mathit{𝑏𝑐},\mathit{𝑏𝑑}).& \hfill & \end{array}$$

Therefore $\star$ is commutative.

(e)

The operation $\star$ on $\mathbb{Q}-\{0\}$ is defined by $a\star b=\frac{a}{b}$.

Then

$$\begin{array}{llll}\hfill 1\star 2& =\frac{1}{2},& \hfill & \\ \hfill 2\star 1& =\frac{2}{1}=2\ne \frac{1}{2}.& \hfill & \end{array}$$

Therefore $\star$ is not commutative.