Exercise 1.1.1 (Associative laws)

Proof. If the operation star is no associative, we give a counterexample.

(a)

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

Then

$$\begin{array}{llll}\hfill (0\star 1)\star 2& =-1\star 2=-3,& \hfill & \\ \hfill 0\star (1\star 2)& =0\star -1=1\ne -3.& \hfill & \end{array}$$

Therefore $\star$ is not associative.

(b)

The operation $\star$ on $ℝ$ is defined by $a\star b=a+b+\mathit{𝑎𝑏}=(1+a)(1+b)-1$.

Then, for all $a,b,c\in ℝ$,

$$\begin{array}{llll}\hfill (a\star b)\star c& =[(1+a)(1+b)-1]\star c& \hfill & \\ \hfill & =(1+a)(1+b)(1+c)-1& \hfill & \\ \hfill a\star (b\star c)& =a\star ((1+b)(1+c)-1)& \hfill & \\ \hfill & =(1+a)(1+b)(1+c)-1.& \hfill & \end{array}$$

Therefore $\star$ is associative.

(c)

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

Then

$$\begin{array}{llll}\hfill (0\star 1)\star 2& =\frac{1}{5}\star 2=\frac{\frac{1}{5}+2}{5}=\frac{11}{25}& \hfill & \\ \hfill 0\star (1\star 2)& =0\star \frac{3}{5}=\frac{3}{25}\ne \frac{11}{25}.& \hfill & \end{array}$$

Therefore $\star$ is not associative.

(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 $(a,b),(c,d),(e,f)$ in ${\mathbb{Z}}^2$,

$$\begin{array}{llll}\hfill ((a,b)\star (c,d))\star (e,f)& =(\mathit{𝑎𝑑}+\mathit{𝑏𝑐},\mathit{𝑏𝑑})\star (e,f)& \hfill & \\ \hfill & =((\mathit{𝑎𝑑}+\mathit{𝑏𝑐})f+\mathit{𝑏𝑑𝑒},\mathit{𝑏𝑑𝑓})& \hfill & \\ \hfill & =(\mathit{𝑎𝑑𝑓}+\mathit{𝑏𝑐𝑓}+\mathit{𝑏𝑑𝑒},\mathit{𝑏𝑑𝑓}),& \hfill & \\ \hfill (a,b)\star ((c,d)\star (e,f))& =(a,b)\star (\mathit{𝑐𝑓}+\mathit{𝑑𝑒},\mathit{𝑑𝑓})& \hfill & \\ \hfill & =(\mathit{𝑎𝑑𝑓}+b(\mathit{𝑐𝑓}+\mathit{𝑑𝑒}),\mathit{𝑏𝑑𝑓})& \hfill & \\ \hfill & =(\mathit{𝑎𝑑𝑓}+\mathit{𝑏𝑐𝑓}+\mathit{𝑏𝑑𝑒},\mathit{𝑏𝑑𝑓}).& \hfill & \end{array}$$

Therefore $\star$ is associative.

(e)

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

Then

$$\begin{array}{llll}\hfill (1\star 2)\star 3& =\frac{1}{2}\star 3=\frac{\left(\frac{1}{2}\right)}{3}=\frac{1}{6},& \hfill & \\ \hfill 1\star (2\star 3)& =1\star \frac{2}{3}=\frac{1}{\left(\frac{2}{3}\right)}=\frac{3}{2}\ne \frac{1}{6}.& \hfill & \end{array}$$

Therefore $\star$ is not associative.