Question

Check the commutativity and associativity of each of the following binary operations:
(i) '*'. on Z defined by a * b = a + b + ab for all a, b ∈ Z
(ii) '*'. on N defined by a * b = 2^ab for all a, b ∈ N
(iii) '*'. on Q defined by a * b = a − b for all a, b ∈ Q
(iv) '⊙' on Q defined by a ⊙ b = a² + b² for all a, b ∈ Q
(v) 'o' on Q defined by $aob=\frac{ab}{2}$ for all a, b ∈ Q
(vi) '*' on Q defined by a * b = ab² for all a, b ∈ Q
(vii) '*' on Q defined by a * b = a + ab for all a, b ∈ Q
(viii) '*' on R defined by a * b = a + b − 7 for all a, b ∈ R
(ix) '*' on Q defined by a * b = (a − b)² for all a, b ∈ Q
(x) '*' on Q defined by a * b = ab + 1 for all a, b ∈ Q
(xi) '*' on N, defined by a * b = a^b for all a, b ∈ N
(xii) '*' on Z defined by a * b = a − b for all a, b ∈ Z
(xiii) '*' on Q defined by $a*b=\frac{ab}{4}$ for all a, b ∈ Q
(xiv) '*' on Z defined by a * b = a + b − ab for all a, b ∈ Z
(xv) '*' on N defined by a * b = gcd(a, b) for all a, b ∈ N

Answer 1

(i) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Z}.\mathrm{Then}, \\ a*b=a+b+ab \\ =b+a+ba \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{Z} \end{gathered}$$

Thus, * is commutative on Z.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Z}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b+c+bc\right) \\ =a+\left(b+c+bc\right)+a\left(b+c+bc\right) \\ =a+b+c+bc+ab+ac+abc \\ \left(a*b\right)*c=\left(a+b+ab\right)*c \\ =a+b+ab+c+\left(a+b+ab\right)c \\ =a+b+ab+c+ac+bc+abc \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in \mathbf{Z} \end{gathered}$$

Thus, * is associative on Z.

(ii) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{N}.\mathrm{Then}, \\ a*b=2^{ab} \\ =2^{ba} \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{N} \end{gathered}$$

Thus, * is commutative on N.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{N}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(2^{bc}\right) \\ =2^{a*2^{bc}} \\ \left(a*b\right)*c=\left(2^{ab}\right)*c \\ =2^{ab*2^c} \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on N.

(iii) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ a*b=a-b \\ b*a=b-a \\ \text{Therefore,} \\ a*b\ne b*a \end{gathered}$$

Thus, * is not commutative on Q.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b-c\right) \\ =a-\left(b-c\right) \\ =a-b+c \\ \left(a*b\right)*c=\left(a-b\right)*c \\ =a-b-c \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on Q.

(iv) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ a\odot b=a^2+b^2 \\ =b^2+a^2 \\ =b\odot a \\ \text{Therefore,} \\ a\odot b=b\odot a,\forall a,b\in \mathbf{Q} \end{gathered}$$

Thus, $\odot$ is commutative on Q.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ a\odot \left(b\odot c\right)=a\odot \left(b^2+c^2\right) \\ =a^2+{\left(b^2+c^2\right)}^2 \\ =a^2+b^4+c^4+2b^2{c}^2 \\ \left(a\odot b\right)\odot c=\left(a^2+b^2\right)\odot c \\ ={\left(a^2+b^2\right)}^2+c^2 \\ =a^4+b^4+2a^2{b}^2+c^2 \\ \text{Therefore,} \\ a\odot \left(b\odot c\right)\ne \left(a\odot b\right)\odot c \end{gathered}$$

Thus, $\odot$ is not associative on Q.

(v) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ aob=\frac{ab}{2} \\ =\frac{ba}{2} \\ =boa \\ \text{Therefore,} \\ aob=boa,\forall a,b\in \mathbf{Q} \end{gathered}$$

Thus, o is commutative on Q.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ ao\left(boc\right)=ao\left(\frac{bc}{2}\right) \\ =\frac{a\left({\displaystyle \frac{bc}{2}}\right)}{2} \\ =\frac{abc}{4} \\ \left(aob\right)oc=\left(\frac{ab}{2}\right)oc \\ =\frac{\left({\displaystyle \frac{ab}{2}}\right)c}{2} \\ =\frac{abc}{4} \\ \text{Therefore,} \\ ao\left(boc\right)=\left(aob\right)oc,\forall a,b,c\in \mathbf{Q} \end{gathered}$$

Thus, o is associative on Q.

(vi) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ a*b=a{b}^2 \\ b*a=b{a}^2 \\ \text{Therefore,} \\ a*b\ne b*a \end{gathered}$$

Thus, * is not commutative on Q.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b{c}^2\right) \\ =a{\left(b{c}^2\right)}^2 \\ =a{b}^2{c}^4 \\ \left(a*b\right)*c=\left(a{b}^2\right)*c \\ =a{b}^2{c}^2 \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on Q.

(vii) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ a*b=a+ab \\ b*a=b+ba \\ =b+ab \\ \text{Therefore,} \\ a*b\ne b*a \end{gathered}$$

Thus, * is not commutative on Q.

Associativity:
$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b+bc\right) \\ =a+a\left(b+bc\right) \\ =a+ab+abc \\ \left(a*b\right)*c=\left(a+ab\right)*c \\ =\left(a+ab\right)+\left(a+ab\right)c \\ =a+ab+ac+abc \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on Q.

(viii) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{R}.\mathrm{Then}, \\ a*b=a+b-7 \\ =b+a-7 \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{R} \end{gathered}$$

Thus, * is commutative on R.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{R}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b+c-7\right) \\ =a+b+c-7-7 \\ =a+b+c-14 \\ \left(a*b\right)*c=\left(a+b-7\right)*c \\ =a+b-7+c-7 \\ =a+b+c-14 \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in \mathbf{R} \end{gathered}$$

Thus, * is associative on R.

(ix) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ a*b={\left(a-b\right)}^2 \\ ={\left(b-a\right)}^2 \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{Q} \end{gathered}$$

Thus, * is commutative on Q.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ a*\left(b*c\right)=a*{\left(b-c\right)}^2 \\ =a*\left(b^2+c^2-2bc\right) \\ ={\left(a-b^2-c^2+2bc\right)}^2 \\ \left(a*b\right)*c={\left(a-b\right)}^2*c \\ =\left(a^2+b^2-2ab\right)*c \\ ={\left(a^2+b^2-2ab-c\right)}^2 \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on Q.

(x) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ a*b=ab+1 \\ =ba+1 \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{Q} \end{gathered}$$

Thus, * is commutative on Q.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(bc+1\right) \\ =a\left(bc+1\right)+1 \\ =abc+a+1 \\ \left(a*b\right)*c=\left(ab+1\right)*c \\ =\left(ab+1\right)c+1 \\ =abc+c+1 \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on Q.

(xi) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{N}.\mathrm{Then}, \\ a*b=a^b \\ b*a=b^a \\ \text{Therefore,} \\ a*b\ne b*a \end{gathered}$$

Thus, * is not commutative on N.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{N}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b^c\right) \\ =a^{b^c} \\ \left(a*b\right)*c=\left(a^b\right)*c \\ ={\left(a^b\right)}^c \\ =a^{bc} \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on N.

(xii) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Z}.\mathrm{Then}, \\ a*b=a-b \\ b*a=b-a \\ \text{Therefore,} \\ a*b\ne b*a \end{gathered}$$

Thus, * not is commutative on Z.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Z}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b-c\right) \\ =a-\left(b-c\right) \\ =a-b+c \\ \left(a*b\right)*c=\left(a-b\right)-c \\ =a-b-c \\ \text{Therefore,} \\ a*\left(b*c\right)\ne \left(a*b\right)*c \end{gathered}$$

Thus, * is not associative on Z.

(xiii) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Q}.\mathrm{Then}, \\ a*b=\frac{ab}{4} \\ =\frac{ba}{4} \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{Q} \end{gathered}$$

Thus, * is commutative on Q.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Q}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(\frac{bc}{4}\right) \\ =\frac{a\left({\displaystyle \frac{bc}{4}}\right)}{4} \\ =\frac{abc}{16} \\ \left(a*b\right)*c=\left(\frac{ab}{4}\right)*c \\ =\frac{\left({\displaystyle \frac{ab}{4}}\right)c}{4} \\ =\frac{abc}{16} \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in \mathbf{Q} \end{gathered}$$

Thus, * is associative on Q.

(xiv) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{Z}.\mathrm{Then}, \\ a*b=a+b-ab \\ =b+a-ba \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{Z} \end{gathered}$$

Thus, * is commutative on Z.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{Z}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left(b+c-bc\right) \\ =a+b+c-bc-a\left(b+c-bc\right) \\ =a+b+c-bc-ab-ac+abc \\ \left(a*b\right)*c=\left(a+b-ab\right)*c \\ =a+b-ab+c-\left(a+b-ab\right)c \\ =a+b+c-ab-ac-bc+abc \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in \mathbf{Z} \end{gathered}$$

Thus, * is associative on Z.

Disclaimer : The answer given in the textbook is incorrect. The same has been corrected here.

(xv) Commutativity:

$$\begin{gathered} \text{Let}a,b\in \mathbf{N}.\mathrm{Then}, \\ a*b=\gcd \left(a,b\right) \\ =\gcd \left(b,a\right) \\ =b*a \\ \text{Therefore,} \\ a*b=b*a,\forall a,b\in \mathbf{N} \end{gathered}$$

Thus, * is commutative on N.

Associativity:

$$\begin{gathered} \text{Let}a,b,c\in \mathbf{N}.\mathrm{Then}, \\ a*\left(b*c\right)=a*\left[\gcd \left(a,b\right)\right] \\ =\gcd \left(a,b,c\right) \\ \left(a*b\right)*c=\left[\gcd \left(a,b\right)\right]*c \\ =\gcd \left(a,b,c\right) \\ \text{Therefore,} \\ a*\left(b*c\right)=\left(a*b\right)*c,\forall a,b,c\in \mathbf{N} \end{gathered}$$

Thus, * is associative on N.