Q: $A=N \times N$ and * be the binary operation on A defined by $(a, b) *(c, d)=(a+c, b+d)$. Show that * is commutative and associative. Find the identity element for * on $A$, if any.

$A=N \times N$ and * be the binary operation on A defined by $$ \begin{aligned} & (a, b) *(c, d)=(a+c, b+d) \\ & (a, b) *(c, d) \in A \\ & a, b, c, d \in N \\ & (a, b) *(c, d)=(a+c, b+d) \\ & (c, d) *(a, b)=(c+a, d+b)=(a+c, b+d) \\ & \therefore(a, b) *(c, d)=(c, d) *(a, b) \end{aligned} $$ Operation * is commutative. Now, let $(a, b),(c, d),(e, f) \in A$ $a, b, c, d, e, f \in N$ $[(a, b) *(c, d)] *(e, f)=(a+c, b+d) *(e, f)=(a+c+e, b+d+f)$ $(a, b) *[(c, d) *(e, f)]=(a, b) *(c+e, d+f)=(a+c+e, b+d+f)$ $\therefore[(a, b) *(c, d)] *(e, f)=(a, b) *[(c, d) *(e, f)]$ Operation * is associative. An element $e=\left(e_{1}, e_{2}\right) \in A$ will be an identity element for the operation * if $a+e=a=e^{*} a$ for all $a=\left(a_{1}, a_{2}\right) \in A$ i.e., $\left(a_{1}+e_{1}, a_{2}+e_{2}\right)=\left(a_{1}, a_{2}\right)=\left(e_{1}+a_{1}, e_{2}+a_{2}\right)$, which is not true for any element in A . Therefore, the operation * does not have any identity element.

Question 1

Determine whether or not each of the definition of * given below gives a binary operation. In the event that $*$ is not a binary operation, give justification for this.

i. On $\mathbf{Z}^{+}$, define * by $a^{*} b=a-b$

ii. On $\mathbf{Z}^{+}$, define * by $a^{*} b=a b$

iii. On $\mathbf{R}$, define ${ }^{*}$ by $a^{*} b=a b^{2}$

iv. On $\mathbf{Z}^{+}$, define * by $a^{*} b=|a-b|$

v. On $\mathbf{Z}^{+}$, define * by $a^{*} b=a$

Accepted Answer

i. On $\mathbf{Z}^{+}$, define * by $a^{*} b=a-b$

It is not a binary operation as the image of $(1,2)$ under * is

$$

\begin{aligned}

& 1 * 2=1-2 \

& \Rightarrow-1 
otin \mathbf{Z}^{+} .

\end{aligned}

$$

Therefore, * is not a binary operation.

ii. On $\mathbf{Z}^{+}$, define * by $a^{*} b=a b$

It is seen that for each $a, b \in \mathbf{Z}^{+}$, there is a unique element $a b$ in $\mathbf{Z}^{+}$.

This means that * carries each pair $(a, b)$ to a unique element $a^{*} b=a b$ in $\mathbf{Z}^{+}$.

Therefore, * is a binary operation.

iii. On $\mathbf{R}$, define ${ }^{*} a^{*} b=a b^{2}$

It is seen that for each $a, b \in \mathbf{R}$, there is a unique element $a b^{2}$ in $\mathbf{R}$. This means that * carries each pair $(a, b)$ to a unique element $a^{*} b=a b^{2}$ in $\mathbf{R}$.

Therefore, *is a binary operation.

iv. On $\quad \mathbf{Z}^{+}$, define * by $\quad a^{*} b=|a-b|$

It is seen that for each $a, b \in \mathbf{Z}^{+}$,there is a unique element $|a-b|$ in $\mathbf{Z}^{+}$. This means that * carries each pair $(a, b)$ to a unique element $a^{*} b=|a-b|$ in $\mathbf{Z}^{+}$. Therefore, *is a binary operation.

v. On $\mathbf{Z}^{+}$, define * by $a^{*} b=a$

* carries each pair ( $a, b$ ) to a unique element in $a^{*} b=a$ in $\mathbf{Z}^{+}$.

Therefore, * is a binary operation.

Question 2

For each binary operation *defined below, determine whether * is commutative or associative.

i. On $\mathbf{Z}^{+}$, define $a^{*} b=a-b$

ii. On $\mathbf{Q}$, define $a^{*} b=a b+1$

iii. On $\mathbf{Q}$, define $a^{*} b=\frac{a b}{2}$

iv. On $\mathbf{Z}^{+}$, define $a^{*} b=2^{a b}$

v. On $\mathbf{Z}^{+}$, define $a^{*} b=a^{b}$

vi. On $\mathbf{R}-\{-1\}$, define $a^{*} b=\frac{a}{b+1}$

Accepted Answer

i. On $\mathbf{Z}^{+}$, define $a^{*} b=a-b$

It can be observed that $1 * 2=1-2=-1$ and $2 * 1=2-1=1$.

$	herefore 1 * 2 
eq 2 * 1$; where $1,2 \in \mathbf{Z}$

Hence, the operation * is not commutative.

Also,

$(1 * 2) * 3=(1-2) * 3=-1 * 3=-1-3=-4$

$1 *(2 * 3)=1 *(2-3)=1 *-1=1-(-1)=2$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

where $1,2,3 \in \mathbf{Z}$

Hence, the operation * is not associative.

ii. On $\mathbf{Q}$, define $a^{*} b=a b+1$

$a b=b a \quad$ for all $a, b \in Q$

$\Rightarrow a b+1=b a+1 \quad$ for all $a, b \in Q$

$\Rightarrow a^{*} b=b^{*} a \quad$ for all $a, b \in Q$

Hence, the operation * is commutative.

$(1 * 2) * 3=(1 	imes 2+1) * 3=3 * 3=3 	imes 3+1=10$

$1 *(2 * 3)=1 *(2 	imes 3+1)=1 * 7=1 	imes 7+1=8$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

where $1,2,3 \in \mathbf{Q}$

Hence, the operation * is not associative.

iii. On $\mathbf{Q}$, define $a^{*} b=\frac{a b}{2}$

$a b=b a \quad$ for all $a, b \in Q$

$\Rightarrow \frac{a b}{2}=\frac{a b}{2} \quad$ for all $a, b \in Q$

$\Rightarrow a^{*} b=b^{*} a \quad$ for all $a, b \in Q$

Hence, the operation * is commutative.

$(a * b) * c=\left(\frac{a b}{2}ight) * c=\frac{\left(\frac{a b}{2}ight) c}{2}=\frac{a b c}{4}$

And

$a *(b * c)=a *\left(\frac{b c}{2}ight)=\frac{a\left(\frac{b c}{2}ight)}{2}=\frac{a b c}{4}$

$	herefore\left(a^{*} bight) * c=a^{*}\left(b^{*} cight)$

where $a, b, c \in \mathbf{Q}$

Hence, the operation * is associative.

iv. On $\mathbf{Z}^{+}$, define $a^{*} b=2^{a b}$

$a b=b a \quad$ for all $a, b \in Z$

$\Rightarrow 2^{a b}=2^{b a} \quad$ for all $a, b \in Z$

$\Rightarrow a^{*} b=b^{*} a \quad$ for all $a, b \in Z$

Hence, the operation * is commutative.

$(1 * 2) * 3=2^{1 	imes 2} * 3=4 * 3=2^{4 	imes 3}=2^{12}$

$1 *(2 * 3)=1 * 2^{2 	imes 3}=1 * 2^{6}=1 * 64=2^{64}$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

where $1,2,3 \in \mathbf{Z}^{+}$

Hence, the operation * is not associative.

v. On $\mathbf{Z}^{+}$, define $a^{*} b=a^{b}$

$1 * 2=1^{2}=1$

$2 * 1=2^{1}=2$

$	herefore 1 * 2 
eq 2 * 1$

where $1,2, \in \mathbf{Z}^{+}$

Hence, the operation * is not commutative.

$(2 * 3) * 4=2^{3} * 4=8 * 4=8^{4}=2^{12}$

$2 *(3 * 4)=2 * 3^{4}=2 * 81=2^{81}$

$	herefore(2 * 3) * 4 
eq 2 *(3 * 4)$

where $2,3,4 \in \mathbf{Z}^{+}$

Hence, the operation * is not associative.

vi. On $\mathbf{R}-\{-1\}$, define $a^{*} b=\frac{a}{b+1}$

$1 * 2=\frac{1}{2+1}=\frac{1}{3}$

$2 * 1=\frac{2}{1+1}=\frac{2}{2}=1$

$	herefore 1 * 2 
eq 2 * 1$

where $1,2, \in \mathbf{R}-\{-1\}$

Hence, the operation * is not commutative.

$(1 * 2) * 3=\frac{1}{3} * 3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12}$

$1 *(2 * 3)=1 * \frac{2}{3+1}=1 * \frac{2}{4}=1 * \frac{1}{2}=\frac{1}{\frac{1}{2}+1}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

where $1,2,3 \in \mathbf{R}-\{-1\}$

Hence, the operation * is not associative.

Question 3

Consider the binary operation $\wedge$ on the set $\{1,2,3,4,5\}$ defined by $a \wedge b=\min \{a, b\}$. Write the operation table of the operation $\wedge$.

Accepted Answer

The binary operation $\wedge$ on the set $\{1,2,3,4,5\}$ is defined by $a \wedge b=\min \{a, b\}$ for all $a, b \in\{1,2,3,4,5\}$.

The operation table for the given operation $\wedge$ can be given as:

\begin{tabular}{|l|l|l|l|l|l|}

\hline$\wedge$ & 1 & 2 & 3 & 4 & 5 \

\hline 1 & 1 & 1 & 1 & 1 & 1 \

\hline 2 & 1 & 2 & 2 & 2 & 2 \

\hline 3 & 1 & 2 & 3 & 3 & 3 \

\hline 4 & 1 & 2 & 3 & 4 & 4 \

\hline 5 & 1 & 2 & 3 & 4 & 5 \

\hline

\end{tabular}

Question 4

Consider a binary operation * on the set $\{1,2,3,4,5\}$ given by the following multiplication table.

i. Compute $(2 * 3) * 4$ and $2 *(3 * 4)$

ii. Is *commutative?

iii. Compute $(2 * 3) *(4 * 5)$.

(Hint: Use the following table)

\begin{tabular}{|l|l|l|l|l|l|}

\hline$*$ & 1 & 2 & 3 & 4 & 5 \

\hline 1 & 1 & 1 & 1 & 1 & 1 \

\hline 2 & 1 & 2 & 1 & 2 & 1 \

\hline

\end{tabular}

\begin{tabular}{|l|l|l|l|l|l|}

\hline 3 & 1 & 1 & 3 & 1 & 1 \

\hline 4 & 1 & 2 & 1 & 4 & 1 \

\hline 5 & 1 & 1 & 1 & 1 & 5 \

\hline

\end{tabular}

Accepted Answer

$$

(2 * 3) * 4=1 * 4=1

$$

i. $\quad 2 *(3 * 4)=2 * 1=1$

ii. For every $a, b \in\{1,2,3,4,5\}$, we have $a^{*} b=b^{*} a$. Therefore, * is commutative.

iii. $\quad(2 * 3) *(4 * 5)$

$$

\begin{aligned}

& (2 * 3)=1 	ext { and }(4 * 5)=1 \

& 	herefore(2 * 3) *(4 * 5)=1 * 1=1

\end{aligned}

$$

Question 5

Let ${ }^{* \prime}$ be the binary operation on the set $\{1,2,3,4,5\}$ defined by $a^{* \prime} b=$ H.C.F. of $a$ and $b$. Is the operation *' same as the operation * defined in Exercise 4 above? Justify your answer.

Accepted Answer

The binary operation on the set $\{1,2,3,4,5\}$ is defined by $a^{* \prime} b=$ H.C.F. of $a$ and $b$. The operation table for the operation *' can be given as:

\begin{tabular}{|l|l|l|l|l|l|}

\hline${ }^{* \prime}$ & 1 & 2 & 3 & 4 & 5 \

\hline 1 & 1 & 1 & 1 & 1 & 1 \

\hline 2 & 1 & 2 & 1 & 2 & 1 \

\hline 3 & 1 & 1 & 3 & 1 & 1 \

\hline 4 & 1 & 2 & 1 & 4 & 1 \

\hline 5 & 1 & 1 & 1 & 1 & 5 \

\hline

\end{tabular}

The operation table for the operations ${ }^{* \prime}$ and ${ }^{*}$ are same. operation *' is same as operation *.

Question 6

Let * be the binary operation on N defined by $a^{*} b=$ L.C.M. of $a$ and $b$. Find

i. $5 * 7,20 * 16$

ii. Is *commutative?

iii. Is *associative?

iv. Find the identity of *in N

v. Which elements of N are invertible for the operation *?

Accepted Answer

The binary operation on N is defined by $a^{*} b=$ L.C.M. of $a$ and $b$.

i. $\quad 5 * 7=$ L.C.M of 5 and $7=35$

$20 * 16=\mathrm{LCM}$ of 20 and $16=80$

ii. L.C.M. of $a$ and $b=\mathrm{LCM}$ of $b$ and $a$ for all $a, b \in N$

$	herefore a^{*} b=b^{*} a$

Operation *is commutative.

iii. For $a, b, c \in N$

$\left(a^{*} bight)^{*} c=(	ext { L.C.M. of } a 	ext { and } b)^{*} c=$ L.C.M. of $a, b, c$

$a^{*}\left(b^{*} cight)=a^{*}($ L.C.M. of $b$ and $c)=$ L.C.M. of $a, b, c$

$	herefore\left(a^{*} bight)^{*} c=a^{*}\left(b^{*} cight)$

Operation * is associative.

iv. L.C.M. of $a$ and $1=a=$ L.C.M. of 1 and $a$ for all $a \in N$

$a * 1=a=1 * a$ for all $a \in N$

Therefore, 1 is the identity of *in N .

v. An element a in N is invertible with respect to the operation * if there exists an element b in N , such that $a^{*} b=e=b^{*} a$

$e=1$

L.C.M. of $a$ and $b=1=\mathrm{LCM}$ of $b$ and $a$ possible only when $a$ and $b$ are equal to 1.

1 is the only invertible element of N with respect to the operation *.

Question 7

Is * defined on the set $\{1,2,3,4,5\}$ by $a^{*} b=$ LCM of $a$ and $b$ a binary operation? Justify your answer.

Accepted Answer

The operation *on the set $\{1,2,3,4,5\}$ is defined by $a^{*} b=\mathrm{LCM}$ of $a$ and $b$.

The operation table for the operation *' can be given as:

\begin{tabular}{|l|l|l|l|l|l|}

\hline$*$ & 1 & 2 & 3 & 4 & 5 \

\hline 1 & 1 & 2 & 3 & 4 & 5 \

\hline 2 & 2 & 2 & 6 & 4 & 10 \

\hline 3 & 3 & 6 & 3 & 12 & 15 \

\hline 4 & 4 & 4 & 12 & 4 & 20 \

\hline 5 & 5 & 10 & 15 & 20 & 5 \

\hline

\end{tabular}

$3 * 2=2 * 3=6 
otin A$,

$5 * 2=2 * 5=10 
otin A$,

$3 * 4=4 * 3=12 
otin A$,

$3 * 5=5 * 3=15 
otin A$,

$4 * 5=5 * 4=20 
otin A$

The given operation *is not a binary operation.

Question 8

Let * be the binary operation on N defined by $a^{*} b=$ H.C.F. of $a$ and $b$. Is * commutative? Is * associative? Does there exist identity for this binary operation on N ?

Accepted Answer

The binary operation * on N defined by $a^{*} b=$ H.C.F. of $a$ and $b$.

$	herefore a^{*} b=b^{*} a$

Operation * is commutative.

For all $a, b, c \in N$,

$\left(a^{*} bight)^{*} c=(	ext { HCF of } a 	ext { and } b)^{*} c=$ HCF of $a, b, c$

$a^{*}\left(b^{*} cight)=a^{*}($ HCF. of $b$ and $c)=$ HCF of $a, b, c$

$	herefore\left(a^{*} bight)^{*} c=a^{*}\left(b^{*} cight)$

Operation * is associative.

$e \in N$ will be the identity for the operation*if $a^{*} e=a=e^{*} a$ for all $a \in N$. But this relation is not true for any $a \in N$.

Operation * does not have any identity in N .

Question 9

Let * be the binary operation on Q of rational numbers as follows:

i. $\quad a^{*} b=a-b$

ii. $\quad a^{*} b=a^{2}+b^{2}$

iii. $\quad a^{*} b=a+a b$

iv. $\quad a^{*} b=(a-b)^{2}$

v. $a+b=\frac{a b}{4}$

vi. $a * b=a b^{2}$

Find which of the binary operations are commutative and which are associative.

Accepted Answer

i. On Q, the operation * is defined as $a^{*} b=a-b$

$\frac{1}{2} * \frac{1}{3}=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{3}=\frac{1}{6}$

And

$\frac{1}{3} * \frac{1}{2}=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6}$

$	herefore\left(\frac{1}{2} * \frac{1}{3}ight) 
eq\left(\frac{1}{3} * \frac{1}{2}ight)$

where $\frac{1}{2}, \frac{1}{3} \in Q$

Operation * is not commutative.

$\left(\frac{1}{2} * \frac{1}{3}ight) * \frac{1}{4}=\left(\frac{1}{2}-\frac{1}{3}ight) * \frac{1}{4}=\frac{1}{6} * \frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{2-3}{12}=\frac{-1}{12}$

$\frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}ight)=\frac{1}{2} *\left(\frac{1}{3}-\frac{1}{4}ight)=\frac{1}{2} * \frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{6-1}{12}=\frac{5}{12}$

$	herefore\left(\frac{1}{2} * \frac{1}{3}ight) * \frac{1}{4} 
eq \frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}ight)$

where $\frac{1}{2}, \frac{1}{3}, \frac{1}{4} \in Q$

Operation * is not associative.

ii. On Q , the operation * is defined as $a^{*} b=a^{2}+b^{2}$

For $a, b \in Q$

$a^{*} b=a^{2}+b^{2}=b^{2}+a^{2}=b^{*} a$

$	herefore a^{*} b=b^{*} a$

Operation * is commutative.

$(1 * 2) * 3=\left(1^{2}+2^{2}ight) * 3=(1+4) * 3=5 * 3=5^{2}+3^{2}=25+9=34$

$1 *(2 * 3)=1 *\left(2^{2}+3^{2}ight)=1 *(4+9)=1 * 13=1^{2}+13^{2}=1+169=170$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

where $1,2,3 \in Q$

Operation * is not associative.

iii. On Q, the operation * is defined as $a^{*} b=a+a b$

$1 * 2=1+1 	imes 2=1+2=3$

2*1 = ? + 2 $	imes 1=2+2=4$

$	herefore 1 * 2 
eq 2 * 1$

where $1,2 \in Q$

Operation * is not commutative.

$(1 * 2) * 3=(1+1 	imes 2) * 3=3 * 3=3+3 	imes 3=3+9=12$

$1 *(2 * 3)=1 *(2+2 	imes 3)=1 * 8=1+1 	imes 8=1+8=9$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

where $1,2,3 \in Q$

Operation * is not associative.

iv. On Q , the operation * is defined as $a * b=(a-b)^{2}$

For $a, b \in Q$

$$

\begin{aligned}

& a^{*} b=(a-b)^{2} \

& b^{*} a=(b-a)^{2}=[-(a-b)]^{2}=(a-b)^{2}

\end{aligned}

$$

$	herefore a * b=b * a$

Operation * is commutative.

$$

\begin{aligned}

& (1 * 2) * 3=(1-2)^{2} * 3=(-1)^{2} * 3=1 * 3=(1-3)^{2}=(-2)^{2}=4 \

& 1 *(2 * 3)=1 *(2-3)^{2}=1 *(-1)^{2}=1 * 1=(1-1)^{2}=0 \

& 	herefore(1 * 2) * 3 
eq 1 *(2 * 3) \quad 	ext { where } 1,2,3 \in Q

\end{aligned}

$$

Operation * is not associative.

v. On Q, the operation * is defined as $a+b=\frac{a b}{4}$

For $a, b \in Q$

$$

\begin{aligned}

& a * b=\frac{a b}{4}=\frac{b a}{4}=b * a \

& 	herefore a * h=b * a

\end{aligned}

$$

Operation * is commutative.

For $a, b, c \in Q$

$$

\begin{aligned}

& (a * b) * c=\frac{a b}{4} * c=\frac{\frac{a b}{4} \cdot c}{4}=\frac{a b c}{16} \

& a *(b * c)=a * \frac{a b}{4}=\frac{a \cdot \frac{a b}{4}}{4}=\frac{a b c}{16} \

& 	herefore(a * b) * c=a *(b * c)

\end{aligned}

$$

where $a, b, c \in Q$

Operation * is associative.

vi. On Q, the operation * is defined as $a * b=a b^{2}$

$$

\begin{aligned}

& \frac{1}{2} * \frac{1}{3}=\frac{1}{2} \cdot\left(\frac{1}{3}ight)^{2}=\frac{1}{2} \cdot \frac{1}{9}=\frac{1}{18} \

& \frac{1}{3} * \frac{1}{2}=\frac{1}{3} \cdot\left(\frac{1}{2}ight)^{2}=\frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}

\end{aligned}

$$

where $\frac{1}{2}, \frac{1}{3} \in Q$

Operation * is not commutative.

$$

\begin{aligned}

& \left(\frac{1}{2} * \frac{1}{3}ight) * \frac{1}{4}=\left(\frac{1}{2} \cdot\left(\frac{1}{3}ight)^{2}ight) * \frac{1}{4}=\frac{1}{18} * \frac{1}{4}=\frac{1}{18} \cdot\left(\frac{1}{4}ight)^{2}=\frac{1}{18 	imes 16} \

& \frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}ight)=\frac{1}{2} *\left(\frac{1}{3} \cdot\left(\frac{1}{4}ight)^{2}ight)=\frac{1}{2} * \frac{1}{48}=\frac{1}{2} \cdot\left(\frac{1}{48}ight)^{2}=\frac{1}{2 	imes(48)^{2}} \

& 	herefore\left(\frac{1}{2} * \frac{1}{3}ight) * \frac{1}{4} 
eq \frac{1}{2} *\left(\frac{1}{3} * \frac{1}{4}ight) \quad 	ext { where } \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \in Q

\end{aligned}

$$

Operation * is not associative.

Operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

Question 10

Find which of the operations given above has identity.

Accepted Answer

An element $e \in Q$ will be the identity element for the operation * if

$a^{*} e=a=e^{*} a$, for all $a \in Q$

$a^{*} b=\frac{a b}{4}$

$\Rightarrow a^{*} e=a$

$\Rightarrow \frac{a e}{4}=a$

$\Rightarrow e=4$

Similarly, it can be checked for $e * a=a$, we get $e=4$ is the identity.

Question 11

$A=N 	imes N$ and * be the binary operation on A defined by $(a, b) *(c, d)=(a+c, b+d)$. Show that * is commutative and associative. Find the identity element for * on $A$, if any.

Accepted Answer

$A=N 	imes N$ and * be the binary operation on A defined by

$$

\begin{aligned}

& (a, b) *(c, d)=(a+c, b+d) \

& (a, b) *(c, d) \in A \

& a, b, c, d \in N \

& (a, b) *(c, d)=(a+c, b+d) \

& (c, d) *(a, b)=(c+a, d+b)=(a+c, b+d) \

& 	herefore(a, b) *(c, d)=(c, d) *(a, b)

\end{aligned}

$$

Operation * is commutative.

Now, let $(a, b),(c, d),(e, f) \in A$

$a, b, c, d, e, f \in N$

$[(a, b) *(c, d)] *(e, f)=(a+c, b+d) *(e, f)=(a+c+e, b+d+f)$

$(a, b) *[(c, d) *(e, f)]=(a, b) *(c+e, d+f)=(a+c+e, b+d+f)$

$	herefore[(a, b) *(c, d)] *(e, f)=(a, b) *[(c, d) *(e, f)]$

Operation * is associative.

An element $e=\left(e_{1}, e_{2}ight) \in A$ will be an identity element for the operation * if $a+e=a=e^{*} a$ for all $a=\left(a_{1}, a_{2}ight) \in A$ i.e., $\left(a_{1}+e_{1}, a_{2}+e_{2}ight)=\left(a_{1}, a_{2}ight)=\left(e_{1}+a_{1}, e_{2}+a_{2}ight)$, which is not true for any element in A .

Therefore, the operation * does not have any identity element.

Question 12

State whether the following statements are true or false. Justify.

i. For an arbitrary binary operation * on a set $\mathrm{N}, a^{*} a=a$ for all $a \in N$.

ii. If * is a commutative binary operation on N , then $a^{*}\left(b^{*} c\right)=\left(c^{*} b\right)^{*} a$

Accepted Answer

i. Define operation * on a set N as $a^{*} a=a$ for all $a \in N$.

In particular, for $a=3$,

$$

3 * 3=9 
eq 3

$$

Therefore, statement (i) is false.

ii. R.H.S. $=(c * b)^{*} a$

$=\left(b^{*} cight)^{*} a$ [* is commutative]

$=a^{*}\left(b^{*} cight)$ [Again, as * is commutative]

= L.H.S.

$	herefore a *(b * c)=(c * b) * a$

Therefore, statement (ii) is true.

Question 13

Consider a binary operation * on N defined as $a^{*} b=a^{3}+b^{3}$. Choose the correct answer.

A. Is * both associative and commutative?

B. Is * commutative but not associative?

C. Is * associative but not commutative?

D. Is * neither commutative nor associative?

Accepted Answer

On N, operation *is defined as $a^{*} b=a^{3}+b^{3}$.

For all $a, b \in N$

$a^{*} b=a^{3}+b^{3}=b^{3}+a^{3}=b^{*} a$

Operation * is commutative.

$(1 * 2) * 3=\left(1^{3}+2^{3}ight) * 3=(1+8) * 3=9 * 3=9^{3}+3^{3}=729+27=756$

$1 *(2 * 3)=1 *\left(2^{3}+3^{3}ight)=1 *(8+27)=1 * 35=1^{3}+35^{3}=1+42875=42876$

$	herefore(1 * 2) * 3 
eq 1 *(2 * 3)$

Operation *is not associative.

Therefore, Operation * is commutative, but not associative.

The correct answer is B.

\section*{MISCELLANEOUS EXERCISE}