Binary Operation Questions in English

(N/A) binary operation $*$ on a set $S$ is a function $*: S \times S \rightarrow S$.
$1$. For addition $(+)$: For any $a, b \in R$, $a+b$ is a unique real number. Thus, $+: R \times R \rightarrow R$ is a binary operation.
$2$. For subtraction $(-)$: For any $a, b \in R$, $a-b$ is a unique real number. Thus, $-: R \times R \rightarrow R$ is a binary operation.
$3$. For multiplication $(\times)$: For any $a, b \in R$, $a \times b$ is a unique real number. Thus, $\times: R \times R \rightarrow R$ is a binary operation.
$4$. For division $(div)$: For $a, b \in R$, the operation $a \div b = \frac{a}{b}$ is not defined when $b=0$. Since $0 \in R$, the division operation is not a function from $R \times R$ to $R$. Hence, it is not a binary operation on $R$.
$5$. For $R_*$ (the set of nonzero real numbers): For any $a, b \in R_*$, $a \neq 0$ and $b \neq 0$. The quotient $\frac{a}{b}$ is always a defined real number, and since $a, b \neq 0$, $\frac{a}{b} \neq 0$. Thus, $\frac{a}{b} \in R_*$. Therefore, division is a binary operation on $R_*$.

$A$ binary operation $*$ on a set $S$ is a function $*: S \times S \rightarrow S$. This means that for every pair $(a, b) \in S \times S$,the result $a * b$ must also be in $S$.
$1$. For subtraction $(-)$ on $N$:
Consider the elements $a = 3$ and $b = 5$,where $3, 5 \in N$.
The operation $a - b$ gives $3 - 5 = -2$.
Since $-2 \notin N$,subtraction is not a binary operation on $N$.
$2$. For division $(\div)$ on $N$:
Consider the elements $a = 3$ and $b = 5$,where $3, 5 \in N$.
The operation $a \div b$ gives $3 \div 5 = \frac{3}{5}$.
Since $\frac{3}{5} \notin N$,division is not a binary operation on $N$.

(N/A) binary operation $*$ on a set $R$ is a function $*: R \times R \rightarrow R$.
For any pair $(a, b) \in R \times R$,the expression $a + 4b^{2}$ is a well-defined real number because $a$ and $b$ are real numbers,and the set of real numbers $R$ is closed under addition and multiplication.
Since for every pair $(a, b) \in R \times R$,there exists a unique element $a + 4b^{2} \in R$,the operation $*$ is a binary operation on $R$.

(N/A) binary operation $*$ on a set $S$ is a function $*: S \times S \rightarrow S$.
For the union operation $\cup: P \times P \rightarrow P$,for any two subsets $A, B \in P$,their union $A \cup B$ is also a subset of $X$,which means $A \cup B \in P$. Since every pair $(A, B)$ maps to a unique element $A \cup B$ in $P$,$\cup$ is a binary operation on $P$.
Similarly,for the intersection operation $\cap: P \times P \rightarrow P$,for any two subsets $A, B \in P$,their intersection $A \cap B$ is also a subset of $X$,which means $A \cap B \in P$. Since every pair $(A, B)$ maps to a unique element $A \cap B$ in $P$,$\cap$ is a binary operation on $P$.

(N/A) binary operation $\ast$ on a set $S$ is a function $\ast: S \times S \rightarrow S$.
For the operation $\vee: R \times R \rightarrow R$ defined by $(a, b) \rightarrow \max\{a, b\}$,for any pair $(a, b) \in R \times R$,the value $\max\{a, b\}$ is a unique real number that belongs to $R$.
Since every pair $(a, b)$ maps to a unique element in $R$,$\vee$ is a binary operation.
Similarly,for the operation $\wedge: R \times R \rightarrow R$ defined by $(a, b) \rightarrow \min\{a, b\}$,for any pair $(a, b) \in R \times R$,the value $\min\{a, b\}$ is a unique real number that belongs to $R$.
Since every pair $(a, b)$ maps to a unique element in $R$,$\wedge$ is also a binary operation.

(N/A) binary operation $\ast$ on a set $S$ is commutative if $a \ast b = b \ast a$ for all $a, b \in S$.
For addition $(+)$ on $R$: Since $a + b = b + a$ for all $a, b \in R$,the operation $+$ is commutative.
For multiplication $(\times)$ on $R$: Since $a \times b = b \times a$ for all $a, b \in R$,the operation $\times$ is commutative.
For subtraction $(-)$ on $R$: Since $a - b \neq b - a$ in general (e.g.,$3 - 4 = -1$ and $4 - 3 = 1$),the operation $-$ is not commutative.
For division $(\div)$ on $R_*$: Since $a \div b \neq b \div a$ in general (e.g.,$3 \div 4 = 0.75$ and $4 \div 3 = 1.33$),the operation $\div$ is not commutative.

(N/A) To check if the operation $*$ is commutative,we need to verify if $a * b = b * a$ for all $a, b \in \mathbb{R}$.
Given $a * b = a + 2b$.
Let us take two real numbers,$a = 3$ and $b = 4$.
Then,$a * b = 3 * 4 = 3 + 2(4) = 3 + 8 = 11$.
Now,calculate $b * a = 4 * 3 = 4 + 2(3) = 4 + 6 = 10$.
Since $a * b \neq b * a$ (because $11 \neq 10$),the operation $*$ is not commutative.

For any $a, b, c \in R$,addition is associative because $(a + b) + c = a + (b + c)$.
Multiplication is associative because $(a \times b) \times c = a \times (b \times c)$.
Subtraction is not associative because $(8 - 5) - 3 = 3 - 3 = 0$,while $8 - (5 - 3) = 8 - 2 = 6$. Since $0 \neq 6$,subtraction is not associative.
Division is not associative on $R_*$ (where $R_* = R \setminus \{0\}$) because $(8 \div 5) \div 3 = \frac{8}{5} \div 3 = \frac{8}{15}$,while $8 \div (5 \div 3) = 8 \div \frac{5}{3} = \frac{24}{5}$. Since $\frac{8}{15} \neq \frac{24}{5}$,division is not associative.

(N/A) An operation $*$ is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in R$.
First,calculate $(a * b) * c$:
$(a * b) * c = (a + 2b) * c = (a + 2b) + 2c = a + 2b + 2c$.
Next,calculate $a * (b * c)$:
$a * (b * c) = a * (b + 2c) = a + 2(b + 2c) = a + 2b + 4c$.
Since $a + 2b + 2c \neq a + 2b + 4c$ for all $a, b, c \in R$,the operation is not associative.
For example,let $a = 8, b = 5, c = 3$:
$(8 * 5) * 3 = (8 + 2(5)) * 3 = 18 * 3 = 18 + 2(3) = 18 + 6 = 24$.
$8 * (5 * 3) = 8 * (5 + 2(3)) = 8 * (5 + 6) = 8 * 11 = 8 + 2(11) = 8 + 22 = 30$.
Since $24 \neq 30$,the operation is not associative.

(N/A) For addition,$a+0 = 0+a = a$ for all $a \in R$. Thus,$0$ is the identity element for addition.
For multiplication,$a \times 1 = 1 \times a = a$ for all $a \in R$. Thus,$1$ is the identity element for multiplication.
For subtraction,an identity element $e$ must satisfy $a-e = a$ and $e-a = a$ for all $a \in R$. The first implies $e=0$,but the second implies $e=2a$,which is not constant for all $a$. Thus,no identity exists.
For division,an identity element $e$ must satisfy $a \div e = a$ and $e \div a = a$ for all $a \in R_*$. The first implies $e=1$,but the second implies $e=a^2$,which is not constant for all $a$. Thus,no identity exists.

(N/A) For the addition operation '$+$' on the set of real numbers $R$,the identity element is $0$.
Since $a + (-a) = 0$ and $(-a) + a = 0$,it follows that $-a$ is the additive inverse of $a$.
For the multiplication operation '$\times$' on the set of non-zero real numbers $R \setminus \{0\}$,the identity element is $1$.
Since $a \times \frac{1}{a} = 1$ and $\frac{1}{a} \times a = 1$ for $a \neq 0$,it follows that $\frac{1}{a}$ is the multiplicative inverse of $a$.

(A) For the addition operation $+$ on $N$,the identity element is $0$. However,$0 \notin N$. Even if we consider the condition $a + (-a) = 0$,the element $-a$ must belong to the set $N$ to be an inverse. Since for any $a \in N$,$-a$ is a negative integer,$-a \notin N$. Therefore,$-a$ is not the inverse of $a$ in $N$.
For the multiplication operation $\times$ on $N$,the identity element is $1$. For an element $a \in N$ to have an inverse $b \in N$,we must have $a \times b = 1$. This implies $b = \frac{1}{a}$. For any $a \in N$ where $a \neq 1$,$\frac{1}{a}$ is not an integer,so $\frac{1}{a} \notin N$. Thus,no element $a \in N$ except $a = 1$ has a multiplicative inverse in $N$.

(B) On the set of positive integers $Z^{+}$,the operation $*$ is defined by $a * b = a - b$.
$A$ binary operation $*$ on a set $S$ is a function $*: S \times S \to S$.
For the operation to be binary,for every pair $(a, b) \in Z^{+} \times Z^{+}$,the result $a * b$ must belong to $Z^{+}$.
Consider the elements $a = 1$ and $b = 2$,where $1, 2 \in Z^{+}$.
Then $a * b = 1 * 2 = 1 - 2 = -1$.
Since $-1 \notin Z^{+}$,the operation $*$ is not a binary operation on $Z^{+}$.

(A) On $Z^{+}$,$*$ is defined by $a * b = ab$.
It is observed that for any two elements $a, b \in Z^{+}$,their product $ab$ is also a positive integer,meaning $ab \in Z^{+}$.
Since the product of two positive integers is always a unique positive integer,the operation $*$ maps every pair $(a, b)$ to a unique element $a * b = ab$ in $Z^{+}$.
Therefore,$*$ is a binary operation on $Z^{+}$.

(A) On $R$,the operation $*$ is defined by $a * b = ab^2$.
For any two real numbers $a, b \in R$,the product $ab^2$ is also a real number because the set of real numbers $R$ is closed under multiplication.
Since for every pair $(a, b) \in R \times R$,there exists a unique element $ab^2 \in R$,the operation $*$ satisfies the definition of a binary operation.
Therefore,$*$ is a binary operation on $R$.

(B) On $Z^+$,the operation $*$ is defined by $a * b = |a - b|$.
For any two elements $a, b \in Z^+$,the result $|a - b|$ must be an element of $Z^+$ for $*$ to be a binary operation.
Consider $a = 1$ and $b = 1$. Both $1, 1 \in Z^+$.
Then $a * b = |1 - 1| = 0$.
Since $0 \notin Z^+$ (as $Z^+$ is the set of positive integers ${1, 2, 3, ...}$),the operation $*$ does not map every pair $(a, b)$ to an element in $Z^+$.
Therefore,$*$ is not a binary operation on $Z^+$.

(A) On $Z^{+}$,the operation $*$ is defined by $a * b = a$.
For any two elements $a, b \in Z^{+}$,the result of the operation $a * b$ is $a$.
Since $a$ is an element of $Z^{+}$,it follows that for every pair $(a, b) \in Z^{+} \times Z^{+}$,there exists a unique element $a * b$ in $Z^{+}$.
By definition,a binary operation on a set $S$ is a function from $S \times S$ to $S$.
Since $a * b = a \in Z^{+}$ for all $a, b \in Z^{+}$,the operation $*$ satisfies the condition of being a binary operation.
Therefore,$*$ is a binary operation on $Z^{+}$.

On $Z$,$^*$ is defined by $a ^* b = a - b$.
It can be observed that $1 ^* 2 = 1 - 2 = -1$ and $2 ^* 1 = 2 - 1 = 1$.
$\therefore 1 ^* 2 \neq 2 ^* 1$,where $1, 2 \in Z$.
Hence,the operation $^*$ is not commutative.
Also,we have:
$(1 ^* 2) ^* 3 = (1 - 2) ^* 3 = -1 ^* 3 = -1 - 3 = -4$.
$1 ^* (2 ^* 3) = 1 ^* (2 - 3) = 1 ^* (-1) = 1 - (-1) = 2$.
$\therefore (1 ^* 2) ^* 3 \neq 1 ^* (2 ^* 3)$,where $1, 2, 3 \in Z$.
Hence,the operation $^*$ is not associative.

(A) On $Q$,the operation $^*$ is defined by $a ^* b = ab + 1$.
For commutativity:
We know that $ab = ba$ for all $a, b \in Q$.
Therefore,$ab + 1 = ba + 1$ for all $a, b \in Q$.
This implies $a ^* b = b ^* a$ for all $a, b \in Q$.
Thus,the operation $^*$ is commutative.
For associativity:
We check if $(a ^* b) ^* c = a ^* (b ^* c)$ for all $a, b, c \in Q$.
Consider $(1 ^* 2) ^* 3 = (1 \times 2 + 1) ^* 3 = 3 ^* 3 = 3 \times 3 + 1 = 10$.
Consider $1 ^* (2 ^* 3) = 1 ^* (2 \times 3 + 1) = 1 ^* 7 = 1 \times 7 + 1 = 8$.
Since $(1 ^* 2) ^* 3 \neq 1 ^* (2 ^* 3)$,the operation $^*$ is not associative.

(N/A) On $Q$,the operation $^*$ is defined by $a ^* b = \frac{ab}{2}$.
Commutativity:
We know that $ab = ba$ for all $a, b \in Q$.
Therefore,$\frac{ab}{2} = \frac{ba}{2}$,which implies $a ^* b = b ^* a$ for all $a, b \in Q$.
Thus,the operation $^*$ is commutative.
Associativity:
For all $a, b, c \in Q$,we have:
$(a ^* b) ^* c = \left(\frac{ab}{2}\right) ^* c = \frac{(\frac{ab}{2})c}{2} = \frac{abc}{4}$.
Also,$a ^* (b ^* c) = a ^* (\frac{bc}{2}) = \frac{a(\frac{bc}{2})}{2} = \frac{abc}{4}$.
Since $(a ^* b) ^* c = a ^* (b ^* c)$ for all $a, b, c \in Q$,the operation $^*$ is associative.

(A) On $Z^+$,the operation $^*$ is defined by $a ^* b = 2^{ab}$.
For commutativity:
We check if $a ^* b = b ^* a$ for all $a, b \in Z^+$.
$a ^* b = 2^{ab}$
$b ^* a = 2^{ba}$
Since $ab = ba$ for all $a, b \in Z^+$,it follows that $2^{ab} = 2^{ba}$.
Thus,$a ^* b = b ^* a$.
Therefore,the operation $^*$ is commutative.
For associativity:
We check if $(a ^* b) ^* c = a ^* (b ^* c)$ for all $a, b, c \in Z^+$.
Consider $a=1, b=2, c=3$:
$(1 ^* 2) ^* 3 = (2^{1 \times 2}) ^* 3 = 4 ^* 3 = 2^{4 \times 3} = 2^{12} = 4096$.
$1 ^* (2 ^* 3) = 1 ^* (2^{2 \times 3}) = 1 ^* 2^6 = 1 ^* 64 = 2^{1 \times 64} = 2^{64}$.
Since $2^{12} \neq 2^{64}$,the operation is not associative.

(NONE) On $Z^+$,the operation $^*$ is defined by $a ^* b = a^b$.
$1$. Commutativity:
We check if $a ^* b = b ^* a$ for all $a, b \in Z^+$.
Consider $a = 1$ and $b = 2$.
$1 ^* 2 = 1^2 = 1$
$2 ^* 1 = 2^1 = 2$
Since $1 \neq 2$,$1 ^* 2 \neq 2 ^* 1$.
Therefore,the operation $^*$ is not commutative.
$2$. Associativity:
We check if $(a ^* b) ^* c = a ^* (b ^* c)$ for all $a, b, c \in Z^+$.
Consider $a = 2, b = 3, c = 4$.
$(2 ^* 3) ^* 4 = (2^3) ^* 4 = 8 ^* 4 = 8^4 = (2^3)^4 = 2^{12}$.
$2 ^* (3 ^* 4) = 2 ^* (3^4) = 2 ^* 81 = 2^{81}$.
Since $2^{12} \neq 2^{81}$,$(2 ^* 3) ^* 4 \neq 2 ^* (3 ^* 4)$.
Therefore,the operation $^*$ is not associative.

The binary operation is defined as $a ^* b = \frac{a}{b+1}$ for $a, b \in R - \{-1\}$.
$1$. Commutativity:
To check if $^*$ is commutative,we compare $a ^* b$ and $b ^* a$.
$a ^* b = \frac{a}{b+1}$
$b ^* a = \frac{b}{a+1}$
Since $\frac{a}{b+1} \neq \frac{b}{a+1}$ in general (e.g.,let $a=1, b=2$,then $1 ^* 2 = \frac{1}{3}$ and $2 ^* 1 = \frac{2}{2} = 1$),the operation $^*$ is not commutative.
$2$. Associativity:
To check if $^*$ is associative,we compare $(a ^* b) ^* c$ and $a ^* (b ^* c)$.
$(a ^* b) ^* c = (\frac{a}{b+1}) ^* c = \frac{\frac{a}{b+1}}{c+1} = \frac{a}{(b+1)(c+1)}$
$a ^* (b ^* c) = a ^* (\frac{b}{c+1}) = \frac{a}{\frac{b}{c+1} + 1} = \frac{a}{\frac{b+c+1}{c+1}} = \frac{a(c+1)}{b+c+1}$
Since $\frac{a}{(b+1)(c+1)} \neq \frac{a(c+1)}{b+c+1}$ in general (e.g.,let $a=1, b=2, c=3$,then $(1 ^* 2) ^* 3 = \frac{1}{(3)(4)} = \frac{1}{12}$ and $1 ^* (2 ^* 3) = \frac{1(4)}{2+3+1} = \frac{4}{6} = \frac{2}{3}$),the operation $^*$ is not associative.

(N/A) The binary operation $\wedge$ on the set $S = \{1, 2, 3, 4, 5\}$ is defined as $a \wedge b = \min\{a, b\}$ for all $a, b \in S$.
To construct the operation table,we calculate $a \wedge b$ for every pair $(a, b)$ where $a, b \in \{1, 2, 3, 4, 5\}$. The value in each cell is the minimum of the corresponding row and column headers.

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

(N/A) To compute $(2 \,^* \,3) \,^* \,4$:
From the table,$2 \,^* \,3 = 2$.
Then,$(2 \,^* \,3) \,^* \,4 = 2 \,^* \,4 = 2$.
To compute $2 \,^* \,(3 \,^* \,4)$:
From the table,$3 \,^* \,4 = 3$.
Then,$2 \,^* \,(3 \,^* \,4) = 2 \,^* \,3 = 2$.

(A) binary operation $*$ on a set $S$ is commutative if $a \,^*\, b = b \,^*\, a$ for all $a, b \in S$.
To check if the operation is commutative,we check if the table is symmetric with respect to the main diagonal.
Looking at the table:
- $1 \,^*\, 2 = 1$ and $2 \,^*\, 1 = 1$. Thus,$1 \,^*\, 2 = 2 \,^*\, 1$.
- $1 \,^*\, 3 = 1$ and $3 \,^*\, 1 = 1$. Thus,$1 \,^*\, 3 = 3 \,^*\, 1$.
- $2 \,^*\, 3 = 2$ and $3 \,^*\, 2 = 2$. Thus,$2 \,^*\, 3 = 3 \,^*\, 2$.
- $2 \,^*\, 4 = 2$ and $4 \,^*\, 2 = 2$. Thus,$2 \,^*\, 4 = 4 \,^*\, 2$.
- $3 \,^*\, 4 = 3$ and $4 \,^*\, 3 = 3$. Thus,$3 \,^*\, 4 = 4 \,^*\, 3$.
- $1 \,^*\, 5 = 1$ and $5 \,^*\, 1 = 1$. Thus,$1 \,^*\, 5 = 5 \,^*\, 1$.
- $2 \,^*\, 5 = 2$ and $5 \,^*\, 2 = 2$. Thus,$2 \,^*\, 5 = 5 \,^*\, 2$.
- $3 \,^*\, 5 = 3$ and $5 \,^*\, 3 = 3$. Thus,$3 \,^*\, 5 = 5 \,^*\, 3$.
- $4 \,^*\, 5 = 4$ and $5 \,^*\, 4 = 4$. Thus,$4 \,^*\, 5 = 5 \,^*\, 4$.
Since $a \,^*\, b = b \,^*\, a$ for all $a, b \in \{1,2,3,4,5\}$,the operation $^*$ is commutative.

(B) From the given table:
$1$. Find the value of $(2 \,^* \,3)$. Looking at the row for $2$ and column for $3$,we get $(2 \,^* \,3) = 2$.
$2$. Find the value of $(4 \,^* \,5)$. Looking at the row for $4$ and column for $5$,we get $(4 \,^* \,5) = 4$.
$3$. Now,compute the final expression: $(2 \,^* \,3) \,^* \,(4 \,^* \,5) = 2 \,^* \,4$.
$4$. Looking at the row for $2$ and column for $4$ in the table,we get $2 \,^* \,4 = 2$.
Thus,the final result is $2$.

(B) The binary operation $*^{\prime}$ on the set $\{1, 2, 3, 4, 5\}$ is defined as $a *^{\prime} b = \text{H.C.F. of } a \text{ and } b$.
The operation table for the operation $*^{\prime}$ is given below:

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

In Exercise $4$,the operation $*$ is defined as $a * b = \min\{a, b\}$.
Comparing the tables,we observe that the operation tables for the operations $*$ and $*^{\prime}$ are different.
Thus,the operation $*^{\prime}$ is not the same as the operation $*$.

(A) The binary operation $^*$ on $N$ is defined as $a \,^* \,b = \text{L.C.M. of } a \text{ and } b$.
For $5 \,^* \,7$:
$5 \,^* \,7 = \text{L.C.M. of } 5 \text{ and } 7 = 35$.
For $20 \,^* \,16$:
$20 \,^* \,16 = \text{L.C.M. of } 20 \text{ and } 16$.
Since $20 = 2^2 \times 5$ and $16 = 2^4$,the $L$.$C$.$M$. is $2^4 \times 5 = 16 \times 5 = 80$.
Thus,$5 \,^* \,7 = 35$ and $20 \,^* \,16 = 80$.

(A) The binary operation $^*$ is defined on the set of natural numbers $N$ as $a \, ^* \, b = \text{L.C.M. of } a \text{ and } b$.
For any $a, b \in N$,we know that the least common multiple of $a$ and $b$ is the same as the least common multiple of $b$ and $a$.
Therefore,$a \, ^* \, b = \text{L.C.M. of } a \text{ and } b = \text{L.C.M. of } b \text{ and } a = b \, ^* \, a$.
Since $a \, ^* \, b = b \, ^* \, a$ for all $a, b \in N$,the operation $^*$ is commutative.

(A) For $a, b, c \in N,$
$(a \,^*\, b) \,^*\, c = (\text{L.C.M. of } a \text{ and } b) \,^*\, c = \text{L.C.M. of } a, b, \text{ and } c$
$a \,^*\, (b \,^*\, c) = a \,^*\, (\text{L.C.M. of } b \text{ and } c) = \text{L.C.M. of } a, b, \text{ and } c$
$\therefore (a \,^*\, b) \,^*\, c = a \,^*\, (b \,^*\, c)$
Thus,the operation $^*$ is associative.

(A) The identity element $e$ for a binary operation $^*$ on a set $N$ is an element such that $a \, ^* \, e = a = e \, ^* \, a$ for all $a \in N$.
Given that $a \, ^* \, b = \text{L.C.M. of } a \text{ and } b$.
We know that the $\text{L.C.M.}$ of any natural number $a$ and $1$ is $a$,i.e.,$\text{L.C.M.}(a, 1) = a$ and $\text{L.C.M.}(1, a) = a$.
Therefore,$a \, ^* \, 1 = a = 1 \, ^* \, a$ for all $a \in N$.
Thus,$1$ is the identity element of $^*$ in $N$.

(A) An element $a$ in $N$ is invertible with respect to the operation $^*$ if there exists an element $b$ in $N$ such that $a ^* b = e = b ^* a$,where $e$ is the identity element.
For the operation of $L.C.M.$ on the set of natural numbers $N$,the identity element $e$ must satisfy $L.C.M.(a, e) = a$ for all $a \in N$. This holds true for $e = 1$.
Thus,we need to find $b \in N$ such that $L.C.M.(a, b) = 1$.
Since $L.C.M.(a, b) \geq a$ and $L.C.M.(a, b) \geq b$ for all $a, b \in N$,the condition $L.C.M.(a, b) = 1$ implies that $a = 1$ and $b = 1$.
Therefore,$1$ is the only invertible element of $N$ with respect to the operation $^*$.

(D) binary operation $^*$ on a set $A$ is a function $^*: A \times A \to A$. This means that for all $a, b \in A$,the result $a \,^*\, b$ must also be an element of $A$.
Let $A = \{1, 2, 3, 4, 5\}$. The operation is defined as $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$.
For the operation to be a binary operation,$a \,^*\, b$ must belong to $A$ for all $a, b \in A$.
Let us test this with some elements from the set $A$:
Consider $a = 2$ and $b = 3$. Both $2, 3 \in A$.
$2 \,^*\, 3 = \text{L.C.M. of } 2 \text{ and } 3 = 6$.
Since $6 \notin \{1, 2, 3, 4, 5\}$,the result of the operation is not in the set $A$.
Similarly,$2 \,^*\, 5 = 10 \notin A$,$3 \,^*\, 4 = 12 \notin A$,etc.
Since there exist elements $a, b \in A$ such that $a \,^*\, b \notin A$,the operation $^*$ is not a binary operation on the set $\{1, 2, 3, 4, 5\}$.

(N/A) The binary operation $^*$ on $N$ is defined as: $a \,^*\, b = \text{H.C.F. of } a \text{ and } b$.
$1$. Commutativity:
We know that the $\text{H.C.F.}$ of $a$ and $b$ is the same as the $\text{H.C.F.}$ of $b$ and $a$ for all $a, b \in N$.
Therefore,$a \,^*\, b = b \,^*\, a$.
Thus,the operation $^*$ is commutative.
$2$. Associativity:
For $a, b, c \in N$,we have:
$(a \,^*\, b) \,^*\, c = (\text{H.C.F. of } a \text{ and } b) \,^*\, c = \text{H.C.F. of } a, b, \text{ and } c$.
$a \,^*\, (b \,^*\, c) = a \,^*\, (\text{H.C.F. of } b \text{ and } c) = \text{H.C.F. of } a, b, \text{ and } c$.
Since $(a \,^*\, b) \,^*\, c = a \,^*\, (b \,^*\, c)$,the operation $^*$ is associative.
$3$. Identity Element:
An element $e \in N$ is the identity for $^*$ if $a \,^*\, e = a = e \,^*\, a$ for all $a \in N$.
This implies $\text{H.C.F.}(a, e) = a$,which means $a$ must be a divisor of $e$ for all $a \in N$. Since there is no such fixed element $e \in N$ that is a multiple of every natural number,there is no identity element for this operation.

On $Q,$ the operation $^*$ is defined as $a ^* b = a - b.$
$1$. Commutativity:
For the operation to be commutative,$a ^* b = b ^* a$ must hold for all $a, b \in Q.$
Consider $a = \frac{1}{2}$ and $b = \frac{1}{3}.$
$\frac{1}{2} ^* \frac{1}{3} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6}$
$\frac{1}{3} ^* \frac{1}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6}$
Since $\frac{1}{6} \neq -\frac{1}{6},$ the operation $^*$ is not commutative.
$2$. Associativity:
For the operation to be associative,$(a ^* b) ^* c = a ^* (b ^* c)$ must hold for all $a, b, c \in Q.$
Consider $a = \frac{1}{2}, b = \frac{1}{3}, c = \frac{1}{4}.$
$(a ^* b) ^* c = (\frac{1}{2} - \frac{1}{3}) ^* \frac{1}{4} = \frac{1}{6} ^* \frac{1}{4} = \frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12}$
$a ^* (b ^* c) = \frac{1}{2} ^* (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} ^* \frac{1}{12} = \frac{1}{2} - \frac{1}{12} = \frac{6 - 1}{12} = \frac{5}{12}$
Since $-\frac{1}{12} \neq \frac{5}{12},$ the operation $^*$ is not associative.

(B) On $Q$,the operation $^*$ is defined as $a \,^* \,b = a^{2} + b^{2}$.
For $a, b \in Q$,we have:
$a \,^* \,b = a^{2} + b^{2} = b^{2} + a^{2} = b \,^* \,a$.
Therefore,$a \,^* \,b = b \,^* \,a$,which means the operation $^*$ is commutative.
Now,check for associativity:
$(1 \,^* \,2) \,^* \,3 = (1^{2} + 2^{2}) \,^* \,3 = (1 + 4) \,^* \,3 = 5 \,^* \,3 = 5^{2} + 3^{2} = 25 + 9 = 34$.
$1 \,^* \,(2 \,^* \,3) = 1 \,^* \,(2^{2} + 3^{2}) = 1 \,^* \,(4 + 9) = 1 \,^* \,13 = 1^{2} + 13^{2} = 1 + 169 = 170$.
Since $(1 \,^* \,2) \,^* \,3 \neq 1 \,^* \,(2 \,^* \,3)$,the operation $^*$ is not associative.
Thus,the operation $^*$ is commutative but not associative.

(D) On $Q$,the operation $^*$ is defined as $a * b = a + ab$.
To check for commutativity:
$1 * 2 = 1 + (1 \times 2) = 1 + 2 = 3$
$2 * 1 = 2 + (2 \times 1) = 2 + 2 = 4$
Since $1 * 2 \neq 2 * 1$,the operation $^*$ is not commutative.
To check for associativity:
$(1 * 2) * 3 = (1 + 1 \times 2) * 3 = 3 * 3 = 3 + (3 \times 3) = 3 + 9 = 12$
$1 * (2 * 3) = 1 * (2 + 2 \times 3) = 1 * 8 = 1 + (1 \times 8) = 1 + 8 = 9$
Since $(1 * 2) * 3 \neq 1 * (2 * 3)$,the operation $^*$ is not associative.
Thus,the operation $^*$ is neither commutative nor associative.

(B) On the set $Q$,the operation $^*$ is defined by $a * b = (a - b)^2$.
For commutativity,we check if $a * b = b * a$ for all $a, b \in Q$:
$a * b = (a - b)^2$
$b * a = (b - a)^2 = [-(a - b)]^2 = (a - b)^2$
Since $a * b = b * a$,the operation $^*$ is commutative.
For associativity,we check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in Q$:
Consider $a = 1, b = 2, c = 3$:
$(1 * 2) * 3 = (1 - 2)^2 * 3 = (-1)^2 * 3 = 1 * 3 = (1 - 3)^2 = (-2)^2 = 4$
$1 * (2 * 3) = 1 * (2 - 3)^2 = 1 * (-1)^2 = 1 * 1 = (1 - 1)^2 = 0$
Since $(1 * 2) * 3 \neq 1 * (2 * 3)$,the operation $^*$ is not associative.

(C) On the set $Q$,the operation $*$ is defined as $a * b = \frac{ab}{4}$.
For commutativity,consider $a, b \in Q$:
$a * b = \frac{ab}{4} = \frac{ba}{4} = b * a$.
Since $a * b = b * a$,the operation is commutative.
For associativity,consider $a, b, c \in Q$:
$(a * b) * c = (\frac{ab}{4}) * c = \frac{(\frac{ab}{4})c}{4} = \frac{abc}{16}$.
$a * (b * c) = a * (\frac{bc}{4}) = \frac{a(\frac{bc}{4})}{4} = \frac{abc}{16}$.
Since $(a * b) * c = a * (b * c)$,the operation is associative.
Thus,the operation is both commutative and associative.

On the set $Q$,the operation $^*$ is defined as $a \,^*\, b = a b^{2}$.
To check for commutativity,we compare $a \,^*\, b$ and $b \,^*\, a$.
Consider $a = \frac{1}{2}$ and $b = \frac{1}{3}$ in $Q$.
$\frac{1}{2} \,^*\, \frac{1}{3} = \frac{1}{2} \times (\frac{1}{3})^{2} = \frac{1}{2} \times \frac{1}{9} = \frac{1}{18}$.
$\frac{1}{3} \,^*\, \frac{1}{2} = \frac{1}{3} \times (\frac{1}{2})^{2} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.
Since $\frac{1}{18} \neq \frac{1}{12}$,the operation $^*$ is not commutative.
To check for associativity,we compare $(a \,^*\, b) \,^*\, c$ and $a \,^*\, (b \,^*\, c)$.
Consider $a = \frac{1}{2}, b = \frac{1}{3}, c = \frac{1}{4}$ in $Q$.
$(\frac{1}{2} \,^*\, \frac{1}{3}) \,^*\, \frac{1}{4} = (\frac{1}{2} \times (\frac{1}{3})^{2}) \,^*\, \frac{1}{4} = \frac{1}{18} \,^*\, \frac{1}{4} = \frac{1}{18} \times (\frac{1}{4})^{2} = \frac{1}{18 \times 16} = \frac{1}{288}$.
$\frac{1}{2} \,^*\, (\frac{1}{3} \,^*\, \frac{1}{4}) = \frac{1}{2} \,^*\, (\frac{1}{3} \times (\frac{1}{4})^{2}) = \frac{1}{2} \,^*\, \frac{1}{48} = \frac{1}{2} \times (\frac{1}{48})^{2} = \frac{1}{2 \times 2304} = \frac{1}{4608}$.
Since $\frac{1}{288} \neq \frac{1}{4608}$,the operation $^*$ is not associative.

(N/A) An element $e \in Q$ is the identity element for a binary operation $^*$ if $a * e = a = e * a$ for all $a \in Q$.
For the six operations typically defined on the set of rational numbers $Q$ (such as $a * b = a + b + ab$,$a * b = a - b$,etc.),we test the condition $a * e = a$.
If we solve $a * e = a$ for each operation,we find that the resulting value of $e$ depends on $a$ or does not exist within the set $Q$ for all $a$.
Since there is no single element $e \in Q$ that satisfies the condition $a * e = a = e * a$ for all $a \in Q$ across these operations,none of the operations has an identity element.

(A) Given $A = N \times N$ and the binary operation $^*$ defined by $(a, b) \,^*\, (c, d) = (a + c, b + d)$.
$1$. Commutativity:
For any $(a, b), (c, d) \in A$,we have:
$(a, b) \,^*\, (c, d) = (a + c, b + d)$
$(c, d) \,^*\, (a, b) = (c + a, d + b) = (a + c, b + d)$
Since addition is commutative in $N$,$(a, b) \,^*\, (c, d) = (c, d) \,^*\, (a, b)$. Thus,$^*$ is commutative.
$2$. Associativity:
For any $(a, b), (c, d), (e, f) \in A$,we have:
$[(a, b) \,^*\, (c, d)] \,^*\, (e, f) = (a + c, b + d) \,^*\, (e, f) = (a + c + e, b + d + f)$
$(a, b) \,^*\, [(c, d) \,^*\, (e, f)] = (a, b) \,^*\, (c + e, d + f) = (a + c + e, b + d + f)$
Since addition is associative in $N$,the operation $^*$ is associative.
$3$. Identity Element:
Let $e = (e_1, e_2) \in A$ be the identity element. Then $(a, b) \,^*\, (e_1, e_2) = (a, b)$,which implies $(a + e_1, b + e_2) = (a, b)$.
This requires $a + e_1 = a$ and $b + e_2 = b$,so $e_1 = 0$ and $e_2 = 0$.
Since $0 \notin N$,there is no identity element in $A$.

(B) The statement is false.
To justify this,consider a counterexample.
Let the binary operation $^*$ on the set of natural numbers $N$ be defined as $a \,^* \,b = a + b$ for all $a, b \in N$.
Now,consider an element $a = 3 \in N$.
According to the given condition,we should have $3 \,^* \,3 = 3$.
However,using our defined operation,we get $3 \,^* \,3 = 3 + 3 = 6$.
Since $6 \neq 3$,the condition $a \,^* \,a = a$ does not hold for all binary operations on $N$.
Thus,the statement is false.

(A) Given that $^*$ is a commutative binary operation on $N$,which means $x ^* y = y ^* x$ for all $x, y \in N$.
Consider the right-hand side $(RHS)$:
$RHS = (c ^* b) ^* a$
Since $^*$ is commutative,we can write $(c ^* b) = (b ^* c)$.
So,$RHS = (b ^* c) ^* a$.
Now,by the commutative property,we can swap the elements around the operation $^*$:
$(b ^* c) ^* a = a ^* (b ^* c)$.
Thus,$RHS = a ^* (b ^* c) = LHS$.
Therefore,the statement is true.

(C) On $N$,the operation $*$ is defined as $a * b = a^{3} + b^{3}$.
For $a, b \in N$,we have:
$a * b = a^{3} + b^{3} = b^{3} + a^{3} = b * a$ (Since addition is commutative in $N$).
Therefore,the operation $*$ is commutative.
Now,check for associativity:
$(1 * 2) * 3 = (1^{3} + 2^{3}) * 3 = (1 + 8) * 3 = 9 * 3 = 9^{3} + 3^{3} = 729 + 27 = 756$.
$1 * (2 * 3) = 1 * (2^{3} + 3^{3}) = 1 * (8 + 27) = 1 * 35 = 1^{3} + 35^{3} = 1 + 42875 = 42876$.
Since $(1 * 2) * 3 \neq 1 * (2 * 3)$,the operation $*$ is not associative.
Thus,the operation $*$ is commutative but not associative. The correct answer is $C$.

(A) $1$. Commutativity: $A$ binary operation $\ast$ on a set $N$ is commutative if $a \ast b = b \ast a$ for all $a, b \in N$. Here,$a \ast b = 1$ and $b \ast a = 1$. Since $1 = 1$,the operation is commutative.
$2$. Associativity: $A$ binary operation $\ast$ on a set $N$ is associative if $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a, b, c \in N$. Here,$(a \ast b) \ast c = 1 \ast c = 1$. Also,$a \ast (b \ast c) = a \ast 1 = 1$. Since $1 = 1$,the operation is associative.
Therefore,the operation is both associative and commutative.

(N/A) For commutativity,we check if $a * b = b * a$ for all $a, b \in N$.
$a * b = \frac{a+b}{2} = \frac{b+a}{2} = b * a$.
Since $a * b = b * a$,the operation is commutative.
For associativity,we check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in N$.
$(a * b) * c = \left(\frac{a+b}{2}\right) * c = \frac{\frac{a+b}{2} + c}{2} = \frac{a+b+2c}{4}$.
$a * (b * c) = a * \left(\frac{b+c}{2}\right) = \frac{a + \frac{b+c}{2}}{2} = \frac{2a+b+c}{4}$.
Since $\frac{a+b+2c}{4} \neq \frac{2a+b+c}{4}$ in general,the operation is not associative.

(A) binary operation $^*$ on $\{1, 2\}$ is a function from $\{1, 2\} \times \{1, 2\}$ to $\{1, 2\}$,i.e.,a function from $\{(1, 1), (1, 2), (2, 1), (2, 2)\}$ to $\{1, 2\}$.
Since $1$ is the identity for the binary operation $^*$,we must have $1 * 1 = 1$,$1 * 2 = 2$,and $2 * 1 = 2$.
This determines the values for the pairs $(1, 1)$,$(1, 2)$,and $(2, 1)$.
For the pair $(2, 2)$,we are given that $2$ is the inverse of $2$. By definition of inverse,$2 * 2$ must equal the identity element,which is $1$.
Thus,$2 * 2 = 1$.
Since all values of the function $^*$ are uniquely determined,there is exactly one such binary operation.

(A) It is given that the binary operation $^*: P(X) \times P(X) \rightarrow P(X)$ is defined by $A \,^*\, B = A \cap B$ for all $A, B \in P(X)$.
We know that for any set $A \in P(X)$,$A \cap X = A$ and $X \cap A = A$.
This implies $A \,^*\, X = A$ and $X \,^*\, A = A$ for all $A \in P(X)$.
Thus,$X$ is the identity element for the given binary operation $^*$.
Now,an element $A \in P(X)$ is invertible if there exists an element $B \in P(X)$ such that $A \,^*\, B = X$ and $B \,^*\, A = X$ (since $X$ is the identity element).
This means $A \cap B = X$ and $B \cap A = X$.
Since $A \subseteq X$ and $B \subseteq X$,the intersection $A \cap B$ can only be equal to $X$ if $A = X$ and $B = X$.
Therefore,$X$ is the only invertible element in $P(X)$ with respect to the given operation $^*$.
Hence,the result is proved.