Binary Operation Questions in English

(A) Given $a \,^*\, b = |a-b|$ and $a \,o\, b = a$ for all $a, b \in R$.
For $^*$,$a \,^*\, b = |a-b|$ and $b \,^*\, a = |b-a| = |-(a-b)| = |a-b|$. Since $a \,^*\, b = b \,^*\, a$,$^*$ is commutative.
For associativity of $^*$,consider $(1 \,^*\, 2) \,^*\, 3 = |1-2| \,^*\, 3 = 1 \,^*\, 3 = |1-3| = 2$,while $1 \,^*\, (2 \,^*\, 3) = 1 \,^*\, |2-3| = 1 \,^*\, 1 = |1-1| = 0$. Since $2 \neq 0$,$^*$ is not associative.
For $o$,$1 \,o\, 2 = 1$ and $2 \,o\, 1 = 2$. Since $1 \neq 2$,$o$ is not commutative.
For associativity of $o$,$(a \,o\, b) \,o\, c = a \,o\, c = a$ and $a \,o\, (b \,o\, c) = a \,o\, b = a$. Since both sides equal $a$,$o$ is associative.
For distributivity of $^*$ over $o$,$a \,^*\, (b \,o\, c) = a \,^*\, b = |a-b|$ and $(a \,^*\, b) \,o\, (a \,^*\, c) = |a-b| \,o\, |a-c| = |a-b|$. Thus,$^*$ distributes over $o$.
For distributivity of $o$ over $^*$,consider $1 \,o\, (2 \,^*\, 3) = 1 \,o\, |2-3| = 1 \,o\, 1 = 1$,while $(1 \,o\, 2) \,^*\, (1 \,o\, 3) = 1 \,^*\, 1 = |1-1| = 0$. Since $1 \neq 0$,$o$ does not distribute over $^*$.

(A) It is given that $^*: P(X) \times P(X) \rightarrow P(X)$ is defined as $A \,^*\, B = (A - B) \cup (B - A)$ for all $A, B \in P(X)$.
$1$. Identity Element:
For any $A \in P(X)$,we have:
$A \,^*\, \Phi = (A - \Phi) \cup (\Phi - A) = A \cup \Phi = A$
$\Phi \,^*\, A = (\Phi - A) \cup (A - \Phi) = \Phi \cup A = A$
Since $A \,^*\, \Phi = A = \Phi \,^*\, A$ for all $A \in P(X)$,the empty set $\Phi$ is the identity element for the operation $^*$.
$2$. Invertible Elements:
An element $A \in P(X)$ is invertible if there exists $B \in P(X)$ such that $A \,^*\, B = \Phi = B \,^*\, A$,where $\Phi$ is the identity element.
Consider $A \,^*\, A$:
$A \,^*\, A = (A - A) \cup (A - A) = \Phi \cup \Phi = \Phi$
Since $A \,^*\, A = \Phi$,every element $A \in P(X)$ is its own inverse.
Thus,all elements $A$ of $P(X)$ are invertible with $A^{-1} = A$.

(A) Let $X = \{0, 1, 2, 3, 4, 5\}$.
The operation $^*$ on $X$ is defined as $a \,^* \, b = \begin{cases} a+b, & \text{If } a+b < 6 \\ a+b-6, & \text{If } a+b \geq 6 \end{cases}$.
An element $e \in X$ is the identity element for the operation $^*$ if $a \,^* \, e = a = e \,^* \, a$ for all $a \in X$.
For $a \in X$,we have:
$a \,^* \, 0 = a + 0 = a$ (since $a \in X \Rightarrow a+0 < 6$)
$0 \,^* \, a = 0 + a = a$ (since $a \in X \Rightarrow 0+a < 6$)
Therefore,$a \,^* \, 0 = a = 0 \,^* \, a$ for all $a \in X$.
Thus,$0$ is the identity element for the given operation $^*$.
An element $a \in X$ is invertible if there exists $b \in X$ such that $a \,^* \, b = 0 = b \,^* \, a$.
If $a \neq 0$,let $b = 6-a$. Since $a \in \{1, 2, 3, 4, 5\}$,$b \in \{5, 4, 3, 2, 1\} \subset X$.
Then $a \,^* \, b = a + (6-a) - 6 = 0$ (since $a+b = 6 \geq 6$).
Similarly,$b \,^* \, a = (6-a) + a - 6 = 0$.
Thus,$6-a$ is the inverse of $a$ for all $a \in X, a \neq 0$.

(C) binary operation $^*$ on a set $S$ is a function from $S \times S$ to $S$.
Here,the set is $S = \{a, b\}$,so $S \times S = \{(a, a), (a, b), (b, a), (b, b)\}$.
The number of elements in $S \times S$ is $n(S \times S) = 2 \times 2 = 4$.
The number of elements in the codomain $S$ is $n(S) = 2$.
The total number of functions from $S \times S$ to $S$ is given by $(n(S))^{n(S \times S)} = 2^4$.
Calculating this,we get $2^4 = 16$.
Thus,the total number of binary operations on the set $\{a, b\}$ is $16$.
The correct answer is $C$.

(A) Let $e$ be the identity element for the operation $*$.
By definition,$a * e = a = e * a$.
Given $a * b = 10ab$,we have $10ae = a$.
Since $a \in Q^{+}$,$a \neq 0$,so $10e = 1$,which gives $e = \frac{1}{10}$.
Let $a^{\prime}$ be the inverse of $0.01$.
By definition,$0.01 * a^{\prime} = e$.
Substituting the values,$10 \times 0.01 \times a^{\prime} = \frac{1}{10}$.
$10 \times \frac{1}{100} \times a^{\prime} = \frac{1}{10}$.
$0.1 \times a^{\prime} = 0.1$.
Therefore,$a^{\prime} = 1$.

(C) The operation $a * b$ is defined as the remainder when $ab$ is divided by $7$,where $a, b \in A$.
First,we find the identity element $e \in A$ such that $a * e = a$ for all $a \in A$.
This means the remainder of $ae$ divided by $7$ is $a$.
For $a \in \{1, 2, 3, 4, 5, 6\}$,$a * 1 = \text{remainder of } (a \times 1) / 7 = a$.
Thus,the identity element $e = 1$.
To find the inverse of $2$,we need an element $x \in A$ such that $2 * x = e = 1$.
This means the remainder of $(2 \times x)$ divided by $7$ is $1$.
Testing values from $A$:
$2 * 1 = 2$
$2 * 2 = 4$
$2 * 3 = 6$
$2 * 4 = 8 \equiv 1 \pmod{7}$
Since $2 * 4 = 1$,the inverse of $2$ is $4$.

(B) binary operation $*$ on a set $S$ is commutative if $a * b = b * a$ for all $a, b \in S$.
Given the operation $a * b = a - b$ for all $a, b \in Q$.
To check for commutativity,we calculate $b * a$:
$b * a = b - a$.
Now,compare $a * b$ and $b * a$:
$a * b = a - b$ and $b * a = b - a$.
Since $a - b \neq b - a$ in general (for example,let $a = 5$ and $b = 3$,then $5 - 3 = 2$ while $3 - 5 = -2$,and $2 \neq -2$),the operation is not commutative.
Therefore,the binary operation $*$ is not commutative.

(A) The binary operation is defined as $a * b = a^{2} + b^{2}$.
To check for commutativity,we need to verify if $a * b = b * a$ for all $a, b \in Q$.
Given $a * b = a^{2} + b^{2}$.
Since addition is commutative in the set of rational numbers,we have $a^{2} + b^{2} = b^{2} + a^{2}$.
By the definition of the operation,$b^{2} + a^{2} = b * a$.
Therefore,$a * b = b * a$ for all $a, b \in Q$.
Thus,the binary operation $*$ is commutative.

(B) The binary operation is defined as $a * b = a + ab$.
To check for commutativity,we compare $a * b$ with $b * a$.
By definition,$a * b = a + ab$.
Similarly,$b * a = b + ba = b + ab$.
For the operation to be commutative,we must have $a * b = b * a$ for all $a, b \in Q$.
This implies $a + ab = b + ab$.
Subtracting $ab$ from both sides,we get $a = b$.
Since $a = b$ is not true for all $a, b \in Q$,the operation is not commutative.
Example: Let $a = 3$ and $b = 2$.
$a * b = 3 * 2 = 3 + 3(2) = 3 + 6 = 9$.
$b * a = 2 * 3 = 2 + 2(3) = 2 + 6 = 8$.
Since $9 \neq 8$,$a * b \neq b * a$.

(A) binary operation $*$ is commutative if $a * b = b * a$ for all $a, b \in Q$.
Given $a * b = (a - b)^{2}$.
Now,calculate $b * a$:
$b * a = (b - a)^{2}$.
Since $(b - a)^{2} = [-(a - b)]^{2} = (-1)^{2}(a - b)^{2} = (a - b)^{2}$.
Therefore,$b * a = (a - b)^{2} = a * b$.
Since $a * b = b * a$ for all $a, b \in Q$,the operation $*$ is commutative.

(A) Given the binary operation $a * b = 1 + ab$ for all $a, b \in R$.
$1$. Commutativity:
$a * b = 1 + ab$
$b * a = 1 + ba = 1 + ab$
Since $a * b = b * a$,the operation $*$ is commutative.
$2$. Associativity:
Consider $a * (b * c) = a * (1 + bc) = 1 + a(1 + bc) = 1 + a + abc$.
Now consider $(a * b) * c = (1 + ab) * c = 1 + (1 + ab)c = 1 + c + abc$.
Since $1 + a + abc \neq 1 + c + abc$ for all $a, b, c \in R$,the operation $*$ is not associative.
Therefore,the operation $*$ is commutative but not associative.

(B) Given the binary operation $a * b = \frac{2ab}{5}$.
First,find the identity element $e$ such that $a * e = a$:
$\frac{2ae}{5} = a \implies e = \frac{5}{2}$.
Next,find the inverse $a^{-1}$ such that $a * a^{-1} = e$:
$\frac{2a(a^{-1})}{5} = \frac{5}{2} \implies a^{-1} = \frac{25}{4a}$.
For $a = 3$,the inverse is $3^{-1} = \frac{25}{4(3)} = \frac{25}{12}$.
Now,solve $2 * x = 3^{-1}$:
$\frac{2(2x)}{5} = \frac{25}{12} \implies \frac{4x}{5} = \frac{25}{12}$.
$x = \frac{25 \times 5}{12 \times 4} = \frac{125}{48}$.

(C) Given the binary operation $a * b = \frac{a}{b+1}$.
Check for commutativity:
$a * b = \frac{a}{b+1}$
$b * a = \frac{b}{a+1}$
Since $\frac{a}{b+1} \neq \frac{b}{a+1}$ for all $a, b \in R - \{-1\}$,the operation is not commutative.
Check for associativity:
$(a * b) * c = \left(\frac{a}{b+1}\right) * c = \frac{\frac{a}{b+1}}{c+1} = \frac{a}{(b+1)(c+1)}$
$a * (b * c) = a * \left(\frac{b}{c+1}\right) = \frac{a}{\frac{b}{c+1} + 1} = \frac{a(c+1)}{b+c+1}$
Since $\frac{a}{(b+1)(c+1)} \neq \frac{a(c+1)}{b+c+1}$,the operation is not associative.
Therefore,the operation $*$ is neither commutative nor associative.

(D) Given the operation $ a \oplus b = a^{2} + b^{2} $.
First,calculate $ (2 \oplus 3) $:
$ 2 \oplus 3 = 2^{2} + 3^{2} = 4 + 9 = 13 $.
Now,substitute this result into the expression $ (2 \oplus 3) \oplus 4 $:
$ 13 \oplus 4 = 13^{2} + 4^{2} $.
$ 13^{2} + 4^{2} = 169 + 16 = 185 $.
Thus,the final result is $ 185 $.

(B) Given that,$a * b = \frac{a+b}{4}$.
For commutativity,we check $b * a = \frac{b+a}{4} = \frac{a+b}{4} = a * b$. Thus,the operation is commutative.
For associativity,we check $a * (b * c)$ and $(a * b) * c$.
$a * (b * c) = a * \left(\frac{b+c}{4}\right) = \frac{a + \frac{b+c}{4}}{4} = \frac{4a + b + c}{16}$.
$(a * b) * c = \left(\frac{a+b}{4}\right) * c = \frac{\frac{a+b}{4} + c}{4} = \frac{a + b + 4c}{16}$.
Since $\frac{4a + b + c}{16} \neq \frac{a + b + 4c}{16}$,the operation is not associative.
Therefore,the operation is commutative but not associative.

(A) Given that $a^{2} = e$ in the group $(G, *)$.
Multiplying both sides by $a^{-1}$ (the inverse of $a$):
$a^{-1} * (a * a) = a^{-1} * e$
$(a^{-1} * a) * a = a^{-1}$
$e * a = a^{-1}$
$a = a^{-1}$

(D) Given,in a group $(Z, *)$,the operation is defined as $a * b = a + b - n$ for all $a, b \in Z$.
First,we find the identity element $e$ such that $a * e = a$.
$a + e - n = a \implies e = n$.
Now,to find the inverse of $(-n)$,let the inverse be $x$ such that $(-n) * x = e$.
Since $e = n$,we have $(-n) * x = n$.
Using the definition of the operation:
$(-n) + x - n = n$.
$x - 2n = n$.
$x = 3n$.
Thus,the inverse of $(-n)$ is $3n$.

(A) set $G$ with a binary operation $\cdot$ is a group if it satisfies closure,associativity,identity,and inverse properties.
For $(N, \cdot)$,where $N$ is the set of natural numbers,the multiplicative inverse of an element $a \in N$ (where $a \neq 1$) is $1/a$,which is not in $N$.
Thus,$(N, \cdot)$ fails the inverse property and is not a group.
$(N, +)$ is a semi-group because it is closed and associative under addition.
$(Z, +)$ is a group as it satisfies all group axioms.
The set of even integers $2Z = \{..., -4, -2, 0, 2, 4, ...\}$ forms a group under addition because it contains the identity $0$,inverses for all elements,and is closed and associative.
Therefore,the statement $(N, \cdot)$ is a group is false.

(D) subset $H$ of a group $G$ is a subgroup if it is itself a group under the same operation.
For $H = \{4^{n} \mid n \in \mathbb{Z}\}$,we note that $4^{n} = (2^{2})^{n} = 2^{2n}$. Since $2n \in \mathbb{Z}$ for all $n \in \mathbb{Z}$,$H \subset G$.
$1.$ Closure: $4^{n} \cdot 4^{m} = 4^{n+m} \in H$.
$2.$ Identity: $4^{0} = 1 = 2^{0} \in H$.
$3.$ Inverse: For $4^{n} \in H$,the inverse is $4^{-n} \in H$.
Thus,$\{4^{n} \mid n \in \mathbb{Z}\}$ is a subgroup of $G$.

(A) We are given the equation $4 \otimes_{7} x = 5$ in the group $G = \{1, 2, 3, 4, 5, 6\}$ under multiplication modulo $7$.
To find $x$,we test the elements of the group:
$4 \otimes_{7} 1 = 4$
$4 \otimes_{7} 2 = 8 \equiv 1 \pmod{7}$
$4 \otimes_{7} 3 = 12 \equiv 5 \pmod{7}$
$4 \otimes_{7} 4 = 16 \equiv 2 \pmod{7}$
$4 \otimes_{7} 5 = 20 \equiv 6 \pmod{7}$
$4 \otimes_{7} 6 = 24 \equiv 3 \pmod{7}$
Comparing this with the given equation $4 \otimes_{7} x = 5$,we find that $x = 3$.

(C) Given binary operation is $a * b = \frac{3ab}{2}$.
First,we find the identity element $e$ such that $a * e = a$.
$\frac{3ae}{2} = a \implies e = \frac{2}{3}$.
Now,we find the inverse $3^{-1}$ such that $3 * 3^{-1} = e = \frac{2}{3}$.
$\frac{3 \cdot 3 \cdot 3^{-1}}{2} = \frac{2}{3} \implies \frac{9 \cdot 3^{-1}}{2} = \frac{2}{3} \implies 3^{-1} = \frac{4}{27}$.
Also,$2 * x = \frac{3 \cdot 2 \cdot x}{2} = 3x$.
Given equation: $(2 * x) * 3^{-1} = 4^{-1}$.
Note that $4^{-1}$ is the inverse of $4$,so $4 * 4^{-1} = \frac{2}{3} \implies \frac{3 \cdot 4 \cdot 4^{-1}}{2} = \frac{2}{3} \implies 6 \cdot 4^{-1} = \frac{2}{3} \implies 4^{-1} = \frac{1}{9}$.
Substituting these into the equation: $(3x) * \frac{4}{27} = \frac{1}{9}$.
$\frac{3 \cdot (3x) \cdot (4/27)}{2} = \frac{1}{9}$.
$\frac{36x}{54} = \frac{1}{9} \implies \frac{2x}{3} = \frac{1}{9}$.
$x = \frac{3}{18} = \frac{1}{6}$.

(B) The binary operation is defined as $A * B = A \cup B$ for all $A, B \in P(X)$.
$1$. Commutative Law: $A * B = A \cup B = B \cup A = B * A$. Thus,it is commutative.
$2$. Associative Law: $(A * B) * C = (A \cup B) \cup C = A \cup (B \cup C) = A * (B * C)$. Thus,it is associative.
$3$. Identity Law: The identity element $E$ must satisfy $A * E = A$,which means $A \cup E = A$. This holds for $E = \phi$. Thus,the identity element exists.
$4$. Inverse Law: For an element $A$ to have an inverse $B$,we must have $A * B = E$,where $E = \phi$. This implies $A \cup B = \phi$. This is only possible if $A = \phi$ and $B = \phi$. For any non-empty set $A \neq \phi$,there is no $B \in P(X)$ such that $A \cup B = \phi$. Therefore,the inverse law is not satisfied.

(D) Let $e$ be the identity element in $Q^{+}$ such that $a * e = a$ for all $a \in Q^{+}$.
$\frac{a \times e}{2010} = a \implies e = 2010$.
Thus,the identity element is $2010$.
Let $x$ be the inverse of $2010$. By definition,$2010 * x = e$.
$\frac{2010 \times x}{2010} = 2010$.
$x = 2010$.
Therefore,the inverse of $2010$ is $2010$.

(C) binary operation $*$ on a set $N$ is a function $*: N \times N \to N$. This means for any $a, b \in N$,the result $a * b$ must also be in $N$.
$(A)$ $a * b = \sqrt{ab}$: If $a=1, b=2$,then $a * b = \sqrt{2} \notin N$.
$(B)$ $a * b = \frac{a-b}{a+b}$: If $a=1, b=2$,then $a * b = \frac{-1}{3} \notin N$.
$(C)$ $a * b = a + 3b$: Since $a, b \in N$,$a + 3b$ is always a natural number. Thus,this is a binary operation.
$(D)$ $a * b = 3a - 4b$: If $a=1, b=2$,then $a * b = 3(1) - 4(2) = -5 \notin N$.
Hence,option $C$ is correct.

(B) In the group $G = \{0, 1, 2, 3, 4, 5\}$ under addition modulo $6$,the identity element is $0$.
First,we find the inverse of $3$ under addition modulo $6$. Since $3 +_{6} 3 = 6 \equiv 0 \pmod{6}$,the inverse $3^{-1} = 3$.
Now,calculate the expression inside the parentheses: $2 +_{6} 3^{-1} +_{6} 4 = 2 +_{6} 3 +_{6} 4 = 9 \pmod{6} = 3$.
Finally,we find the inverse of the result $3$ under addition modulo $6$. Since $3 +_{6} 3 = 0$,the inverse of $3$ is $3$.
Thus,$(2 +_{6} 3^{-1} +_{6} 4)^{-1} = 3^{-1} = 3$.

(B) The set of fourth roots of unity is given by $S = \{1, -1, i, -i\}$.
For a set to form an additive group,it must satisfy the closure property under addition.
Consider $1 + (-1) = 0$. Since $0 \notin S$,the set is not closed under addition.
Therefore,the statement that the fourth roots of unity form an additive abelian group is false.

(D) Given the binary operation $a * b = a + b - 5$.
First,evaluate the inner expression $(x * 3)$:
$x * 3 = x + 3 - 5 = x - 2$.
Now,substitute this into the equation $2 * (x * 3) = 5$:
$2 * (x - 2) = 5$.
Applying the definition of the operation again:
$2 + (x - 2) - 5 = 5$.
Simplify the expression:
$x - 5 = 5$.
$x = 10$.

(C) For the set of natural numbers $N$ with respect to addition,the operation is commutative $(a+b = b+a)$ and associative $((a+b)+c = a+(b+c))$.
For the set of natural numbers $N$ with respect to multiplication,the operation is associative $((a \times b) \times c = a \times (b \times c))$.
For the operation $a * b = a^{b}$,we check for commutativity: $a * b = a^{b}$ and $b * a = b^{a}$.
Since $a^{b} \neq b^{a}$ for all $a, b \in N$ (e.g.,$2 * 3 = 2^{3} = 8$ while $3 * 2 = 3^{2} = 9$),the operation $*$ is not commutative.
Therefore,option $C$ is false.

(D) Let $G = \{-1, 0, 1\}$.
For a set to be a multiplicative group,every element must have a multiplicative inverse.
The multiplicative inverse of an element $a$ is an element $b$ such that $a \times b = 1$.
For the element $0$,there is no such element $b$ in the set such that $0 \times b = 1$.
Therefore,the inverse law fails for the element $0$.

(B) The identity element $e$ in a group $(G, \times_{15})$ must satisfy $a \times_{15} e = a$ for all $a \in G$.
From the composition table,we observe the row corresponding to $6$ reproduces the elements of $G$ in the same order.

$\times_{15}$	$3$	$6$	$9$	$12$
$3$	$9$	$3$	$12$	$6$
$6$	$3$	$6$	$9$	$12$
$9$	$12$	$9$	$6$	$3$
$12$	$6$	$12$	$3$	$9$

Since $3 \times_{15} 6 = 3$,$6 \times_{15} 6 = 6$,$9 \times_{15} 6 = 9$,and $12 \times_{15} 6 = 12$,the identity element is $6$.

(B) In any group $(G, *)$,the identity element $e$ is always its own inverse,because $e * e = e$.
Since the group $G$ must contain at least the identity element,the minimum number of elements that are their own inverses is $1$.

(B) set $G$ with a binary operation $*$ is a group if it satisfies closure,associativity,existence of identity,and existence of inverse.
For the set of odd integers under addition,let $a = 1$ and $b = 3$. Then $a + b = 4$,which is an even integer.
Since the sum of two odd integers is always even,the set of odd integers is not closed under addition.
Furthermore,the identity element for addition is $0$,which is not an odd integer.
Therefore,the set of odd integers under addition is not a group.

(D) binary operation on a set $A$ is a function from $A \times A$ to $A$.
If the number of elements in set $A$ is $n$,then the number of elements in $A \times A$ is $n^{2}$.
The number of binary operations on set $A$ is given by the formula $n^{(n^{2})}$.
Here,the set $A = \{a, b, c\}$,so the number of elements $n = 3$.
Substituting $n = 3$ into the formula,we get the number of binary operations as $3^{(3^{2})} = 3^{9}$.

(A) Given that,$a * b = 1 + ab$ $\rightarrow (1)$
For commutativity,check if $a * b = b * a$:
$b * a = 1 + ba = 1 + ab = a * b$
Since $a * b = b * a$,the operation is commutative.
For associativity,check if $(a * b) * c = a * (b * c)$:
$(a * b) * c = (1 + ab) * c = 1 + (1 + ab)c = 1 + c + abc$
$a * (b * c) = a * (1 + bc) = 1 + a(1 + bc) = 1 + a + abc$
Since $1 + c + abc \neq 1 + a + abc$,the operation is not associative.
Therefore,the operation is commutative but not associative.

(D) Let the relation be defined as $R = \{(A, B) : B = P A Q^{-1}\}$ where $P$ and $Q$ are invertible matrices.
For reflexive: Since $A = I A I^{-1}$ where $I$ is the identity matrix (which is invertible),$(A, A) \in R$. Thus,$R$ is reflexive.
For symmetric: Let $(A, B) \in R$. Then $B = P A Q^{-1}$. Multiplying by $P^{-1}$ on the left and $Q$ on the right,we get $P^{-1} B Q = A$. Since $P^{-1}$ and $Q$ are invertible,let $P' = P^{-1}$ and $Q' = Q^{-1}$. Then $A = P' B (Q')^{-1}$,so $(B, A) \in R$. Thus,$R$ is symmetric.
For transitive: Let $(A, B) \in R$ and $(B, C) \in R$. Then $A = P_1 B Q_1^{-1}$ and $B = P_2 C Q_2^{-1}$. Substituting $B$,we get $A = P_1 (P_2 C Q_2^{-1}) Q_1^{-1} = (P_1 P_2) C (Q_1 Q_2)^{-1}$. Since the product of invertible matrices is invertible,$(A, C) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.