MCQ 11 Mark
The relation $S$ defined on the set $R$ of all real number by the rule $aSb$ iff $a ≥ b$ is:
- A
- ✓
Reflexive, transitive but not symmetric.
- C
Symmetric, transitive but not reflexive.
- D
Neither transitive nor reflexive but symmetric.
AnswerCorrect option: B. Reflexive, transitive but not symmetric.
The relation $S$ is reflexive, since for any $(\text{a, a})\in\text{S}$ the condition $a2b$ holds,
The relation $S$ is not symmetric since, for any $(\text{a, b}]\in\text{S}$ but $(\text{b, a})\notin\text{S}$
The relation $S$ is transitive since, for any $(\text{a, b}]\in\text{S}$ and $(\text{b, c})\in\text{S}$
Therefore, $(\text{a, c})\notin\text{S}$
View full question & answer→MCQ 21 Mark
$S$ is a relation over the set $R$ of all real numbers and it is given by $(\text{a, b})\in\text{S}\Leftrightarrow\text{ab}\geq0.$ Then, $S$ is:
- A
Symmetric and transitive only.
- B
Reflexive and symmetric only.
- C
- ✓
AnswerReflexivity: Let $\text{a}\in\text{R}$
Then,
$aa = a^2 > 0$ $\Rightarrow\ \text{a, }\forall$
So, $S$ is reflexive on $R$.
Symmetry: Let $(\text{a, b})\in\text{S}$
Then,
$\text{a, b}\in\text{S}\Rightarrow\ \text{ab}\geq0$ $\Rightarrow\ \text{ba}\geq0\Rightarrow\ \text{ba}\geq0\Rightarrow\ \text{b, a}\in\text{S}\ \forall\ \text{a, b}\in\text{R}$
So, $S$ is symmetric on $R.$
Transitivity: If $\text{a, b, b, c}\in\text{S}\Rightarrow\ \text{ab}\geq0$ and $\text{bc}\geq0\Rightarrow\ \text{ab}\times\text{bc}\geq0$ $\Rightarrow\ \text{ac}\geq0$
$\text{b}^2\geq0\Rightarrow\ \text{a, c}\in\text{S}$ for all $\text{a, b, c}\in\text{set R}$
Hence, $S$ is an equivalence relation on $R.$
View full question & answer→MCQ 31 Mark
The relation $R = \{(1, 1), (2, 2), (3, 3)\}$ on the set $\{1, 2, 3\}$ is:
Answer$R = \{(a, b): a = b$ and $a, b \in\text{A}\}$
Reflexivity: Let $\text{a}\in\text{A}$
Here,
$a = a$
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$
So, $R$ is reflexive on $A.$
Symmetry: Let $\text{a, b}\in\text{A}$ such that $ (\text{a, b})\in\text{R}.$ Then,
$ (\text{a, b})\in\text{R}$
$\Rightarrow\ \text{a}=\text{b}$
$\Rightarrow\ \text{b}=\text{a}$
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a}\in\text{A}$
So, $R$ is symmetric on $A.$
Transitive: Let $\text{a, b, c}\in\text{A}$ such that $ (\text{a, b})\in\text{R}$ and $ (\text{b, c})\in\text{R}.$ Then,
$ (\text{a, b})\in\text{R}$
$\Rightarrow\ \text{a}=\text{b}$
and $ (\text{b, c})\in\text{R}$
$\Rightarrow\ \text{b}=\text{c}$
$\Rightarrow\ \text{a}=\text{c}$
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a}\in\text{A}$
So, $R$ is transitive on $A.$
Hence, $R$ is an equivalence relation on $A.$
View full question & answer→MCQ 41 Mark
Let $L$ denote the set of all straight lines in a plane. Let a relation $R$ be defined by $lRm$ if $l$ is perpendicular to $m$ for all $l, m \in L.$ Then, $R$ is:
AnswerGiven that $L$ denote the set of all straight lines in a plane.
A relation $R$ be defined by $lRm$ if is perpendicular to m for all $l, m \in L.$
$R$ is not reflexive. $R$ is symmetric as we can say $\text{l}\bot\text{m}$ or $\text{m}\bot\text{l}.$
View full question & answer→MCQ 51 Mark
In the set $Z$ of all integers, which of the following relation $R$ is not an equivalence relation?
- ✓
$xRy :$ if $\text{x}\leq\text{y}$
- B
$xRy :$ if $x = y$
- C
$xRy :$ if $x - y$ is an even integer
- D
$xRy :$ if $\text{x}\equiv\text{y}\ (\text{mod 3})$
AnswerCorrect option: A. $xRy :$ if $\text{x}\leq\text{y}$
In the set of $Z$ of all integers $xRy :$ if $\text{x}\leq\text{y}$ is not an equivalence relation.
For the relation $\text{x}\leq\text{y}(\text{x, y})\in\text{R}$ but $(y, x)$ not belongs to $y$ as $\text{y}\geq\text{x}$ given.
Hence, it is not an equivalence relation.
View full question & answer→MCQ 61 Mark
If A = $\{1, 2, 3\},$ then a relation $R = \{(2, 3)\}$ on $A$ is:
- A
Symmetric and transitive only.
- B
- ✓
- D
AnswerThe relation $R$ is not reflexive because every element of $A$ is not related to itself.
Also, $R$ is not symmetric since on interchanging the elements,
the ordered pair in $R$ is not contained in it.
$R$ is transitive by default because there is only one element in it.
View full question & answer→MCQ 71 Mark
$R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3$. Then, $R^{-1}$ is:
- ✓
${(8, 11), (10, 13)}$
- B
${(11, 8), (13, 10)}$
- C
${(10, 13), (8, 11)}$
- D
AnswerCorrect option: A. ${(8, 11), (10, 13)}$
Given that $R$ is a relation from $\{11, 12, 13\}$ to ${8, 10, 12}$ defined by $y = x - 3.$
$R = \{(8, 11), (10, 13)\}$
$R^{-1} = \{(8, 11), (10, 13)\}$
As inverse function of $R$ is,
$y + 3 = x$
$\Rightarrow y = x + 3$
View full question & answer→MCQ 81 Mark
If $R$ is a relation on the set $A = \{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3)\},$ then $R$ is:
Answer$R = a, b : a = b$ and $\text{a, b}\in\text{A}$
Reflexivity: Let $\text{a}\in\text{A}$
Then, $a = a$
$\Rightarrow\ \text{a, a}\in\text{R}$ for all $\text{a}\in\text{A}$
So, $R$ is reflexive on $A.$
Symmetry: Let $\text{a, b}\in\text{A}$ such that $\text{a, b}\in\text{R.}$
Then, $\text{a, b}\in\text{R}$
$\Rightarrow a = b \Rightarrow b = a \Rightarrow b, \text{a}\in\text{R}$ for all $\text{a}\in\text{A}$
So, $R$ is symmetric on $A$.
View full question & answer→MCQ 91 Mark
If $A = \{a, b, c, d\},$ then a relation $R = \{(a, b), (b, a), (a, a)\}$ on $A$ is:
- ✓
Symmetric and transitive only.
- B
Reflexive and transitive only.
- C
- D
AnswerCorrect option: A. Symmetric and transitive only.
Given that $A = \{a, b, c, d\}$ then a relation $R = \{(a, b), (b, a), (a, a)\}$ on $A.$
$(a, b), (b, a) \in\text{R}$
$\Rightarrow R$ is symmetric.
Also for $(a, a) R$ is symmetric.
View full question & answer→MCQ 101 Mark
If $R$ is the largest equivalence relation on a set $A$ and $S$ is any relation on $A$, then:
AnswerCorrect option: B. $\text{S}\subset\text{R}$
Given that $R$ is the largest relation on $A$ and $S$ is any relation on $A.$
We know that $R$ is always subset of $A \times A.$
Hence, $\text{S}\subset\text{R}.$
View full question & answer→MCQ 111 Mark
If a relation $R$ is defined on the set $Z$ of integers as follows: $(a, b) \in R ⇔ a^2 + b^2 = 25$. Then, domain $(R)$ is:
- A
${3, 4, 5}$
- B
${0, 3, 4, 5}$
- ✓
$\{0,\pm3,\pm4,\pm5\}$
- D
AnswerCorrect option: C. $\{0,\pm3,\pm4,\pm5\}$
As $aRb ⇔ a < b$
does not satisfy reflexive and symmetric relation.
View full question & answer→MCQ 121 Mark
If $A = \{1, 2, 3\}, B = \{1, 4, 6, 9\}$ and $R$ is a relation from $A$ to $B$ defined by $'x$ is greater than $y'$. The range of $R$ is:
- A
$\{1, 4, 6, 9\}$
- B
$\{4, 6, 9\}$
- ✓
$\{1\}$
- D
AnswerCorrect option: C. $\{1\}$
Here, $\text{R}=\text{x, y}:\text{x}\in\text{A}$ and $\text{y}\in\text{B}:\text{x}>\text{y}$
$\Rightarrow R = 2, 1, 3, 1$
Thus, Range of $R = \{1\}$
View full question & answer→MCQ 131 Mark
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is:
AnswerThe maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is,
$R_1 = \{(1, 1)\}$
$R_2 = \{(2, 2)\}$
$R_3 = \{(3, 3)\}$
$R_4 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_5 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is $5.$
View full question & answer→MCQ 141 Mark
Let $R$ be the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$ Then, $R$ is:
AnswerGiven $R$ is the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$
It is symmetric relation as we can say either or $\text{l}_2\bot\text{l}_1.$
View full question & answer→MCQ 151 Mark
Let $R$ be the relation on the set $A = \{1, 2, 3, 4\}$ given by $R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}.$ Then,
- A
$R$ is reflexive and symmetric but not transitive.
- ✓
$R$ is reflexive and transitive but not symmetric.
- C
$R$ is symmetric and transitive but not reflexive.
- D
$R$ is an equivalence relation.
AnswerCorrect option: B. $R$ is reflexive and transitive but not symmetric.
Reflexivity: Clearly, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$
So, $R$ is reflexive on $A.$
Symmetry: Since, $1,2\in\text{R},$ but $2,1\notin\text{R,} R$ is not symmetric on $A.$
Transitivity: Since, $1,3,3,2\in\text{R}$ and $1,2\in\text{R}, R$ is transitive on $A.$
View full question & answer→MCQ 161 Mark
Let $R$ be a relation on the set $N$ given by $R = \{(a, b): a = b - 2, b > 6\}$. Then,
- ✓
$(2, 4) \in R$
- B
$(3, 8) \in R$
- C
$(6, 8) \in R$
- D
$(8, 7) \in R$
AnswerCorrect option: A. $(2, 4) \in R$
$a = b - 2$
$\Rightarrow 6 = 8 - 2$ and $b = 8 > 6$
Hence, $(6, 8) \in R$
View full question & answer→MCQ 171 Mark
Consider a non$-$empty set consisting of children in a family and a relation $R$ defined as $aRb$ if a is brother of $b.$ Then, $R$ is:
- A
Symmetric but not transitive.
- B
Transitive but not symmetric.
- C
Neither symmetric nor transitive.
- ✓
Both symmetric and transitive.
AnswerCorrect option: D. Both symmetric and transitive.
We have,
$R = \{(a, b): a$ is brother of $b\}$
Let $(\text{a, b})\in\text{R}.$ Then,
$a$ is brother of $b.$
but $b$ is not necessary brother of a $($ As, $b$ can be sister of $a)$
$\Rightarrow\ (\text{b, a})\notin\text{R}$
So, $R$ is not symmetric.
Also,
Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$
$\Rightarrow a$ is brother of $b$ and $b$ is brother of $c$
$\Rightarrow a$ is brother of $c$
$\Rightarrow\ (\text{a, c})\in\text{R}$
So, $R$ is transitive.
View full question & answer→MCQ 181 Mark
Let $R$ be a relation on $N$ defined by $x + 2y = 8.$ The domain of $R$ is:
- A
$\{2, 4, 8\}$
- B
$\{2, 4, 6, 8\}$
- ✓
$\{2, 4, 6\}$
- D
$\{1, 2, 3, 4\}$
AnswerCorrect option: C. $\{2, 4, 6\}$
The relation $R$ is defined as $R = x, y: \text{x, y}\in\text{N}$ and $x + 2y = 8$
$\Rightarrow R = x, y: \text{x, y}\in\text{N}$ and $\text{y}=\frac{8-\text{x}}{2}$
Domain of $R$ is all values of $\text{x}\in\text{N}$ satisfying the relation $R.$
Also, there are only three values of $x$ that result in $y,$
which is a natural number.
These are $\{2, 6, 4\}.$
View full question & answer→MCQ 191 Mark
The relation $'R\ '$ in $N \times N$ such that $(a, b)R(c, d) ⇔ a + d = b + c$ is:
- A
Reflexive but not symmetric.
- B
Reflexive and transitive but not symmetric.
- ✓
- D
AnswerWe observe the following properties of relation $R.$
Reflexivity: Let $(\text{a, b})\in\text{N}\times\text{N}$
$\Rightarrow\ \text{a, b}\in\text{N}$
$\Rightarrow\ \text{a}+\text{b}=\text{b}+\text{a}$
$\Rightarrow\ (\text{a, b})\in\text{R}$
So, $R$ is reflexive on $N \times N.$
Symmetry: Let $(\text{a, b}),\ (\text{c, d})\in\text{N}\times\text{N}$ such that $(a, b)R(c, d)$
$\Rightarrow\ \text{a}+\text{d}=\text{b}+\text{c}$
$\Rightarrow\ \text{d}+\text{a}=\text{c}+\text{b}$
$\Rightarrow\ (\text{d, c}),\ (\text{b, a})\in\text{R}$
So, $R$ is symmetric on $N \times N.$
Transitivity: Let $(\text{a, b}),\ (\text{c, d}),\ (\text{e, f})\in\text{N}\times\text{N}$ such that $(a, b)R(c, d)$ and $(c, d)R(e, f)$
$\Rightarrow a + d = b + c$ and $c + f = d + e$
$\Rightarrow a + d + c + f = b + c + d + e$
$\Rightarrow a + f = b + e$
$\Rightarrow (a, b)R(e, f)$
So, $R$ is transitive on $N \times N.$
Hence, $R$ is an equivalence relation on $N.$
View full question & answer→MCQ 201 Mark
Let $A = \{1, 2, 3\}$ and $B = \{(1, 2), (2, 3), (1, 3)\}$ be a relation on $A$. Then, $R$ is:
- A
Neither reflexive nor transitive.
- B
Neither symmetric nor transitive.
- ✓
- D
AnswerReflexivity: Since $(1, 1)\notin\text{B,} B$ is not reflexive on $A.$
Symmetry: Since $1,2\in\text{B}$ but $2,1\notin\text{B,} B$ is not symmetric on $A.$
Transitivity: Since $1,2\in\text{B},\ 2,3\in\text{B}$ and $1,3\in\text{B,} B$ is transitive on $A.$
View full question & answer→MCQ 211 Mark
$R$ is a relation on the set $Z$ of integers and it is given by $(x, y) \in R ⇔ | x - y | \leq 1.$ Then, $R$ is:
- A
Reflexive and transitive.
- ✓
- C
Symmetric and transitive.
- D
AnswerReflexivity: Let $\text{x}\in\text{R.}$ Then,
$\text{x}-\text{x}=0<1$
$\Rightarrow\ |\text{x}-\text{x}|\leq1$
$\Rightarrow\ (\text{x, x})\in\text{R}$ for all $\text{x}\in\text{Z}$
So, $R$ is reflexive on $Z.$
Symmetry: Let $\text{x, y}\in\text{R.}$ Then,
$|\text{x}-\text{y}|\leq0$
$\Rightarrow\ |-(\text{y}-\text{x})|\leq1$
$\Rightarrow\ |(\text{y}-\text{x})|\leq1 [$Since $|x - y| = |y - x|]$
$\Rightarrow\ (\text{y, x})\in\text{R}$ for all $\text{x, y}\in\text{Z}$
So, $R$ is symmetric on $Z.$
Transitivity: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}.$ Then,
$|\text{x}-\text{y}|\leq1$ and $|\text{y}-\text{z}|\leq1$
$\Rightarrow$ It is not always true that $|\text{x}-\text{y}|\leq1.$
$\Rightarrow\ (\text{x, z})\notin\text{R}$
So, $R$ is not transitive on $Z.$
View full question & answer→MCQ 221 Mark
If $A = \{a, b, c\},$ then the relation $R = \{(b, c)\}$ on $A$ is:
- A
- B
- ✓
- D
Reflexive and transitive only.
AnswerThe relation $R = \{(b, c)\}$ is neither reflexive nor symmetric because every element of $A$ is not related to itself.
Also, the ordered pair of $R$ obtained by interchanging its elements is not contained in $R.$
We observe that $R$ is transitive on $A$ because there is only one pair.
View full question & answer→MCQ 231 Mark
For real numbers $x$ and $y,$ define $xRy$ if $\text{x}-\text{y}+\sqrt{2}$ is an irrational number. Then the relation $R$ is:
AnswerWe have,
$\text{R} = \big\{(\text{x, y}):\text{x}-\text{y}+\sqrt{2}$ is an irrational number, $\text{x, y}\in\text{R}\big\}$
As, $\text{x}-\text{x}+\sqrt{2}=\sqrt{2},$
which is an irrational number
$\Rightarrow\ (\text{x, x})\in\text{R}$
So, $R$ is reflexive relation.
Since, $\Big(\sqrt{2},2\Big)\in\text{R}$
i.e. $\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,$
which is an irrational number
but $2-\sqrt{2}+\sqrt{2}=2,$
which is a rational number
$\Rightarrow\ \Big(2,\sqrt{2}\Big)\notin\text{R}$
So, $R$ is not symmetric relation.
Also, $\Big(\sqrt{2},2\Big)\in\text{R}$ and $\Big(2,2\sqrt{2}\Big)\in\text{R}$
$\Rightarrow\ \Big(\sqrt{2},2\sqrt{2}\Big)\notin\text{R}$
So, $R$ is not transitive relation.
View full question & answer→MCQ 241 Mark
Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $aRb$ if $a$ is congruent to $b$ for all $a, b \in T.$ Then, $R$ is:
- A
Reflexive but not symmetric.
- B
Transitive but not symmetric.
- ✓
- D
AnswerGiven that $R$ is $T$ be the set of all triangle in the Euclidean plane,
and a relation $R$ on $T$ be defined as $aRb$ if a is congruent to $b$ for all $a, b \in T.$
Here, congruency of triangles follows reflexive, symmetric and transitive property.
Hence, it is an equivalence relation.
View full question & answer→MCQ 251 Mark
Let $R = \{(a, a), (b, b), (c, c), (a, b)\}$ be a relation on set $A = a, b, c.$ Then, $R$ is:
AnswerReflexivity: Since $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}, R$ is reflexive on $A.$
Symmetry: Since $(\text{a, b})\in\text{R}$ but $(\text{b, a})\notin\text{R,}$ is not symmetric on $A.$
$\Rightarrow R$ is not antisymmetric on $A.$
Also, $R$ is not an identity relation on $A.$
View full question & answer→MCQ 261 Mark
Let $A = \{1, 2, 3\}$. Then, the number of equivalence relations containing $(1, 2)$ is:
AnswerGiven that $A = \{1, 2, 3\}$. Then, the number of equivalence relation containing $(1, 2)$ is,
$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)\}$
Then, the number of equivalence relation containing $(1, 2)$ is $2$.
View full question & answer→MCQ 271 Mark
If $R$ is a relation on the set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ given by $xRy ⇔ y = 3x,$ then $R =$
- A
$\{(3, 1), (6, 2), (8, 2), (9, 3)\}$
- B
$\{(3, 1), (6, 2), (9, 3)\}$
- C
$\{(3, 1), (2, 6), (3, 9)\}$
- ✓
AnswerThe relation $R$ is defined as,
$\text{R}=\{(\text{x, y}):\ \text{x, y}\in\text{A}:\text{y}=3\text{x}\}$
$\Rightarrow R = \{(1, 3), (2, 6), (3, 9)\}$
View full question & answer→MCQ 281 Mark
A relation $\phi$ from $C$ to $R$ is defined by $\text{x }\phi\text{ y}\Leftrightarrow|\text{x}|=\text{y.}$ Which one is correct?
- A
$(2+3\text{i})\phi13$
- B
$3\phi(-3)$
- C
$(1+\text{i})\phi2$
- ✓
$\text{i}\phi1$
AnswerCorrect option: D. $\text{i}\phi1$
$\because\ |2+3\text{i}|=\sqrt{13}\neq13$
$|3|\neq-3$
$|1+\text{i}|=\sqrt{2}\neq2$
and $|\text{i}|=1$
So, $(\text{i, }1)\in\phi$
View full question & answer→MCQ 291 Mark
The relation $R$ defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(a, b): |a^2 - b^2| < 16\}$ is given by:
- A
$\{(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)\}$
- B
$\{(2, 2), (3, 2), (4, 2), (2, 4)\}$
- C
$\{(3, 3), (4, 3), (5, 4), (3, 4)\}$
- ✓
Answer$R$ is given by $\{(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4),(1, 3), (3, 1), (1, 4), (4, 1), (2, 4), (4, 2)\}$ which is not mentioned in $(a), (b)$ or $(c)$.
View full question & answer→MCQ 301 Mark
Let $A = \{1, 2, 3\}$ and consider the relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$. Then, $R$ is:
- ✓
Reflexive but not symmetric.
- B
Reflexive but not transitive.
- C
Symmetric and transitive.
- D
Neither symmetric nor transitive.
AnswerCorrect option: A. Reflexive but not symmetric.
We have,
$R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$
As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$
So, $R$ is reflexive relation.
Also, $(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$
So, $R$ is not symmetric relation.
And, $(1,2)\in\text{R},\ (2,3)\in\text{R}$ and $(1,3)\in\text{R}$
So, $R$ is transitive relation.
View full question & answer→MCQ 311 Mark
Let $R$ be a relation on the set $N$ of natural numbers defined by $nRm$ if $n$ divides $m$. Then, $R$ is:
- A
- B
Transitive and symmetric.
- C
- ✓
Reflexive, transitive but not symmetric.
AnswerCorrect option: D. Reflexive, transitive but not symmetric.
We have,
$R = \{(m, n): n$ divides $m; m, n \in N\}$
As, $m$ divides $m$
$\Rightarrow\ (\text{m, m})\in\text{R}\ \forall\ \text{m}\in\text{N}$
So, $R$ is reflexive.
Since, $(2,1)\in\text{R}$
i.e. $1$ divides $2$
but $2$ cannot divide $1$
i.e. $(2,1)\notin\text{R}$
So, $R$ is not symmetric.
Let $(\text{m, n})\in\text{R}$ and $(\text{n, p})\in\text{R.}$ Then,
$n$ divides $m$ and $p$ divides $n$
$\Rightarrow p$ divides $m$
$\Rightarrow\ (\text{m, p})\in\text{R}$
So, $R$ is transitive.
View full question & answer→MCQ 321 Mark
Let $A = \{2, 3, 4, 5, ..., 17, 18\}.$ Let $'\simeq\ '$ be the equivalence relation on $A \times A,$ cartesian product of $A$ with itself, defined by $(\text{a, b})\simeq(\text{c, d)}$ if $ad = bc$. Then, the number of ordered pairs of the equivalence class of $(3, 2)$ is:
AnswerThe ordered pairs of the equivalence class of $(3, 2)$ are $\{(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)\}$.
We observe that these are $6$ pairs.
View full question & answer→MCQ 331 Mark
A relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $: xRy ⇔ x$ is relatively prime to $y.$ Then, domain of $R$ is:
- A
$\{2, 3, 5\}$
- B
$\{3, 5\}$
- C
$\{2, 3, 4\}$
- ✓
$\{2, 3, 4, 5\}$
AnswerCorrect option: D. $\{2, 3, 4, 5\}$
Given that relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $: xRy ⇔ x$ is relatively prime to $y.$
$R$ can be written as,
$\{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)\}$
Here we can see that domain means x element which is $2\leq\text{x}\leq5.$
Hence, $\{2, 3, 4, 5\}$
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Let $A = \{1, 2, 3\}.$ Then, the number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is:
AnswerGiven that $A = {1, 2, 3}$
To find the number of relations containing $(1, 2)$ and $(1, 3)$ then $R$ can be written as $\{(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)\}$
Here, we can see that
$(3, 1)$ and $(1, 2) \Rightarrow (3, 2)$ which is not belongs to $R.$
The number of relations containing $(1, 2)$ and $(1, 3)$
Which are reflexive and symmetric but not transitive is $1.$
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