Question 1: Let A be the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive:

$\displaystyle \text{(i) } R = \{(x, y) : x \text{ and } y \text{ work at the same place } \}$

$\displaystyle \text{(ii) } R = \{(x, y): x \text{ and } y \text{ live in the same locality } \}$

$\displaystyle \text{(iii) } R = \{(x, y) : x \text{ is wife of } y \}$

$\displaystyle \text{(iv) } R = \{(x, y) : x \text{ is father of } y \}$

$\displaystyle \text{(i) } \text{A be the set of all human beings in a town at a particular time. } \\ \\ \text{Given } R = \{(x, y) : x \text{ and } y \text{ work at the same place } \}$
Reflexive:

Let $x$ be an arbitrary element of $R$. Then, $x \in R$

$\Rightarrow x$ and $x$ work at the same place

$\Rightarrow ( x, x) \in R$

Thus, $R$ is reflexive relation.

Symmetric:

Let $x$ and $y$ be an arbitrary element of $R$. Then, $(x, y) \in R$

$\Rightarrow x$ and $y$  work at the same place

$\Rightarrow y$ and $x$  work at the same place

$\Rightarrow ( y, x) \in R$

Thus, $R$ is a symmetric relation.

Transitive:

Let $x, y$  and $z$ be an arbitrary element of $R$. Then, $(x, y) \in R$ and  $(y, z) \in R$

$\Rightarrow x$ and $y$  work at the same place

$\Rightarrow y$ and $z$   work at the same place

$\Rightarrow x, y, z$ work at the same place

$\therefore x$ and  $z$   work at the same place

$\Rightarrow ( x, z) \in R$

Thus, $R$ is a transitive relation.

$\displaystyle \text{(ii) } \\ \\ \text{Given } R = \{(x, y): x \text{ and } y \text{ live in the same locality } \}$

Reflexive:

Let $x$ be an arbitrary element of $R$. Then, $x \in R$

$\Rightarrow x$ and $x$ live in the same locality

$\Rightarrow ( x, x) \in R$

Thus, $R$ is a reflexive relation.

Symmetric:

Let $x$ and $y$ be an arbitrary element of $R$. Then, $(x, y) \in R$

$\Rightarrow x$ and $y$  live in the same locality

$\Rightarrow y$ and $x$ live in the same locality

$\Rightarrow ( y, x) \in R$

Thus, $R$ is a symmetric relation.

Transitive:

Let $x, y$  and $z$ be an arbitrary element of $R$. Then, $(x, y) \in R$ and  $(y, z) \in R$

$\Rightarrow x$ and $y$  live in the same locality

$\Rightarrow y$ and $z$  live in the same locality

$\Rightarrow x, y, z$ live in the same locality

$\therefore x$ and  $z$  live in the same locality

$\Rightarrow ( x, z) \in R$

Thus, $R$ is a transitive relation.

$\displaystyle \text{(iii) } \text{Given } R = \{(x, y) : x \text{ is wife of } y \}$

Reflexive:

Let $x$ be an arbitrary element of $R$. Let, $x \in R$

$\Rightarrow x$ is wife of $x$ is not possible

$\Rightarrow ( x, x) \notin R$

Thus, $R$ is not a reflexive relation.

Symmetric:

Let $x$ and $y$ be an arbitrary element of $R$. Let, $(x, y) \in R$

$\Rightarrow x$ is wife of  $y$

$\text{but } y$ is husband of $x$

$\Rightarrow ( x, y) \notin R$

Thus, $R$ is not a symmetric relation.

Transitive:

Let $x, y$  and $z$ be an arbitrary element of $R$. Let, $(x, y) \in R$ and  $(y, z) \in R$

$\Rightarrow x$ is wife of  $y$

$\Rightarrow y$ is not husband of  $z$

$\Rightarrow ( x, z) \notin R$

Thus, $R$ is not a transitive relation.

$\displaystyle \text{(iv) } \\ \\ \text{Given } R = \{(x, y) : x \text{ is father of } y \}$

Reflexive:

Let $x$ be an arbitrary element of $R$. Let, $x \in R$

$\Rightarrow x$ if father of $x$ is not possible

$\Rightarrow ( x, x) \notin R$

Thus, $R$ is not a reflexive relation.

Symmetric:

Let $x$ and $y$ be an arbitrary element of $R$. Let, $(x, y) \in R$

$\Rightarrow x$ is father of  $y$

$\text{but } y$ cannot be father of $x$

$\Rightarrow ( y, x) \notin R$

Thus, $R$ is not a symmetric relation.

Transitive:

Let $x, y$  and $z$ be an arbitrary element of $R$. Let, $(x, y) \in R$ and  $(y, z) \in R$

$\Rightarrow x$ is father of  $y$

$\Rightarrow y$ is father of  $z$

$\Rightarrow ( x, z) \notin R$

Thus, $R$ is not a transitive relation.

$\\$

$\displaystyle \text{Question 2: Relations } R_1, R_2 , R_3 \text{ and } R_4 \text{ are defined on a set } A = \{ a, b, c \} \text{ as follows: }$

$\displaystyle R_1 = \{(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c) \}$

$\displaystyle R_2 = \{(a, a) \}$

$\displaystyle R_3 = \{(b, c) \}$

$\displaystyle R_4 = \{(a,b), (b, c), (c, a) \}$

$\displaystyle \text{Find whether or not each of the relations } R_1, R_2, R_3, R_4 \text{ on } \\ \\ A \text{ is (i) reflexive (ii) symmetric (iii) transitive. }$

$\displaystyle R_1 = \{(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c) \}$

$R_1 \text{ is reflexive as } (a, a) \in R_1,(b, b) \in R_1,(c, c) \in R_1$

$R_1 \text{ is not symmetric as } (a, b) \in R_1 \text{ but } (b, a) \notin R_1$

$R_1 \text{ is not transitive as } (b, c) \in R_1, (c, a) \in R_1 \text{ but } (b, a) \notin R_1$

$\displaystyle R_2 = \{(a, a) \}$

$R_2 \text{ is reflexive as } (a, a) \in R_2$

$R_2 \text{ is symmetric as } (a, a) \in R_2$

$R_2 \text{ is transitive, since there is only one element in it. }$

$\displaystyle R_3 = \{(b, c) \}$

$R_3 \text{ is not reflexive as } (b, b) \notin R_3 \text{ nor } (c, c) \notin R_3$

$R_3 \text{ is not symmetric }$

$R_3 \text{ is not transitive as it has only two elements }$

$\displaystyle R_4 = \{(a,b), (b, c), (c, a) \}$

$R_4 \text{ is not reflexive as } (a, a) \notin R_4$

$R_4 \text{ is not symmetric as } (a, b) \in R_4 \text{ but } (b, a) \notin R_4$

$R_4 \text{ is not transitive as } (a, b) \in R_4 \text{ and } (b, c) \in R_4 \text{ but } (a, c) \notin R_4$

$\\$

$\displaystyle \text{Question 3: Test whether the following relations } R_1, R_2, \text{ and } R_3 \\ \\ \text{ are (i) reflexive (ii) symmetric and (iii) transitive: }$

$\displaystyle \text{(i) } R_1 \text{ on } Q_0 \text{ defined by } ( a, b) \in R_1 \Leftrightarrow a = \frac{1}{b}$

$\displaystyle \text{(ii) } R_2 \text{ on } Z \text{ defined by } ( a, b) \in R_2 \Leftrightarrow |a-b| \leq 5$

$\displaystyle \text{(iii) } R_3 \text{ on } R \text{ defined by } ( a, b) \in R_3 \Leftrightarrow a^2 - 4ab + 3b^2 = 0$

$\displaystyle \text{(i) } R_1 \text{ on } Q_0 \text{ defined by } ( a, b) \in R_1 \Leftrightarrow a = \frac{1}{b}$

Reflexivity:

Let $a$ be an arbitrary element of $R_1$.

Then, $a \in R_1$

$\displaystyle \Rightarrow a \neq \frac{1}{a} \text{ for all } a \in Q_0$

Clearly, $R_1$ is not reflexive.

Symmetry:

Let $(a, b) \in R_1$

Then, $(a, b) \in R_1$

$\displaystyle \text{Therefore we can write 'a' as } a = \frac{1}{b}$

$\displaystyle \Rightarrow b = \frac{1}{a} \Rightarrow (b, a) \in R_1$

Hence, $R_1$ is symmetric.

Transitivity:

Here, $(a, b) \in R_1 \text{ and } (b, c) \in R_2$

$\displaystyle \Rightarrow a = \frac{1}{b} \text{ and } b = \frac{1}{c}$

$\displaystyle \Rightarrow a = \frac{1}{\frac{1}{c}} = c$

$\displaystyle \Rightarrow a \neq \frac{1}{c} \Rightarrow (a, c) \notin R1$

So, $R_1$ is not transitive.

$\displaystyle \text{(ii) } R_2 \text{ on } Z \text{ defined by } ( a, b) \in R_2 \Leftrightarrow |a-b| \leq 5$

Reflexivity:

Let $a$ be an arbitrary element of $R_2$.

Then, $a \in R_2$

On applying the given condition we get,

$\Rightarrow |a - a| = 0 \nleq 5$

So, $R_2$ is reflexive.

Symmetry:

Let $(a, b) \in R_2$

$\Rightarrow |a - b| \nleq 5 \ \ \ \ \ \ \ [ \text{ Since, } |a - b| = |b - a|]$

$\Rightarrow |b - a| \nleq 5$

$\Rightarrow (b, a) \in R_2$

So, $R_2$ is symmetric.

Transitivity:

Let $(1, 3) \in R_2 \text{ and } (3, 7) \in R_2$

$\Rightarrow |1 - 3| \nleq 5 \text{ and } |3 - 7| \nleq 5 \text{ But } |1 - 7| \nleq 5$

$\Rightarrow (1, 7) \notin R_2$

Clearly, $R_2$ is not transitive.

$\displaystyle \text{(iii) } R_3 \text{ on } R \text{ defined by } ( a, b) \in R_3 \Leftrightarrow a^2 - 4ab + 3b^2 = 0$

Reflexivity:

Let $a$ be an arbitrary element of $R_3$.

Then, $a \in R_3 \Rightarrow a^2 - 4a \times a + 3a^2 = 0$

Clearly, $R_3$ is reflexive

Symmetry:

Let $(a, b) \in R_3 \Rightarrow a^2 - 4ab + 3b^2 = 0$

But $b^2 - 4ba + 3a^2 \neq 0 \text{ for all } a, b \in R$

So, $R_3$ is not symmetric.

Transitivity:

Let $(1, 2) \in R_3 \text{ and } (2, 3) \in R_3$

$\Rightarrow 1 - 8 + 6 = 0 \text{ and } 4 - 24 + 27 = 0$

But $1 - 12 + 9 \neq 0$

Clearly, $R_3$ is not transitive.

$\\$

$\displaystyle \text{Question 4: Let }A= \{1, 2, 3\} , \text{ and let } R_1 =\{(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3) \}, \\ \\ R_2 = \{(2, 2), (3, 1), (1, 3) \}, R_3 = \{(1, 3), (3, 3)\}. \text{ Find whether or not each of the } \\ \\ \text{ relations } R_1, R_2, R_3 \text{ on } A \text{ is (i) reflexive (ii) symmetric (iii) transitive }$

Given that $R_1 = \{(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)\}$

Reflexivity:

Here, $(1, 1), (2, 2), (3, 3) \in R_1$

Clearly, $R_1$ is reflexive.

Symmetry:

Here, $(2, 1) \in R_1$, But $(1, 2) \notin R_1$

So, $R_1$ is not symmetric.

Transitivity:

Here, $(2, 1) \in R1 \text{ and } (1, 3) \in R_1,$ But $(2, 3) \notin R_1$

So, $R_1$ is not transitive.

Now we consider $R_2$

Given that $R_2 = \{(2, 2), (3, 1), (1, 3)\}$

Reflexivity:

Clearly, $(1, 1) \text{ and } (3, 3) \notin R_2$

So, $R_2$ is not reflexive.

Symmetry:

Here, $(1, 3) \in R_2 \text{ and } (3, 1) \in R_2$

Clearly, $R_2$ is symmetric.

Transitivity:

Here, $(1, 3) \in R_2 \text{ and } (3, 1) \in R_2$

But $(3, 3) \notin R_2$

So, $R_2$ is not transitive.

Consider as $R_3$

Given that $R_3 = \{(1, 3), (3, 3) \}$

Reflexivity:

Clearly, $(1, 1) \notin R_3$

So, $R_3$ is not reflexive.

Symmetry:

Here, $(1, 3) \in R_3, \text{ but } (3, 1) \notin R_3$

So, $R_3$ is not symmetric.

Transitivity:

Here, $(1, 3) \in R_3 \text{ and } (3, 3) \in R_3 \text{ Also, } (1, 3) \in R_3$

Clearly, $R_3$ is transitive.

$\\$

Question 5: The following relations are defined on the set of real numbers:

$\displaystyle \text{(i) } aRb \text{ if } a-b > 0 \hspace{1.0cm} \text{(ii) } aRb \text{ iff } 1 + ab > 0 \hspace{1.0cm} \text{(iii) } aRb \text{ if } |a| \leq b.$

Find whether these relations are reflexive, symmetric or transitive.

$\displaystyle \text{(i) } aRb \text{ if } a-b > 0$

Reflexivity:

Let $a$ be an arbitrary element of $R.$

Then, $a \in R \text{ but } a - a = 0 \ngtr 0$

Thus, this relation is not reflexive.

Symmetry:

Let $(a, b) \in R \Rightarrow a - b > 0$

$\Rightarrow - (b - a) > 0 \Rightarrow b - a < 0$

Therefore, the given relation is not symmetric.

Transitivity:

Let $(a, b) \in R$ and $(b, c) \in R.$

Then, $a - b > 0$ and $b - c > 0$

Adding the two, we get $a - b + b - c > 0$

$\Rightarrow a - c > 0 \Rightarrow (a, c) \in R.$

Clearly, the given relation is transitive.

$\text{(ii) } aRb \text{ iff } 1 + ab > 0$

Reflexivity:

Let $a$ be an arbitrary element of $R.$

Then, $a \in R$

$\Rightarrow 1 + a \times a > 0$

i.e. $+ a^2 > 0$            [Since, square of any number is positive]

So, the given relation is reflexive.

Symmetry:

Let $(a, b) \in R \Rightarrow 1 + ab > 0$

$\Rightarrow 1 + ba > 0 \Rightarrow (b, a) \in R$

So, the given relation is symmetric.

Transitivity:

Let $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow 1 + ab > 0 \text{ and } 1 + bc >0$

$\text{But } 1 + ac \ngtr 0 \Rightarrow (a, c) \notin R$

Thus, the given relation is not transitive.

$\text{(iii) } aRb \text{ if } |a| \leq b.$

Reflexivity:

Let $a$ be an arbitrary element of $R.$

Then, $a \in R \ \ \ \ [Since, |a| = a] \Rightarrow |a| \nless a$

Clearly, $R$ is not reflexive.

Symmetry:

Let $(a, b) \in R \Rightarrow |a| \leq b$

$\Rightarrow |b| \nleq a$ for all $a, b \in R$

$\Rightarrow (b, a) \notin R$

Thus, $R$ is not symmetric.

Transitivity:

Let $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow |a| \leq b$  and $|b| \leq c$

Multiplying the corresponding sides, we get

$|a| \times |b| \leq b c \Rightarrow |a| \leq c \Rightarrow (a, c) \in R$

So, $R$ is transitive.

$\\$

Question 6: Check whether the relation $R$ defined on the set $A=\{ 1, 2, 3, 4, 5, 6 \}$ as $R = \{ ( a, b): b = a+1 \}$ is reflexive, symmetric or transitive.

Let $A = \{ 1, 2, 3, 4, 5, 6 \}$

$A$ relation $R$ is defined on set $A$ as $: R \{ (a, b) : b = a+1 \}$

$\therefore R = \{ (1, 2), ( 2, 3), (3, 4), (4, 5), (5, 6) \}$

We observe that $( a, a ) \notin R,$ where $a \in A$

For instance, $(1, 1), (2, 2), (3, 3), (4, 4) , (5, 5, ) , (6, 6) \notin R$

Therefore $R$ is not reflexive.

It is also observed that $(1, 2) \in R$ but $( 2, 1 ) \notin R$

Therefore $R$ is not symmetric.

Not, $(1, 2) , ( 2, 3) \in R$ but $( 1, 3) \notin R$

Therefore $R$ is not transitive.

Hence, $R$ is neither reflexive, not symmetric, nor transitive.

$\\$

Question 7: Check whether the relation $R$ on R defined by $R = \{ (a, b): a \leq b^3 \}$ is reflexive, symmetric or transitive.

$R = \{ (a, b) : a \leq b^3 \}$

$\displaystyle \text{It is observed that } \Big(\frac{1}{2},\frac{1}{2} \Big) \notin R, \text{ since, } \frac{1}{2} > \Big(\frac{1}{2} \Big)^3$

Therefore, $R$ is not reflexive.

Now, $(1, 2) \in R ( \text{ as } 1 < 2^3 = 8 )$

But $(2, 1) \notin R ( \text{ as } 2^3 > 1)$

Therefore R is not symmetric.

$\displaystyle \text{We have } \Big(3,\frac{3}{2} \Big), \Big(\frac{3}{2},\frac{6}{5} \Big) \in R, \text{ since } 3 < \Big(\frac{3}{2} \Big)^3 \text{ and } \frac{3}{2} < \Big(\frac{6}{5} \Big)^3$

$\displaystyle \text{But } \Big(3,\frac{6}{5} \Big) \notin R \text{ as } 3 > \Big(\frac{6}{5} \Big)^3$

Therefore $R$ is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

$\\$

Question 8: ‘ Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Let us first understand what ‘Reflexive Relation’ is and what ‘Identity Relation’ is.

Reflexive Relation: A binary relation $R$ over a set $A$ is reflexive if every element of $A$ is related to itself. Formally, this may be written as $\forall \ x \in A: x R x.$

Identity Relation: Let $A$ be any set.

Then the relation $R= \{ (x, x): x \in A \}$ on $A$ is called the identity relation on $A.$ Thus, in an identity relation, every element is related to itself only.

Let $A = \{ a, b, c \}$ be a set.

Let $R$ be a binary relation defined on $A.$

Let $R_A = \{ (a, a): a \in A \}$ is the identity relation on $A.$

Hence, every identity relation on set $A$ is reflexive by definition.

Converse: Let $A = \{ a, b, c \}$ is the set.

Let $R_A= \{ (a, a), (b, b), (c, c), (a, b), (c, a) \}$ be a relation defined on $A.$

$R$ is reflexive as per definition.

$[ \because (a, a) \in R, (b, b) \in R \ \ \& \ \ (c, c) \in R]$

But, $(a, b) \in R$

$(c, a) \in R \Rightarrow R$ is not identity relation by definition.

Hence, proved that every identity relation on a set is reflexive, but the converse is not necessarily true.

$\\$

Question 9: . If $A = \{ 1, 2, 3, 4 \}$, define relations on A which have properties of being

(i) reflexive, transitive but not symmetric.

(ii) symmetric but neither reflexive nor transitive.

(iii) reflexive, symmetric and transitive.

(i) The relation on $A$ having properties of being reflexive, transitive, but not symmetric is

$R = \{(1, 1), (2, 2), (3, 3), (4, 4), (2, 1) \}$

Relation $R$ satisfies reflexivity and transitivity.

$\Rightarrow (1, 1), (2, 2), (3, 3) \in R$

And $(1, 1), (2, 1) \in R \Rightarrow (1, 1) \in R$

However, $(2, 1) \in R, \text{ but } (1, 2) \notin R$

(ii)  The relation on $A$ having properties of being reflexive, transitive, but not symmetric is

$R = \{(1, 1), (2, 2), (3, 3), (4, 4), (2, 1) \}$

Relation $R$ satisfies reflexivity and transitivity.

$\Rightarrow (1, 1), (2, 2), (3, 3) \in R$

And $(1, 1), (2, 1) \in R \Rightarrow (1, 1) \in R$

However, $(2, 1) \in R, \text{ but } (1, 2) \notin R$

(iii) The relation on $A$ having properties of being symmetric, reflexive and transitive is

$R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1) \}$

So, the $R$ is an equivalence relation on $A.$

$\\$

Question 10: Let R be a relation defined on the set of natural numbers N as

$R = \{ (x, y): x, y \in N , 2x + y = 41 \}$

Find the domain and range of R. Also, verify whether R is (i) reflexive, (ii) symmetric (iii) transit

We have, $R = \{ (x, y): x, y \in N , 2x + y = 41 \}$

$\displaystyle \text{The Domain of } R \text{ is } x \in N, \text{ such that } x = \frac{41-y}{2}$

Since $y \in N$, largest value that $x$ can take corresponds to the smallest value that $y$ can take.

$\therefore x = \{ 1, 2, 3, \ldots , 20 \}$

Range of $R$ is $y \in N$ such that

$2x + y = 41$

$\Rightarrow y = 41 - 2x$

Since $x = \{ 1, 2, 3, \ldots , 20 \}$

$\therefore y = \{ 39, 27, 35, 33, \ldots , 7, 5, 3, 1 \}$

Since, $( 2, 2) \notin R, R$ is not reflexive.

Also, since $( 1, 39) \in R \text{ but } ( 39, 1) \notin R, \text{ therefore } R$ is not symmetric.

$\text{Finally, since } (15, 11) \in R \text{ and } ( 11, 19) \in R \text{ but } ( 15, 19) \notin R.$

Therefore $R$ is not transitive.

$\\$

Question 11: Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.

It is not true that every relation which is symmetric and transitive is also reflexive.

Take for example:

Take a set $A = \{1, 2, 3, 4\}$

And define a relation $R$ on $A$.

Symmetric relation:

$R = \{(1, 2), (2, 1)\}$, is symmetric on set $A$.

Transitive relation:

$R = \{(1, 2), (2, 1), (1, 1)\}$, is the simplest transitive relation on set $A$.

$\Rightarrow R = \{(1, 2), (2, 1), (1, 1)\}$ is symmetric as well as transitive relation.

But $R$ is not reflexive here.

If only $(2, 2) \in R,$ had it been reflexive.

Thus, it is not true that every relation which is symmetric and transitive is also reflexive.

$\\$

Question 12: An integer $m$ is said to be related to another integer $n$ if $m$ is a multiple of $n$. Check if the relation is symmetric, reflexive and transitive.

Given, $R = \{ (m, n) : m, n \in Z , m = kn, \text{ where } k \in N \}$

Reflexivity:

$m = km \text{ is true for } k = 1$

$\Rightarrow (m, m) \in R$

Thus, $R$ is reflexive relation.

Symmetry:

Let $(m, n) \in R$

$\Rightarrow m = kn \text{ for some } k \in N$

$\displaystyle \Rightarrow n = \frac{1}{k} m$

$\Rightarrow (n, m) \notin R$

Thus $R$ is not symmetric.

Transitivity:

Let $(m, n) \text{ and } ( n, o) \in R$

$\Rightarrow m = kn \text{ and } n = lo \text{ for some } k, l \in N$

$\Rightarrow m = (kl)o$

Here $kl \in R$

$\Rightarrow ( m, o) \in R$

Thus, $R$ is transitive.

$\\$

Question 13: Show that the relation $" \geq "$ on the set $R$ of all real numbers is reflexive and transitive but not symmetric.

Let $R$ be the set such that $R = \{ (a, b) : a, b \in R ; a \geq b \}$

Reflexivity:

Let $a$ be an arbitrary element of $R.$

$\Rightarrow a \in R$

$\Rightarrow a \geq a \text{ is true as } a = a$

$\Rightarrow (a, b) \in R$

Hence, $R$ is reflexive.

Symmetry:

Let $(a, b) \in R$

$\Rightarrow a \geq b \text{ is the same as } b \leq a , \text{ but not } b \geq a$

Hence, $( b, a ) \notin R$

Hence, $R$ is not symmetric.

Transitivity:

$\text{Let } (a, b) \text{ and } (b, c) \in R$

$\Rightarrow a \geq b \text{ and } b \geq c$

$\Rightarrow a \geq b \geq c$

$\Rightarrow a \geq c$

$\Rightarrow(a, c) \in R$

Hence, $R$ is transitive.

$\\$

Question 14: Give an example of a relation which is

(i) reflexive and symmetric but not transitive.

(ii) reflexive and transitive but not symmetric.

(iii) symmetric and transitive but not reflexive.

(iv) symmetric but neither reflexive nor transitive.

(v) transitive but neither reflexive nor symmetric

(i) reflexive and symmetric but not transitive.

Let $A = \{ 1, 2, 3 \}$

Let $R$ be a relation on $A$ such that $R = \{ (1, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2) \}$

Relation $R$ is reflexive if for every $a \in A (a, a) \in R.$ $\text{ Here, } (1, 1), (2, 2), (3, 3) \in R \text{ Hence Reflexive.}$

Relation $R$ is symmetric if $(a, b) \in R \Rightarrow (b, a) \in R$ $\text{ Here, } (1, 3), (3, 1), (2, 3), (3, 2) \in R \text{. Hence Symmetric.}$

Relation $R$ is transitive if $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ $\text{ Here, } (2, 3) , (3, 1) \in R \text{ but } ( 2, 1) \notin R \text{. Hence not Transitive.}$

(ii) reflexive and transitive but not symmetric.

Let $A = \{ 1, 2, 3 \}$

Let $R$ be a relation on $A$ such that $R = \{ (1, 1), (2, 2), (3, 3), (1, 2) , ( 1, 3), ( 2, 3) \}$

Relation $R$ is reflexive if for every $a \in A (a, a) \in R.$ $\text{ Here, } (1, 1), (2, 2), (3, 3) \in R \text{ Hence Reflexive.}$

Relation $R$ is symmetric if $(a, b) \in R \Rightarrow (b, a) \in R$ $\text{ Here, } (1, 3) \in R \text{ but } (3, 1) \notin R \text{. Hence not Symmetric.}$

Relation $R$ is transitive if $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ $\text{ Here, } (1, 2) , (2, 3) , (1, 3) \in R \text{. Hence Transitive.}$

(iii) symmetric and transitive but not reflexive.

Let $A = \{ 1, 2, 3 \}$

Let $R$ be a relation on $A$ such that $R = \{ (1,3), (3,1),(1,1), (3,3) \}$

Relation $R$ is reflexive if for every $a \in A (a, a) \in R.$ $\text{ Here, } (1, 1) \in R \text{ but } (2, 2), (3, 3) \notin R. \text{ Hence not Reflexive.}$

Relation $R$ is symmetric if $(a, b) \in R \Rightarrow (b, a) \in R$ $\text{ Here, } (1, 3) , (3, 1) \in R \text{. Hence Symmetric.}$

Relation $R$ is transitive if $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ $\text{ Here, } (1, 3) , (3, 1), (1,1) \in R \text{. Hence Transitive.}$

(iv) symmetric but neither reflexive nor transitive.

Let $A = \{ 1, 2, 3 \}$

Let $R$ be a relation on $A$ such that $R = \{ (1, 1), (1,3), (3, 1), (2,3), (3, 2) \}$

Relation $R$ is reflexive if for every $a \in A (a, a) \in R.$ $\text{ Here, } (1, 1) \in R \text{ but } (2, 2), (3, 3) \notin R. \text{ Hence not Reflexive.}$

Relation $R$ is symmetric if $(a, b) \in R \Rightarrow (b, a) \in R$ $\text{ Here, } (1, 3) , (3, 1), (2, 3), (3, 2) \in R \text{. Hence Symmetric.}$

Relation $R$ is transitive if $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ $\text{ Here, } (2, 3) , (3, 1) \in R \text{ but } (2,1) \notin R \text{. Hence not Transitive.}$

(v) transitive but neither reflexive nor symmetric

Let $A = \{ 1, 2, 3 \}$

Let $R$ be a relation on $A$ such that $R = \{ (1, 2), (2,3), (1, 3) \}$

Relation $R$ is reflexive if for every $a \in A (a, a) \in R.$ $\text{ Here, } (1, 1), (2, 2), (3, 3) \notin R. \text{ Hence not Reflexive.}$

Relation $R$ is symmetric if $(a, b) \in R \Rightarrow (b, a) \in R$ $\text{ Here, } (1, 2) , (2, 3), (1, 3) \in R \text{ but } (2, 1), (3, 2), (3, 1) \notin R \text{. Hence not Symmetric.}$

Relation $R$ is transitive if $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ $\text{ Here, } (1, 2) , (2, 3), ( 1, 3) \in R \text{. Hence Transitive.}$

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Question 15: Given the relation $R = \{ (1, 2), (2, 3) \}$ on the set $A = \{ 1, 2, 3 \}$ add a minimum number of ordered pairs so that the enlarged relation is symmetric, transitive and reflexive.

$A = \{ 1, 2, 3 \}$

$R = \{ (1, 2), (2, 3) \}$

To make R an equivalence relation, it should be:

(i) Reflexive: So three more ordered pairs $(1, 1), (2, 2), (3, 3)$ should be added to $R$ to make it reflexive.

(ii) Symmetric : As $R$ contains $(1, 2) \text{ and } (2, 3)$ so two more ordered pairs $(2, 1) \text{ and } (3, 2)$ should be added to make it symmetric.

(iii) Transitive: $(1, 2) \in R, (2, 3) \in R.$ So to make $R$ transitive $(1, 3)$ should be added to $R$.  Also to maintain the symmetric property $(3, 1)$ should then be added to $R.$

So, $R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (2, 1), (3, 2), (1, 3), (3, 1)}$ is an equivalence relation. So minimum 7 ordered pairs are to be added.

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Question 16: Let $A = \{1, 2, 3 \}$ and $R = \{(1,2), (1,1), (2,3) \}$ be a relation on $A.$ What minimum number of ordered pairs may be added to R so that it may become a transitive relation on $A.$

We have the relation $R$ such that  $R = \{(1, 2), (1, 1), (2, 3)\}$

$R$ is defined on set $A.$  $A = \{1, 2, 3\}$

For transitive relation:

Note in $R, (1, 2) \in R \text{ and } (2, 3) \in R. \text{ Then } (1, 3) \in R$

So, add $(1, 3) \text{ in } R.$

$R = \{(1, 2), (1, 1), (2, 3), (1, 3)\}$

Now, we can see that $R$ is transitive. Hence, the ordered pair to be added is $(1, 3).$

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Question 17: Let $A= \{a, b, c \}$ and the relation $R$ be defined on $A$ as follows: $R = \{ ( a, a), (b, c), (a, b) \}.$ Then, write minimum number of ordered pairs to be added in $R$ to make it reflexive and transitive

We have the relation $R$ such that $R = \{ ( a, a), (b, c), (a, b) \}.$

$R$ is defined on set $A.\ A= \{a, b, c \}$

To make $R$ as reflexive we should add $(b, b)$ and $(c, c)$ to $R.$

Also, to make $R$ as transitive we should add $(a, c)$ to $R.$

Hence, the minimum number of ordered pairs to be added in $R$ is $3$.

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Question 18: Each of the following defines a relation on N:

$\displaystyle \text{(i) } x > y , x, y \in N$

$\displaystyle \text{(ii) } x + y = 10, x, y \in N$

$\displaystyle \text{(iii) } xy \text{ is square of an integer} , x, y \in N$

$\displaystyle \text{(iv) } x + 4y = 10, x, y \in N$

Determine which of the above relations are reflexive, symmetric and transitive.

$\displaystyle \text{(i) } x > y , x, y \in N$

If $( x, x) \in R$, then $x > x$  which is not true for any $x \in N.$

Therefore $R$ is not reflexive.

$\text{Let } (x, y) \in R$

$\Rightarrow ( x, y) \in R \Rightarrow x > y$

$\Rightarrow ( y, x) \in R \Rightarrow y > x$ which is not true.

Therefore $R$ is not symmetric.

$\text{Let } (x, y) \in R \text{ and } (y, z) \in R$

$\Rightarrow x > y \text{ and } y > z$

$\Rightarrow x > z \Rightarrow (x, z) \in R$

Therefore $R$ is transitive.

$\displaystyle \text{(ii) } x + y = 10, x, y \in N$

$\Rightarrow R = \{ (x, y) : x+y = 10 \text{ and } x, y \in N \}$

$\Rightarrow R = \{ ( 1, 9), (2, 8), (3, 7), (4, 6), ( 5, 5), 6, 4), (7, 3), (8, 2), ( 9, 1) \}$

$\Rightarrow ( 1, 1) \notin R$

Therefore $R$ is not reflexive.

Also, $R$ is symmetric since for all $(a, b) \in R, \text{ we have } ( b, a) \in R.$

$\text{Also, } (1,9) \in R \text{ and }(9, 1) \in R \text{ but } ( 1, 1) \notin R.$

Therefore $R$ is not transitive.

$\displaystyle \text{(iii) } xy \text{ is square of an integer} , x, y \in N$

$\Rightarrow R = \{ (x, y) : xy \text{ is square of an integer} , x, y \in N \}$

$\text{Clearly, } (x, x) \in R \ \forall \ x \in N$

Therefore $R$ is reflexive.

$\text{If } ( x, y) \in R \Rightarrow (y, x) \in R$

Therefore $R$ is symmetric.

Now if $xy$ is square of an integer and $yz$ is square of an integer

$\text{Then let } xy = m^2 \text{ and } yz = n^2$

$\displaystyle \Rightarrow x = \frac{m^2}{y} \text{ and } z = \frac{n^2}{y}$

$\displaystyle \Rightarrow xz = \frac{m^2n^2}{y^2}, \text{ which is square of an integer. }$

Therefore $R$ is transitive.

$\displaystyle \text{(iv) } x + 4y = 10, x, y \in N$

$\text{Given } R = \{ (x, y) : x+4y = 10; x, y \in N \}$

$\Rightarrow R = \{ (2, 2), (6, 1) \}$

$\text{Clearly, } (1, 1) \notin R.$

Therefore $R$ is not reflexive.

$(6, 1) \in R \text{ but } (1, 6) \notin R$

Therefore $R$ is not symmetric.

$(x, y) \in R \Rightarrow x+4y = 10$

$\text{and } ( y, z) \in R \Rightarrow y + 4z = 10$

$\Rightarrow x - 16z = - 30$

$\Rightarrow ( x, z) \notin R$

Therefore $R$ is not transitive.