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Relations Excercise: 1.1
Relations Exercise 1.1 Question 1 (i)
Answer : Reflexive, symmetric, transitive.
Hint :If
$R$ is reflexive
$\Rightarrow(a, a) \in R \text { for all } a \in A$If
$R$ is symmetric
$\Rightarrow(a, b) \in R \Rightarrow(b, a) \in R_{\text {for all }} a, b \in A$If
$R$ is transitive
$\Rightarrow(a, b) \in R \text { and }(b, c) \in R \Rightarrow(a, c) \in R \text { for all } a, b, c \in A$Given : $R=\{(x, y): x \text { and } y \text { work at same place }\}$Solution : A relation
$R$ on set A is said to be reflexive if every element of A is related to itself.
Thus,
$R$ is reflexive
$\Rightarrow(a, a) \in R \text { for all } a \in A$A relation
$R$ on set A is said to be symmetric relation if
$(a, b) \in R \Rightarrow(b, a) \in R$ for all
$a, b \in A$i.e.
$a R b \Rightarrow b R a \text { for all } a, b \in A$A relation
$R$ on set A is said to be transitive relation
If
$(a, b) \in R \; {\text {and }}\; (b, a) \in R \Rightarrow(c, a) \in R\; {\text {for all }} \; a, b, c \in A$i.e
$a R b$ and
$b R c \Rightarrow a R c$ for all
$a, b, c \in A$Now, let's get back to the actual problem
For Reflexive :$x$ and
$x$ work at same place
Similarly,
$y$ and
$y$ work at the same place.
So,
$R$ is reflexive.
For Symmetric :It is given that
$x$ and
$y$ work at the same place.
So, we can say that
$y$ and
$x$ work at the same place.
So,
$R$ is symmetric.
For Transitive:Let
$z$ be a person such that
$y$ and
$z$ work at the same place.
And we know that
$x$ and
$y$ work at the same place.
So, we can say that
$x$ and
$z$ work at the same place.
So,
$R$ is Transitive.
Relations Exercise 1.1 Question 1 (ii)
Answer : Reflexive, symmetric, transitive.
Hint :If
$R$ is reflexive
$\Rightarrow(a, a) \in R \text { for all } a \in A$If
$R$ is symmetric
$\Rightarrow(a, b) \in R \Rightarrow(b, a) \in R_{\text {for all }} a, b \in A$If
$R$ is transitive
$\Rightarrow(a, b) \in R \text { and }(b, c) \in R \Rightarrow(a, c) \in R \text { for all } a, b, c \in A$Given :$\mathrm{R}=\{(x, y): x \text { and } y \text { live in same locality\} }$Solution : A relation
$R$ on set A is said to be reflexive if every element of A is related to itself.
Thus,
$R$ is reflexive
$\Leftrightarrow(a, a) \in R \; {\text {for all }} a \in A$A relation
$R$ on set A is said to be symmetric relation if
$(a, b) \in R \Rightarrow(b, a) \in R$for all
$a, b \in A$I.e.
$a R b \Rightarrow b R a$ for all
$a, b \in A$A relation
$R$ on set A is said to be transitive relation if
$(a, b) \in R \; {\text {and }}(b, c) \in R \Rightarrow(c, a) \in R$ for all
$a, b, c \in A$ i.e
$a R b \text { and } b R c \Rightarrow a R c \text { for all } a, b, c \in A$For Reflexive:$x$ and
$x$ live in the same locality.
Similarly,
$y$ and
$y$ live in same locality
So,
$R$ is reflexive.
For Symmetric:$x$ and
$y$ live in same locality.
So, we can easily say that
$y$ and
$x$ live in same locality.
So,
$R$ is symmetric.
For Transitive:Let
$z$ be a person;
$z \in A$ and
$z$ and
$y$ live in same locality
And it is given that
$x$ and
$y$ live in same locality
So, we can say that
$x$ and
$z$ live in the same locality.
So,
$R$ is Transitive.
Relations Exercise 1.1 Question 1 (iii)
Answer:Neither reflexive, nor symmetric, not transitive.
Hints :If
$R$ is reflexive
$\Rightarrow(a, a) \in R \; {\text {for all }} a \in A$If
$R$ is symmetric
$\Rightarrow(a, b) \in R \Rightarrow(b, a) \in R \text { for all } a, b \in A$If
$R$ is transitive
$\Rightarrow(a, b) \in R,(b, c) \in R \Rightarrow(a, c) \in R \text { for all } a, b, c \in A$Given :$R=\left\{(x, y): x\; {\text {is wife of }} y\right\}$Solution :A relation
$R$ on set
$A$ is said to be reflexive if every element of
$A$ is related to itself.
Thus,
$R$ is reflexive
$\Leftrightarrow(a, a) \in R\; {\text {for all }} a \in A$.
A relation
$R$ on set
$A$ is said to be symmetric relation if
$(a, b) \in R \Rightarrow(b, a) \in R$for all
$a, b \in A$i.e;
$a R b \Rightarrow b R a \text { for all } a, b \in A$A relation
$R$ on set
$A$ is said to be transitive relation if
$(a, b) \in R$ and
$(b, c) \in R \Rightarrow$ $(c, a) \in R$ for all
$a, b, c \in A$i.e;
$a R b$ and
$b R c \Rightarrow a R c$ for all
$a, b, c \in A$ For Reflexive:$x$ is not wife of
$x$ and
$y$ is not wife of
$y$So,
$R$ is not reflexive.
For Symmetric:$x$ is the wife of
$y$ but
$y$ is not the wife of
$x$.
So,
$R$ is not symmetric.
For Transitive:Let
$z$ be a person;
$\mathrm{z} \in A$ such that
$y$ is the wife of
$z$ .
And it is given that
$x$ is the wife of
$y$ but this case is not possible. Also, here we can’t show x is the wife of z.
So, R is not transitive.
Relations Exercise 1.1 Question 1 (iv)
Answer:Neither reflexive, nor symmetric nor transitive.
Hints :If
$R$ is reflexive
$\Rightarrow(a, a) \in R\; {\text {for all }} a \in A$If
$R$ is symmetric
$\Rightarrow(a, b) \in R \Rightarrow(b, a) \in R\; {\text {for all }} a, b \in A$If
$R$ is transitive
$\Rightarrow(a, b) \in R,(b, c) \in R \Rightarrow(a, c) \in R \text { for all } a, b, c \in A$Given :$R=\{(x, y): x \text { is father of } \mathrm{y}\}$Solution :A relation
$R$ on set
$A$ is said to be reflexive if every element of
$A$ is related to itself.
Thus,
$R$ is reflexive
$\Leftrightarrow(a, a) \in R_{\text {for all }} a \in A$A relation
$R$ on set
$A$ is said to be symmetric relation if
$(a, b) \in R \Rightarrow(b, a) \in R$ for all
$a, b \in A$i.e
$a R b \Rightarrow b R a$ for all
$a, b \in A$A relation
$R$ on set
$A$ is said to be transitive relation if
$(a, b) \in R$ and
$(b, c) \in R \Rightarrow$ $(c, a) \in R$ for all
$a, b, c \in A$i.e
$a R b$ and
$b R c \Rightarrow a R c$ for all
$a, b, c \in A$ For Reflexive:$x$ is not father of
$x$ and
$y$ is not father of
$y$So,
$R$ is not reflexive
For Symmetric:It is given that
$x$ is the father of
$y$.
But we can say that
$y$ is not the father of
$x$.
So
$R$ is not symmetric.
For Transitive:Let
$z$ be a person;
$z \in A$ such that
$y$ is father of
$z$ and it is given that x is a father of y.
Then
$x$ is grandfather of
$z$So,
$R$ is not Transitive.
Relations Exercise 1.1 Question 2
Answer: $R_{1}$ is reflexive and transitive but not symmetric
$R_{2}$ is reflexive, symmetric, and transitive.
$R_{3}$ is transitive but neither reflexive nor symmetric
$R_{4}$ is neither reflexive nor symmetric nor transitive
Hint :
A relation R on set A is
Reflexive relation: If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:If
$(a, b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:If
$(a, b)$and
$(b,c)$ $\in\; R$, then
$(a, c)$$\in\; R$ for every
$a, b \in A$Given :
Set
$A=\{a, b, c\}$$\begin{aligned} &R_{1}=\{(a, a),(a, b),(a, c),(b, b),(b, c),(c, a),(c, b),(c, c)\} \\ &R_{2}=\{(a, a)\} \\ &R_{3}=\{(b, c)\} \\ &R_{4}=\{(a, b),(b, c),(c, a)\} \end{aligned}$Consider
$R_{1}=\{(a, a),(a, b),(a, c),(b, b),(b, c),(c, a),(c, b),(c, c)\}$Reflexive :
Given
$(a, a),(b, b)$ and
$(c, c) \in R_{1}$So,
$R_{1}$ is reflexive.
For Symmetric:
We see that the ordered pairs obtained by interchanging the components of
$R_{1}$ are not in
$R_{1}$.
For ex :
$(a, b) \in R_{1} \text { but }(b, a) \notin R_{1}$So,
$R_{1}$ is not symmetric.
For Transitive:
Here,
$(a, b) \in R_{1} \text { and }(b, c) \in R_{1}$ but
$(a, c) \in R_{1}$So,
$R_{1}$ is transitive
(ii) Consider
$R_{2}$$R_{2}=\{(a, a)\}$Reflexive:
clearly
$(a, a) \in R_{2}$So,
$R_{2}$ is reflexive.
Symmetric:
Clearly
$(a,a)\in \; R_{2}$So,
$R_{2}$ is symmetric.
Transitive:
$R_{2}$ is a transitive relation, since there is only one element in it.
(iii) Consider
$R_{3}$$R_{3}=\{(b, c)\}$Reflexive:
Here neither
$(b, b) \notin R_{3}$ nor
$(c, c) \notin R_{3}$So,
$R_{3}$ is not reflexive
Symmetric:
Here neither
$(b, c) \in R_{3}$ nor
$(c, b) \notin R_{3}$So,
$R_{3}$ is not symmetric.
Transitive:
$R_{3}$ has only one element
Hence
$R_{3}$ is transitive.
(iv) Consider
$R_{4}=\{(a, b),(b, c),(c, a)\}$Reflexive:
Here
$(a, b) \in R_{4} \text { but }(b, a) \notin R_{4}$So,
$R_{4}$ is not reflexive
Symmetric:
Here
$(a, b) \in R_{4} \text { but }(b, a) \notin R_{4}$So,
$R_{4}$ is not symmetric
Transitive:
Here
$(a, b) \in R_{4},(b, c) \in R_{4}$ but
$(a, c) \notin R_{4}$Hence
$R_{4}$ is not transitive.
Relations Exercise 1.1 Question 3 (i)
Answer: $R_{1}$is symmetric but neither reflexive nor transitive.
Hint : A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b, a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$a, b, c \in A$Given :$R_{1} \text { on } \mathrm{Q}_{0}$ defined by
$(a, b) \in R_{1} \Leftrightarrow a=\frac{1}{b}$Solution :
Reflexivity:
Let
$a$ be an arbitrary element of
$R_{1}$Then,
$a \in R_{1}$$a \neq \frac{1}{a} \text { for all } a \in Q_{0}$So,
$R_{1}$ is not reflexive
Symmetry:
Let
$(a, b) \in R_{1}$Therefore, we can write 'a' as
$a=\frac{1}{b}$$b=\frac{1}{a}$Then
$(b, a) \in R_{1}$So,
$R_{1}$ is symmetric.
For Transitive:
Let
$(a, b) \in R_{1} \text { and }(b, c) \in R_{1}$$\begin{aligned} &a=\frac{1}{b} \text { and } b=\frac{1}{c} \\ &a=\frac{1}{\left(\frac{1}{c}\right)} \Rightarrow c \\ &a \neq \frac{1}{c} \\ &(a, c) \notin R_{1} \end{aligned}$So,
$R_{1}$ is not transitive.
Relations Exercise 1.1 Question 3 (ii)
Answer:$R_{2}$ is reflexive and symmetric but not transitive.
Hint : A relation R on set A is
Reflexive relation:
If $(a, a) \in R$ for every $a \in A$
Symmetric relation:
if $(a, b)$ is true then $(b,a)$ is also true for every $a, b \in A$
Transitive relation:
If $(a, b)$ and $(b, c) \in R$, then $(a, c) \in R$ for every $a, b, c \in A$
Given:
$R_{2} \text { on } Z$ defined by $(a, b) \in R_{2} \Leftrightarrow|a-b| \leq 5$
Solution :
Reflexivity:
Let a be an arbitrary element of $R_{2}$
Then, $a \in R_{2}$
On applying the given condition, we get
$|a-a|=0 \leq 5$
So, $R_{2}$ is reflexive
Symmetry :
Let $(a, b) \in R_{2} \quad|a-b| \leq 5$
[Since $|a-b|=|b-a|$ ]
Then $|b-a| \leq 5$
$(b, a) \in R_{2}$
So, $R_{2}$ is symmetric.
Transitivity :
Let $(1,3) \in R_{2} \text { and }(3,7) \in R_{2}$
$|1-3| \leq 5 \text { and }|3-7| \leq 5$
But, $|1-7| \leq 5$
$(1,7) \neq R_{2}$
So, $R_{2}$ is not transitive.
Relations Exercise 1.1 Question 3 (iii)
Answer:
$R_{3}$ is reflexive but neither symmetric nor transitive
Hint :
A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in R$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
if
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{A}$Given :
$R_{3}$ on R defined by
$(a, b) \in R_{3} \Leftrightarrow a^{2}-4 a b+3 b^{2}=0$Solution :
Reflexivity:
Let a be an arbitrary element of
$R_{3}$Then,
$a \in R_{3}$$a^{2}-4 a \times a+3 a^{2}=0$So,
$R_{3}$ is reflexive
Symmetry :
Let :
$(a, b) \in R_{3}$$a^{2}-4 a b+3 b^{2}=0$But
$b^{2}-4 b a+3 a^{2} \neq 0$ for all
$a, b \in R$So,
$R_{3}$ is not symmetric.
Transitivity:
Let
$(a, b) \in R_{3} \text { and }(b, c) \in R_{3}$$\begin{aligned} &a^{2}-4 a b+3 b^{2}=0\; \; \; \; \; \; ...(i)\\ &\text { and } b^{2}-4 b c+3 c^{2}=0 \; \; \; \; \; \; \; ....(ii) \end{aligned}$Adding (i) and (ii) we get,
$\begin{aligned} &\qquad a^{2}-4 a b+3 b^{2}+b^{2}-4 b c+3 c^{2}=0 \\ &\Rightarrow a^{2}-4 a b+4 b^{2}-4 b c+3 c^{2}=0 \\ &\text { But } a^{2}-4 a c+3 c^{2}=-4 a c-4 a b+4 b^{2}-4 b c \neq 0 \\ &\Rightarrow a^{2}-4 a c+3 c^{2} \neq 0 \end{aligned}$$(a, c) \notin R_{3}$So,
$R_{3}$ is not transitive.
Relations Exercise 1.1 Question 4
Answer:
$R_{1}$ is reflexive but neither symmetric nor transitive
$R_{2}$ is symmetric but neither reflexive nor transitive
$R_{3}$ is transitive but neither reflexive nor symmetric
Hint :
A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in R$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
if
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{A}$Given :
$\begin{aligned} &R_{1}=\{(1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\} \\ &R_{2}=\{(2,2),(3,1),(1,3)\} \\ &R_{3}=\{(1,3),(3,3)\} \end{aligned}$Solution :
Consider
$R_{1}$$R_{1}=\{(1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\}$Reflexivity:
Here
$(1,1),(2,2),(3,3) \in R$So,
$R_{1}$ is Reflexive
Symmetric :
Here
$(2,1) \in R_{1}$But,
$(1,2) \notin R_{1}$So,
$R_{1}$ is not symmetric
Transitivity:
$\begin{aligned} &\text { Here }(2,1) \in R_{1} \text { and }(1,3) \in R_{1} \\ &\text { But }(2,3) \notin R_{1} \end{aligned}$So,
$R_{1}$ is not transitive.
Now consider
$R_{2}$$R_{2}=\{(2,2),(3,1),(1,3)\}$Reflexivity :
Clearly,
$(1,1) \text { and }(3,3) \notin R_{2}$So,
$R_{2}$ is not reflexive.
Symmetric :
$\text { Here }(1,3) \in R_{2} \text { and }(3,1) \in R_{2}$So,
$R_{2}$ is symmetric.
Transitivity:
$\begin{aligned} &\text { Here }\\ &(1,3) \in R_{2} \text { and }(3,1) \in R_{2}\\ &\text { But }(3,3) \notin R_{2} \end{aligned}$So,
$R_{2}$ is not transitive.
Now consider
$R_{3}$$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
$(1,3) \in R_{3} \text { and }(3,3) \in R_{3}$But
$(1,3) \in R_{3}$So,
$R_{3}$ is transitive.
Relation Exercise 1.1 Question 5 (i)
Answer: Given relation is transitive
Hint: A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$a, b, c \in A$Given :
$a R b \text { if } a-b>0$Solution :Reflexivity:
Let
$a$ be an arbitrary element of
$R$Then,
$a \in A$But
$a-a=0 \ngtr 0$So, this relation is not reflexive.
Symmetry :
Let
$\begin{aligned} &(a, b) \in R \\ &a-b>0 \\ &-(b-a)>0 \\ &b-a<0 \end{aligned}$So, the given relation is not symmetric.
Transitivity :
$\begin{aligned} &\text { Let }(a, b) \in R \; {\text {and }}(b, c) \in R \\ &\text { Then, } a-b>0 \; \; \; \; \; \; \; ...(i)\\ &\; \; \; \; \qquad b-c>0 \; \; \; \; \; \; \;...(ii) \end{aligned}$Adding eq (i) & (ii), we get
$\begin{aligned} &a-b+b-c>0 \\ &a-c>0 \\ &(a, c) \in R \end{aligned}$So, the given relation is transitive.
Relation Exercise 1.1 Question 5 (ii)
Answer:Reflexive and symmetric but not transitive.
Hint:A relation R on set A is
Reflexive relation:
If
$(a,a)\in \; R$ for every
$a\in \; R$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$\mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given :$a R b \text { if } 1+a b>0$Solution :Reflexivity:
Let
$a$ be an arbitrary element of
$R$Then,
$a\in \; R$$\begin{aligned} &1+a \times a>0 \\ &1+a^{2}>0 \end{aligned}$Since, Square of any number is positive.
So, the given relation is reflexive.
Symmetry:
Let
$(a, b) \in R$$\begin{aligned} &1+a b>0 \\ &1+b a>0 \\ &(b, a) \in R \end{aligned}$So, the given relation is symmetric.
Transitivity:
$\begin{aligned} &\text { Let }(a, b) \in R \text { and }(b, c) \in R \\ &\text { Let } \mathrm{a}=-8, \mathrm{~b}=-2, \mathrm{c}=\frac{1}{4} \\ &\text { Then } 1+a b>0 \text { i.e; } 1+(-8)(-2)=17>0 \\ &\text { and } 1+b c>0 \text { i.e; } 1+(-2) \frac{1}{4}=\frac{1}{2}>0 \text { But } 1+a c \neq 0 \text { i.e; } 1+(-8) \frac{1}{4}=-1 \ngtr 0 \\ &(a, c) \epsilon R \end{aligned}$So, the given relation is not transitive.
Relation Exercise 1.1 Question 5 (iii)
Answer: Transitive neither reflexive nor symmetric.
Hint:A relation R on set A is
Reflexive relation:
If
$(a,a)\in \; R$ for every
$a\in \; R$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$\mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given :
$a R b \text { if }|a| \leq b$Solution :
Reflexivity:
Let -a be an arbitrary element of R
Then
$-a \in R$$\Rightarrow|-a| \neq-a$So,
$R$ is not reflexive
Symmetry:
Let
$(a, b) \in R$$\begin{aligned} &|a| \leq b \\ &|b| \lessgtr a \text { for all } a, b \in R \\ &(b, a) \notin R \end{aligned}$So,
$R$ is not symmetric.
Transitivity:
$\begin{aligned} &\text { Let }(a, b) \in R_{\text {and }}(b, c) \in R \\ &|a| \leq b \text { and }|b| \leq c \text { for } a, b, c \in R \end{aligned}$Multiplying the corresponding sides, we get
$\begin{aligned} &|a| \times|b| \leq b c \\ &|a| \leq c \\ &(a, c) \in R \end{aligned}$Thus,
$R$ is transitive
Relation Exercise 1.1 Question 6
Answer:$R$ is neither reflexive nor symmetric nor transitive.
Hint:A relation R on set A is
Reflexive relation:
If
$(a,a)\in \; R$ for every
$a\in \; R$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$\mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given :
$R=\{(a, b): b=a+1\}$Solution :Let
$a$ be an arbitrary element of set A.
Then,
$a=a+1$ cannot be true for all
$a\; \in\; A$$(a, a) \notin R$So,
$R$ is not reflexive on
$A$Symmetry :
$\begin{aligned} &\text { Let }(a, b) \in R\\ &b=a+1\\ &a=b-1\\ &-a=-b+1\\ &\text { Thus }\\ &(b, a) \notin R \end{aligned}$So,
$R$ is not symmetric on
$A$Transitivity :
Let
$(1,2) \text { and }(2,3) \in R$$\begin{aligned} &2=1+1 \text { and } \\ &3=2+1 \text { is true } \\ &\text { But } 3 \neq 1+1 \\ &(1,3) \notin R \end{aligned}$So,
$R$ is not transitive on
$A$Relation Exercise 1.1 Question 7
Answer:$R$ is neither reflexive nor symmetric nor transitive.
Hint:A relation R on set A is
Reflexive relation:
If
$(a,a)\in \; R$ for every
$a\in \; R$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$\mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Solution :
Reflexive :
It is observed that
$(1 / 2,1 / 2) \text { is } R \text { as } 1 / 2>(1 / 2)^{3}=\frac{1}{8}$$\therefore R \text { is not reflexive. }$Symmetric :
Now
$(1,2) \in R\left(\text { as } 1<2^{3}=8\right)$But
$(2,1) \notin R\left(\text { as } 2>1^{3}=1\right)$$R$ is not symmetric
Transitive:
We have
$\left(3, \frac{3}{2}\right),\left(\frac{3}{2}, \frac{6}{5}\right) \operatorname{in} R \text { as } 3<\left(\frac{3}{2}\right)^{3}$and
$\left(\frac{3}{2}\right)<\left(\frac{6}{5}\right)^{3}$but
$\left(3, \frac{6}{5}\right) \notin R \text { as } 3>\left(\frac{6}{5}\right)^{3}$$R$ is not transitive
Hence,
$R$ is neither reflexive, nor symmetric nor transitive
Relations Exercise 1.1 Question 8
Answer:Hence, prove every identity relation on a set is reflexive but the converse is not necessarily true.
Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$ is true then
$\left ( b,a\right )$ is also true for every
$a, b \in A$Transitive relation:
$\text { If }(a, b) \text { and }(b, c) \in R, \text { then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given: Every Identity relation is reflexive.
Solution:Let
$A$ be a set
$A=\{1,2,3\}$Then,
$I_{A}=\{(1,1),(2,2),(3,3)\}$Identity relation is reflexive, since
$(a, a) \in A \quad \forall a$The converse of it need not be necessarily true.
Counter example:
Consider the set
$A=\{1,2,3\}$Here,
Relation
$R=\{(1,1),(2,2),(3,3),(2,1),(1,3)\}$ is reflexive on
$A$However,
$R$ is not an identity relation.
Relations Exercise 1.1 Question 9(i)
Answer: $R=\{(1,1),(2,2),(3,3),(4,4),(2,1)\}$Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$is true then
$\left ( b,a \right )$ is also true for every
$a, b \in A$Transitive relation:
$\text { If }(a, b) \text { and }(b, c) \in R, \text { then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given: $A=\{1,2,3,4\}$Solution:
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
$(1,1),(2,2),(3,3) \in R$And
$(2,2),(2,1) \in R \Rightarrow(2,1) \in R$However,
$(2,1) \in R_{\text {but }}(1,2) \notin R$Relations Exercise 1.1 Question 9 (ii)
Answer: $R=\{(1,2),(2,1)\}$Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$ is true then
$\left ( b,a \right )$ is also true for every
$a, b \in A$Transitive relation:
$\text { If }(a, b) \text { and }(b, c) \in R, \text { then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given: $A=\{1,2,3\}$Solution: The relation on A having properties of being symmetric but neither transitive nor reflexive.
Let
$R=\{(1,2),(2,1)\}$Now,
$(1,2) \in R,(2,1) \in R$So, it is symmetric.
Clearly
$R$ is not transitive
$(1,2) \in R,(2,1) \in R \text { but }(1,1) \notin R$Relations Exercise 1.1 Question 9(iii)
Answer: $R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)\}$The relation
$R$ is an equivalence relation on
$A$Hint:A relation
$R$ on set
$A$ is
Reflexive relation:
If
$(a, a) \in R$for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$ is true then
$\left ( b,a \right )$ is also true for every
$a, b \in A$Transitive relation:
$\text { If }(a, b) \text { and }(b, c) \in R, \text { then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given:$A=\{1,2,3,4\}$Solution:
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)\}$The relation
$R$ is an equivalence relation on
$A$Relations Exercise 1.1 Question 10
Answer:Domain of
$R$ is x ∈ N where x∈
$\left \{ 1,2,3............20 \right \}$and
range of
$R$ is y ∈ N where
$\left \{ 39,37,35,......3,1 \right \}$The relation having properties of being neither symmetric nor transitive nor reflexive.
Hint: A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$is true then
$\left (b,a\right )$ is also true for every
$a \in A$Transitive relation:
$\text { If }(a, b) \text { and }(b, c) \in R, \text { then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given: $R=\{(x, y): x, y \in N, 2 x+y=41\}$Solution:The domain of
$R \text { is } x \in N, \text { such that } 2 x+y=41$$x=(41-y) / 2$Since
$y \in N$, largest value that
$x$ can take corresponds to the smallest value that
$y$ can take.
$\therefore \quad x=\{1,2,3 \ldots .20\}$Range
$R$ of is
$y \in N$such that
Since
$\begin{gathered} 2 x+y=41 \\ y=41-2 x \\ x=\{1,2,3 \ldots 20\} \end{gathered}$$y=\{39,37,35, \ldots . .3,1\}$Since
$(2,2) \notin R, R$,is not reflexive.
Also, since
$(1,39) \in R(39,1) \notin R, R$ is not symmetric.
Finally, since
$(15,11) \in R \text { and }(11,19) \in R \text { but }(15,19) \notin R, R$ is not transitive.
Relations Exercise 1.1 Question 11
Answer:No, it is not true that every relation which is symmetric and transitive is also reflexive.
Hint: A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a, b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a, b)$and
$(b,a)$ $\in R$, then
$(a, c) \in R$ for every
$a, b \in A$Given: Answer that, whether every relation which is symmetric and transitive is also reflexive.
Solution:No, it is not necessary that a relation that is symmetric and transitive is reflexive as well.
For example:
$R=\{(1,1),(1,2),(2,1)\} \text { on } A=\{1,2\}$ is symmetric and transitive but not reflexive.
Because
$(2,2) \notin R$Relations Exercise 1.1 Question 13
Answer:Hence proved, the relation “
$\geq$” on the set
$R$ of all real numbers is reflexive and transitive but not symmetric.
Hint:A relation
$R$ on set
$A$ is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a, b)$ is true then
$(b, a)$is also true for every
$a, b \in A$Transitive relation:
If
$(a, b) \text { and }(b, c) \in R$ then
$(a, c) \in R$ for every
$a, b, c \in A$Given: Relation is ”
$\geq$”on the
$R$ of all real numbers.
Solution:Reflexivity:
Let
$a \in R$$a \geq a$"
$\geq$ "is reflexive
.Transitive:
Let
$a, b, c \in R$Such that
$a \geq b \text { and } b \geq c$Then
$a \geq c$$" \geq "$ is transitive.
Symmetric: Let
$a, b \in R$Such that
$a \geq b \text { but } b \ngeqslant a$$" \geq "$not symmetric
Relations Exercise 1.1 Question 14(ii)
Answer:$R=\left\{(a, b): a^{3} \geq b^{3}\right\}$Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$ is true then
$\left (b,a\right )$ is also true for every
$a, b \in A$Transitive relation:
$\text { If }(a, b) \text { and }(b, c) \in R \text { , then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given:We have to give the example of a relation which is reflexive and transitive but not symmetric.
Solution: The relation having properties of being reflexive and transitive but not symmetric.
Define a relation
$R$ in
$R$ as:
$R=\left\{(a, b): a^{3} \geq b^{3}\right\}$Clearly
$(a, a) \in R \text { as } a^{3}=a^{3}$Therefore
$R$ is reflexive.
Now
$(2,1) \in R\left(\text { as } 2^{3} \geq 1^{3}\right)$But
$(1,2) \notin R\left(\text { as } 1^{3}<2^{3}\right)$Therefore
$R$ is not symmetric.
Now let
$\begin{aligned} &(a, b), \quad(b, c) \in R \\ &a^{3} \geq b^{3} \text{and }b^{3} \geq c^{3} \text { then } a^{3} \geq c^{3} \quad(a, c) \in R \end{aligned}$$R$ is transitive.
Hence, relation $R$ is reflexive and transitive but not symmetric.
Relations Exercise 1.1 Question 14(iii)
Answer: $R=\{(-5,-6),(-6,-5),(-5,-5)\}$Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$ is true then
$\left (b,a\right )$ is also true for every
$a, b \in A$Transitive relation:
$\text { If }(a, b) \text { and }(b, c) \in R \text { , then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given: We have to give the example of a relation which is symmetric and transitive but not reflexive.
Solution:The relation having properties of being symmetric and transitive but not reflexive.
Let
$A=\{-5,-6\}$Define a relation
$R$ on
$A$ as
$R=\{(-5,-6),(-6,-5),(-5,-5)\}$Relation
$R$ is not reflexive as
$(-6,-6) \notin R$Relation
$R$ is symmetric as
$(-5,-6),(-6,-5) \in R$It is seen that
$(-5,-6),(-6,-5) \in R$Also
$(-5,-5) \in R$The relation
$R$ is transitive.
Hence relation
$R$ is symmetric and transitive but not reflexive.
Relations Exercise 1.1 Question 14(iv)
Answer: $R=\{(5,6),(6,5)\}$Hint:
A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$ is true then
$\left ( b,a \right )$is also true for every
$a, b \in A$Transitive relation:
If
$(a, b) \text { and }(b, c) \in R$, then
$(a, c) \in R$ for every
$a, b, c \in A$Given:$\text { Let } A=\{5,6,7\} \text { . }$Solution:$\\\text{Define a relation R on A as R}=\{(5,6),(6,5)\}. \\\text{Relation R is not reflexive as }(5,5),(6,6),(7,7) \notin \mathrm{R}.$$\begin{aligned} &\text { Now, as }(5,6) \in \mathrm{R} \text { and also }(6,5) \in \mathrm{R}, \mathrm{R} \text { is symmetric. }\\ &\Rightarrow(5,6),(6,5) \in \mathrm{R}, \text { but }(5,5) \notin \mathrm{R} \end{aligned}$$\\\text{Therefore, R is not transitive. }\\ \text{Hence, relation R is symmetric but not reflexive or transitive.}$Relations Exercise 1.1 Question 14(v)
Answer:$R=\{(a, b): a<b\}$Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$for every
$a \in A$Symmetric relation:
If
$\left ( a,b \right )$ is true then
$\left ( b,a\right )$is also true for every
$a, b \in A$Transitive relation:
If
$(a, b) \text { and }(b, c) \in R, \text { then }(a, c) \in R \text { for every } \mathrm{a}, \mathrm{b}, \mathrm{c} \in A$Given:We have to give the example of a relation which is transitive but neither symmetric nor reflexive.
Solution:The relation having properties of being transitive but neither symmetric nor reflexive.
Consider a relation R in
$R$ defined as:
$R=\{(a, b): a<b\}$For any
$a \in R$ we have
$(a, a) \notin R$, since
$a$ can’t be strictly less than
$a$ itself.
Infact
$a=a$∴ Relation
$R$ is not reflexive.
Now,
$(1,2) \in R(\text { as } 1<2)$But 2 is not less than 1
∴
$(2,1) \notin R$∴
$R$ is not symmetric
Now let
$(a, b),(b, c) \in R$$\begin{aligned} &a<b \text { and } b<c \\ &a<c \\ &(a, c) \in R \end{aligned}$$R$ is transitive.
Relation Exercise 1.1 Question 15
Answer :$R=\{(1,2),(2,3),(1,1),(2,2),(3,3),(3,2),(2,1),(1,3),(3,1)\}$Hint : A relation R on set A is
Reflexive relation:
$a, b, c \in A$If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and , then
$(b, c) \in R$ for every
$(a, c) \in R$Given :
$\begin{aligned} &\text { Relation } R=\{(1,2),(2,3)\} \text { on the set }\\ &A=\{1,2,3\} \end{aligned}$Solution :To make
$R$ reflexive we will add
$(1,1),(2,2),(3,3)$ to get
$R^{\prime}=\{(1,2),(2,3),(1,1),(2,2),(3,3)\}$ is reflexive.
Again, to make
$R$ symmetric we will add
$(3,2)$ and
$(2,1)$ $R^{\prime \prime}=\{(1,2),(2,3),(1,1),(2,2),(3,3),(3,2),(2,1)\}$is reflexive and symmetric .
To make
$R$ transitive we will add
$(1,3)$ and
$(3,1)$ $R^{\prime \prime \prime}=\{(1,2),(2,3),(1,1),(2,2),(3,3),(3,2),(2,1),(1,3),(3,1)\}$ is reflexive and symmetric and transitive.
Relation Exercise 1.1 Question 16
Answer:only 1 ordered pair maybe added to
$R$ so that it may become a transitive relation on
$A$Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$a, b, c \in A$Given :$A=\{1,2,3\}$Solution :$R=\{(1,2),(1,1),(2,3)\}$ be a relation on
$A$To make
$R$ transitive we shall add
$(1,3)$ only
$R^{\prime}=\{(1,2),(1,1),(2,3),(1,3)\}$As we know,
Transitive relation
$\\\text {x=y} \; and \; \text {y=z}\\ \text {Then} \; \text {x=z}$Note: for
$R$ to be transitive
$(a,c)$ must be in
$R$ because
$(a, b) \in R \text { and }(b, c) \in R$ So,
$(a,c)$ must be in
$R$Relation Exercise 1.1 Question 17
Answer:At least 3 ordered pairs must be added for
$R$ to be reflexive and transitive
Hint:A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$, then
$(a, c) \in R$ for every
$a, b, c \in A$Given: Minimum number of order pair that makes R reflexive and transitive.
Solution:A relation
$R$ in
$A$ is said to be reflexive if
$aRa$ for all
$a \in A$$R$ is said to be transitive if
$aRb$ and bRc then aRc for all
$a, b, c \in A$Hence for
$R$ to be reflexive
$(b,b)$ and
$(c,c)$must be there in set
$R$Also for
$R$ to be transitive
$(a,c)$ must be in
$R$ because
$(a, b) \in R(b, c) \in R$So,
$(a,c)$ must be in
$R$So, at least 3 ordered pairs must be added for
$R$ to be reflexive and transitive.
Relation Exercise 1.1 Question 18 (i)
Answer : Transitive
Hint : A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$ then
$(a, c) \in R$ for every
$a, b, c \in A$Given :$x>y, x, y \in N(x, y) \in\{(2,1),(3,1) \ldots \ldots(3,2),(4,2) \ldots\}$Solution :This is not reflexive as
$(1,1),(2,2) \ldots \ldots$are absents.
This is not symmetric as
$(2,1)$ is present but
$(1,2)$ is absent
This is transitive as
$(3,2) \in R \text { and }(2,1) \in R \text { also }(3,1) \in R$Hence, this relation is not satisfying reflexivity and symmetricity.
Relation Exercise 1.1 Question 18 (ii)
Answer : Symmetric
Hint : A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$ then
$(a, c) \in R$ for every
$a, b, c \in A$Given :$x+y=10, x, y \in N(x, y) \in\{(9,1),(1,9),(2,8),(8,2),(3,7),(7,3),(4,6),(6,4),(5,5)\}$Solution :This is not reflexive as
$(1,1),(2,2) \ldots \ldots$ are absent.
This only follows the condition of symmetry as
$(1,9) \in R \operatorname{also}(9,1) \in R$This is not transitive because
$\{(1,9),(9,1)\} \in R \text { but }(1,1)$ is absent.
Hence, this relation is not satisfying reflexivity and transitivity.
Relation Exercise 1.1 Question 18 (iii)
Answer : Symmetric, Reflexive and transitive
Hint : A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$ then
$(a, c) \in R$ for every
$a, b, c \in A$Given :$x y \text { is square of an integer } \ x, y \in N$$(x, y) \in\{(1,1),(2,2),(4,1),(1,4)(3,3),(9,1),(1,9),(4,4),(2,8),(8,2),(16,1),(1,16) \ldots\}$Solution :This is reflexive as
$(1,1),(2,2) \ldots \ldots$are present.
This is also symmetric because
$a R b \Leftrightarrow b R a \text { for all } a, b \in N$This is transitive because if
$a R b \text { and } b R c \Leftrightarrow a R c \ \ \ a, b, c \in N$This relation is reflexive, symmetric, and transitive.
Relation Exercise 1.1 Question 18 (iv)
Answer : This relation is neither symmetric nor reflexive nor transitive.
Hint : A relation R on set A is
Reflexive relation:
If
$(a, a) \in R$ for every
$a \in A$Symmetric relation:
If
$(a,b)$ is true then
$(b,a)$ is also true for every
$a, b \in A$Transitive relation:
If
$(a,b)$ and
$(b, c) \in R$ then
$(a, c) \in R$ for every
$a, b, c \in A$Given :$\\x+4 y=10, x, y \in N \\ \\ (x, y) \in\{(6,1),(2,2)\}$ x
Solution :
This is not reflexive as $(1,1),(6,6)$ are absent.
This is not symmetric as $(6,1)$is present but $(1,6)$ is absent.
This is not transitive as there are only two elements in the set having no element in common.
This relation is neither symmetric nor reflexive nor transitive.
With such a vast number of topics, studying all of them at once gets pretty confusing. Therefore, the questions from RD Sharma Class 12th Exercise 1.1 are divided into two parts: Level 1 and Level 2. The best approach for this chapter it's to complete 20 Level 1 questions and 10 Level 2 questions per day to cover the entire chapter systematically. This will ensure that you cover a major part of the portion without the burden.
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