# NCERT Solutions of Class 12 Maths | CBSE Textbook Solutions | Chapter 1 | Relations and Functions | Exercise 1.1 | Question 1 |

Question Details
 Board CBSE Book NCERT Textbook Class 12 Subject Mathematics Chapter 1 [Relations and Functions] Exercise 1.1 Question No. 1 Question Type Exercise

Question 1. Determine whether each of the following relations are reflexive, symmetric and transitive :
(i) Relation $$R$$ in the set $$A=\{1,2,3, \ldots \ldots, 13,14\}$$ defined as $$R=\{(x, y): 3 x-y=0\}$$
(ii) Relation $$R$$ in set $$N$$ of natural numbers defined as $$R=\{(x, y): y=x+5$$ and $$x<4\}$$
(iii) Relation $$R$$ in the set $$A=\{1,2,3,4,5,6\}$$ as $$R=\{(x, y): y$$ is divisible by $$x\}$$
(iv) Relation $$R$$ in the set $$Z$$ of all integers defined as $$R=\{(x, y): x-y$$ is an integer $$\}$$
(v) Relation $$R$$ in set $$A$$ of human beings in a town at a particular time given by
(a) $$R=\{(x, y): x$$ and $$y$$ work to the same place $$\}$$
(b) $$R=\{(x, y): x$$ and $$y$$ live in the same locality $$\}$$
(c) $$R=\{(x, y): x$$ is exactly $$7 \mathrm{~cm}$$ taller than $$y\}$$
(d) $$R=\{(x, y): x$$ wife of $$y\}$$
(e) $$R=\{(x, y): x$$ is father of $$y\}$$

(i) (a) Given that, $$A=\{1,2,3,4, \ldots \ldots, 12,13,14\}$$ and $$R$$ is
$$=\{(\mathrm{x}, \mathrm{y}): 3 \mathrm{x}-\mathrm{y}=0\}$$ for reflexive relation $$(\mathrm{x}, \mathrm{x}) \in R \forall \mathrm{x} \in A$$
$$\therefore$$ If $$y=x, 3 x-y=0 \Rightarrow 2 x \neq 0, R$$ is not reflexive.
$$\Rightarrow \quad(x, x) \notin R$$

(b) Given $$R=\{(\mathrm{x}, \mathrm{y}): 3 \mathrm{x}-\mathrm{y}=0\}$$ for symmetric relation $$(\mathrm{x}, \mathrm{y}) \in R$$ $$\Rightarrow(\mathrm{y}, \mathrm{x}) \in R$$
If $$(x, y) \in R \Rightarrow 3 x-y=0$$, then $$3 y-x \neq 0 \Rightarrow(y, x) \notin R$$, so $$R$$ is not symmetric.

(c) For transitive, if $$(x, y) \in R \Rightarrow 3 x-y=0,(y, z) \in R \Rightarrow 3 y-z=0$$, then $$3 \mathrm{x}-\mathrm{z} \neq 0$$. So, $$R$$ is not transitive
For example, when, $$x=1, y=3, z=9$$, then $$3 \times 1-3=0,3 \times 3-9=0,3 \times 1-9 \neq 0$$

(ii) Here, $$A=N$$, the set of natural numbers and
$R=\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=\mathrm{x}+5, \mathrm{x}<4\}$
$=\{(x, x+5): x \in N \text { and } x<4\}=\{(1,6),(2,7),(3,8)\}$
(a) For reflexive $$(\mathrm{x}, \mathrm{x}) \in R \forall \mathrm{x}$$ putting $$\mathrm{y}=\mathrm{x}, \mathrm{x} \neq \mathrm{y}+5 \Rightarrow(1,1) \notin R$$. So, $$R$$ is not reflexive.

(b) For symmetrical $$(x, y) \in R \Rightarrow(y, x) \in R$$ putting $$y=x+5$$, then $$\mathrm{x} \neq \mathrm{y}+5 \Rightarrow(1,6) \in R$$ but $$(6,1) \notin R$$. So, $$R$$ is not symmetric.

(c) For transitivity $$(\mathrm{x}, \mathrm{y}) \in R,(\mathrm{y}, \mathrm{z}) \in R \Rightarrow(\mathrm{x}, \mathrm{z}) \in R$$ if $$\mathrm{y}=\mathrm{x}+5$$, $$z=y+5$$, then $$z \neq x+5$$. Since, $$(1,6) \in R$$ and there is no order pair in $$R$$ which has 6 as the first element same in the case for $$(2,7)$$ and $$(3,8)$$. So, $$R$$ is not transitive.

(iii) (a) For reflexive, we know that $$\mathrm{x}$$ is divisible by $$\mathrm{x}$$ for all $$\mathrm{x} \in A$$. $$\therefore \quad(\mathrm{x}, \mathrm{x}) \in R$$ for all $$\mathrm{x} \in R$$. So, $$R$$ is reflexive.

(b) For symmetry, we observe that 6 is divisible by 2 . This means that $$(2,6) \in R$$ but $$(6,2) \notin R$$. So, $$R$$ is not symmetric.

(c) For transitivity, let $$(x, y) \in R$$ and $$(y, z) \in R$$, then $$z$$ is divisible by $$\mathrm{X}$$
$$\Rightarrow \quad(\mathrm{x}, \mathrm{z}) \in R$$
For example, 2 is divisible by 1,4 is divisible by 2
So, 4 is divisible by 1 . So, $$R$$ is transitive.

(iv) (a) For reflexive put $$y=x, x-x=0$$ which is an integer for all $$x \in Z . \mathrm{So}^{\prime} R$$ is reflexive on $$Z$$.

(b) For symmetry let $$(x, y) \in R$$, then $$(x-y)$$ is an integer $$\lambda$$ and also $$y-x=-\lambda \quad[\because \lambda \in Z \Rightarrow-\lambda \in Z]$$ $$\therefore y-x$$ is an integer $$\Rightarrow(y, x) \in R$$. So, $$R$$ is symmetric.

(c) For transitivity let $$(x, y) \in R$$ and $$(y, z) \in R \therefore x-y=$$ integer and $$y-z=$$ integers, then $$x-z$$ is also an integer $$\therefore \quad(\mathrm{x}, \mathrm{z}) \in R$$. So, $$R$$ is transitive.

(v) (a) $$R$$ is reflexive, symmetric and transitive obviously.

(b) $$R$$ is reflexive, symmetric and transitive obviously.

(c) Here, $$R$$ is not reflexive as $$\mathrm{x}$$ is not $$7 \mathrm{~cm}$$ taller than $$\mathrm{x}$$
$$R$$ is not symmetric as if $$\mathrm{x}$$ is exactly $$7 \mathrm{~cm}$$ taller than $$\mathrm{y}$$, then $$\mathrm{y}$$ cannot be $$7 \mathrm{~cm}$$ taller than $$\mathrm{x}$$ and $$R$$ is not transitive as if $$\mathrm{x}$$ is exactly $$7 \mathrm{~cm}$$ taller than $$y$$ and $$y$$ is exactly $$7 \mathrm{~cm}$$ taller than $$z$$, then $$x$$ is exactly $$14 \mathrm{~cm}$$ taller than $$\mathrm{z}$$.

(d) Here, $$R$$ is not reflexive; as $$\mathrm{x}$$ is not wife of $$\mathrm{x}, R$$ is not symmetric, as if $$x$$ is wife of $$y$$, then $$y$$ is husband (not wife) of $$x$$ and $$R$$ is transitive as transitivity is not contradicted in this case. Whenever $$(x, y) \in R$$, then $$(y, z) \notin R$$ for any $$z$$ as if $$x$$ is wife of $$y_1$$ then $$y$$ is a male and a male cannot be a wife.

(e) Here, $$R$$ is not reflexive; as $$\mathrm{x}$$ cannot be father of $$\mathrm{x}$$. For any $$\mathrm{x}, R$$ is not symmetric as if $$x$$ is father of $$y$$, then $$y$$ cannot be father of $$x . R$$ is not transitive as if $$x$$ is father of $$y$$ and $$y$$ is father of $$z$$, then $$x$$ is grandfather (not father) of $$z$$.

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