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

Question Details
BookNCERT Textbook
Chapter1 [Relations and Functions]
Question No.1
Question TypeExercise

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 \(y\}\)

Expert Answer

(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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