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Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

Relation and Function Exercise 1.1 Solutions

1. Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set A = {1, 2, 3…13, 14} defined as

R = {(x, y): 3x − y = 0}

(ii) Relation R in the 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 as integer}

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) R = {(x, y): x and y work at the same place}

(b) R = {(x, y): x and y live in the same locality}

(c) R = {(x, y): x is exactly 7 cm taller than y}

(d) R = {(x, y): x is wife of y}

(e) R = {(x, y): x is father of y}

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Vishal Baghel | Contributor-Level 10

4 months ago

(i) We have, $R = {(x, y) : 3 x - y = 0}$ a relation in set A= ${1, 2, 3 . . . . . . . . . . 14}$

For $x \in A, y = 3 x$ or $y \neq x$ i.e.,

$(x, x)$ does not exist in R

$∴$ R is not reflexive.

For $(x, y) \in R, y = 3 x$

Then $(y, x) x \neq 3 y$

So $(y, x) \notin R$

$∴$ R is not symmetric

For $(x, y) \in R$ and $(y, z) \in R$ . We have

$y = 3 x$ and $z = 3 y$

Then $z = 3 (3 x) = 9 x$

i.e., $(x, z) \notin R$

$∴$ R is not Transitive

(ii) We have,

R= ${(x, y) : y = x + 5 & x < 4}$ is a relation in N

= ${(1, 1 + 5), (2, 2 + 5), (3, 3 + 5)}$

= ${(1, 6), (2, 7), (3, 8)}$

Clearly, R is not reflexive as $(x, x) \notin R$ and $x < 4 & x \in N$

Also, R is not symmetric as $(1, 6) \in R$ but $(6, 1) \notin R$

And for $(x, y) \in R (y, z) \notin R$ . Hence, R is not Transitive.

(iii) R= ${(x, y); y$ is divisible by x $}$ is a relation in set

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

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

Hence, R is reflexive because $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) \in R$ i.e., $(x, x) \in R$

R is not sy

...more

(i) We have, $R = {(x, y) : 3 x - y = 0}$ a relation in set A= ${1, 2, 3 . . . . . . . . . . 14}$

For $x \in A, y = 3 x$ or $y \neq x$ i.e.,

$(x, x)$ does not exist in R

$∴$ R is not reflexive.

For $(x, y) \in R, y = 3 x$

Then $(y, x) x \neq 3 y$

So $(y, x) \notin R$

$∴$ R is not symmetric

For $(x, y) \in R$ and $(y, z) \in R$ . We have

$y = 3 x$ and $z = 3 y$

Then $z = 3 (3 x) = 9 x$

i.e., $(x, z) \notin R$

$∴$ R is not Transitive

(ii) We have,

R= ${(x, y) : y = x + 5 & x < 4}$ is a relation in N

= ${(1, 1 + 5), (2, 2 + 5), (3, 3 + 5)}$

= ${(1, 6), (2, 7), (3, 8)}$

Clearly, R is not reflexive as $(x, x) \notin R$ and $x < 4 & x \in N$

Also, R is not symmetric as $(1, 6) \in R$ but $(6, 1) \notin R$

And for $(x, y) \in R (y, z) \notin R$ . Hence, R is not Transitive.

(iii) R= ${(x, y); y$ is divisible by x $}$ is a relation in set

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

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

Hence, R is reflexive because $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) \in R$ i.e., $(x, x) \in R$

R is not symmetric as $(1, 2) \in R$ but $(2, 1) \notin R$

And for $x, y, z \in A$ and $(x, y) \in R & (y, z) \in R$

$\frac{y}{z} = n$ and $\frac{z}{y} = m$ where $n & m \in N$

Then, $z = m y = m (n x) = m n . x$

$\frac{z}{x} = m . n, m, m n \in N$

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

$∴$ R is transitive

(iv) R= ${(x, y) : x - y$ is an integer $}$ is a relation in set Z

For $x \in Z$

$x - x = 0$ is an integer

So, $(x, x) \in R$ i.e., R is reflexive

For $x, y \in Z$ and $(x, y) \in R$

$x - y$ is an integer

$- (y - x)$ is an integer

$(y - x)$ is an integer

So, $(y, x) \in R$ i.e., R is symmetric

For $x, y, z \in R$ and $(x, y) \in R & (y, z) \in R$ We have,

$x - y$ is an integer

$y - z$ is an integer

So, $(x - y) + (y - z)$ is also an integer

$x - z$ is an integer

So, $(x, z) \in R$ i.e., R is transitive.

(v) (a) R= ${(x, y) : x$ and $y$ work at same place $}$ in set A of human being.

For $x \in A$ we get

$x & x$ work at same place

$(x, x) \in R$ So, R is reflexive.

For $x, y \in A$ and $(x, y) \in R$ We get,

$x & y$ work at same place

$y & x$ work at same place

$(y, x) \in R$ . So, R is symmetric.

For $x, y, z \in A$ and $(x, y)$ and $(y, z) \in R$ . We have,

$x & y$ work at same place

$y & z$ work at same place

$x & z$ work at same place

i.e., $(x, z) \in R$ . So, R is Transitive.

(b) R= ${(x, y) : x & y$ live in same locality $}$

For $x \in A$ , we get,

$x & x$ live in same locality

$(x, x) \in R$ So, R is reflexive.

For $x, y \in A$ and $(x, y) \in R$ we get,

$x & y$ live in same locality

$y & x$ live in same locality

i.e., $(y, x) \in R$ . So, R is symmetric.

For $x, y, z \in A$ and $(x, y)$ & $(y, z) \in R$ . Then

$x, y$ live in same locality

$y, z$ live in same locality

$x, z$ live in same locality

So, $(x, z) \in R$ i.e., R is transitive.

(c) R= ${(x, y) : x$ is exactly 7 cm taller than y $}$

For $x \in A$ ,

Height $(x) \neq 7 +$ height of $(x)$

So, $(x, x) \notin R$ . i.e., R is not reflexive.

For, $x, y \in A$ and $(x, y) \in R$ we have,

Height $(x)$ $= 7 + h e i g h t (y)$

But $h e i g h t (y) \neq 7 + h e i g h t (x)$

So, $(y, x) \in R$ i.e., R is not symmetric.

For $x, y, z \in A$ and $(x, y) \in R$ and $(y, z) \in R$ we have,

Height $(x) = 7 + h e i g h t (y)$

And $h e i g h t (y) = 7 + h e i g h t (z)$

So, $h e i g h t (x) = 7 + 7 + h e i g h t (z) = 14 + h e i g h t (z)$

i.e., $(x, z) \notin R$

So, R is not transitive.

(d) R= ${(x, y) : x$ is wife of y $}$

For $x \in A,$

X is not wife of x.

So, $(x, x) \notin R$ i.e., R is not reflexive.

For $x, y \in A$ and $(x, y) \in R$ ,

X is wife of y but y is not wife of x

So, $(y, x) \notin R$ . i.e., R is not symmetric.

For $x, y, z \in A$ and $(x, y)$ and $(y, z) \in R$

X is wife of y

Y is wife of z

But y can never be husband & wife simultaneously

So, R is not transitive.

(e) R= ${(x, y) : x$ is father of y $}$

For $x \in A$ ,

X is not a father of x

So, $(x, x) \notin R$ i.e., R is not reflexive.

For $x, y \in A$ and $(x, y) \in R$

X is father of y

Y is father of z

But x is not father of z

So, $(x, z) \notin R$ i.e., R is not transitive.

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$?$ R is symmetric relation

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