NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1 - Relations and Functions

Question1(i) . 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 $R = \{(x,y): 3x - y = 0\}$

Answer:

$A = \{1, 2, 3, \ldots, 13, 14\}$

$R = \{(x, y) : 3x - y = 0\} = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$

Since $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), \ldots, (14, 14) \notin R$, so $R$ is not reflexive.

Since $(1, 3) \in R$ but $(3, 1) \notin R$, so $R$ is not symmetric.

Since $(1, 3), (3, 9) \in R$ but $(1, 9) \notin R$, so $R$ is not transitive.

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

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

(ii) Relation R in the set N of natural numbers defined as
$R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\}$

Answer:

$R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\}$ $= \left \{ \left ( 1,6 \right ),\left ( 2,7 \right ),\left ( 3,8 \right ) \right \}$

Since, $\left ( 1,1 \right ) \notin R$

so $R$ is not reflexive.

Since, $\left ( 1,6 \right )\in R$ but $\left ( 6,1 \right )\notin R$

so $R$ is not symmetric.

Since there is no pair in $R$ such that $\left ( x,y \right ),\left ( y,x \right )\in R$ so this is not transitive.

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

Question1(iii) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iii) Relation R in the set $A = \{1,2,3,4,5,6\}$ as $R = \{(x,y) : y \; \textup{is} \; divisible \; by\; x\}$

Answer:

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

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

Any number is divisible by itself and $\left ( x,x \right ) \in R$ .So it is reflexive.

$\left ( 2,4 \right ) \in R$ but $\left ( 4,2 \right ) \notin R$ .Hence,it is not symmetric.

$\left ( 2,4 \right ),\left ( 4,4 \right ) \in R$ and 4 is divisible by 2 and 4 is divisible by 4.

Hence, it is transitive.

Hence, it is reflexive and transitive but not symmetric.

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

(iv). Relation R in the set Z of all integers defined as $R = \{(x,y): x - y \;is\;an\;integer\}$

Answer:

$R = \{(x,y): x - y \;is\;an\;integer\}$

For $x \in Z$ , $\left ( x,x \right ) \in R$ as $x-x = 0$ which is an integer.

So,it is reflexive.

For $x,y \in Z$ , $\left ( x,y \right ) \in R$ and $\left ( y,x \right ) \in R$ because $x-y \, \, and \, \, y-x$ are both integers.

So, it is symmetric.

For $x,y,z \in Z$ , $\left ( x,y \right ),\left ( y,z \right ) \in R$ as $x-y \, \, and \, \, y-z$ are both integers.

Now, $x-z = \left ( x-y \right )+\left ( y-z \right )$ is also an integer.

So, $\left ( x,z \right ) \in R$ and hence it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

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

$\left ( x,x \right )\in R$ ,so it is reflexive

$\left ( x,y \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ .

$y \;and\; x\;work\;at\;the\;same\;place$ i.e. $\left ( y,x \right )\in R$ so it is symmetric.

$\left ( x,y \right ),\left ( y,z \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ also $y \;and\; z\;work\;at\;the\;same\;place$ .It states that $x \;and\; z\;work\;at\;the\;same\;place$ i.e. $\left ( x,z \right )\in R$ .So, it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question 1 (v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

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

$\left ( x,x \right )\in R$ as $x$ and $x$ is same human being.So, it is reflexive.

$\left ( x,y \right )\in R$ means $x\;and\;y\;live\;in\;the\;same\;locality$ .

It is same as $y\;and\;x\;live\;in\;the\;same\;locality$ i.e. $\left ( y,x \right )\in R$ .

So,it is symmetric.

$\left ( x,y \right ),\left ( y,z \right )\in R$ means $x\;and\;y\;live\;in\;the\;same\;locality$ and $y\;and\;z\;live\;in\;the\;same\;locality$ .

It implies that $x\;and\;z\;live\;in\;the\;same\;locality$ i.e. $\left ( x,z \right )\in R$ .

Hence, it is reflexive, symmetric and transitive.

Question 1 (v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

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

$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ but $x\;is\;not\;\;taller\;than\;x$ i.e. $\left ( x,x \right )\notin R$ .So, it is not reflexive.

$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ but $y\;is\;not\;\;taller\;than\;x$ i.e $\left ( y,x \right )\notin R$ .So, it is not symmetric.

$\left ( x,y\right ),\left ( y,z \right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ and $y\;is\;exactly\;7\;cm\;taller\;than\;z$ .

$x\;is\;exactly\;14\;cm\;taller\;than\;z$ i.e. $\left ( x,z \right )\notin R$ .

Hence, it is not reflexive, symmetric, or transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

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

$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but $x\;is\;not\, wife\;of\;x$ i.e. $\left ( x,x \right ) \notin R$ .

So, it is not reflexive.

$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but $y\;is\;not\, wife\;of\;x$ i.e. $\left ( y,x \right ) \notin R$ .

So, it is not symmetric.

Let, $\left ( x,y \right ),\left ( y,z \right ) \in R$ means $x\;is\;wife\;of\;y$ and $y\;is\;wife\;of\;z$ .

This case is not possible, so it is not transitive.

Hence, it is not reflexive, symmetric or transitive.

Question 1 (v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

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

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

Answer:

If \( (x, y) \in R \), then \( x \) is the father of \( y \). \\

But a person cannot be their own father, so \( (x, x) \notin R \). \\

\(\Rightarrow R \text{ is not reflexive.} \)

If \( (x, y) \in R \), then \( x \) is the father of \( y \). \\

This does not imply that \( y \) is the father of \( x \), so \( (y, x) \notin R \). \\

\(\Rightarrow R \text{ is not symmetric.} \)

Suppose \( (x, y), (y, z) \in R \). \\

Then \( x \) is the father of \( y \), and \( y \) is the father of \( z \). \\

This means \( x \) is the grandfather of \( z \), not the father. \\

So, \( (x, z) \notin R \). \\

\(\Rightarrow R \text{ is not transitive.} \)

So, it is not transitive.

Hence, it is neither reflexive nor symmetric nor transitive.

Question:2 Show that the relation R in the set R of real numbers defined as
$R = \{(a, b) : a \leq b^2 \}$ is neither reflexive nor symmetric nor transitive.

Answer:

$R = \{(a, b) : a \leq b^2 \}$

Taking

$\left ( \frac{1}{2},\frac{1}{2} \right )\notin R$

and

$\left ( \frac{1}{2} \right )> \left ( \frac{1}{2} \right )^{2}$

So, R is not reflexive.

Now,

$\left ( 1,2 \right )\in R$ because $1< 4$ .

But, $4\nless 1$ i.e. 4 is not less than 1

So, $\left ( 2,1 \right )\notin R$

Hence, it is not symmetric.

$\left ( 3,2 \right )\in R\, \, and \, \, \left ( 2,1.5 \right )\in R$ as $3< 4\, \, and \, \, 2< 2.25$

Since $\left ( 3,1.5 \right )\notin R$ because $3\nless 2.25$

Hence, it is not transitive.

Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive.

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

Answer:

$R$ defined on the set $\{1, 2, 3, 4, 5, 6\}$

$R = \{(a, b) : b = a + 1\}$

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

Since $\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\} \notin R$, so it is not reflexive.

$\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\} \in R$ but $\{(2, 1), (3, 2), (4, 3), (5, 4), (6, 5)\} \notin R$, so it is not symmetric.

$\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\} \in R$ but $\{(1, 3), (2, 4), (3, 5), (4, 6)\} \notin R$, so it is not transitive.

So, it is not transitive.

Hence, it is neither reflexive, nor symmetric, nor transitive

Question:4 Show that the relation R in R defined as $R = \{(a, b) : a \leq b\}$ , is reflexive and

transitive but not symmetric.

Answer:

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

As $\left ( a,a \right )\in R$ so it is reflexive.

Now we take an example

$\left ( 2,3 \right )\in R$ as $2< 3$

But $\left ( 3,2 \right )\notin R$ because $2 \nless 3$ .

So,it is not symmetric.

Now if we take, $\left ( 2,3 \right )\in R\, \, and\, \, \left ( 3,4 \right )\in R$

Than, $\left ( 2,4 \right )\in R$ because $2< 4$

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

Question:5 Check whether the relation R in R defined by $R = \{(a, b) : a \leq b^3 \}$ is reflexive,
symmetric or transitive.

Answer:

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

$\left ( \frac{1}{2},\frac{1}{2} \right )\notin R$ because $\frac{1}{2}\nleqslant (\frac{1}{2}) ^{3}$

So, it is not symmetric

Now, $\left ( 1,2 \right ) \in R$ because $1< 2^{3}$

but $\left ( 2,1 \right )\notin R$ because $2\nleqslant 1^{3}$

It is not symmetric

$\left ( 3,1.5 \right ) \in R\, \, and \, \, \left ( 1.5,1.2 \right ) \in R$ as $3< 1.5^{3} \, \, and \, \, 1.5< 1.2^{3}$ .

But, $\left ( 3,1.2 \right )\notin R$ because $3 \nleqslant 1.2^{3}$

So it is not transitive

Thus, it is neither reflexive, nor symmetric, nor transitive.

Question:6 Show that the relation R in the set $\{1, 2, 3\}$ given by $R = \{(1, 2), (2, 1)\}$ is
symmetric but neither reflexive nor transitive.

Answer:

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

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

We can see $\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right )\notin R$ so it is not reflexive.

As $\left ( 1,2 \right )\in R \, and \, \left ( 2,1 \right )\in R$ so it is symmetric.

$(1, 2) \in R \, and\, (2, 1)\in R$

But $(1, 1)\notin R$ so it is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question:7 Show that the relation R in the set A of all the books in a library of a college,
given by $R = \{(x, y) : x \;and\;y\;have\;same\;number\;of\;pages\}$ is an equivalence
relation.?

Answer:

A = all the books in a library of a college

$R = \{(x, y) : x \;and\;y\;have\;same\;number\;of\;pages\}$

$(x,x) \in R$ because x and x have the same number of pages, so it is reflexive.

Let $(x,y) \in R$ means x and y have the same number of pages.

Since y and x have the same number of pages, so $(y,x) \in R$.

Hence, it is symmetric.

Let $(x,y) \in R$ means x and y have the same number of pages.

and $(y,z) \in R$ means y and z have the same number of pages.

This states,x and z also have the same number of pages i.e. $(x,z) \in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive, i.e. it is an equivalence
relation.

Question:8 Show that the relation R in the set $A = \{1, 2, 3, 4, 5\}$ given by $R = \{(a, b) : |a - b| \;is\;even\}$ , is an equivalence relation. Show that all the elements of $\{1, 3, 5\}$ are related to each other and all the elements of $\{2, 4\}$ are related to each other. But no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$ .

Answer:

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

$R = \{(a, b) : |a - b| \;is\;even\}$

$R=\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 3,1 \right ),\left ( 5,1 \right ),\left ( 4,2 \right ),\left ( 5,3 \right )\right \}$

Let there be $a\in A$ then $(a,a)\in R$ as $\left | a-a \right |=0$ which is even number. Hence, it is reflexive

Let $(a,b)\in R$ where $a,b\in A$ then $(b,a)\in R$ as $\left | a-b \right |=\left | b-a \right |$

Hence, it is symmetric

Now, let $(a,b)\in R \, and\, (b,c)\in R$

$\left | a-b \right | \, and \, \left | b-c \right |$ are even number i.e. $(a-b)\, and\,(b-c)$ are even

then, $(a-c)=(a-b)+(b-c)$ is even (sum of even integer is even)

So, $(a,c)\in R$ . Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive, i.e. it is an equivalence relation.

The elements of $\{1, 3, 5\}$ are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.

The elements of $\{2, 4\}$ are related to each other because the difference of even number is even number and in this set, all numbers are even.

The element of $\{1, 3, 5\}$ is not related to $\{2, 4\}$ because a difference of odd and even number is not even.

Question:9(i) Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$ , given by

(i) $R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

$A = \{x \in Z : 0 \leq x \leq 12\}$

$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$

$R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$

For $a\in A$ , $(a,a)\in R$ as $\left | a-a \right |=0$ which is multiple of 4.

Henec, it is reflexive.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4.

then $\left | b-a \right |$ is also multiple of 4 because $\left | a-b \right |$ = $\left | b-a \right |$ i.e. $(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4 and $(b,c)\in R$ i.e. $\left | b-c \right |$ is multiple of 4 .

$( a-b )$ is multiple of 4 and $(b-c)$ is multiple of 4

$(a-c)=(a-b)+(b-c)$ is multiple of 4

$\left | a-c \right |$ is multiple of 4 i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is $\left \{1,5,9 \right \}$

$\left | 1-1 \right |=0$ is multiple of 4.

$\left | 5-1 \right |=4$ is multiple of 4.

$\left | 9-1 \right |=8$ is multiple of 4.

Question:9(ii) Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$ , given by

(ii) $R = \{(a, b) : a = b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

$A = \{x \in Z : 0 \leq x \leq 12\}$

$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$

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

For $a\in A$ , $(a,a)\in R$ as $a=a$

Henec, it is reflexive.

Let, $(a,b)\in R$ i.e. $a=b$

$a=b$ $\Rightarrow$ $b=a$ i.e. $(b,a)\in R$

Hence, it is symmetric.

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

$\therefore$ $a=b=c$

$a=c$ i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is {1}

Question:10(i) Give an example of a relation.

(i) Which is Symmetric but neither reflexive nor transitive.

Answer:

Let

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

$R = \left \{ \left ( 1,2 \right ),\left ( 2,1 \right )\right \}$

$\left ( 1,1 \right ),\left ( 2,2 \right ),(3,3) \notin R$ so it is not reflexive.

$(1,2)\in R$ and $(2,1)\in R$ so it is symmetric.

$(1,2)\in R \, and\, (2,1)\in R$ but $(1,1)\notin R$ so it is not transitive.

Hence, symmetric but neither reflexive nor transitive.

Question:10(ii) Give an example of a relation.

(ii) Which is transitive but neither reflexive nor symmetric.

Answer:

Let

$R = \left \{ \left ( x,y \right ): x> y \right \}$

Now for $x\in R$ , $(x,x)\notin R$ so it is not reflexive.

Let $(x,y) \in R$ i.e. $x> y$

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

Let $(x,y) \in R$ i.e. $x> y$ and $(y,z) \in R$ i.e. $y> z$

we can write this as $x> y> z$

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

Hence, it is transitive but neither reflexive nor symmetric.

Question:10(iii) Give an example of a relation.

(iii) Which is Reflexive and symmetric but not transitive.

Answer:

Let

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

Define a relation R on A as

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

If $x\in A$ , $(x,x)\in R$ i.e. $\left \{ (1,1),(2,2),(3,3)\right \} \in R$ . So it is reflexive.

If $x,y\in A$ , $(x,y)\in R$ and $(y,x)\in R$ i.e. $\left \{(1,2),(2,1),(2,3),(3,2) \right \}\in R$ . So it is symmetric.

$(x,y)\in R$ and $(y,z)\in R$ i.e. $(1,2)\in R$ . and $(2,3)\in R$

But $(1,3)\notin R$, So it is not transitive.

Hence, it is Reflexive and symmetric but not transitive.

Question 10 (iv) Give an example of a relation.

(iv) Which is Reflexive and transitive but not symmetric.

Answer:

Let there be a relation R in R

$R=\left \{ (a,b):a\leq b \right \}$

$(a,a)\in R$ because $a=a$

Let $(a,b)\in R$ i.e. $a\leq b$

But $(b,a)\notin R$ i.e. $b\nleqslant a$

So it is not symmetric.

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

This can be written as $a\leq b\leq c$ i.e. $a\leq c$ implies $(a,c)\in R$

Hence, it is transitive.

Thus, it is Reflexive and transitive but not symmetric.

Question 10 (v) Give an example of a relation.

(v) Which is Symmetric and transitive but not reflexive.

Answer:

Let there be a relation A in R

$A= \left \{ 1,2 \right \}$

$R=\left \{ (1,2),(2,1),(2,2)\right \}$

$(1,1)\notin R$ So R is not reflexive.

We can see $(1,2)\in R$ and $(2,1)\in R$

So it is symmetric.

Let $(1,2)\in R$ and $(2,1)\in R$

Also $(2,2)\in R$

Hence, it is transitive.

Thus, it Symmetric and transitive but not reflexive.

Question:11 Show that the relation R in the set A of points in a plane given by
$R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}$ , is an equivalence relation. Further, show that the set of
all points related to a point $P \neq (0, 0)$ is the circle passing through P with origin as
centre.

Answer:

$R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}$

The distance of point P from the origin is always the same as the distance of same point P from origin i.e. $(P,P)\in R$

$\therefore$ R is reflexive.

Let $(P,Q)\in R$ i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

this is the same as distance of the point Q from the origin is the same as the distance of the point P from the origin i.e. $(Q,P)\in R$

$\therefore$ R is symmetric.

Let $(P,Q)\in R$ and $(Q,S)\in R$

i.e. the distance of point P from the origin is the same as the distance of point Q from the origin, and also the distance of point Q from the origin is the same as the distance of the point S from the origin.

We can say that the distance of point P, Q, S from the origin is the same. Means distance of point P from the origin is the same as the distance of point S from origin i.e. $(P,S)\in R$

$\therefore$ R is transitive.

Hence, R is an equivalence relation.

The set of all points related to a point $P \neq (0, 0)$ are points whose distance from the origin is the same as the distance of point P from the origin.

In other words, we can say there be a point O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.

Hence, these sets of points form a circle with the centre as the origin and this circle passes through the point.

Question:12 Show that the relation R defined in the set A of all triangles as $R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$ , is equivalence relation. Consider three right angle triangles T 1 with sides 3, 4, 5, T 2 with sides 5, 12, 13 and T 3 with sides 6, 8, 10. Which triangles among T 1 , T 2 and T 3 are related?

Answer:

$R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$

All triangles are similar to itself, so it is reflexive.

Let,

$(T_1,T_2) \in R$ i.e.T 1 is similar to T2

T 1 is similar to T2 is the same asT2 is similar to T 1 i.e. $(T_2,T_1) \in R$

Hence, it is symmetric.

Let,

$(T_1,T_2) \in R$ and $(T_2,T_3) \in R$ i.e. T 1 is similar to T2 and T2 is similar toT 3 .

$\Rightarrow$ T 1 is similar toT 3 i.e. $(T_1,T_3) \in R$

Hence, it is transitive,

Thus, $R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$ , is equivalence relation.

Now, we see the ratio of sides of triangle T 1 andT 3 are as shown

$\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$

i.e. ratios of sides of T 1 and T 3 are equal.Hence, T 1 and T 3 are related.

Question:13 Show that the relation R defined in the set A of all polygons as $R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$ , is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer:

$R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$

The same polygon has the same number of sides with itself,i.e. $(P_1,P_2) \in R$ , so it is reflexive.

Let,

$(P_1,P_2) \in R$ i.e.P 1 have same number of sides as P 2

P 1 have the same number of sides as P 2 is the same as P 2 have same number of sides as P 1 i.e. $(P_2,P_1) \in R$

Hence,it is symmetric.

Let,

$(P_1,P_2) \in R$ and $(P_2,P_3) \in R$ i.e. P 1 have the same number of sides as P 2 and P 2 have same number of sides as P 3

$\Rightarrow$ P 1 have same number of sides as P 3 i.e. $(P_1,P_3) \in R$

Hence, it is transitive,

Thus, $R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$ , is an equivalence relation.

The elements in A related to the right angle triangle T with sides 3, 4 and 5 are those polygons which have 3 sides.

Hence, the set of all elements in A related to the right angle triangle T is set of all triangles.

Question:14 Let L be the set of all lines in XY plane and R be the relation in L defined as $R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$ . Show that R is an equivalence relation. Find the set of all lines related to the line $y = 2x + 4.$

Answer:

$R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$

All lines are parallel to itself, so it is reflexive.

Let,

$(L_1,L_2) \in R$ i.e.L 1 is parallel to L 2 .

L1 is parallel to L 2 is same as L 2 is parallel to L 1 i.e. $(L_2,L_1) \in R$

Hence, it is symmetric.

Let,

$(L_1,L_2) \in R$ and $(L_2,L_3) \in R$ i.e. L1 is parallel to L 2 and L 2 is parallel to L 3 .

$\Rightarrow$ L 1 is parallel to L 3 i.e. $(L_1,L_3) \in R$

Hence, it is transitive,

Thus, $R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$ , is equivalence relation.

The set of all lines related to the line $y = 2x + 4.$ are lines parallel to $y = 2x + 4.$

Here, Slope = m = 2 and constant = c = 4

It is known that the slope of parallel lines are equal.

Lines parallel to this ( $y = 2x + 4.$ ) line are $y = 2x + c$ , $c \in R$

Hence, set of all parallel lines to $y = 2x + 4.$ are $y = 2x + c$ .

Question:15 Let R be the relation in the set A= {1,2,3,4}

given by $R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}$ . Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

Answer:

A = {1,2,3,4}

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

For every $a \in A$ there is $(a,a) \in R$ .

$\therefore$ R is reflexive.

Given, $(1,2) \in R$ but $(2,1) \notin R$

$\therefore$ R is not symmetric.

For $a,b,c \in A$ there are $(a,b) \in R \, and \, (b,c) \in R$ $\Rightarrow$ $(a,c) \in R$

$\therefore$ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is option B.

Question 16 Let R be the relation in the set N given by $R = \{(a, b) : a = b - 2, b > 6\}$ . Choose the correct answer.

(A) $(2, 4) \in R$
(B) $(3,8) \in R$
(C) $(6,8) \in R$
(D) $(8,7) \in R$

Answer:

$R = \{(a, b) : a = b - 2, b > 6\}$

(A) Since, $b< 6$ so $(2, 4) \notin R$

(B) Since, $3\neq 8-2$ so $(3,8) \notin R$

(C) Since, $8> 6$ and $6=8-2$ so $(6,8) \in R$

(d) Since, $8\neq 7-2$ so $(8,7) \notin R$

The correct answer is option C.