Exercise 1.2 - Chapter 1 Sets Relations and Functions 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 1.2
Question 1.
Discuss the following relations for reflexivity, symmetricity and transitivity:
(i) The relation $\mathrm{R}$ defined on the set of all positive integers by " $m R n$ if $m$ divides $n$ ".

Solution:
$\mathrm{S}=\{$ set of all positive integers $\}$
(a) $\mathrm{mRm} \Rightarrow$ ' $\mathrm{m}$ ' divides' $\mathrm{m}$ ' $\Rightarrow$ reflexive
(b) $\mathrm{mRn} \Rightarrow \mathrm{m}$ divides $\mathrm{n}$ but $\mathrm{nRm} \Rightarrow \mathrm{n}$ does not divide $\mathrm{m}$ (i.e.,) $m R n \neq n R m$ It is not symmetric
(c) $\mathrm{mRn} \Rightarrow \mathrm{nRr}$ as $\mathrm{n}$ divides $\mathrm{r}$
It is transitive
(ii) Let $\mathrm{P}$ denote the set of all straight lines in a plane. The relation $\mathrm{R}$ defined by "IRm if 1 is perpendicular to $\mathrm{m}$ ".
Solution:
$P=\{$ set of all straight lines in a plane $\}$
$\mathrm{IRm} \Rightarrow 1$ is perpendicular to $\mathrm{m}$
(a) $\mathrm{lRl} \Rightarrow 1$ is not perpendicular to $\mathrm{l}$
$\Rightarrow$ It is not reflexive
(b) $\mathrm{lRm} \Rightarrow 1$ is perpendicular to $\mathrm{m}$ $\mathrm{mRl} \Rightarrow \mathrm{m}$ is perpendicular to $\mathrm{l}$
It is symmetric
(c) I perpendicular to $\mathrm{m} \Rightarrow \mathrm{m}$ perpendicular to $\mathrm{n} \Rightarrow 1$ is parallel to $\mathrm{n}$ It is not transitive 

(iii) Let $\mathrm{A}$ be the set consisting of all the members of a family. The relation $\mathrm{R}$ defined by "aRb if $a$ is not a sister of $\mathrm{b}$ ".
Solution:
$A=\{$ set of all members of the family $\}$
$a R b$ is $a$ is not a sister of $b$
(a) $\mathrm{aRa} \Rightarrow \mathrm{a}$ is not a sister of a It is reflexive
(b) $a R b \Rightarrow a$ is not a sister of $b$.
$\mathrm{bRa} \Rightarrow \mathrm{b}$ is not a sister of $\mathrm{a}$.
It is symmetric
(c) $\mathrm{aRb} \Rightarrow \mathrm{a}$ is not a sister of $\mathrm{b}$.
$\mathrm{bRc} \Rightarrow \mathrm{b}$ is not a sister of $\mathrm{c}$.
$\Rightarrow \mathrm{aRc} \Rightarrow \mathrm{a}$ can be a sister of $\mathrm{c}$
It is not transitive.
(iv) Let $\mathrm{A}$ be the set consisting of all the female members of a family. The relation $\mathrm{R}$ defined by " $a R b$ if $a$ is not a sister of $b$ ".
Solution:
$A=\{$ set of all female members of a family $\}$
(a) $\mathrm{aRa} \Rightarrow \mathrm{a}$ is a sister of a
It is reflexive
(b) $\mathrm{aRb} \Rightarrow \mathrm{a}$ is a sister of $b$
$\mathrm{bRa} \Rightarrow \mathrm{b}$ is a sister of a
$\Rightarrow$ It is symmetric
(c) $\mathrm{aRb} \Rightarrow \mathrm{a}$ is a sister of $\mathrm{b} b R \mathrm{bRc} \mathrm{b}$ is a sister of $\mathrm{c} a R \mathrm{c} \Rightarrow \mathrm{a}$ can be sister of $\mathrm{c}$ It is not transitive.
(v) On the set of natural numbers the relation R defined by " $x R y$ if $x+2 y=1$ ".
Solution:
$
\mathrm{N}=\{1,2,3,4,5, \ldots\}
$
$x R y$ if $x+2 y=1 R$ is an empty set

(a) $x R x \Rightarrow x+2 x=1 \Rightarrow x=\frac{1}{3} \notin N$. It is not reflexive $x R y=y R x \Rightarrow x+2 y=1$ It does not imply that $y+2 x=1$ as $y=\frac{1-x}{2}$ It is not symmetric.
(b) $-\mathrm{x}=\mathrm{y} \Rightarrow(-1,1) \notin \mathrm{N}$
It is not transitive.
Question 2.
Let $\mathrm{X}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\}$ and $\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),\{\mathrm{b}, \mathrm{b}),(\mathrm{a}, \mathrm{c})\}$. Write down the minimum number of ordered pairs to be included to $\mathrm{R}$ to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
$\mathrm{X}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\}$
$\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{a}, \mathrm{c})\}$
(i) To make $\mathrm{R}$ reflexive we need to include (c, c) and (d, d)
(ii) To make $\mathrm{R}$ symmetric we need to include (c, a)
(iii) $\mathrm{R}$ is transitive
(iv) To make $\mathrm{R}$ reflexive we need to include (c, c)
To make R symmetric we need to include (c, c) and (c, a) for transitive
$\therefore$ The relation now becomes
$\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{a}, \mathrm{c}),(\mathrm{c}, \mathrm{c}),(\mathrm{c}, \mathrm{a})\}$
$\therefore \mathrm{R}$ is equivalence relation.

Question 3.
Let $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\}$ and $\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{a}, \mathrm{c})\}$. Write down the minimum number of ordered pairs to be included to $R$ to make $i$
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
(i) $(\mathrm{c}, \mathrm{c})$
(ii) $(\mathrm{c}, \mathrm{a})$
(iii) nothing
(iv) (c, c) and (c, a)
Question 4.
Let $P$ be the set of all triangles in a plane and $R$ be the relation defined on $P$ as aRb if a is
similar to $b$. Prove that $\mathrm{R}$ is an equivalence relation.
Solution:
$\mathrm{P}=\{$ set of all triangles in a plane $\}$
$\mathrm{aRb} \Rightarrow \mathrm{a}$ similar to $\mathrm{b}$
(a) $\mathrm{aRa} \Rightarrow$ every triangle is similar to itself
$\therefore \mathrm{aRa}$ is reflexive
(b) $a R b \Rightarrow$ if a is similar to $b \Rightarrow b$ is also similar to $a$.
$\Rightarrow$ It is symmetric
(c) $\mathrm{aRb} \Rightarrow \mathrm{bRc} \Rightarrow \mathrm{aRc}$
$\mathrm{a}$ is similar to $\mathrm{b}$ and $\mathrm{b}$ is similar to $\mathrm{c}$
$\Rightarrow \mathrm{a}$ is similar to $\mathrm{a}$
$\Rightarrow$ It is transitive

$\therefore \mathrm{R}$ is an equivalence relation
Question 5.
On the set of natural numbers let $R$ be the relation defined by $a R b$ if $2 a+3 b=30$. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
$\mathrm{N}=\{$ set of natural numbers $\}$
$\mathrm{R}=\{(3,8),(6,6),(9,4),(12,2)\}$
(a) $(3,3) \notin \mathrm{R} \Rightarrow \mathrm{R}$ is not reflexive
$
\begin{aligned}
& 2 \mathrm{a}+3 \mathrm{~b}=30 \\
& 3 \mathrm{~b}=30-2 \mathrm{a} \\
& \mathrm{b}=\frac{30-2 a}{3}
\end{aligned}
$
(b) $(3,8) \in \mathrm{R}(8,3) \notin \mathrm{R}$
$\Rightarrow \mathrm{R}$ is not symmetric
(c) (a, b) (b, c) $\notin \mathrm{R} \Rightarrow \mathrm{R}$ is transitive
$\therefore$ It is not equivalence relation.
Question 6.
Prove that the relation "friendship" is not an equivalence relation on the set of all people in Chennai.
Solution:
(a) $\mathrm{S}=\mathrm{aRa}$ (i.e. ) a person can be a friend to himself or herself.
So it is reflextive.
(b) $\mathrm{aRb} \Rightarrow \mathrm{bRa}$ so it is symmetric

(c) aRb, bRc does not $\Rightarrow$ aRc so it is not transitive
$\Rightarrow$ It is not an equivalence relation
Question 7.
On the set of natural numbers let $\mathrm{R}$ be the relation defined by $a R b$ if $a+b \leq 6$. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
Set of all natural numbers $a R b$ if $a+b \leq 6$
$\mathrm{R}=\{(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4$, $1),(4,2),(5,1)\}$
(i) $(5,1) \in \mathrm{R}$ but $(5,5) \notin \mathrm{R}$
It is not reflexive
(ii) $\mathrm{aRb} \Rightarrow \mathrm{bRa} \Rightarrow$ It is symmetric
(iii) $(4,2),(2,3) \in \mathrm{R} \Rightarrow(4,3) \notin \mathrm{R}$
$\therefore$ It is not transitive
(iv) $\therefore$ It is not an equivalence relation
Question 8.
Let $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}\}$. What is the equivalence relation of smallest cardinality on $\mathrm{A}$ ? What is the equivalence relation of largest cardinality on $A$ ?
Solution:
$\mathrm{R}=\{\{\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{c}, \mathrm{c})\}$ is this smallest cardinality of $\mathrm{A}$ to make it equivalence relation $\mathrm{n}(\mathrm{R})=3$
(i) $R=\{(a, a),\{a, b),(a, c),(b, c),(b, b),\{b, c),(c, a),(c, b),(c, c)\}$
$\mathrm{n}(\mathrm{R})=9$ is the largest cardinality of $\mathrm{R}$ to make it equivalence.
Question 9.
In the set $\mathrm{Z}$ of integers, define $\mathrm{mRn}$ if $\mathrm{m}-\mathrm{n}$ is divisible by 7 . Prove that $\mathrm{R}$ is an equivalence relation.
Solution:
$\mathrm{mRn}$ if $m-n$ is divisible by 7
(a) $\mathrm{mRm}=\mathrm{m}-\mathrm{m}=0$

0 is divisible by 7
$\therefore$ It is reflexive
(b) $\mathrm{mRn}=(\mathrm{m}-\mathrm{n})$ is divisible by 7
$\mathrm{nRm}=(\mathrm{n}-\mathrm{m})=-\{\mathrm{m}-\mathrm{n})$ is also divisible by 7
It is symmetric
(c) $m \mathrm{R} n \Rightarrow(m-n)$ is divisible by $7=\frac{(m-n)}{7}=\frac{k}{7}$
$n \mathrm{R} r \Rightarrow(n-r)$ is divisible by $7=\frac{(n-r)}{7}=\frac{l}{7}$
$
\begin{aligned}
& \Rightarrow m \mathrm{R} r= m-r=\left(\frac{k}{7}+n\right)-(n- \\
& m-r=\frac{k}{7}+n-n+\frac{l}{7}
\end{aligned}
$
$(m-r)=\frac{1}{7}(k+l)$ is divisible by 7
It is transitive
$\mathrm{mRn}$ if $\mathrm{m}-\mathrm{n}$ is divisible by 7
$\therefore \mathrm{R}$ is an equivalence relation.