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### Course: Class 12 math (India)>Unit 1

Lesson 8: Solutions to select NCERT problems

# Select problems from exercise 1.1

Solutions to few selected problems from exercise 1.1 NCERT class 12.
In this article we will look at solutions of a few selected problems from exercise 1.1 of NCERT.
Problem 1:
Determine whether the following relation is reflexive, symmetric and transitive:
Relation $R$ in the set $A=\left\{1,2,3,4,5,6\right\}$ as
Solution:
Let's check whether the given relation is reflexive, symmetric and transitive one-by-one.
Any number is divisible by itself. So all pairs like $\left(1,1\right)$, $\left(2,2\right)$ etc. belong to $R$. Thus, the relation is reflexive.
$\left(2,4\right)$ belongs to $R$, but $\left(4,2\right)$ does not ($2$ is not divisible by $4$). Thus, the relation is not symmetric.
Let's see an example. The set of pairs $\left(1,3\right)$, $\left(3,6\right)$ and $\left(1,6\right)$ belong to $R$. This hints that $R$ is transitive.
In general, let $\left(x,y\right)$ and $\left(y,z\right)$ belong to $R$. This means $z$ is divisible by $y$ and $y$ is divisible by $x$. So, $z$ is divisible by $x$. Hence the pair $\left(x,z\right)$ will also belong to $R$. The relation is transitive.
Finally, $R$ is reflexive and transitive, but not symmetric.
Problem 2:
Show that the relation $R$ in the set $\mathbb{R}$ of real numbers, defined as $R=\left\{\left(a,b\right):a\le {b}^{2}\right\}$ is neither reflexive nor symmetric nor transitive.
Solution:
To show that the given relation is neither reflexive nor symmetric nor transitive, we just need to find one counter-example for each case.
$0.5\nleqq {0.5}^{2}$. So the pair $\left(0.5,0.5\right)\notin R$. Thus, the relation is not reflexive.
try it out
Can you find the general rule when the inequality $a\le {a}^{2}$ fails?

$\left(2,5\right)$ belongs to $R$, but $\left(5,2\right)$ does not. Thus, the relation is not symmetric.
Consider this example.
$\begin{array}{rl}15\le {4}^{2}& \phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\left(15,4\right)\in R\\ \\ 4\le {3}^{2}& \phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\left(4,3\right)\in R\\ \\ 15\nleqq {3}^{2}& \phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\left(15,3\right)\notin R\end{array}$
From above, we can conclude the relation is not transitive.
Finally, $R$ is neither reflexive nor symmetric nor transitive.
Problem 3:
Show that the relation $R$ in the set $A=\left\{1,2,3,4,5\right\}$ given by is an equivalence relation. Show that all the elements of $\left\{1,3,5\right\}$ are related to each other and all the elements of $\left\{2,4\right\}$ are related to each other. But no element of $\left\{1,3,5\right\}$ is related to any element of $\left\{2,4\right\}$.
A relation is an equivalence relation if it is reflexive, symmetric and transitive. Let's show that the given relation is all three one-by-one.
$|a-a|=0$ and $0$ is even. So all pairs of the form $\left(a,a\right)$ like $\left(1,1\right)$, $\left(2,2\right)$ etc. belong to $R$. Thus, the relation is reflexive.
$|a-b|=|b-a|$. If $|a-b|$ is even, $|b-a|$ is also even. So for every pair $\left(a,b\right)$, its symmetrical counterpart $\left(b,a\right)$ will also belong to $R$. Thus, the relation is symmetric.
Let's see an example. The set of pairs $\left(1,3\right)$, $\left(3,5\right)$ and $\left(1,5\right)$ belong to $R$. This hints that $R$ is transitive.
In general, let $\left(a,b\right)$ and $\left(b,c\right)$ belong to $R$.
Hence the pair $\left(a,c\right)$ will also belong to $R$. The relation is transitive.
Finally, $R$ is an equivalence relation.
For the next part of the question, consider the two sets ${A}_{1}=\left\{1,3,5\right\}$ and ${A}_{2}=\left\{2,4\right\}$.
See that the difference between any two elements of ${A}_{1}$ is even, and same is the case with ${A}_{2}$. However, if we pick one element from ${A}_{1}$ and another from ${A}_{2}$ the difference between them will be odd, and hence they cannot relate to each other.
Problem 4:
Give an example of a relation which is
(i) Symmetric but neither reflexive nor transitive
(ii) Transitive but neither reflexive nor symmetric
(iii) Reflexive and symmetric but not transitive
(iv) Reflexive and transitive but not symmetric
(v) Symmetric and transitive but not reflexive
Solution:
There are two ways to write examples. One is to write mathematical relations, for example, . Another way is to write a relation explicitly with all the pairs in it, for example, $R=\left\{\left(0,0\right),\left(2,3\right),\left(1,3\right)\dots \right\}$.
For each case given in the question, a list of relations are given below. Try and find the right examples on your own from that list. Be careful, there might me multiple correct options.
(i) Symmetric but neither reflexive nor transitive
pick the right examples 1
$\mathbb{Z}$ is the set of integers. $L$ is the set of all lines in a plane.

(ii) Transitive but neither reflexive nor symmetric
pick the right examples 2
$\mathbb{Z}$ is the set of integers. $A=\left\{1,2,3\right\}$.

(iii) Reflexive and symmetric but not transitive
pick the right examples 3
$\mathbb{R}$ is the set of real numbers. $L$ is the set of all lines in a plane.

(iv) Reflexive and transitive but not symmetric
pick the right examples 4
$\mathbb{N}$ is the set of natural numbers.
$\mathbb{N}$ is the set of natural numbers. $A=\left\{1,2,3\right\}$.