Let $n$ be a fixed positive integer. Define a relation $R$ in $Z$ as follows $\forall a$, $b \in Z, a R b$ if and only if $a-b$ is divisible by $n$. Show that $R$ is an equivalence relation.
Given that, $\forall a, b \in Z, a R b$ if and only if $a-b$ is divisible by $n$.
Now,
I. Reflexive
$a R a \Rightarrow(a-a)$ is divisible by $n$, which is true for any integer $a$ as ' $O$ ' is divisible by $n$. Hence, $R$ is reflexive.
II. Symmetric
$$\begin{array}{ll} & a R b \\ \Rightarrow & a-b \text { is divisible by } n . \\ \Rightarrow & -b+a \text { is divisible by } n \\ \Rightarrow & -(b-a) \text { is divisible by } n \\ \Rightarrow & (b-a) \text { is divisible by } n . \\ \Rightarrow & b R a \end{array}$$
Hence, R is symmetric.
III. Transitive
Let $a R b$ and $b R c$
$\Rightarrow\quad (a-b)$ is divisible by $n$ and $(b-c)$ is divisible by $n$
$\Rightarrow\quad(a-b)+(b-c)$ is divisibly by $n$
$\Rightarrow\quad(a-c)$ is divisible by $n$
$\Rightarrow\quad aRc$
Hence, R is transitive.
So, R is an equivalence relation.
If $A=\{1,2,3,4\}$, define relations on $A$ which have properties of being
(i) reflexive, transitive but not symmetric.
(ii) symmetric but neither reflexive nor transitive.
(iii) reflexive, symmetric and transitive.
Given that, $$A=\{1,2,3,4\}$$
(i) Let $$R_1=\{(1,1),(1,2),(2,3),(2,2),(1,3),(3,3)\}$$
$R_1$ is reflexive, since, $(1,1)(2,2)(3,3)$ lie in $R_1$.
Now, $$(1,2) \in R_1,(2,3) \in R_1 \Rightarrow(1,3) \in R_1$$
Hence, $R_1$ is also transitive but $(1,2) \in R_1 \Rightarrow(2,1) \notin R_1$.
So, it is not symmetric.
(ii) Let $$R_2=\{(1,2),(2,1)\}$$
Now, $$(1,2) \in R_2,(2,1) \in R_2$$
So, it is symmetric.
(iii) Let $$R_3=\{(1,2),(2,1),(1,1),(2,2),(3,3),(1,3),(3,1),(2,3)\}$$
Hence, $R_3$ is reflexive, symmetric and transitive.
Let $R$ be relation defined on the set of natural number $N$ as follows, $R=\{(x, y): x \in N, y \in N, 2 x+y=41\}$. Find the domain and range of the relation $R$. Also verify whether $R$ is reflexive, symmetric and transitive.
$$\begin{aligned} \text{Given that,}\quad R & =\{(x, y): x \in N, y \in N, 2 x+y=41\} \\ \text { Domain } & =\{1,2,3, \ldots, 20\} \\ \text { Range } & =\{1,3,5,7, \ldots, 39\} \\ R & =\{(1,39),(2,37),(3,35), \ldots,(19,3),(20,1)\} \end{aligned}$$
$R$ is not reflexive as $(2,2) \notin R$
$$2 \times 2+2 \neq 41$$
So, $R$ is not symmetric.
As $$(1,39) \in R \text { but }(39,1) \notin R $$
So, $R$ is not transitive.
As $$(11,19) \in R,(19,3) \in R$$
But $(11,3) \notin R$
Hence, $R$ is neither reflexive, nor symmetric and nor transitive.
Given, $A=\{2,3,4\}, B=\{2,5,6,7\}$. Construct an example of each of the following
(i) an injective mapping from $A$ to $B$.
(ii) a mapping from $A$ to $B$ which is not injective.
(iii) a mapping from $B$ to $A$.
Given that, $$A=\{2,3,4\}, B=\{2,5,6,7\}$$
(i) Let $f: A \rightarrow B$ denote a mapping
$$\begin{aligned} & f=\{(x, y): y=x+3\} \\ \text{i.e.,}\quad & f=\{(2,5),(3,-6),(4,7)\}, \text { which is an injective mapping. } \end{aligned}$$
(ii) Let $g: A \rightarrow B$ denote a mapping such that $g=\{(2,2),(3,5),(4,5)\}$, which is not an injective mapping.
(iii) Let $h: B \rightarrow A$ denote a mapping such that $h=\{(2,2),(5,3),(6,4),(7,4)\}$, which is a mapping from $B$ to $A$.
Give an example of a map
(i) which is one-one but not onto.
(ii) which is not one-one but onto.
(iii) which is neither one-one nor onto.
(i) Let $f: N \rightarrow N$, be a mapping defined by $f(x)=2 x$ which is one-one.
$$\begin{aligned} \text{For}\quad f\left(x_1\right) & =f\left(x_2\right) \\ \Rightarrow\quad 2 x_1 & =2 x_2 \\ x_1 & =x_2 \end{aligned}$$
Further $f$ is not onto, as for $1 \in N$, there does not exist any $x$ in $N$ such that $f(x)=2 x+1$.
(ii) Let $f: N \rightarrow N$, given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2$ is onto but not one-one. $f$ is not one-one as $f(1)=f(2)=1$. But $f$ is onto.
(iii) The mapping $f: R \rightarrow R$ defined as $f(x)=x^2$, is neither one-one nor onto.