If functions $f: A \rightarrow B$ and $g: B \rightarrow A$ satisfy $g \circ f=I_A$, then show that $f$ is one-one and $g$ is onto.
Given that,
$f: A \rightarrow B$ and $g: B \rightarrow A$ satisfy $g \circ f=I_A$
$$\begin{aligned} \because\quad g \circ f & =I_A \\ \Rightarrow\quad\operatorname{g\circ f}\left\{f\left(x_1\right)\right\} & =g \circ f\left\{f\left(x_2\right)\right\} \\ \Rightarrow\quad g\left(x_1\right) & =g\left(x_2\right) \quad \left[\because g \circ f=I_A\right]\\ \therefore\quad x_1 & =x_2 \end{aligned}$$
Hence, f is one-one and g is onto.
Let $f: R \rightarrow R$ be the function defined by $f(x)=\frac{1}{2-\cos x}, \forall x \in R$. Then, find the range of $f$.
$$\begin{aligned} \text{Given function,}\quad & f(x)=\frac{1}{2-\cos x}, \forall x \in R \\ \text{Let,}\quad & y=\frac{1}{2-\cos x} \\ \Rightarrow\quad & 2 y-y \cos x=1 \\ \Rightarrow\quad & y \cos x=2 y-1 \\ \Rightarrow\quad & \cos x=\frac{2 y-1}{y}=2-\frac{1}{y} \Rightarrow \cos x=2-\frac{1}{y} \\ \Rightarrow\quad & -1 \leq \cos x \leq 1 \quad \Rightarrow \quad-1 \leq 2-\frac{1}{y} \leq 1 \\ \Rightarrow\quad & -3 \leq-\frac{1}{y} \leq-1 \quad \Rightarrow \quad 1 \leq \frac{1}{y} \leq 3 \\ \Rightarrow\quad & \frac{1}{3} \leq \frac{1}{y} \leq 1 \end{aligned}$$
So, $y$ range is $\left[\frac{1}{3}, 1\right]$.
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.