Show that the relation R on ℝ defined as R={(a, b): a ≤ b}, is reflexive, and transitive
but not symmetric.

$\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{OR}$

Prove that the function f: N → N, defined by $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}+1$ is one-one but not onto. Find inverse of f: N → S, where S is range of f.

Given: $\mathrm{R}=\left\{(\mathrm{a},\mathrm{b}): \mathrm{a}\le \mathrm{b}\right\},$

Clearly $(\mathrm{a}, \mathrm{a}) \in \mathrm{R} \mathrm{as} \mathrm{a}=\mathrm{a}.$

$\therefore \mathrm{R}$ is reflexive.

Now, 

$(2, 4)\in \mathrm{R} (\mathrm{as} 2<4)$

But $(4, 2) \notin \mathrm{R}$ as 4 is greater than 2.

$\therefore \mathrm{R}$ is not symmetric.

Now, let $(\mathrm{a}, \mathrm{b}), (\mathrm{b}, \mathrm{c})\in \mathrm{R}$

Then, 

$\mathrm{a}\le \mathrm{b}$ and $\mathrm{b}\le \mathrm{c}$

$$\begin{gathered} \Rightarrow \mathrm{a}\le \mathrm{c} \\ \Rightarrow (\mathrm{a}, \mathrm{c})\in \mathrm{R} \end{gathered}$$

$\therefore \mathrm{R}$ is transitive.

Therefore, R is reflexive and transitive but not symmetric have proven.

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Given:  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}+1$

Let ${\mathrm{x}}_1, {\mathrm{x}}_2 \in \mathrm{N}$

Such that, ${\text{⨍ (x}}_1{\text{)=⨍ (x}}_2\text{)}$

$$\begin{gathered} \therefore {\mathrm{x}}_1^2+{\mathrm{x}}_1+1={\mathrm{x}}_2^2+{\mathrm{x}}_2+1 \\ \therefore {\mathrm{x}}_1^2-{\mathrm{x}}_2^2+{\mathrm{x}}_1-{\mathrm{x}}_2=0 \\ \therefore \left({\mathrm{x}}_1-{\mathrm{x}}_2\right) \left({\mathrm{x}}_1+{\mathrm{x}}_2+1\right)=0 \\ \therefore {\mathrm{x}}_2={\mathrm{x}}_1 \mathrm{or} {\mathrm{x}}_2=-{\mathrm{x}}_{ }-1 \\ \therefore {\mathrm{x}}_1\in \mathrm{N} \\ \therefore -{\mathrm{x}}_1-1\in \mathrm{N} \end{gathered}$$

So ${\mathrm{x}}_2\ne -{\mathrm{x}}_1-1$

$\therefore f\left(x_2\right)=f\left(x_1\right)$ only for ${\mathrm{x}}_1={\mathrm{x}}_2$

So f(x) is one-one function.

$$\begin{gathered} \therefore f\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}+1 \\ \therefore f\left(\mathrm{x}\right)={\left(\mathrm{x}+\frac{1}{2}\right)}^2+\frac{3}{4} \end{gathered}$$ 

Which is an increasing function.

f(1)=3

$\because$ Range of f(x) will be, {3, 7, ....}

Which is a subset of N.

So it is an into function.

i.e., f(x) is not an onto function.

let $\mathrm{y}={\mathrm{x}}^2+\mathrm{x}+1$

$$\begin{gathered} \therefore {\mathrm{x}}^2+\mathrm{x}+1-\mathrm{y}=0 \\ \therefore \mathrm{x}=\frac{-1\pm \sqrt{1-4(1-\mathrm{y})}}{2} \\ \therefore \mathrm{x}=\frac{-1\pm \sqrt{4\mathrm{y}-3}}{2} \end{gathered}$$

So, two possibilities are their for ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)$

$f^{-1}\left(\mathrm{x}\right)=\frac{-1+\sqrt{4\mathrm{x}-3}}{2}, \frac{-1-\sqrt{4\mathrm{x}-3}}{2}$ and we know ${\mathrm{f}}^{-1}\left(3\right)=1$ because $f\left(1\right)=3$

so $f^{-1}\left(\mathrm{x}\right)=\frac{-1+\sqrt{4\mathrm{x}-3}}{2}$

Therefore, proven that the function f: N → N, defined by $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}+1$ is one-one but not onto and inverse of f: N → S, where S is range of f is ${\text{⨍ }}^{-1}\text{(x)=}\frac{-1+\sqrt{4\mathrm{x}-3}}{2}\text{.}$