Question

Complete the proof of Theorem G.2 by proving that

(a) fis injective;

(b) fis surjective

Short answer

1. Itis proved that$f$ is injective.
2. It is proved that$f$ is onto or surjective.

Step-by-step solution

Introduction of polynomial

Consider the polynomial in $\mathit{x}$and $\mathbf{ℝ}$will be express as;

$\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{a}}_{\mathbf{0}}{\mathbf{x}}^{\mathbf{0}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{a}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}$

Polynomial is always finite but sequence can be finite and infinite.

Prove that f  is injective 

(a)

Consider the G.2.

Consider the function $f:R\to R^{*}$and consider $r_1,r_2\in R$and let $f\left(r_1\right)=f\left(r_2\right)$then,

$\begin{array}{c}f\left(r_1\right)=f\left(r_2\right)\\ (r_1,0_R,0_R,\mathrm{...})=(r_2,0_R,0_R,\mathrm{...})\\ r_1=r_2\end{array}$

Therefore,$f$ is one-one mapping.

Hence, it is proved that$f$ is injective.

Prove that f  is surjective

(b)

Consider the function $f:R\to R^{*}$.

$\mathrm{f}\left(\mathrm{r}\right)=(\mathrm{r},0_{\mathrm{R}},0_{\mathrm{R}},\mathrm{...})$

Therefore,$f$ is onto or a surjective because for every $r\in R$their exist an image$f\left(r\right)$ in$R$ . Since, every element in$R^{*}$ is an image of an element in $R$.

Hence, it is proved that$f$ is onto or surjective.