Question

Verify that the Rational Root Test (Theorem 4.21) is valid with$\mathrm{\mathbb{Z}}$ and$\mathrm{\mathbb{Q}}$ replaced by R and F.

Short answer

It is verified that the Rational Root Test is valid with $\mathrm{\mathbb{Z}}$and$\mathrm{\mathbb{Q}}$replaced by R and F.

Step-by-step solution

Theorem 4.21

Let $\mathit{f}\left(x\right)\mathbf{=}{\mathit{a}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}\mathbf{+}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}{\mathit{a}}_{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathit{a}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{a}}_{\mathbf{0}}$be a polynomial with integer coefficients. If $r\ne 0$and the rational number $\mathit{r}\mathbf{/}\mathit{s}$is root of $\mathit{f}\left(x\right)$then, $\mathit{r}\mathbf{/}{\mathit{a}}_{\mathbf{0}}$and$\mathit{s}\mathbf{/}{\mathit{a}}_{\mathbf{n}}$ .

Take the proof of Theorem 4.21

The proof of theorem 4.21 can be directly generalized.

Let $f\left(x\right)=\sum _{i=0}^n{a}_i{x}^i\in R\left[x\right]$.

If $\frac{r}{s}\in F,\frac{r}{s}\ne 0$is a root of $f\left(x\right)$in reduced form then, $r/a_0$and $s/a_n$.

Since $0=\sum _{i=0}^n{a}_i\frac{r^i}{s^i}$, then multiplying by we have, $0=\sum _{i=0}^n{a}_i{r}^i{s}^{n-1}$.

Which implies, $a_n{r}^n=s\left(\sum _{i=0}^{n-1}\left(-a_i{r}^i{s}^{n-i-1}\right)\right)$.

Since s is relatively prime to $r,s/a_n$.

The above equation also give us that, $a_0{s}^n=r\left(\sum _{i=0}^{n-1}\left(-a_i{r}^{i-1}{s}^{n-i}\right)\right)$.

So, $r/a_0$.

Hence, it is verified that the Rational Root Test is valid with$\mathrm{\mathbb{Z}}$ and$\mathrm{\mathbb{Q}}$ replaced by R and F.