$\newcommand{\br}{\\}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\P}{\mathbb{P}}$ $\newcommand{\F}{\mathbb{F}}$ $\newcommand{\L}{\mathcal{L}}$ $\newcommand{\spa}[1]{\text{span}(#1)}$ $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\emptyset}{\varnothing}$ $\newcommand{\otherwise}{\text{ otherwise }}$ $\newcommand{\if}{\text{ if }}$ $\newcommand{\proj}{\text{proj}}$ $\newcommand{\union}{\cup}$ $\newcommand{\intercept}{\cap}$ $\newcommand{\abs}[1]{| #1 |}$ $\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$ $\newcommand{\pare}[1]{\left(#1\right)}$ $\newcommand{\t}[1]{\text{ #1 }}$ $\newcommand{\head}{\text H}$ $\newcommand{\tail}{\text T}$ $\newcommand{\d}{\text d}$ $\newcommand{\limu}[2]{\underset{#1 \to #2}\lim}$ $\newcommand{\inv}[1]{{#1}^{-1}}$ $\newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $\newcommand{\nullity}[1]{\text{nullity}(#1)}$ $\newcommand{\rank}[1]{\text{rank }#1}$ $\newcommand{\var}[1]{\text{var}(#1)}$ $\newcommand{\tr}[1]{\text{tr}(#1)}$ $\newcommand{\oto}{\text{ one-to-one }}$ $\newcommand{\ot}{\text{ onto }}$ $\newcommand{\Re}[1]{\text{Re}(#1)}$ $\newcommand{\Im}[1]{\text{Im}(#1)}$ $\newcommand{\Vcw}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Vce}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Vcr}[4]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \end{bmatrix}}$ $\newcommand{\Vct}[5]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \end{bmatrix}}$ $\newcommand{\Vcy}[6]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \end{bmatrix}}$ $\newcommand{\Vcu}[7]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \br #7 \end{bmatrix}}$ $\newcommand{\vcw}[2]{\begin{matrix} #1 \br #2 \end{matrix}}$ $\newcommand{\vce}[3]{\begin{matrix} #1 \br #2 \br #3 \end{matrix}}$ $\newcommand{\vcr}[4]{\begin{matrix} #1 \br #2 \br #3 \br #4 \end{matrix}}$ $\newcommand{\vct}[5]{\begin{matrix} #1 \br #2 \br #3 \br #4 \br #5 \end{matrix}}$ $\newcommand{\vcy}[6]{\begin{matrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \end{matrix}}$ $\newcommand{\vcu}[7]{\begin{matrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \br #7 \end{matrix}}$ $\newcommand{\Mqw}[2]{\begin{bmatrix} #1 & #2 \end{bmatrix}}$ $\newcommand{\Mqe}[3]{\begin{bmatrix} #1 & #2 & #3 \end{bmatrix}}$ $\newcommand{\Mqr}[4]{\begin{bmatrix} #1 & #2 & #3 & #4 \end{bmatrix}}$ $\newcommand{\Mqt}[5]{\begin{bmatrix} #1 & #2 & #3 & #4 & #5 \end{bmatrix}}$ $\newcommand{\Mwq}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Meq}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Mrq}[4]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \end{bmatrix}}$ $\newcommand{\Mtq}[5]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \end{bmatrix}}$ $\newcommand{\Mqw}[2]{\begin{bmatrix} #1 & #2 \end{bmatrix}}$ $\newcommand{\Mwq}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Mww}[4]{\begin{bmatrix} #1 & #2 \br #3 & #4 \end{bmatrix}}$ $\newcommand{\Mqe}[3]{\begin{bmatrix} #1 & #2 & #3 \end{bmatrix}}$ $\newcommand{\Meq}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Mwe}[6]{\begin{bmatrix} #1 & #2 & #3\br #4 & #5 & #6 \end{bmatrix}}$ $\newcommand{\Mew}[6]{\begin{bmatrix} #1 & #2 \br #3 & #4 \br #5 & #6 \end{bmatrix}}$ $\newcommand{\Mee}[9]{\begin{bmatrix} #1 & #2 & #3 \br #4 & #5 & #6 \br #7 & #8 & #9 \end{bmatrix}}$
Definition: Rational Numbers $\Q$

A rational number is a number of the form $\frac{ m }{ n } $ where $m, n \in \Z $ and $n \neq 0 $.

The space of all rational numbers is denoted by $\Q $.

Definition: Algebraic Number

A number is called an algebraic number if it satisfies a polynomial equation

$$c_nx^n + c_{n-1}x^{n-1} + … + c_1x + c_0 = 0 $$

where the coefficients $c_0, c_1, …, c_n$ are integers, $c_n \neq 0$ and $n \geq 1 $.

Note

Rational numbers are always algebraic numbers, since if $r = \frac{ m }{ n } $ is a rational number ($m, n \in \Z$ and $n \neq 0 $), then it satisfies the equation $nx - m = 0$.

Example

Prove $\frac{ 4 }{ 17 }, \sqrt{ 3 }, \sqrt[ 3 ]{ 17 }, \sqrt[ ]{ 2 + \sqrt[ 3 ]{ 5 }}, $ and $\sqrt[ ]{ \frac{ 4 - 2 \sqrt[ ]{ 3 }}{ 7 }} $ are algebraic numbers.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

$\frac{ 4 }{ 17 }$ is a solution of $17x - 4 = 0 $.

$\sqrt[ ]{ 3 } $ is a solution of $x^2 = 3 $.

$\sqrt[ ]{ 2 + \sqrt[ 3 ]{ 5 }} $ is a solution of $x^6 - 6x^4 + 12 x^2 -13 = 0 $.

$\sqrt[ ]{ \frac{ 4 - 2 \sqrt[ ]{ 3 }}{ 7 }} $ is a solution of $49x^4 - 56x^2 + 4 = 0 $.

Definition: Factor

An integer $k $ is a factor of an integer $m $ or divides $m $ if $\frac{ m }{ k } $ is also an integer.

Theorem : Rational Zeros Theorem

Suppose $c_1, _2, …, _{ n } $ are integers and $r $ is a rational number satisfying the polynomial equation

$$c_nx^n + c_{n-1}x^{n-1} + … + c_1x + c_0 = 0 $$

where $n \geq 1, c_n \neq 0 $ and $c_0 \neq 0 $. Let $r = \frac{ c }{ d } $ where $c, d $ are integeres having no common factors and $d \neq 0 $. Then $c $ divides $c_0 $ and $d $ divdes $c_n $.

In other words, the only rational candidates for solutions of the equation have the form $\frac{ c }{ d } $ where $c $ divides $c_0 $ and $d $ divides $c_n $.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Want to prove that for any rational number $r = \frac{ c }{ d } $ satisfying the polynomial

$$c_nx^n + c_{n-1}x^{n-1} + … + c_1x + c_0 = 0 $$

$c$ divides $c_0 $, $d $ divides $c_n$.

Since $r $ satisfies the equation, we have

$$c_n \pare{\frac{ c }{ d }}^n + c_{n-1} \pare{\frac{ c }{ d }}^{n-1} + … + c_1 \pare{\frac{ c }{ d }} + c_0 = 0 $$

Multiply through by $d^n $,

$$c_nc^n + c_{n-1}c^{n-1}d + … + c_1cd^{n-1} + c_0d^n = 0 $$ Solve for $c_0d^n $, we obtain

$$c_0d^n = -c \times [c_nc^{n-1} + c_{n-1}c^{n-2}d + c_{n-2}c^{n-3}d^2 + … + c_2cd^{n-2} + c_1 d^{n-1}] $$

Since $c $ and $d^n $ have no common factors, so $c $ divides $c_0 $.

Similarly solve for $c_nc^n $ and obtain

$$c_nc^n = -d \times [c_{n-1}c^{n-1} + c_{n-2}c^{n-2}d + … + c_2c^2d^{n-3} + c_1 cd^{n-2} + c_0d^{n-1}] $$

Again, since $c $ and $d^n $ have no common factors, so $d $ divides $c_nc^n $.

Corollary

Consider the polynomial equation

$$x^n + c_{n-1}x^{n-1} + … + c_1x + c_0 = 0 $$

where the coefficients $c_1, c_2, …, c_{ n-1 }$ are integers and $c_0 \neq 0 $. Any rational solution of this equation must be an integer that divides $c_0$.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

In the Ration Zeros Theorem, the denominator of $r$ must divide the coefficient of $x^n$, which is $1$ in this case. Thus $r$ is an integer and it divides $c_0$.

Note

Polynomials where the highest power has coefficient 1, are called monic polynomials.

Example

Prove that $\sqrt[ ]{ 2 } $ is not a rational number.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Let $x = \sqrt{ 2 }$, then we $x^2 - 2 = 0$.

By the corollary, the only rational numbers that could possibly be solutions of $x^2 - 2 = 0$ are $\pm 1, \pm 2$, which does not include $\sqrt[ ]{ 2 } $.

Hence, $x $ is not a rational number.