Arithmetic enzyme root

Definition

Given a real number $a \geq 0$ and an integer $n \geq 2$, there exists a unique real number $b \geq 0$ such that $b^n = a$. The number b we will call the arithmetic enzymatic root of a and we will indicate it by the symbol $\sqrt[n]{a}$ where a is called the radicand and n is the index.

Examples

  1. $\sqrt[5]{32} = 2$ because $2^5 = 32$
  2. $\sqrt[3]{8} = 2$ because $2^3 = 8$
  3. $\sqrt{9} = 3$ because $3^2 = 9$
  4. $\sqrt[7]{0} = 0$ because $0^7 = 0$
  5. $\sqrt[6]{1} = 1$ because $1^6 = 1$

Observations

  1. From the definition follows $(\sqrt[n]{a})^n = a$.

  2. We observe in the given definition that:

$\sqrt{36} = 6$ and not $\sqrt{36} = \pm 6$

$\sqrt{\frac{9}{4}}$ = $\frac{3}{2}$ and not $\sqrt{\frac{9}{4}}$ = $\pm \frac{3}{2}$

but

$-\sqrt[3]{8}$ = -2, $-\sqrt{4}$ = -2, $\pm \sqrt{9}$ = $\pm 3$

are true sentences where the radical “is not causing” the sign that precedes it.

  1. We must be careful when calculating the square root of a square perfect:

$\sqrt{a^2}$ = |a|

Examples

  1. $\sqrt[2]{(-5)^2} = |-5| = 5$ and not $\sqrt[2]{(-5)^2} = -5$
  2. $\sqrt[2]{x^2} = |x|$ and not $\sqrt[2]{x^2} = x$

Properties

If $a\in \mathbb{R}_+$, $b\in \mathbb{R}_+$, $m\in \mathbb{Z}_+$, $n\in \mathbb{N}_+$ and $p\in \mathbb{N}_+$, we have:

R_1. $\sqrt[n]{a^m}$ = $\sqrt[np]{a^{mp}}$

R_2. $\sqrt[n]{a \times b} = \sqrt[n]{a} \times \sqrt[n]{b}$

R_3. $\sqrt[n]{\frac{a}{b}}$ = $\frac {\sqrt[n]{a}} {\sqrt[n]{b}}$

R_4. $(\sqrt[n]{a})^m$ = $\sqrt[n]{a^m}$

R_5. $\sqrt[p]{\sqrt[n]{a}}$ = $\sqrt[p \times n]{a}$

Demos

R$_1$- $\sqrt[n]{a^m}$ = $\sqrt[np]{a^{mp}}$

Façamos $x = \sqrt[n]{a^m}$, então:

$x^{np}$ = $(\sqrt[n]{a^m})^{np}$ = $[(\sqrt[n]{a^m})^n]^p$ = $[\sqrt[n]{a^m}]^p$ $\rightarrow$ $x = \sqrt[np]{a^{mp}}$.

R$_2$- $\sqrt[n]{a}$ $\times$ $\sqrt[n]{b}$ = $\sqrt[n]{ab}$

Façamos x = $\sqrt[n]{a}$ $\times$ $\sqrt[n]{b}$, então:

$x^n$ = $(\sqrt[n]{a}$ $\times$ $\sqrt[n]{b})^n$ = $(\sqrt[n]{a})^n$ $\times$ $(\sqrt[n]{b})^n$ = $a \times b$ $\rightarrow$ x = $\sqrt[n]{a \times b}$

R$_4$- $(\sqrt[n]{a})^m$ = $\sqrt[n]{a^m}$

Considerando n fixo e m > 0, provaremos por indução sobre m.

  1. A propriedade é verdadeira para m = 0 pois

$(\sqrt[n]{a})^0$ = 1 = $\sqrt[n]{1}$ = $\sqrt[n]{a^0}$

  1. Supondo a propriedade verdadeira para m = P. isto é $(\sqrt[n]{a})^p$ = $\sqrt[n]{a^p}$, provemos que é verdadeira para m = p + 1, isto é

$(\sqrt[n]{a})^{p+1}$ = $\sqrt[n]{a^{p+1}}$.

De fato:

$(\sqrt[n]{a})^{p+1}$ = $\sqrt[n]{a^{p+1}}$.

Se m < 0, façamos $-m = q > 0$, então

$(\sqrt[n]{a})^m$ =

Observação Notemos que se $b\in \mathbb{R}$ e $b\in \mathbb{N}^*$, temos:

para $b > 0$, $b \times \sqrt[n]{a}$ = $\sqrt[n]{a \times b^n}$ para $b < 0$, $b \times \sqrt[n]{a}$ = $-\sqrt[n]{a \times |b|^n}$

isto é, O coeficiente do radical (a menos do sinal) pode ser colocado no radicando com expoente igual ao índice do radical.

Exemplos

  1. $2 \times \sqrt[3]{3} = \sqrt[3]{3 \times 2^3} = \sqrt[3]{24}$
  2. $-5 \times \sqrt{2} = -\sqrt{2 \times 5^2} = -\sqrt{50}$
  3. $-2 \times \sqrt[4]{2} = -\sqrt[4]{2 \times 2^4} = -\sqrt[4]{32}$