Powers and Roots.

For any positive integer $ q$ , the function $ f_q: x \mapsto x^q$ is continuous and strictly increasing on $ [0,+\infty )$ . Moreover $ f_q(0)=0$ and $ \underset{x \rightarrow -\infty}{\text{lim}} x^q = +\infty$ . Therefore $ f_q$ is a bijection from $ [0,+\infty )$ onto itself. It is invertible and its inverse $ f_q^{-1}$ is called the $ q^{th}-$ root function.

We denote:

\begin{displaymath}\begin{cases}y=\sqrt[q]{x}  x \geq 0 \end{cases} \Longleftrightarrow \begin{cases}x=y^q  y \geq 0 \end{cases}\end{displaymath}    

Remark 5.7.3   We give here a definition of roots valid only for non negative numbers. Many pocket calculators compute roots for negative numbers; in fact their ``definition'' of roots is different from what we explain here, and coincides with ours for non negative numbers.

Properties 5.7.4       

  1. $ \forall q \in \mathbb{N}-{0}, \forall x \in [0,+\infty), \forall y \in [0,+\infty),
\; \sqrt[q]{xy}=\sqrt[q]{x} \cdot \sqrt[q]{y}$ .
  2. $ \forall q \in \mathbb{N}-{0}, \forall x \in [0,+\infty), \forall y \in (0,+\infty),
\; \sqrt[q]{\frac xy}=\frac{\sqrt[q]{x}}{\sqrt[q]{y}}$ .
  3. $ \forall p \in \mathbb{N}-{0},\forall q \in \mathbb{N}-{0}, \forall x \in [0,+\infty),
\; \sqrt[p]{ \sqrt[q]{x}}=\; \sqrt[pq]{x}$ .

We can now define rational powers of a non negative real number:

Definition 5.7.5   let $ \alpha$ be a rational number; we denote it in reduced form $ \alpha = \frac pq$ , where $ p$ is an integer and $ q$ is a positive integer. Then $ \forall x \in [0,+\infty ), \; x^{\alpha} = \sqrt[q]{x^p}$ .

For example: $ 8^{4/3}=(\sqrt[3]{8})^4=2^4=16$ .

Properties 5.7.6       

  1. $ \forall \alpha \in \mathbb{Q}, \forall x \in [0,+\infty), \forall y \in [0,+\infty),
\quad x^{\alpha} \cdot y^{\alpha} =(xy)^{\alpha}$ .
  2. $ \forall \alpha \in \mathbb{Q}, \forall \beta \in \mathbb{Q}, \forall x \in [0,+\infty),
\quad x^{\alpha} \cdot x^{\beta} = x^{\alpha + \beta}$ .

For example:

$ 25^{5/2} \cdot 125^{2/5}=(5^2)^{5/2} \cdot (5^3)^{5/2}
=5^5 \cdot 5^{15/2} = 5^{5+15/2}=5^{25/2}=\sqrt{5^{25}}=5^{12}\sqrt{5}$ .

Noah Dana-Picard 2007-12-28