#### Chapter 1 Elementary Arithmetic

Section 1.4 Powers and Roots

# 1.4.1 Exponentiation and Roots

The following section deals with expressions of the form ${a}^{s}$, where $a\in ℝ$. But the question is: For which numbers $s$ can this power be reasonably defined?
Powers with positive integer exponents are defined as follows:
##### Info 1.4.1

Let $n\in ℕ$. The $n$-th power of a number $a\in ℝ$ is the $n$-fold product of the number $a$ by itself:

$\begin{array}{ccc}\multicolumn{1}{c}{{a}^{n}}& =\hfill & a·a·a·\dots ·a .\hfill \\ \multicolumn{1}{c}{}& \hfill & \mathrm{ }\mathrm{ }\left(n \text{factors}\right)\hfill \end{array}$

$a$ is called the base and $n$ is called the exponent.

Here, some special cases exist that you should ideally know by heart:
##### Info 1.4.2

For a zero exponent, the value of the power is one, i.e. for example, ${4}^{0}=1$, also ${0}^{0}=1$. But for a zero base, for $n>0$, we have ${0}^{n}=0$. For base $-1$, we have

$\left(-1{\right)}^{n}\mathrm{ }=\mathrm{ }-1\mathrm{ }\text{if the exponent is odd}\mathrm{ }\mathrm{ },\mathrm{ }\mathrm{ }\left(-1{\right)}^{n}\mathrm{ }=\mathrm{ }1\mathrm{ }\text{if the exponent is even} .$

##### Example 1.4.3

${3}^{2}=3·3=9 , \left(-2{\right)}^{3}=\left(-2\right)·\left(-2\right)·\left(-2\right)=-8 , {\left(\frac{1}{2}\right)}^{4}=\frac{1}{2}·\frac{1}{2}·\frac{1}{2}·\frac{1}{2}=\frac{1}{16} .$

Many powers can be calculated using the calculation rule presented above - but what about ${2}^{-2}$?
##### Info 1.4.4

Powers with negative integer exponents are defined by the formula ${a}^{-n}=\frac{1}{{a}^{n}},n\in ℕ,a\ne 0.$
Hence ${2}^{-2}=\frac{1}{{2}^{2}}=\frac{1}{4}$. By analogy, we have $\left(-2{\right)}^{-2}=\frac{1}{\left(-2{\right)}^{2}}=\frac{1}{4}$.
##### Exercise 1.4.5
Calculate the values of the following powers.
1. ${5}^{3}$$=$ .

2. $\left(-1{\right)}^{1001}$$=$ .

3. ${\left(-\frac{1}{2}\right)}^{-3}$$=$ .

4. ${\left(\left(-3{\right)}^{2}\right)}^{3}$$=$ .

However, even for a rational exponent of the form $\frac{1}{n},n\in ℕ$, we need to extend this definition again so that we can calculate ${4}^{\frac{1}{2}}$, for example. This power can be expressed as a root as well, namely ${4}^{\frac{1}{2}}=\sqrt{4}=2$. Generally, we have:
##### Info 1.4.6

Let $n\in ℕ$ and $a\in ℝ$ with $a\ge 0.$ The $n$-th root has the power representation $\sqrt[n]{a}={a}^{\frac{1}{n}}$.
This leads to one inverse operation of the exponentiation: extracting a root.
##### Info 1.4.7

The $n$-th root of a number $a\in ℝ,a\ge 0,$ is the number whose $n$-th power is $a$:

${a}^{\frac{1}{n}}=\sqrt[n]{a}=b⇒{b}^{n}=a .$

$a$ is called the base of the root or radicand, and $n$ is called the exponent of the root or order of the root. We have $\sqrt[1]{a}=a$, and $\sqrt[2]{a}=\sqrt{a}$ is called the square root and $\sqrt[3]{a}$ is called the cube root of $a$.

##### Example 1.4.8

${16}^{\frac{1}{2}}=\sqrt[2]{16}=\sqrt{16}=4 , {27}^{\frac{1}{3}}=\sqrt[3]{27}=3 .$

##### Info 1.4.9

For $n\in ℕ,a,b\in ℝ$ with $a,b\ge 0$ we have the following calculation rules:
1. Two roots with the same exponent are multiplied by multiplying the radicands and extracting the root of the product, leaving the exponent of the root unchanged:

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

2. Two roots with the same exponent are divided by dividing the radicands and extracting the root of the quotient, leaving the exponent of the root unchanged:

$\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}},b\ne 0 .$

But how can we calculate the number ${\left(\sqrt[10]{4}\right)}^{5}$?
##### Info 1.4.10

Let $m,n\in ℕ$ and $a\in ℝ,a\ge 0.$
1. The $m$-th power of a root is calculated by raising the radicand to the power of $m$, leaving the exponent of the root unchanged:

${\left(\sqrt[n]{a}\right)}^{m}=\sqrt[n]{{a}^{m}}={a}^{\frac{m}{n}} .$

2. The $m$-th root of a root is calculated by multiplying the exponents of the roots, leaving the radicand unchanged (root of a root).

$\sqrt[m]{\sqrt[n]{a}}=\sqrt[m·n]{a} .$

Hence

$\sqrt[10]{{4}^{5}}=\left({4}^{5}{\right)}^{\frac{1}{10}}={4}^{\frac{5}{10}}={4}^{\frac{1}{2}}=\sqrt[2]{4}=2 .$

##### Example 1.4.11
A general power with rational exponent is then calculated as follows:

${\left(\frac{1}{4}\right)}^{-\frac{2}{3}}=\sqrt[3]{{\left(\frac{1}{4}\right)}^{-2}}=\sqrt[3]{\frac{1}{{\left(\frac{1}{4}\right)}^{2}}}=\sqrt[3]{\frac{1}{\frac{1}{16}}}=\sqrt[3]{16}=\sqrt[3]{{2}^{3}·2}=2\sqrt[3]{2} .$

The calculation rules for powers with real base and rational exponent are known as exponent rules. The rules vary depending on whether powers of the same base or the same exponent are considered.
##### Exercise 1.4.12
Calculate the following roots (here, the result is ):
1. ${\left(\sqrt[5]{3}\right)}^{5}$$=$ .

2. $\sqrt[4]{256}$$=$ .