Tutorial: Exponents and radicals
Adaptive game version
This tutorial: Part B: Radicals and rational exponents
(This topic is also in Section 0.2 in Finite Mathematics and Applied Calculus)
#[I don't like this new tutorial. Take me back to the older tutorial!][No me gusta este nueve tutorial. ¡Regresame al tutorial más viejo!]#
Radicals
If $a$ is any nonnegative real number, then its square root is the nonnegative number whose square is $a.$ For example, the square root of $16$ is $\color{indianred}{4},$ since $\color{indianred}{4^{\color{black}{2}}} = 16.$
%%Q
What about $-4$? If you square $-4$ you also get $16,$ so is it true that $-4$ is another square root of $16$?
%%A No. The term "square root of $a$" refers only to the nonnegative number whose square is $a$, so each number has only a single square root.
Similarly, the fourth root of the nonnegative number $a$ is the nonnegative number whose fourth power is $a.$ Thus, the fourth root of $16$ is $\color{indianred}{2},$ since $\color{indianred}{2^{\color{black}{4}}} = 16.$ (As with square roots, the fourth root of a number cannot be negative.)We can similarly define sixth roots, eighth root, and so on.
%%A No. The term "square root of $a$" refers only to the nonnegative number whose square is $a$, so each number has only a single square root.
%%Q
What about odd-numbered roots?
%%A There is a slight difference with odd-numbered roots: For example, the cube root of any number $a$ is the number whose cube is $a.$ For example, the cube root of $8$ is $\color{indianred}{2},$ (as $\color{indianred}{2^{\color{black}{3}}} = 8$).Note that we can take the cube root of any number, positive, negative or zero. For instance, the cube root of $-8$ is $\color{indianred}{-2},$ as $(\color{indianred}{-2})^2 = -8$. Unlike square roots, the cube root of a number may be negative. In fact, the cube root of $a$ always has the same sign as $a.$ The other odd-numbered roots are defined in the same way.
We use "radical" notation to designate roots, as shown below:
%%A There is a slight difference with odd-numbered roots: For example, the cube root of any number $a$ is the number whose cube is $a.$ For example, the cube root of $8$ is $\color{indianred}{2},$ (as $\color{indianred}{2^{\color{black}{3}}} = 8$).Note that we can take the cube root of any number, positive, negative or zero. For instance, the cube root of $-8$ is $\color{indianred}{-2},$ as $(\color{indianred}{-2})^2 = -8$. Unlike square roots, the cube root of a number may be negative. In fact, the cube root of $a$ always has the same sign as $a.$ The other odd-numbered roots are defined in the same way.
Radicals
Suggested video for this topic: Video by MrB4math
| Name | Symbol | %%Examples | |
| Square root of $a$ | $\sqrt{a}$ | $\sqrt{16}=4$ | The square root of $16$ is $4.$ |
| $\sqrt{9}=3$ | To get $9,$ you square $\color{indianred}{3.}$ | ||
| $\sqrt{1}=1$ | To get $1,$ you square $\color{indianred}{1.}$ | ||
| $\sqrt{0}=0$ | To get $0,$ you square $\color{indianred}{0.}$ | ||
| $\sqrt{2 \cdot 2}=2$ | To get $2 \cdot 2,$ you square $\color{indianred}{2.}$ | ||
| $\sqrt{3 \cdot 3}=3$ | To get $3 \cdot 3,$ you square $\color{indianred}{3.}$ | ||
| $\sqrt{2}\sqrt{2}=2$ | If you square the number whose square is 2, you get 2. | ||
| $\sqrt{a}\sqrt{a}=a$ | If you square the number whose square is a, you get a. | ||
| $\sqrt{2}$ ${}=1.4142136...$ |
An irrational number. We normally don't write it as a decimal. |
||
| $\sqrt{-1}$ is not a real number. | $-1$ is negative. | ||
| Cube root of $a$ | $\sqrt[3]{a}$ | $\sqrt[3]{8}=2$ | The cube root of $8$ is $2.$ |
| $\sqrt[3]{-8}=-2$ | To get $-8,$ you cube $\color{indianred}{-2.}$ | ||
| $\sqrt[3]{1}=1$ | To get $1,$ you cube $\color{indianred}{1.}$ | ||
| $\sqrt[3]{-1}=-1$ | To get $-1,$ you cube $\color{indianred}{-1.}$ | ||
| $\sqrt[3]{a}\sqrt[3]{a}\sqrt[3]{a}$ $=a$ | If you cube the number whose cube is a, you get a. | ||
| Fourth root of $a$ | $\sqrt[4]{a}$ | $\sqrt[4]{10{,}000}$ $=10$ | To get $10\,000,$ you raise $\color{indianred}{10}$ to the fourth power. |
| $\sqrt[4]{-3}$ is not a real number. | $-3$ is negative. |
Radicals of products and quotients
In the above quiz we saw that the squre root of a sum is not the sum of the individual square roots. However, the square root of a product is the product of the individual square roots, and the same applied to quotients. Here are the rules:
Radicals of Products and Quotients
In the following identities, $a$ and $b$ stand for any real numbers. In the case of even-numbered roots, they must be nonnegative.
#[Rule][Regla]# \t \t #[Example][Ejemplo]#
\\ $\sqrt[n]{ab} = \sqrt[n]{a}\ \sqrt[n]{b}$ \t $\qquad$ \t $\sqrt{8} $$= \sqrt{4 \cdot 2} = \sqrt{4}\ \sqrt{2} = 2\sqrt{2}$
\\ \t \t #[Equivalently,][De manera equivalente,]#
\\ \t \t $\sqrt{8} = \sqrt{2 \cdot 2 \cdot 2} $$= \sqrt{2 \cdot 2}\ \sqrt{2} = \sqrt{2}\ \sqrt{2} \sqrt{2} $$= 2\sqrt{2}$
\\ \t \t (#[Note that][Observa que]#$\sqrt{2}\ \sqrt{2} = 2.$)
\\ \t \t $\sqrt{54} = \sqrt{9 \cdot 6} $$= \sqrt{9}\ \sqrt{6} = 3\sqrt{6}$
\\ \t \t #[Equivalently,][De manera equivalente,]#
\\ \t \t $\sqrt{54} = \sqrt{3 \cdot 3 \cdot 6} $$= \sqrt{3 \cdot 3}\ \sqrt{6} = \sqrt{3}\ \sqrt{3}\ \sqrt{6} $$= 3\sqrt{6}$
\\ $\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ \t $\quad$ \t $\sqrt[3]{\dfrac{8}{27}} = \dfrac{\sqrt[3]{8}}{\sqrt[3]{27}} = \dfrac{2}{3}$
#[Again, the rule does not apply to sums or differences:][Otra vez, esta regla no se aplica a sumas y restas:]#
$\sqrt{2+2} \neq \sqrt{2} + \sqrt{2}$ #[but rather][sino más bien]# $\sqrt{2+2} = \sqrt{4} = 2.$
\\ $\sqrt{9-5} \neq 3 - \sqrt{5}$ #[but rather][sino más bien]# $\sqrt{9-5} = \sqrt{4} = 2.$
Suggested video for this topic: Video by MrB4math
Exponential notation
Rather than working all the time with radical expressions, we can convert all radical notation to exponential notation, as follows. (Throughout, we take $a$ to be positive if the denominator in the exponent is even.)
Raional exponents
We can use rational exponents for expressions involving radicals as follows:
#[Radical form][Forma radical]# \t \t #[Exponent form][Forma exponente]# \t \t %%Example
\\ $\sqrt{a}$ \t $\quad$ \t $a^{1/2} \quad (\text{or } a^{0.5})$ \t $\quad$ \t $64^{1/2} = \sqrt{64} = 8$
\\ $\sqrt[3]{a}$ \t $\quad$ \t $a^{1/3}$ \t $\quad$ \t $64^{1/3} = \sqrt[3]{64} = 4$
\\ $\sqrt[n]{a}$ \t $\quad$ \t $a^{1/n}$ \t $\quad$ \t $64^{1/6} = \sqrt[6]{64} = 2$
So, if we want the exponent identities to continue to work with rational exponents, we can calculate $a^{m/n}$ in two ways:
\t $a^{m/n} = a^{(m)(1/n)} = (a^m)^{1/n} = \sqrt[n]{a^m}$\t $\qquad$ \t #[(See the third line above.)][(Ve la tercera línea arriba.)]#
\\ #[or][o]# $\quad$
\\ \t $a^{m/n} = a^{(1/n)(m)} = (a^{1/n})^{m} = \left(\sqrt[n]{a}\right)^m.$
Suggested video for this topic: Video by ThinkwellVids
%%Examples
$4^{1/2} = \sqrt{4} = 2$
\\ $4^{3/2} = (4^3)^{1/2} = 64^{1/2} = 8$
\\ $4^{3/2} = (4^{1/2})^3 = \left(\sqrt{4}\right)^3 = 2^3 = 8$
Some for you
%%Q
Do all the usual identities for exponents work with fractional exponents?
%%A Yes. Here is a summary of these rules—exactly the same as those we saw in %%prevparttut—but this time we understand that the exponents can be rational numbers (rather than integers as in the last tutorial):
%%A Yes. Here is a summary of these rules—exactly the same as those we saw in %%prevparttut—but this time we understand that the exponents can be rational numbers (rather than integers as in the last tutorial):
Exponent identities and radicals
| Rule | %%Examples | Comments |
| 1. $a^pa^q = a^{p+q}$ | $8^{5/3}8^{-1/3}=8^{4/3}$ | This is equal to $\left(\sqrt[3]{8}\right)^4=2^4=16$ |
|
2. $\dfrac{a^p}{a^q} = a^{p-q}$ (%if $a \neq 0$) |
$\dfrac{9^3}{9^{3/2}} = 9^{3-3/2} = 9^{3/2}$ | This is equal to $\left(\sqrt{9}\right)^3=3^3=27$ |
|
3. $\dfrac{1}{a^q} = a^{-q}$ (%if $a \neq 0$) |
$9^{-1/2} = \dfrac{1}{9^{1/2}} = \dfrac{1}{3}$ | Put $p=0$ in Rule 2 to obtain Rule 3. |
| 4. $(a^p)^q = a^{pq}\ $ | $(16^{1/2})^2 = 16^{1/2 \times 2} = 16^{1} = 16$ | This example tells us why $16^{1/2}$ has to be 4: Squaring it must give 16. |
| 5. $(ab)^p = a^pb^p$ | $16^{2/3} = (8 \cdot 2)^{2/3} = 8^{2/3} \cdot 2^{2/3}$ | This is equal to $\sqrt[3]{8^2}\ \ \sqrt[3]{2^2} = \sqrt[3]{64}\ \ \sqrt[3]{4} = 4\ \sqrt[3]{4}.$ |
| 6. $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$ | $\sqrt{8}=\sqrt{4 \times 2} = \sqrt{4}\sqrt{2} = 2\sqrt{2}$ |
We already know this rule: The radical a product is the product of the radicals. Put $p=\dfrac{1}{n}$ in Rule 5 to obtain Rule 6. |
|
7. $\left(\dfrac{a}{b}\right)^p = \dfrac{a^p}{b^p}$ (%if $b \neq 0$) |
$\left(\dfrac{27}{8}\right)^{2/3} = \dfrac{27^{2/3}}{8^{2/3}}$ | This is equal to $\dfrac{9}{4}.$ |
|
8. $\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ (%if $b \neq 0$) |
$\sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{\sqrt{4}}= \dfrac{\sqrt{3}}{2}$ |
We already know this rule: The radical a quotient is the quotient of the radicals. Put $p=\dfrac{1}{n}$ in Rule 7 to obtain Rule 8. |
\t $4x^{-2/3}$ \t ${}= \dfrac{4}{x^{2/3}}$ \t ${}= \dfrac{4}{\sqrt[3]{x^2}} \qquad = \dfrac{4}{\left(\sqrt[3]{x}\right)^2}$.
\\ \t Power form \t Positive exponent form \t Radical form
\\ \t !2! (See %%prevparttut.)
Now try the exercises in Section 0.2 in Finite Mathematics and Applied Calculus.
or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.
Copyright © 2019 Stefan Waner and Steven R. Costenoble