Tutorial: Exponents and radicals

Game version

This tutorial: Part B: Radicals and rational exponents

Go to Part C: Positive exponent, power, and simplest radical form

(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!]#

Adaptive game tutorial
This adaptive game tutorial will help you master this topic in a way that adapts to your ability as you practice the questions.
If this is your first time studying the material, you may find yourself requiring more help, and as a result your game scores could end up lower, or you could even "die." Do not be discouraged! Just press "New game" to play the game as many times as you need in order to require less help and improve your scores.
You can also try the non-game version (link on the side), which does not score you, is not randomized or adaptive and gives you all the answers, but will not give you additional practice at the level you might need.
To complete this game you must answer all the questions correctly; you will be allowed to have a number of tries for some questions.
The questions are randomized: Expect to see lots of differences differences every time you load the page.
Press "Scores" at any time to show the scores you currently have.
The game is automatically saved. Relaunching the game on the same computer with the same browser will bring up the saved version.
Press "New game" to discard the saved game and start a new game.
Caution: Clicking on the little pictures that appear from time to time on the left can have unexpected consequences! You might want to experiment with them; this is a game after all!

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.

%%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:

Radicals

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.

Suggested video for this topic: Video by MrB4math

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):

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.

We can now write mathematical expressions with exponents and radicals in a number of ways; for example,

\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.