Chapter 2
Exponents and Radicals
Whole Number Exponents
By relieving the brain of all unnecessary work, a good notation sets it free to concentrate
on more advanced problems, and in effect increases the mental power of the race.
— Alfred North Whitehead
Mathematical notations develop over time. For example, what we now write as $ x^{2} $ was once written xx. It’s possible to read xx at a glance, but not $ $ , so a shorthand was adopted:
Definition (Whole Number Exponents). For each whole number n, we define
\[ x^{n}=\overbrace{x x x x x x x x}^{n\;times} \]
This notation has important algebraic properties that we can discover just by playing with it for a bit. For example, if we think about a specific example such as this
\[ \begin{aligned}x^{5}x^{11}&=\overbrace{xxxxxxx}^{5times}\overbrace{xxxxxxxxxxxxx}^{11times}\\&=\overbrace{xxxxxxxxxxxxxxxxxxxx}^{5+11=16times}\\&=x^{16}\end{aligned} \]
(definition of whole number exponents)
(definition of whole number exponents),
from this, we discern a more general principle: $ x{n}x{m}=x^{n+m} $ for any whole number exponents n and m. This is the first of five basic exponent rules. The other four are equally natural and easy to understand. I’ve listed the full set below, which you should learn immediately.
The Exponent Rules
\[ \mathbf{1}.\ x^{n}x^{m}=x^{n+m} \]
\[ \mathbf{2}.\ \frac{x^{n}}{x^{m}}=x^{n-m} \]
\[ \mathbf{3}.~(x^{n})^{m}=x^{nm} \]
\[ \mathbf{4}.\ (xy)^{n}=x^{n}y^{n} \]
\[ \mathbf{5}.\left(\frac{x}{y}\right)^{n}=\frac{x^{n}}{y^{n}} \]
Commit these five rules to memory, but should your memory ever fail you, you should be able to recover any of these five rules in a few seconds by scribbling a specific numerical example on some scratch paper while mumbling to yourself. (Example: “Wait a second, I can’t remember… What exactly is $ (xy)^{n} $ ? Well, let’s see… when n = 3, we get $ (xy)^{3} = xy xy xy = xxx yyy = x{3}y{3} $ . Yes, that’s it: $ (xy)^{n} = x{n}y{n} $ .”)
To convince you that the five rules hold, here’s an illustration/explanation for each of them:
| Exponent Rule | Illustration/Explanation | |
| 1 | $ x{n}x{m} = x^{n+m} \(</td><td>\) x{5}x{3} = xxxxx xxx = xxxxxxxx = x^{8} \(</td></tr><tr><td>2</td><td>\) = x^{n-m} \(</td><td>\) = = = xx = x^{2} \(</td></tr><tr><td>3</td><td>\) (x{n}){m} = x^{nm} \(</td><td>\) (x{3}){2} = x^{3} x^{3} = xxx xxx = xxxxxxx = x^{6} \(</td></tr><tr><td>4</td><td>\) (xy)^{n} = x{n}y{n} \(</td><td>\) (xy)^{3} = xy xy xy = xxx yyy = x{3}y{3} \(</td></tr><tr><td>5</td><td>\) ()^{n} = \(</td><td>\) ()^{3} = = $ |
In the exercises, you’ll use these five rules to simplify ugly expressions such as the following.
Example 1. Simplify as much as possible: $ x $
Solution. $ xxxxx8x^{9} $
In practice, you need not specify which exponent rule you use at each step. After doing many problems, you will have internalized the rules to such a degree that you’ll use them effortlessly. Here’s one more example, written without explicit appeals to the numbered rules. As you read it, be sure that you can explain why each equality holds. (In some of them, multiple rules are used simultaneously.)
Example 2. Simplify as much as possible: $ $ .
Solution. $ =====y^{6}. $
Exercises
Simplify each of the following expressions as much as possible:
$ x{14}x{73} $
$ x{13}/x{7} $
$ (x{4}){5} $
$ x{11}x{10}x^{9} $
$ (x{3}){2}x $
$ (2()x{2}x{3})^{4} $
$ 3(5x{2}){3}(4xx{5}){2} $ 8. $ [p(q{2}r{3})^{4}]pqr $ 9. $ ()() $
$ $
$ $
$ $
Negative (and Zero) Exponents
One cannot escape the feeling that mathematical formulae have an independent existence and intelligence of their own, that they are wiser than we are, wiser even than their discoverers, that we get more out of them than we originally put into them.
- Heinrich Hertz
In the last section, we saw exponent notation, originally devised as shorthand for repeated multiplication, beget five algebraic rules. In this section, the rules will teach us how to improve exponents themselves, and as a result, we’ll be able to extend exponents’ meaning far beyond repeated multiplication. This is a classic instance of “mere” notation assuming a life of its own and then guiding us to deeper mathematics.
We’ll begin by considering the possibility of raising something to the power of 0. What could $ x^{0} $ mean? Our definition from the previous section does not apply, since 0 is not a whole number. Still, one might reasonably argue that since, for whole numbers, $ x^{n} $ is the product we get from multiplying n copies of x, the expression $ x^{0} $ should be what we get by multiplying zero copies of x. Well, zero copies of anything is…nothing… so $ x^{0} $ ought to equal 0, right?
Wrong. If want to preserve the exponent rules (we do!), then we’re compelled to agree that
\[ \boldsymbol{x}^{\mathbf{0}}=x^{1-1}\stackrel{\mathrm{Rule2}}{\underset{\mathrm{}}\stackrel{\downarrow}{=}}\frac{x^{1}}{x^{1}}=\frac{x}{x}=1. \]
I repeat, if we insist on preserving the exponent rules, then we have no choice: We are forced to accept that the expression $ x^{0} $ must – if it is to have any meaning at all – equal 1. Here, our humble intuition (which suggests that $ x^{0} $ should be 0) clashes with the exponent rules (which insist that $ x^{0} $ is 1). Something must give. Mathematicians have decided that the exponent rules should have the upper hand, because they can prove their mettle on far tougher terrain, where our intuition is quite useless, as we’ll now see. What, for instance, can intuition make of $ 2^{-3} $ ? Nothing. On the other hand, the exponent rules, if we will only listen to them, lead us to a perfectly sensible definition of $ 2^{-3} $ , and, more generally, of $ x^{-n} $ :
\[ \boldsymbol{x}^{-n}=x^{0-n}\stackrel{\mathrm{R u l e2}}{\stackrel{\downarrow}{=}}\frac{x^{0}}{x^{n}}=\frac{1}{x^{n}}. \]
Thus, our exponent rules have taught us exactly what it means to raise something to the power of -n: It means raising something to the power of $ n_{…} $ and taking its reciprocal. (In either order.*)
Examples.
\[ 2^{-3}=\frac{1}{2^{3}}=\frac{1}{8}.\qquad\left(\frac{2}{3}\right)^{-2}=\left(\frac{3}{2}\right)^{2}=\frac{9}{4}.\qquad(-2)^{-10}=\left(\frac{1}{-2}\right)^{10}=\frac{1}{(-2)^{10}}=\frac{1}{1024}. \]
One very special case of raising something to a negative power is raising something to the power of -1. Since this means raising it to the power of 1 (which doesn’t change anything) and taking the reciprocal, the net effect of raising something to the power of -1 is simply… taking that thing’s reciprocal:
Examples.
\[ 13^{-1}=\frac{1}{13}.\quad\left(\frac{5}{42}\right)^{-1}=\frac{42}{5}.\quad\left(-\frac{27}{14}\right)^{-1}=-\frac{14}{27}.\quad\left(.01\right)^{-1}=100.\quad\blacklozenge \]
This is already remarkable. Exponents capture not only repeated multiplication, but also reciprocation. We’ll soon see that they also capture square roots, and even nth roots. Thus, if we accept what our pushy (but wise) exponent rules are trying to teach us, we’ll be able to gather repeated multiplication, reciprocation, and the extraction of roots under the single umbrella of exponentiation. Let us do so. This will greatly simplify mathematics in the long run. Here then are our formal definitions:
Definitions (Zero and Negative Exponents).
For any nonzero x, we make the following definitions: $ ^{*} $
\[ x^{0}=\mathbf{1},\quad\mathsf{a n d}\quad x^{-n}=\frac{1}{x^{n}}. \]
(In particular, raising something to the power of -1 is the same thing as taking its reciprocal.)
The five exponent rules still apply to zero and negative exponents, of course. You’ll soon have the opportunity to practice them again with problems such as the following:
Example. Simplify as much as possible, and remove all negative exponents: $ (5a{3}b{6}b{-3}a{-4})^{-2} $
\[ \begin{aligned}(5a^{3}b^{6}b^{-3}a^{-4})^{-2}&=(5a^{-1}b^{3})^{-2}\quad&(exponent rule1)\\&=\left(5\left(\frac{1}{a}\right)b^{3}\right)^{-2}\quad&(definition of negative exponents)\\&=\left(\frac{5b^{3}}{a}\right)^{-2}\\&=\left(\frac{a}{5b^{3}}\right)^{2}\quad&(definition of negative exponents)\\&=\frac{a^{2}}{(5b^{3})^{2}}\quad&(exponent rule5)\\&=\frac{a^{2}}{5^{2}(b^{3})^{2}}\quad&(exponent rule4)\\&=\frac{a^{2}}{25b^{6}}\quad&(exponent rule3)\end{aligned} \]
Here (as almost everywhere in algebra), there are many different paths that lead to the same answer. $ ^{1} $ So long as you apply the rules correctly, any such path is fine – though some may be more efficient than others. The key is to understand what you are doing. Whenever you write down an equals sign, you should be able to explain why the expressions on each side of it are in fact equal.
The following algebraic trick often comes in handy when we work with exponents in fractions.
If a factor (not a term!) of the numerator or the denominator is raised to a power, we can move it across the fraction bar if we also change the exponent’s sign.
Proof. If $ x^{-n} $ represents a factor of the numerator, the fraction must have the form $ (x^{-n}a)/b $ , where a and b can stand for anything. Manipulating this algebraically, we see that
\[ \frac{\boldsymbol{x}^{-n}\boldsymbol{a}}{\boldsymbol{b}}=\frac{\left(\frac{1}{x^{n}}\right)a}{b}=\frac{\frac{a}{x^{n}}}{b}=\frac{\boldsymbol{a}}{\boldsymbol{x}^{n}\boldsymbol{b}}. \]
Thus, the net effect is clear: We can kick that factor of $ x^{-n} $ downstairs if we change it to $ x^{n} $ . (Moreover, if we read those equalities backwards, we see that a factor of $ x^{n} $ downstairs can be pushed upstairs if we change it to $ x^{-n} $ .) A similar argument, whose details you should verify, shows that if we start with a factor $ x^{-n} $ downstairs, we can bring it upstairs if we change it to $ x^{n} $ , or vice-versa.
\[ Examples.\quad\frac{x^{-5}}{y}=\frac{1}{x^{5}y}.\qquad\quad\frac{5a^{-2}}{17b^{-3}}=\frac{5b^{3}}{17a^{2}}.\qquad\quad\frac{9a^{-2}}{a^{3}}=\frac{9}{a^{3}a^{2}}=\frac{9}{a^{5}}.\qquad\quad\frac{x^{2}y^{-3}z^{-1}}{x^{-2}y^{5}z^{6}}=\frac{x^{4}}{y^{8}z^{7}}.\quad\blacklozenge \]
Please bear in mind that this trick concerns factors of the numerator or denominator – not terms!
Exercises
- Rewrite the following expressions without any negative exponents, and simplify them as much as possible:
\[ \begin{aligned}&a)3x^{-3}\qquad b)(3x)^{-3}\qquad c)a^{3}b^{3}(a^{-3}+b^{-3})\qquad d)8\left(\frac{x^{2}y^{3}z^{-1}}{x^{4}x^{17}y^{-2}zz^{-3}}\right)^{0}\qquad e)\frac{x^{13}y^{-9}}{y^{2}x^{10}x^{-2}}\\&f)(7t^{3}p^{-2})(3p^{2}t^{-2})\left(\frac{1}{7}t^{-1}\right)\qquad g)(6x^{n}y^{n-4})(2x^{2}y^{n+4})\qquad h)\left(\frac{-2a^{3}b^{-2}}{-8(a^{2}b)^{3}}\right)^{-1}\left(\frac{a^{-2}ab^{-1}}{a^{3}b^{4}}\right)^{-2}\\&i)(x+y)(x^{-1}+y^{-1})\qquad j)\frac{\left((2a^{-1})^{-2}\right)^{-3}}{32a^{-5}}+5^{0}-5^{0}\end{aligned} \]
Find the value of $ ((((2/5){-1}){-1}){-1})){-1} $ if this expression contains 1001 exponents, all of which are -1.
Decide whether each statement is true or false. Also, for each false statement, rewrite the left-hand side so that it has neither negative exponents nor fractions within fractions.
\[ a)\frac{a+b^{-5}}{2c}=\frac{a}{2c+b^{5}}.\quad b)\frac{5(x+y)^{-2}a^{2}}{2(x+y)a^{-3}}=\frac{5a^{5}}{2(x+y)^{3}}.\quad c)\frac{a+b}{a^{-1}+d}=\frac{a^{2}+b}{d}. \]
Explain why we defined $ x^{0} $ to be 1 (for $ x $ ).
Explain why we defined $ x^{-n} $ to be $ $ (for $ x $ ).
Provide those details that I asked you to verify in the proof above.
Is the expression $ 0^{-1} $ defined? If so, what is its value? If not, why not?
Radical Review
Before discussing fractional exponents, we must review radicals. The most common ones are square roots. By definition, a square root of c is any number whose square is c.
Negative numbers don’t have square roots, since you can’t get a negative by squaring a real number. (Do you see why?) In contrast, positive numbers have two square roots: one is positive, one is negative. The two square roots of 9, for example, are 3 and -3. The symbol $ $ denotes $ c’ $ s positive square root. $ ^{*} $ Thus, the two square roots of any c > 0 are $ $ and $ - $ . (For example, 5’s square roots are $ $ and $ - $ .)
A particularly simple (and important) fact to keep in mind when working with square roots is that
\[ \left(\sqrt{c}\right)^{2}=c \]
This is obvious. After all, the defining property of any square root of c is that… its square is c.
★
A cube root of c is a number whose cube is c. Cube roots are better behaved than square roots, since every number (including every negative number) c has exactly one cube root, which we denote $ $ .
Examples. $ =6 $ (since $ 6^{3}=216 $ ), $ =-2 $ (since $ (-2)^{3}=-8 $ ), $ =0 $ .
★
Like cube roots, other odd roots are clean: Every number c has one $ 5^{th} $ root, one $ 7^{th} $ root, and so on, which we denote by $ $ , $ $ , etc. (Example: $ = 2 $ , since $ 2^{5} = 32 $ .)
Like square roots, other even roots are messy – but in the same ways that square roots are messy. Negative numbers have no even roots. (Do you see why?) Positive numbers have two of each even root: one positive, one negative. The symbols $ $ , $ $ , etc. specifically refer to the positive even roots.
Exercises
- True or false? (And explain your answers.)
\[ \begin{aligned}&a)-5is a square root of25\qquad b)\sqrt{25}=-5.\qquad c)\sqrt{25}=5\qquad d)\sqrt{-25}=-5\qquad e)\sqrt{0}=0\\&f)-\sqrt{25}=-5\qquad g)\sqrt{8^{2}}=8\qquad h)\sqrt{(-8)^{2}}=-8\qquad i)\left(\sqrt{8}\right)^{2}=8\qquad j)\left(\sqrt{-8}\right)^{2}=-8\\&k)\left(\sqrt{8}\right)^{2}=8\qquad l)\left(\sqrt{3}\right)\left(\sqrt{3}\right)=3\qquad m)\sqrt{9/16}=3/4\qquad n)-\frac{1}{2}is a square root of\frac{1}{4}\qquad o)\sqrt{\frac{1}{4}}=-\frac{1}{2}\end{aligned} \]
- We can add multiples of the same square root like multiples of apples: Just as 5 apples plus 4 apples is 9 apples, so $ 5 + 4 = 9 $ . We can justify this by factoring out the square root: $ 5 + 4 = (5 + 4) = 9 $ . You should be able to justify all your algebraic moves. I’ve seen students claim (wrongly) that $ + $ is $ $ , and then ask “Why can’t I do that?” A better question to ask yourself prior to any move is “Why can I do this?” In this spirit, identify each of the following statements as true or false, and explain your answers.
\[ \begin{aligned}&a)\;2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\qquad b)\;\left(2\sqrt{5}\right)\left(3\sqrt{5}\right)=6\sqrt{5}\qquad c)\;\left(2\sqrt{5}\right)\left(3\sqrt{5}\right)=30\qquad d)\frac{6\sqrt{7}}{2\sqrt{7}}=3\sqrt{7}\\ &e)\;\left(3\sqrt{5}\right)^{2}=9\sqrt{5}\qquad f)\;\left(3\sqrt{5}\right)^{2}=45\qquad g)\;\left(-\sqrt{3}\right)^{2}=3\qquad h)\;\left(-2\sqrt{6}\right)^{2}=24\qquad i)-\left(2\sqrt{2}\right)^{2}=-8\\ \end{aligned} \]
- Identify the logical flaw in this argument: 2 apples times 3 apples is 6 apples, so $ (2)(3)=6 $ .
[Note well: The wrong “answer” is not the argument’s flaw; it’s just a consequence of the flaw. Find the flaw!]
- Simplify as much as possible:
\[ \begin{aligned}a)11\sqrt{11}-\sqrt{11}\qquad b)3\sqrt{121}+4\sqrt{121}+2\sqrt{11}-5\sqrt{11}\qquad c)(3\sqrt{2})(4\sqrt{2})(5\sqrt{1})\qquad d)\frac{2\sqrt{5}}{6\sqrt{5}}\end{aligned} \]
- Goldilocks knows how to approximate square roots by hand. Since $ $ is the positive number whose square is 3, she’ll try to guess its value, square her guess, and depending on the result, she’ll declare “This one is too big” or “This one is too small”. Then she’ll make an adjustment and try again. (Alas, no guess will ever be just right.) Suppose that her first guess is 2. Well, $ 2^{2} $ is greater than 3, so Goldilocks now knows that 2 is too big to be $ $ . Hence, she tries 1. Since $ 1^{2} $ is less than 3, she knows that 1 is too small. She has now shown that $ 1 < < 2 $ . How about 1.5? Well, $ (1.5)^{2} = 2.25 $ , so 1.5 is too small. Thus $ 1.5 < < 2 $ . And so on and so forth.
Your problem: using paper and pencil alone, determine the first two decimal places of $ $ .
- (Food for thought) Decimal approximations are useful; they let us compare numbers’ magnitudes at a glance. (E.g. Which is bigger: $ $ or 120/17? The answer is clear once we know that $ $ and 120/17 $ $ .) However, decimal approximations can also be sources of error if we introduce them into a problem. $ ^{*} $ To see this, contrast the following two calculations:
What is 3 times 2/3? (Obviously, the answer is 2).
What’s 3 times 0.667? (The answer here is not obvious. Nor is it 2.)
We know that $ ()^{2}=a $ . What if we reverse the order of the operations? Is $ $ always equal to a?
Explain why 64 has two square roots, but only one cube root.
Explain why -27 has a cube root, but no square root.
Explain why every positive number has two square roots, but only one cube root.
Explain why every negative number has a cube root, but no square root.
Explain why $ ()^{3}=a $ .
Simplify where possible:
\[ \begin{aligned}&a)\sqrt[3]{216}\quad&b)\sqrt[3]{-1}\quad&c)10\sqrt[3]{1000}\quad&d)\left(\sqrt[3]{7}\right)^{3}\quad&e)-\sqrt[3]{-8}\quad&f)\left(\sqrt[3]{6}\right)\left(\sqrt[3]{6}\right)\left(\sqrt[3]{6}\right)\\ &g)\sqrt[3]{(711)^{3}}\quad&h)\sqrt[3]{2}+\sqrt[3]{3}\quad&i)\sqrt[3]{\frac{8}{-125}}\quad&j)\sqrt[3]{\frac{2}{16}}\quad&k)\sqrt[3]{3\sqrt[3]{1}+\sqrt[3]{-64}}\quad&l)\sqrt[3]{0}(\sqrt{2}+\sqrt[3]{3})\\ &m)\sqrt{\left(\sqrt[3]{-343}+\sqrt[3]{27}\right)^{2}}\quad&n)\left(5\sqrt[3]{2}-3\sqrt[3]{2}\right)^{3}\quad&o)\left(2\sqrt[3]{2}\right)\left(-3\sqrt[3]{2}\right)\left(\frac{\sqrt[3]{2}}{3}\right)\left(\sqrt[3]{-1}\right)+\sqrt[3]{1}\\ \end{aligned} \]
- Simplify where possible:
\[ a)\sqrt[5]{32}\quad b)\sqrt[7]{-1}\quad c)\sqrt{10\sqrt[4]{10000}}\quad d)\left(\sqrt[13]{-22}\right)^{13}\quad e)-\sqrt[4]{81}\quad f)\sqrt[5]{\frac{\sqrt[3]{27a^{3}}}{-3a}} \]
Fractional Exponents
The exponent rules can teach us how to define fractional exponents. We will first consider “unit fraction” exponents (fractions with a numerator of 1). Because one of our exponent rules tells us that $ (x{1/n}){n}=x $ , it follows that $ x^{1/n} $ is an nth root of x. Hence, we define $ x^{1/n} $ to be $ $ . (Example: $ 16^{1/2}==4 $ .) With this preliminary definition in hand, we can now determine the meaning of any fractional exponent:
\[ \begin{aligned}\boldsymbol{x}^{m/n}&=\left(x^{1/n}\right)^{m}\quad&(exponent rule3)\\&=\left(\sqrt[n]{x}\right)^{m}\quad&(definition of unit fraction exponents).\end{aligned} \]
That is, raising something to the power of m/n means raising its $ n^{th} $ root to the $ m^{th} $ power.
Definition (Fractional Exponents).
\[ x^{m/n}=\left(\sqrt[n]{x}\right)^{m}. \]
(Note: $ x^{m/n} $ also equals $ $ , as explained below.)
\[ Examples.8^{2/3}=\left(\sqrt[3]{8}\right)^{2}=2^{2}=4;\qquad(32)^{3/5}=\left(\sqrt[5]{32}\right)^{3}=2^{3}=8;\qquad81^{1/4}=\sqrt[4]{81}=3. \]
The link between radicals and fractional exponents lets us convert any expression involving radicals into one involving exponents; these are usually easier to work with, thanks to the exponent rules’ simplicity.
Problem. Simplify as much as possible: $ ()()()^{5} $ .
Solution.
\[ \begin{aligned}\big(\sqrt[5]{x^{13}}\big)\big(\sqrt[3]{x^{8}}\big)\big(\sqrt{x}\big)^{5}&=\big(x^{13/5}\big)\big(x^{8/3}\big)\big(x^{5/2}\big)\quad&(defn.of fractional exponents)\\&=x^{(13/5)+(8/3)+(5/2)}\quad&(exponent rule1)\\&=x^{233/30}\end{aligned} \]
Usually, the order in which we perform operations matters; changing the order changes the outcome. (First you open the window, and then you put your head through.) So it may come as a surprise that taking a root and raising something to a power can be done in either order without changing the outcome. Proving this curious – and sometimes useful – fact will reinforce the link between radicals and exponents.
Claim. If x is positive, then $ ()^{m} = $ .
Proof. To prove this, we’ll compute $ x^{m/n} $ two ways:
On one hand, $ x^{m/n} = (x{1/n}){m} = ()^{m} $ . On the other, $ x^{m/n} = (x{m}){1/n} = $ .
Equating these two expressions for $ x^{m/n} $ , we conclude that $ = ()^{m} $ as claimed.
Exercises
- Rewrite each of the following radical expressions in terms of exponents, and simplify if possible.
$ $
$ $
$ $
$ $
$ ()^{3} $
$ (){4}(\sqrt[3]{a{2}})^{6} $
$ $
- Rewrite each of the following in terms of radicals.
- $ x^{1/5} $ b) $ y^{2/3} $ c) $ 3^{-1/4} $ d) $ (a + bc)^{3/8} $ e) $ 2^{0.5} $ f) $ w^{-1.5} $
Explain why $ = $ . [Hint: Rewrite this equation in terms of exponents.]
We use the previous exercise’s identity to pull factors out of radicals. (Ex: $ = = = 2 $ .) Rewrite the following expressions, making the number under the radical as small as possible:
- $ $ b) $ $ c) $ $ d) $ $ e) $ $
Explain why $ = $ .
- Compute $ $ , then compute $ + $ . Can radicals be distributed over addition?
- True or false: $ = a + 5 $ .
- Compute $ $ , then compute $ - $ . Can radicals be distributed over subtraction?
- True or false: $ =x-y $ .
- Simplify the following as much as possible:
\[ \sqrt[3]{\frac{64}{125}} \]
\[ \sqrt[3]{-1000} \]
\[ \mathsf{d})\sqrt{2}+\sqrt{8} \]
\[ \sqrt{\frac{200}{144}} \]
\[ )\sqrt{3-\left(\frac{\sqrt{3}}{2}\right)^{2}} \]
\[ 36^{3/2} \]
\[ 32^{4/5} \]
- 216^{2/3}
\[ 100^{5/2} \]
\[ \mathrm{j)}\;7^{2/5}\cdot7^{8/5} \]
\[ (49a^{8}b^{-4})^{1/2} \]
\[ \sqrt[5]{x^{2}y}\cdot\sqrt[5]{x^{3}y^{4}} \]
\[ \mathsf{m})\left(x^{1/2}+y^{1/2}\right)\left(x^{1/2}-y^{1/2}\right) \]
\[ \mathsf{n})a^{-1/6}\left[a^{2/3}\left(\frac{a^{2/3}}{a^{1/4}}\right)^{6}\right]^{1/3} \]
\[ 0)\sqrt{y}\left(\frac{x^{2}y^{-3}}{y^{3}}\right)\left(\frac{y^{13/2}}{x}\right) \]
- A simple trick for removing square roots from a fraction’s denominator is called rationalizing the denominator. It comes in two basic versions. Both rely on the old “multiply by one” trick. I’ll explain each with an example.
Example
\[ \frac{5}{\sqrt{7}}=\left(\frac{5}{\sqrt{7}}\right)\left(\frac{\sqrt{7}}{\sqrt{7}}\right)=\frac{5\sqrt{7}}{7}. \]
The second version is a bit more elaborate, and requires the difference of squares identity.
\[ \frac{3}{1-\sqrt{5}}=\left(\frac{3}{1-\sqrt{5}}\right)\left(\frac{1+\sqrt{5}}{1+\sqrt{5}}\right)=\frac{3+3\sqrt{5}}{1^{2}-\left(\sqrt{5}\right)^{2}}=\frac{3+3\sqrt{5}}{-4} \]
In the second version, we always multiply the top and bottom by the so-called conjugate of the bottom. Rationalize the denominator in the following expressions:
\[ a)\frac{2}{\sqrt{2}}\qquad b)\frac{30}{\sqrt{6}}\qquad c)\frac{14}{\sqrt{7}}\qquad d)\frac{3x}{3+\sqrt{x}}\qquad e)\frac{2}{\sqrt{7}+\sqrt{5}}\qquad f)\frac{\sqrt{6}+2}{\sqrt{6}-2} \]
Rationalize the numerator in the following expressions.
\[ g)\frac{5\sqrt{3}}{9}\qquad h)\frac{2-\sqrt{x}}{5\sqrt{x}} \]