appendix b precalculus

8/13/2019 Appendix B PreCalculus

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REVIEW OF INTEGER EXPONENTS

I write a1, a2, a3, etc., for etc. Isaac Newton (June 13, 1676)

In basic algebra you learned the exponential notation an, where a is a real num

n is a natural number. Because that definition is basic to all that follows in tthe next two sections of this appendix, we repeat it here.

Definition 1 Base and Exponent

Given a real number a and a natural number n, we define an by

n factors

In the expression an the number a is the base, and n is the exponent or po

which the base is raised.

an a # a # a p a

1

aaa,

1

aa,

1

a,

APPENDIX

BB.1 Review of Integer

Exponents

B.2 Review ofnth Roots

B.3 Review of Rational

Exponents

B.4 Review of FactoringB.5 Review of Fractional

Expressions

B.6 Properties of the Real

Numbers

B.1

EXAMPLE

SOLUTION

1 Using Algebraic Notation

Rewrite each expression using algebraic notation:

(a) xto the fourth power, plus five;

(b) the fourth power of the quantityxplus five;

(c) xplusy, to the fourth power;

(d) xplusy to the fourth power.

(a) x4 5 (b) (x 5)4 (c) (xy)4 (d) xy4

Caution: Note that parts (c) and (d) of Example 1 differ only in the use of the c

So for spoken purposes the idea in (c) would be more clearly conveyed by

the fourth power of the quantityxplusy. In the case of (d), if you read it a

another student or your instructor, chances are youll be asked, Do you mean

or do you mean (xy)4?

Moral:Algebra is a precise language; use it carefully. Learn to ask yourself w

what youve written will be interpreted in the manner you intended.


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B-2 Appendix B

In basic algebra the four properties in the following box are developed for work-

ing with exponents that are natural numbers. Each of these properties is a direct con-

sequence of the definition of an. For instance, according to the first property, we have

a2a3 a5. To verify that this is indeed correct, we note that

a2a3 (aa)(aaa) a5

PROPERTY SUMMARY Properties of Exponents

Property Examples

1. aman amn a5a6 a11; (x 1)(x 1)2 (x 1)3

2. (am)n amn (23)4 212; [(x 1)2]3 (x 1)6

3. a4; 1

4. (ab)m

am

bm

; (2x2)3 23 (x2)3 8x6;

x8

y12ax2

y3b4

#

am

bmaa

b bm

1

a4;

a5

a5a2

a6

a6

a2am

an damn if m n1

anmif m n

1 if m n

Of course, we assume that all of these expressions make sense; in particular, we as-

sume that no denominator of a fraction is zero.

Now we want to extend our definition of an to allow for exponents that are inte-

gers but not necessarily natural numbers. We begin by defining a0.

Definition 2 Zero Exponent

For any nonzero real number a, EXAMPLES

(a) 20 1(b) (p)0 1

(00 is not defined.)(c)

Its easy to see the motivation for defining a0 to be 1. Assuming that the exponent zero

is to have the same properties as exponents that are natural numbers, we can write

That is,

Now we divide both sides of this last equation by an to obtain a0 1, which agrees

with our definition.

Our next definition (see the box that follows) assigns a meaning to the expression

an when n is a natural number. Again, its easy to see the motivation for this definition.

We would like to have

anan an(n) a0 1

a0an an

a0an a0n

a 31 a2 b2

b 0 1a0 1


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That is,

Now we divide both sides of this last equation by an to obtain an 1an, in

ment with Definition 3.

Definition 3 Negative Exponent

For any nonzero real number a

and natural number n,

an 1

an

anan 1

B.1 Review of Integer Exponents

EXAMPLE

SOLUTION

EXAMPLE

EXAMPLES

(a) 21

(b) 10

(c) x2

(d) (a2b)3

(e) 23 81

23

1

123

1

(a2b)3

1

a6b3

1

x2

a 110

b1 11 110 21

1

21

1

2

It can be shown that the four properties of exponents that we listed earli

tinue to hold now for all integer exponents. We make use of this fact in the ne

examples.

2 Using Properties of Exponents

Simplify the following expression. Write the answer in such a way that only p

exponents appear.

First Method Alternative Method

3 Using Properties of Exponents

Simplify the following expression, writing the answer so that negative expon

not used.

a a5b2c0a3b1

b 3

as obtained

previously

b5

a

(a2b3)2(a5b)1 a45b61 a1b5

(a2b3)2(a5b)1 a4b6

a5b

b61

a54

b5

a

(a2b3)2(a5b)1 (a4b6)(a5b1)(a2b3)2(a5b)1 (a2b3)2 #1

a5b

(a2b3)2(a5b)1


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B-4 Appendix B

We show two solutions. The first makes immediate use of the property (am)n amn.

In the second solution we begin by working within the parentheses.

First Solution Alternative Solution

( a5b2c0 )3 b9a24

( a5b2c0 )3 b6(3)a9 (15)

b9

a24

a a5b2c0a3b1

b 3 a b3a8b 3a a5b2c0

a3b1b 3 a15b6

a9b3

SOLUTION

EXAMPLE

SOLUTION

EXAMPLE

SOLUTION

4 Simplifying with Variable Exponents

Simplify the following expressions, writing the answers so that negative exponents

are not used. (Assume thatp and q are natural numbers.)

(a) (b) (apbq)2(a3pbq)1

(a)

(b)

b3q

ap

apb3q(apbq)2(a3pbq)1 (a2pb2q)(a3pbq)

a4pqpq a3p2q

a4pq

apq

a4pq(pq)

a4pq

apq

5 Simplifying Numerical Quantities Involving Exponents

Use the properties of exponents to compute the quantity

As an application of some of the ideas in this section, we briefly discuss scientific

notation, which is a convenient form for writing very large or very small numbers. Such

numbers occur often in the sciences. For instance, the speed of light in a vacuum is

As written, this number would be awkward to work with in calculating. In fact, thenumber as written cannot even be displayed on some hand-held calculators, because

there are too many digits. To write the number 29,979,000,000 (or any positive num-

ber) in scientific notation, we express it as a number between 1 and 10, multiplied by

an appropriate power of 10. That is, we write it in the form

where 1 a 10 and nis an integer.

a 10n

29,979,000,000 cm/sec

1

(31513)(21210)

1

32 # 22

1

36

210 # 313

33 # 312 # 212

210 # 313

315 # 212

210 #

313

27# 612 2

10 #3

13

(33)(3# 2)12

210 # 313

27# 612.


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According to this convention, the number 4.03 106 is in scientific notation,

same quantity written as 40.3 105 is not in scientific notation. To convert

number into scientific notation, well rely on the following two-step procedu

1. First move the decimal point until it is to the immediate right of the first n

digit.

2. Then multiply by 10n or 10n, depending on whether the decimal po

moved n places to the left or to the right, respectively.

For example, to express the number 29,979,000,000 in scientific notatio

we move the decimal point 10 places to the left so that its located betwee

and the 9, then we multiply by 1010. The result is

or, more simply,

As additional examples we list the following numbers expressed in both o

and scientific notation. For practice, you should verify each conversion for y

using our two-step procedure.

All scientific calculators have keys for entering numbers in scientific no

(Indeed, many calculators will convert numbers into scientific notation for y

you have questions about how your own calculator operates with respect to sc

notation, you should consult the users manual. Well indicate only one ehere, using a Texas Instruments graphing calculator. The keystrokes on other

are quite similar. Consider the following number expressed in scientific no

3.159 1020. (This gargantuan number is the diameter, in miles, of a galaxy

center of a distant star cluster known as Abell 2029.) The sequence of keystro

entering this number on the Texas Instruments calculator is

3.159 20EE

0.0000002 2 1070.000099 9.9 105

55,708 5.5708 104

29,979,000,000 2.9979 1010

29,979,000,000 2.9979000000 1010

To Express a Number Using Scientific Notation

B.1 Review of Integer Exponents

EXERCISE SET B.1

AIn Exercises 14, evaluate each expression using the givenvalue of x.

1. 2x3 x 4;x 2 2. 1 x 2x2 3x3;x 1

3. x 4. x 11 (x 1)2

1 (x 1)2;

1

2

1 2x2

1 2x3;

In Exercises 516, use the properties of exponents to simeach expression.

5. (a) a3a12 6. (a) (32)3 (23)2

(b) (a 1)3(a 1)12 (b) (x2)3 (x3)2

(c) (a 1)12(a 1)3 (c) (x2)a (xa)2

7. (a) yy2y8

(b) (y 1)(y 1)2(y 1)8

(c) [(y 1)(y 1)8]2


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B-6 Appendix B

33. 34. (2x2)3 2(x2)3

35. 36.

37. (xp)(xp) 38. (2pq)(2qp1)

In Exercises 3942, use the properties of exponents in comput-

ing each quantity, as in Example 5. (The point here is to do as

little arithmetic as possible.)

39. 40.

41. 42.

For Exercises 4350, express each number in scientific notation.

43. The average distance (in miles) from Earth to the Sun:

44. The average distance (in miles) from the planet Pluto to the

Sun:

45. The average orbital speed (in miles per hour) of Earth:

46. The average orbital speed (in miles per hour) of the planet

Mercury:

47. The average distance (in miles) from the Sun to the nearest

star:

48. The equatorial diameter (in miles) of

(a) Mercury: 3031 (c) Jupiter: 88,733

(b) Earth: 7927 (d) the Sun: 865,000

49. The time (in seconds) for light to travel

(a) one foot:

(b) across an atom:

(c) across the nucleus of an atom:

50. The mass (in grams) of

(a) a proton:

(b) an electron:

0.000000000000000000000000000911

0.00000000000000000000000167

0.000000000000000000000001

0.000000000000000001

0.000000001

25,000,000,000,000,000,000

107,300

66,800

3,666,000,000

92,900,000

a 144 # 12523 # 32

b124532 # 124

212 # 513

101228 # 315

9 # 310 # 12

bp2q

(b2)qbp1

bp

x2

y3

x2

y38. (a) 9. (a)

(b) (b)

(c) (c)

10. (a) (y2y3)2 11. (a)

(b) y2(y3)2 (b)

(c) 2m(2n)2 (c)

12. (a) 13. (a)

(b) (b)

(c) (c)

14. (a) (x2y3z)4 15. (a) 4(x3)2

(b) 2(x2y3z)4 (b) (4x3)2

(c) (2x2y3z)4

(c)

16. (a) 2(x 1)7 (x 1)7

(b) [2(x 1)]7 (x 1)7

(c) 2(x 1)7 [2(x 1)]7

For Exercises 1738, simplify each expression. Write the

answers in such a way that negative exponents do not appear.

In Exercises 3538, assume that the letters p and q represent

natural numbers.

17. (a) 640 18. (a) 20 30

(b) (643)0 (b) (20 30)3

(c) (640)3 (c) (23 33)0

19. (a) 101 102 20. (a) 42 41

(b) (101 102)1 (b)

(c) [(101)(102)]1 (c)

21. 22.

23. (a) 52 102 24. (a)

(b) (5 10)2 (b)

25. (a

2

bc

0

)

3

26. (a

3

b)

3

(a

2

b

4

)

1

27. (a2b1c3)2 28. (22 21 20)2

29. 30.

31. 32.a a3b9c2a5b2c4

b 0ax4y8z2xy2z6

b2ax4y8z2

xy2z6b 2ax3y2z

xy2z3b3

1 14 34 221 14 22 1 34 22

1 13 14 211 13 21 1 14 21[ 1 14 21 1 14 22]1[ 1 14 21 1 14 22]1

(4x2)3

(4x3)2

ax2

y

20

x6y15b2

(x2

3x 2)6

x2 3x 2

x2y20

x6y15x

x6

x6y15

x2y20x6

x

(t2 3)15

(t2 3)9

t9

t15

t15

t9

1210

129

210

212

(x2 3)9

(x2 3)10x10

x12

(x2 3)10

(x2 3)9x12

x10


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REVIEW OF nTH ROOTS

In this section we generalize the notion of square root that you studied in elem

algebra. The new idea is that of an nth root; the definition and some basic ex

are given in the box that follows.

B.2 Review of nth Roots

EXAMPLES

B5

132

12

B41

16

1

2;13 8 2

125 5;125 5; 125

Hindu mathematicians first recognized

negative roots, and the two square roots

of a positive number, . . . though they

were suspicious also. David Wells in

The Penguin Dictionary of Curious andInteresting Numbers (Harmondsworth,

Middlesex, England: Penguin Books,

Ltd., 1986)

B.2

Definition 2 Principal nth Root

1. Let n be a natural number. If a and b

are nonnegative real numbers, then

if and only if b an

The number a is the principalnth

root of b.

2. If a and b are negative and n is anodd natural number, then

if and only if b an1n b a

1n b a

Definition 1 nth Roots

Let n be a natural number. If a

and b are real numbers and

an b

then we say that a is an nth root of b.

When n 2 and when n 3, we

refer to the roots as square roots

and cube roots, respectively.

As the examples in the box suggest, square roots, fourth roots, and all eve

of positive numbers occur in pairs, one positive and one negative. In these ca

use the notation to denote the positive, or principal, nth root of b. As ex

of this notation, we can write

The symbol is called a radical sign, and the number within the radical sig

radicand. The natural number n used in the notation is called the index

radical. For square roots, as you know from basic algebra, we suppress the in

simply write rather than So, for example, 5.

On the other hand, as we saw in the examples, cube roots, fifth roots, andall oddroots occur singly, not in pairs. In these cases, we again use the notat

for the nth root. The definition and examples in the following box summar

discussion up to this point.

12512 .11n 1

14 81 3 (The negative of the principal fourth root of 81 is 3.14

81 3 (The principal fourth root of 81 is not 3.)

14 81 3 (The principal fourth root of 81 is 3.)2n b

EXAMPLESBoth 3 and 3 are square roots of

because 32 9 and (3)2 9.

Both 2 and 2 are fourth roots of

because 24 16 and (2)4 16.

2 is a cube root of 8 because 23

3 is a fifth root of 243 becaus

(3)5 243.

There are five properties of nth roots that are frequently used in simplifyi

tain expressions. The first four are similar to the properties of square roots t

developed in elementary algebra. For reference we list these properties side


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(a)

(b) B162

491162149

181127

912

7

713 213 513 (2 5)131413 12513

112

175

1(4)(3)

1(25)(3)SOLUTION

B-8 Appendix B

in the following box. Property 5 is listed here only for the sake of completeness;

well postpone discussing it until Appendix B.3.

PROPERTY SUMMARY Properties of nth Roots

Suppose thatxandy are real numbers and that m and n are

natural numbers. Then each of the following properties holds, CORRESPONDINGprovided only that the expressions on both sides of the PROPERTIES FORequation are defined (and so represent real numbers). SQUARE ROOTS

1.

2.

3.

4. n even:

n odd:

5. m

21nx mn

1x

2n xn x2x2 0x02n xn 0x0B

x

y1x1yBn

x

y1n x1n y

1xy 1x1y1n xy 1n x1n y11x22 x11n x2n x

Our immediate use for these properties will be in simplifying expressions in-

volving nth roots. In general, we try to factor the expression under the radical so

that one factor is the largest perfect nth power that we can find. Then we apply

Property 2 or 3. For instance, the expression is simplified as follows:

In this procedure we began by factoring 72 as (36)(2). Note that 36 is the largest fac-

tor of 72 that is a perfect square. If we were to begin instead with a different factor-

ization, for example, 72 (9)(8), we could still arriveat thesame answer, but it would

take longer. (Check this for yourself.) As another example, lets simplify First,what (if any) is the largest perfect-cube factor of 40? The first few perfect cubes are

so we see that 8 is a perfect-cube factor of 40, and we write

13 40 13 (8)(5) 13 813 5 213 513 1 23 8 33 27 43 64 1

3

40.

172 1(36)(2) 13612 612172

EXAMPLE 1 Simplifying Square Roots

Simplify: (a) (b) B162

49.112 175;


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EXAMPLE

SOLUTION

EXAMPLE

SOLUTION


EXAMPLE

SOLUTION

2 Simplifying nth Roots

Simplify:

3

13

2

213 2 513 2 413 213 16 13 250 13 128 13 813 2 13 12513 2 13 6413 2

13 16 13 250 13 128.

3 Simplifying Radicals Containing Variables

Simplify each of the following expressions by removing the largest possible p

square or perfect-cube factor from within the radical:

(a)

(b) wherex 0;

(c) where a 0;

(d)

(a)

(b)

(c)

(d) 23 16y5 23 8y323 2y2 2y23 2y2 3a312a

218a7 2(9a6)(2a) 29a612a 212x 2x2 xbecausex 028x

214122x2

28x2 2(4)(2)(x2) 14122x2 212 0x023 16y5.2

18a7,

28x2,28x2;

4 Simplifying nth Roots Containing Variables

Simplify each of the following, assuming that a, b, and c are positive:

(a) (b)

(a)

(b)

24 16a4b424 2a2b

c2

2ab24 2a2bc2

B432a6b5

c8

24 32a6b524 c8

2bc2

12ac

28ab2c5 2(4b2c4)(2ac) 24b2c412acB4

32a6b5

c8.28ab2c5;


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B-10 Appendix B

In Examples 1 through 4 we used the definitions and propertiesofnth roots to sim-

plify certain expressions. The box that follows shows some common errors to avoid

in working with roots. Use the error box to test yourself: Cover up the columns labeled

Correction and Comment, and try to decide for yourself where the errors lie.

Error Correction Comment

2 Although 2 is one of the fourth

roots of 16, it is not theprincipal

fourth root. The notation isreserved for the principal fourth root.

5 Although 5 is one of the square

roots of 25, it is not theprincipal

square root.

The expression The properties of roots differ with

cannot respect to addition versus multi-

be simplified. plication. For we do have thesimplification

(assuming that a and b are

nonnegative).

The expression For we do have

cannot

be simplified.

There are times when it is convenient to rewrite fractions involving radicals in al-

ternative forms. Suppose, for example, that we want to rewrite the fraction in

an equivalent form that does not involve a radical in the denominator. This is calledrationalizing the denominator. The procedure here is to multiply by 1 in this way:

That is, as required.

To rationalize a denominator of the form a we need to multiply the fraction

not by but rather by 1a 21a 2( 1). To see why this is necessary,notice that

which still contains a radical, whereas

which is free of radicals. Note that we multiply a by a to obtain an

expression that is free of square roots by taking advantage of the form of a difference

1b1b a2 b

1a 1b 2 1a 1b 2 a2 a1b a1b b1a 1b

21b a1b b

1b1b1b,1b 1b,5

13 513

3,

5

13 5

13 # 1 5

13 #1313

5133

513

13 a b13 ab 13 a13 b.13 ab13 a b 13 a 13 b

1ab 1a1b1ab

1a b1a b 1a 1b

125125 514 16

14 1614 16 2Errors to Avoid


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of two squares. Similarly, to rationalize a denominator of the form a , w

tiply the fraction by 1a 21a 2. (The quantities a and a to be conjugates of each other.) The next two examples make use of these ide

1b1b1b1b 1b


EXAMPLE

5 Rationalizing a Square Root Denominator

Simplify: 3150.112SOLUTION First, we rationalize the denominator in the fraction 1

Next, we simplify the expression

Now, putting things together, we have

2912

2

12 3012

2

(1 30)122

1

12 3150 12

2 1512 1

2

2

30

12

2

3150 31(25)(2) 312512 (3)(5)12 15123150:

1

12 1

12 # 1 1

12 #1212

122

12:

EXAMPLE 6 Using the Conjugate to Rationalize a Denominator

Rationalize the denominator in the expression 42 13

.

SOLUTION We multiply by 1, writing it as

Note: Check for yourself that multiplying the original fraction by d

eliminate radicals in the denominator.

2

13

2 13

8 413

4 12 13 24 113 22

4 12 13 24 3

8 413

1

4

2 13 # 1 4

2 13 #2 132 13

2 132 13.

In the next example we are asked to rationalize the numerator rather th

denominator. This is useful at times in calculus.


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B-12 Appendix B

EXAMPLE 7 Rationalizing a Numerator Using the Conjugate

Rationalize the numeratorof: wherex 0,x 3.1x 13

x 3,

SOLUTION

1

1x13

11x22 113 22

(x 3)11x13 2x 3

(x 3)11x13 2

1x13

x 3

# 1 1x13

x 3

#1x13

1x

13

The strategy for rationalizing numerators or denominators involving nth roots is

similar to that used for square roots. To rationalize a numerator or a denominator in-

volving an nth root, we multiply the numerator or denominator by a factor that yields

a product that itself is a perfect nth power. The next example displays two instances

of this.

EXAMPLE 8 Rationalizing Denominators Containing nth Roots

(a) Rationalize the denominator of:

(b) Rationalize the denominator of: where a 0, b 0.ab

24 a2b3,

6

13 7.

Finally, just as we used the difference of squares formula to motivate the method

of rationalizing certain two-term denominators containing a square root term, we can

use the difference or sum of cubes formula (see Appendix B.4) to rationalize a

denominator involving a cube root term. In particular, in the sum of cubes formula

x3 y3 (xy)(x2 xy y2), the expression (x2 xy y2) is the conjugate of

the expressionxy.

SOLUTION (a)

(b)

ab24 a2b24 a4b4

ab24 a2bab

24 a2b

ab

24 a2b3 # 1 ab

24 a2b3 #24 a2b24 a2b

613 49

7 as required

62

372

23 73

6

13 7 # 1 6

13 7 #23 7223 72


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EXAMPLE 9 Using the Conjugate to Rationalize a Denominator Involving a Cube Ro

Rationalize the denominator in the expression1

23 a 23 b .SOLUTION If we letx andy the conjugate of the denominatorxy

is

where, for example, Then

23 a2 23 ab 23 b2

123 a 23 123 b 23 23 a2 23 ab 23

a b

1

23 a 23 b 1

23 a 23 b #23 a2 23 ab 23 b223 a2 23 ab 23 b2

123 a 22 23 a23 a 23 a2.x2 xy y2 123 a 22 23 a23 b 123 b 22 23 a2 23 ab 23 b2

23 a23 b,

23 a

EXERCISE SET B.2

AIn Exercises 18, determine whether each statement is TRUE

or FALSE.

1. 2.

3. 4.

5. 6.

7. 8.

For Exercises 918, evaluate each expression. If the expression

is undefined (i.e., does not represent a real number), say so.

9. (a) 10. (a)

(b) (b)

11. (a) 12. (a)

(b) (b)

13. (a) 14. (a)

(b) (b)

15. (a) 16. (a)

(b) (b)

17. (a) 18. (a)

(b) (b)

In Exercises 1944, simplify each expression. Unless otherwise

specified, assume that all letters in Exercises 3544 represent posi-

tive numbers.

19. (a) 20. (a)

(b) (b)13 37513 54 1150118

23 (10)3

15 32

24 (10)415 3216 64322712516 64422568114 1614 1614 1611626 1100023 812523 1100023 812515 3214 64 15

3213 64

2(5)2 5113 10 23 10110 6 110 16243 213 149 71100 101256 16181 9

21. (a) 22. (a)

(b) (b)

23. (a) 24. (a)

(b) (b)

25. (a) 26. (a)

(b) (b)

27. (a)

(b)

28. (a)

(b)

29. (a)

(b)

30. (a)

(b)

31. 32.

33. 34.

35. (a) 36. (a)

(b) wherey 0 (b) where

37. (a) 38. (a)

(b) (b) where

39. 40.

41. 42.

43. 44.24 ab324 a3b23 16a12b2c923 8a4b624 16a4b52(a b)5(16a2b272a3b4c5 2

464y4,2ab32a3b 2

3 125x6

2ab2

2a2b

24 16a4,236y2,2225x4y3236x2

23 14096216413 192 13 81 4124 8154 216

281121 23 8133113 2 13 213 0.00810.0915 2 15 64 15 4861

3 112 14814 32

14 162

4150 31128213 81 313 213 2 13 16413 212712 1825 2562434216625222549225413 10815 64127198


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15/30

By replacing 5 and 3 with b and n, respectively, we can see that we want to

b1n to mean Also, by thinking of bmn as (b1n)m, we see that the defini

bmn ought to be These definitions are formalized in the box that follo11n b 2m.1n

b.

B.3 Review of Rational Exponents

Definition Rational Exponents

1. Let b denote a real number and n a

natural number. We define b1n by

(If n is even, we require that b 0.)2. Let mn be a rational number reduced

to lowest terms. Assume that n ispositive and that exists. Then

or, equivalently,

bmn 2n bmbmn 11n b 2m

1n b

b1n 1n b

It can be shown that the four properties of exponents that we listed in Appen

(on page B-2) continue to hold for rational exponents in general. In fact, we

this for granted rather than follow the lengthy argument needed for its verifi

We will also assume that these properties apply to irrational exponents. For in

we have

(The definition of irrational exponents is discussed in Section 5.1.) In the nex

examples we display the basic techniques for working with rational exponen

1215 215 25 32

EXAMPLES

412 2

(8)13

823 22 4

or, equivalently,

823 413 6423 82113 8 2213 8 21

4

EXAMPLE 1 Evaluating Rational Exponents

Simplify each of the following quantities. Express the answers using positiv

nents. If an expression does not represent a real number, say so.

(a) 4912 (b) 4912 (c) (49)12 (d) 4912

SOLUTION (a) 4912 7

(b) 4912 (4912) 7

(c) The quantity (49)12 does not represent a real number because there i

real numberxsuch thatx2

49.

(d) 4912

Alternatively, we have

4912 (4912)1 71 1

7

1

711

14921492491

149149


16/30

B-16 Appendix B

EXAMPLE 2 Simplifying Expressions Containing Rational Exponents

Simplify each of the following. Write the answers using positive exponents. (Assume

that a 0.)

(a) (5a23)(4a34) (b) (c) (x2 1)15(x2 1)45B516a13

a14

SOLUTION (a)

because

(b)

because

(c) (x2 1)15(x2 1)45 (x2 1)1 x2 1

1615a160

13

14

112 (16a

112)15

B516a13

a14 a 16a13

a14b 15

23

34

1712 20a

1712

(5a23)(4a34) 20a(23)(34)

EXAMPLE3 Evaluating Rational Exponents

Simplify: (a) 3225; (b) (8)43.

SOLUTION (a) 3225 22

Alternatively, we have

3225 (25)25 22

(b) (8)43 (2)4 16

Alternatively, we can write

(8)43 [(2)3]43 (2)4 16

113 8 241

4

1

22

1

4

1

22115 32 22

Rational exponents can be used to simplify certain expressions containing radi-

cals. For example, one of the properties of nth roots that we listed but did not discuss

in Appendix B.2 is Using exponents, it is easy to verify this property.

We have

x1mm mn1x as we wished to show

m21nx (x1n)1mm21nx mn1x.

EXAMPLE 4 Using Rational Exponents to Combine Radicals

Consider the expression wherexandy are positive.

(a) Rewrite the expression using rational exponents.

(b) Rewrite the expression using only one radical sign.

1x23 y2,

SOLUTION (a)

(b) rewriting the fractions usinga common denominator

26x326y4 26x3y4x36y46

2x23y2 x12y232x23y2 x12y23


17/30


18/30

B-18 Appendix B

REVIEW OF FACTORING

There are many cases in algebra in which the process offactoring simplifies the work

at hand. To factor a polynomial means to write it as a product of two or more non-

constant polynomials. For instance, a factorization ofx2 9 is given by

In this case,x 3 andx 3 are the factors ofx2 9.

There is one convention that we need to agree on at the outset. If the polynomial

or expression that we wish to factor contains only integer coefficients, then the fac-

tors (if any) should involve only integer coefficients. For example, according to this

convention, we will not consider the following type of factorization in this section:

because it involves coefficients that are irrational numbers. (We should point out,

however, that factorizations such as this are useful at times, particularly in calculus.)

As it happens,x2 2 is an example of a polynomial that cannot be factored using in-

teger coefficients. In such a case we say that the polynomial is irreducible over theintegers.

If the coefficients of a polynomial are rational numbers, then we do allow factors

with rational coefficients. For instance, the factorization of y2 over the rational

numbers is given by

Well consider five techniques for factoring in this section. These techniques will

be applied in the next two sections and throughout the text. In Table 1 we summarize

these techniques. Notice that three of the formulas in the table are just restatements

y2 1

4ay 1

2b ay 1

2b

14

x2 2 1x 12 2 1x 12 2

x2

9 (x 3)(x 3)

B.4fac # tor (fakt r), n. . . . 2. Math.one of two or more numbers, algebraic

expressions, or the like, that when

multiplied together produce a given

product; . . . v.t. 10. Math. to express

(a mathematical quantity) as a product

of two or more quantities of like kind,

as 30 2 # 3 # 5, or x2 y2 (xy) (xy). The Random House

Dictionary of the English Language,

2nd ed. (New York: Random House,

1987)

0

TABLE 1 Basic Factoring Techniques

Technique Example or Formula Remark

Common factor 3x4 6x3 12x2 3x2(x2 2x 4) In any factoring problem, the first step always is

4(x2 1) x(x2 1) (x2 1)(4 x) to look for the common factor of highest degree.

Difference of x2 a2 (x a)(x a) There is no corresponding formula for a sum of

squares squares;x2 a2 is irreducible over the integers.

Trial and error x2 2x 3 (x 3)(x 1) In this example the only possibilities, or trials, are

(a) (x 3)(x 1) (c) (x 3)(x 1)

(b) (x 3)(x 1) (d) (x 3)(x 1)

By inspection or by carrying out the indicated

multiplications, we find that only case (c) checks.Difference of cubes x3 a3 (x a)(x2 ax a2) Verify these formulas for yourself by carrying out

Sum of cubes x3 a3 (x a) (x2 ax a2) the multiplications. Then memorize the formulas.

Grouping This is actually an application of the common

factor technique.

(x 1)(x2 1)

x2(x 1) (x 1)# 1x3 x2 x 1 (x3 x2) (x 1)


19/30

of Special Products formulas. Remember that it is the form or pattern in the f

that is important, not the specific choice of letters.

The idea in factoring is to use one or more of these techniques until each of

tors obtained is irreducible. Theexamples that followshow howthis works in p

B.4 Review of Factoring

EXAMPLE 1 Factoring Using the Difference of Squares Technique

Factor: (a) x2 49; (b) (2a 3b)2 49.

SOLUTION (a)

difference of squares

(b) Notice that the pattern or form is the same as in part (a): Its a differenc

squares. We have

difference of squ

(2a 3b 7)(2a 3b 7)

[(2a 3b) 7][(2a 3b) 7]

(2a 3b)2 49 (2a 3b)2 72

(x 7)(x 7)

x2 49 x2 72

EXAMPLE 2 Factoring Using Common Factor and Difference of Squares

Factor: (a) 2x3 50x; (b) 3x5 3x.

SOLUTION (a) common factor


(b)

3x(x 1)(x 1)(x2 1) difference of squares, again 3x(x2 1)(x2 1) difference of squares 3x[(x2)2 12]

3x5 3x 3x(x4 1) common factor

2x(x 5)(x 5)

2x3 50x 2x(x2 25)

EXAMPLE 3 Factoring by Trial and Error

Factor: (a) x2 4x 5; (b) (a b)2 4(a b) 5.

SOLUTION (a) x2 4x 5 (x 5)(x 1) trial and error

(b) Note that the form of this expression is

This is the sameform as the expression in part (a), so we need only replac

the quantity a b in the solution for part (a). This yields

(a b 5)(a b 1)

(a b)2 4(a b) 5 [(a b) 5][(a b) 1]

( )2 4( ) 5

EXAMPLE 4 Factoring by Trial and Error

Factor: 2z4 9z2 4.

SOLUTION We use trial and error to look for a factorization of the form (2z2 ?)(z2 ?)

are three possibilities:

(i) (2z2 2)(z2 2) (ii) (2z2 1)(z2 4) (iii) (2z2 4)(z2 1)


20/30

B-20 Appendix B

Each of these yields the appropriate first term and last term, but (after checking) only

possibility (ii) yields 9z2 for the middle term. The required factorization is then

2z4 9z2 4 (2z2 1)(z2 4).

Question: Why didnt we consider any possibilities with subtraction signs in placeof addition signs?

EXAMPLE 5 Two Irreducible Expressions

Factor: (a) x2 9; (b) x2 2x 3.

SOLUTION (a) The expressionx2 9 is irreducible over the integers. (This can be discovered

by trial and error.) If the given expression had instead beenx2 9, then it could

have been factored as a difference of squares. Sums of squares, however, cannot

in general be factored over the integers.

(b) The expression x2 2x 3 is irreducible over the integers. (Check this for

yourself by trial and error.)

EXAMPLE 6 Factoring Using Grouping and a Special Product

Factor:x2 y2 10x 25.

SOLUTION Familiarity with the special product (a b)2 a2 2ab b2 suggests that we try

grouping the terms this way:

Then we have


(xy 5)(xy 5)

[(x 5) y][(x 5) y]

(x 5)2 y2x2 y2 10x 25 (x2 10x 25) y2

(x2 10x 25) y2

EXAMPLE 7 Factoring Using Grouping

Factor: ax ay2 bx by2.

SOLUTION We factor a from the first two terms and b from the second two to obtain

We now recognize the quantityxy2 as a common expression that can be factored

out. We then have

The required factorization is therefore

Alternatively,

(a b)(xy2)

x(a b) y2(a b)

(ax bx) (ay2 by2)

ax ay2 bx by2

ax ay2 bx by2 (x y2)(a b)

a(xy2) b(xy2) (xy2)(a b)

ax ay2 bx by2 a(xy2) b(xy2)


21/30

B.4 Review of Factoring

EXAMPLE 8 Factoring Sum and Difference of Cubes

Factor: (a) t3 125; (b) 8 (a 2)3.

SOLUTION (a)

difference of cubes

(b)

sum of c

As you can check now, the expression a2 6a 12 is irreducible over the in

Therefore the required factorization is a(a2 6a 12).

For some calculations (particularly in calculus) its helpful to be able to fa

expression involving fractional exponents. For instance, suppose that we wan

tor the expression

The common expression to factor out here is (2x 1)12; the technique is to

the expression with the smaller exponent.

x(2x 1)12 (2x 1)32

a(a2 6a 12)

a(4 2a 4 a2 4a 4)

[2 (a 2)][22 2(a 2) (a 2)2]

8 (a 2)3 23 (a 2)3

(t 5)(t2 5t 25)

t3 125 t3 53

EXAMPLE 9 Factoring with Rational Exponents

Factor:x(2x 1)12 (2x 1)32.

SOLUTION

We conclude this section with examples of four common errors to avoid in

ing. The first two errors in the box that follows are easy to detect; simply multiplythe supposed factorizations shows that they do not check. The third error ma

from a lack of familiarity with the basic factoring techniques listed on page B-

fourth error indicates a misunderstanding of what is required in factoring; the fin

tity in a factorization must be expressed as aproduct, not a sum, of terms or expr

(2x 1)12(4x2 3x 1)

(2x 1)12[x (2x 1)2]

x(2x 1)12 (2x 1)32 (2x 1)12[x (2x 1)3212]

Error Correction Comment

x2 6x 9 (x 3)2 x2 6x 9 (x 3)2 Check the middle term.

x2 64 (x 8)(x 8) x2 64 is irreducible A sum of squares is, in general, irreducible ove

over the integers. the integers. (A difference of squares, howeveralways be factored.)

x3 64 is irreducible x3 64 (x 4)(x2 4x 16) Although a sum of squares is irreducible, a sum

over the integers of cubes can be factored.

x2 2x 3 factors as x2 2x 3 is irreducible over The polynomialx2 2x 3 is thesum of the e

x(x 2) 3 the integers. sionx(x 2) and the constant 3. By definition,

ever, the factored form of a polynomial must b

product(of two or more nonconstant polynom

Errors to Avoid


22/30

B-22 Appendix B

EXERCISE SET B.4

AIn Exercises 166, factor each polynomial or expression. If a

polynomial is irreducible, state this. (In Exercises 16 the fac-

toring techniques are specified.)

1. (Common factor and difference of squares)

(a) x2 64 (c) 121z z3

(b) 7x4 14x2 (d) a2b2 c2

2. (Common factor and difference of squares)

(a) 1 t4 (c) u2v2 225

(b) x6 x5 x4 (d) 81x4 x2

3. (Trial and error)

(a) x2 2x 3 (c) x2 2x 3

(b) x2 2x 3 (d) x2 2x 3

4. (Trial and error)

(a) 2x2 7x 4 (c) 2x2 7x 4

(b) 2x2 7x 4 (d) 2x2 7x 4

5. (Sum and difference of cubes)

(a) x3 1 (c) 1000 8x6

(b) x3 216 (d) 64a3x3 125

6. (Grouping)

(a) x4 2x3 3x 6

(b) a2x bx a2z bz

7. (a) 144 x2 8. (a) 4a2b2 9c2

(b) 144 x2 (b) 4a2b2 9c2

(c) 144 (y 3)2 (c) 4a2b2 9(ab c)2

9. (a) h3 h5

(b) 100h3 h5

(c) 100(h 1)3 (h 1)5

10. (a) x4 x2

(b) 3x4 48x2

(c) 3(x h)4 48(x h)2

11. (a) x2 13x 40 12. (a) x2 10x 16

(b) x2 13x 40 (b) x2 10x 16

13. (a) x2 5x 36 14. (a) x2 x 6

(b) x2 13x 36 (b) x2 x 6

15. (a) 3x2 22x 16 16. (a) x2 4x 1

(b) 3x2 x 16 (b) x2 4x 3

17. (a) 6x2 13x 5 18. (a) 16x2 18x 9

(b) 6x2 x 5 (b) 16x2 143x 9

19. (a) t4 2t2 1 20. (a) t4 9t2 20

(b) t4 2t2 1 (b) t4 19t2 20

(c) t4 2t2 1 (c) t4 19t2 20

21. (a) 4x3 20x2 25x 22. (a) x3 x2 x

(b) 4x3 20x2 25x (b) x3 2x2 x

23. (a) ab bc a2 ac

(b) (u v)xxy (u v)2 (u v)y

24. (a) 3(x 5)3 2(x 5)2

(b) a(x 5)3 b(x 5)2

25. x2z2 xztxyz yt 26. a2t2 b2t2 cb2 ca2

27. a4 4a2b2c2 4b4c4 28. A2 B2 16A 64

29. A2 B2 30. x2 64

31. x3 64 32. 27 (a b)3

33. (xy)3 y3 34. (a b)3 8c3

35. x3 y3 xy 36. 8a3 27b3 2a 3b

37. (a) p4 1 38. (a) p4 4

(b) p8 1 (b) p4 439. x3 3x2 3x 1 40. 1 6x 12x2 8x3

41. x2 16y2 42. 4u2 25v2

43. 44.

45. 46.

47. 48.

49. xy y2 50. x2 x 1

51. 64(x a)3 x a 52. 64(x a)4 x a

53. x2 a2 y2 2xy 54. a4 (b c)4

55. 21x3 82x2 39x

56. x3a2 8y3a2 4x3b2 32y3b2

57. 12xy 25 4x2 9y2

58. ax2 (1 ab)xy by2

59. ax2 (a b)x b

60. (5a2 11a 10)2 (4a2 15a 6)2

61. (x 1)12 (x 1)32

62. (x2 1)32 (x2 1)72

63. (x 1)12 (x 1)32

64. (x2 1)23 (x2 1)53

65. (2x 3)12 (2x 3)32

66. (ax b)12

In Exercises 67 and 68, evaluate the given expressions usingfactoring techniques. (The point here is to do as little actual

arithmetic as possible.)

67. (a) 1002 992 68. (a) 103 93

(b) 83 63 (b) 502 492

(c) 10002 9992(c)

BIn Exercises 6973, factor each expression.

69. A3 B3 3AB(A B)

70. p3 q3 p(p2 q2) q(p q)2

71. 2x(a2

x2

)12

x3

(a2

x2

)32

72. (x a)12(x a)12 (x a)12(x a)32

73. y4 (p q)y3 (p2q pq2)y p2q2

74. (a) Factorx4 2x2y2 y4.

(b) Factorx4 x2y2 y4. Hint: Add and subtract a

term. [Keep part (a) in mind.]

(c) Factorx6 y6 as a difference of squares.

(d) Factorx6 y6 as a difference of cubes. [Use the result

in part (b) to obtain the same answer as in part (c).]

1

2

1

2

153 103

152 102

1ax bb13

14x

2

x3

8

512

x3125

m3n3 1

(a b)2

4

a2b2

9z4

81

16

81

4y2

25

16 c2


23/30

But if you do use a rule involving

mechanical calculation, be patient,

accurate, and systematically neat in

the working. It is well known to

mathematical teachers that quite half the

failures in algebraic exercises arise from

arithmetical inaccuracy and slovenly

arrangement. George Chrystal (1893)

B.5 Review of Fractional Expressions

REVIEW OF FRACTIONAL EXPRESSIONS

The rules of basic arithmetic that you learned for working with simple fracti

also used for fractions involving algebraic expressions. In algebra, as in arit

we say that a fraction is reduced to lowest terms or simplifiedwhen the num

and denominator contain no common factors (other than 1 and 1). The fa

techniques that we reviewed in Appendix B.4 are used to reduce fractions.

ample, to reduce the fraction we write

In the box that follows, we review two properties of negatives and fractio

will be useful in this section. Notice that Property 2 follows from Property 1 b

a b

b a

a b

(a b) 1

x2 9

x2 3x

(x 3)(x 3)

x(x 3)

x 3

x

x2 9

x2 3x,

B.5

PROPERTY SUMMARY Negatives and Fractions

Property Example

1. b a (a b) 2 x (x 2)

2.2x 5

5 2x 1

a b

b a 1

EXAMPLE 1 Using Property 2 to Simplify a Fractional Expression

Simplify:4 3x

15x 20.

SOLUTION

using Property 2

In Example 2 we display two more instances in which factoring is used to

a fraction. After that, Example 3 indicates how these skills are used to multip

divide fractional expressions.

1

5a 4 3x

3x 4b 1

5

4 3x

15x 20

4 3x

5(3x 4)

EXAMPLE 2 Using Factoring to Simplify a Fractional Expression

Simplify: (a) (b)x2 6x 8

a(x 2) b(x 2).

x3 8

x2 2x;

SOLUTION (a)

(b)x2 6x 8

a(x 2) b(x 2)

(x 2)(x 4)

(x 2)(a b)

x 4

a b

x3 8

x2 2x

(x 2)(x2 2x 4)

x(x 2)

x2 2x 4

x


24/30

B-24 Appendix B

EXAMPLE 3 Using Factoring to Simplify Products and Quotients

Carry out the indicated operations and simplify:

(a) (b)

(c) x2

144x2 4

x 12x 2

.

2x2 x 6

x2 x 1#x3 1

4 x2;

x3

2x2 3x#12x 18

4x2 6x;

SOLUTION (a)

(b)

using the fact

(c)

As in arithmetic, to combine two fractions by addition or subtraction, the fractions

must have a common denominator, that is, the denominators must be the same. The

rules in this case are

For example, we have

Fractions with unlike denominators are added or subtracted by first converting to

a common denominator. For instance, to add 9a and 10a2, we write

Notice that the common denominator used was a2. This is the least commondenominator. In fact, other common denominators (such as a3 or a4) could be used

here, but that would be less efficient. In general, the least common denominator for a

given group of fractions is chosen as follows. Write down a product involving the

irreducible factors from each denominator. The power of each factor should be equal to

(but not greater than) the highest power of that factor appearing in any of the individ-

ual denominators. For example, the least common denominator for the two fractions

9a

a2

10

a2

9a 10

a2

9

a

10

a2

9

a#a

a

10

a2

4x 1

x 1

2x 1

x 1

4x 1 (2x 1)

x 1

4x 1 2x 1

x 1

2x

x 1

a

b

c

b

a c

b

and a

b

c

b

a c

b

(x 12)(x 12)

(x 2)(x 2)#x 2

x 12

x 12

x 2

x2 144

x2 4

x 12

x 2

x2 144

x2 4#x 2

x 12

thatx 22 x

1

2x2 x 3

2 x

(2x 3)(x 1)

2 x

2x2 x 6

x2 x 1#x3 1

4 x2

(2x 3)(x 2)

(x2 x 1)#(x 1)(x2 x 1)

(2 x)(2 x)

6x3

2x2(2x 3)

3x

2x 3

x3

2x2 3x#12x 18

4x2 6x

x3

x(2x 3)#

6(2x 3)

2x(2x 3)


25/30

and is (x 1)2(x 2). In the example that follows

that the denominators must be in factored form before the least common denom

can be determined.

1

(x 1)(x 2)

1

(x 1)2


EXAMPLE 4 Using the Least Common Denominator

Combine into a single fraction and simplify:

(a) (b) (c)15

x2 x 6

x

2

x

x2 9

1

x 3;

3

4x

7x

10y2 2;

SOLUTION (a) Denominators: 22 #x; 2# 5#y2; 1

Least common denominator; 22 # 5xy2 20xy2

Note: The final numerator is irreducible over the integers.

(b) Denominators:x2 9 (x 3)(x 3);x 3Least common denominator: (x 3)(x 3)

(c)

Notice in the last line that we were able to simplify the answer by factor

numerator and reducing the fraction. (This is why we prefer to leave thcommon denominator in factored form, rather than multiplying it out, in th

of problem.)

The next four examples illustrate using the least common denomin

simplify compound fractions, or complex fractions, which are fractions

fractions.

(x 2)(x 6)

(x 2)(x 3)

x 6

x 3

x 6

x 3

15 (x2 4x 3)

(x 2)(x 3)

x2 4x 12

(x 2)(x 3)

15

(x 2)(x 3)

x 1

x 2#x 3

x 3

using the fact tha

2 x (x 2

15

x2 x 6

x 1

2 x

15

(x 2)(x 3)

x 1

x 2

x (x 3)

(x 3)(x 3)

3

(x 3)(x 3)

x

(x 3)(x 3)

1

x 3#x 3

x 3

x

x2 9

1

x 3

x

(x 3)(x 3)

1

x 3

15y2

14x2

40xy2

20xy2

3

4x

7x

10y2

2

1

3

4x#5y2

5y2

7x

10y2#2x

2x

2

1#20xy2

20xy2


26/30

B-26 Appendix B

EXAMPLE 5 Simplifying a Compound Fraction

Simplify:

1

3a

1

4b

5

6a2

1

b

.

SOLUTION The least common denominator for the four individual denominators 3a, 4b, 6a2, and

b is 12a2b. Multiplying the given expression by 12a2b12a2b, which equals 1, yields

12a2b

12a2b#

1

3a

1

4b

5

6a2

1

b

4ab 3a2

10b 12a2

EXAMPLE 6 Simplifying a Fraction Containing Negative Exponents

Simplify: (x1 y1)1. (The answer is not xy.)

SOLUTION After applying the definition of negative exponent to rewrite the given expression,

well use the method shown in Example 5.

multiplying by 1 xy

xy

xy

xy#

1

1

x

1

y

xy

y x

(x1 y1)1 a 1x

1

yb1 1

1

x

1

y


Simplify: (This type of expression occurs in calculus.)

1

x h

1

x

h.

SOLUTION

multiplying by 1

x (x h)

(x h)xh

h

(x h)xh

1

x(x h)

(x h)x

(x h)x

1

x h

1

x

h

(x h)x

(x h)x#

1

x h

1

x

h


Simplify:

x1

x2

1

x2 1

.


27/30


SOLUTION

multiplying by 1

(x 1)(x2 x 1)

(x 1)(x 1)

x2 x 1

x 1

x3 1

1 x2

x3 1

x2 1

x2

x2

x1

x2

1

x2 1

x2

x2#x

1

x2

1

x2 1

EXERCISE SET B.5

AIn Exercises 112, reduce the fractions to lowest terms.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11.

12.

In Exercises 1355, carry out the indicated operations and

simplify where possible.

13. 14.

15.

16.

17.

18.

19. 20.

21. 22.1

a

1

b

1

c

6

a

a

6

1

3x

1

5x2

1

30x34

x

2

x2

a2 a 42

a4 216a

a2 49

a3 6a2 36a

x3 y3

x2

4xy 3y2

(xy)3

x2

2xy 3y2

(3t2 4txx2) 3t2 2txx2

t2 x2

x2 x 2

x2 x 12#

x2 3x

x2 4x 4

ax 3

2a 1

a2x2 3ax

4a2 1

2

x 2#x2 4

x 2

x4 y4

(x4y x2y3 x3y2 xy4)(xy)2

x3 y3

(xy)3

(xy)2(a b)

(x2 y2)(a2 2ab b2)

a3 a2 a 1

a2 1

a3b2 27b5

(ab 3b2)29ab 12b2

6a2 8ab

a b

ax2 bx2x2 2x 4

x3 8

x2

x 202x2 7x 4

x 2x4 16

25 x2

x 5

x2 9

x 3

23. 24.

25. 26.

27.

28.

29. 30.

31.

32.

33.

34.

35.

36.

37. 38. 39.

40. 41. 42.

43. 44. 45.

a

x2

a2 a

3

x2 h

3

x2

h

1

2 h

1

2

h

1x2

1y

1

x

1

y

a 1a

1 1

a

1a

1b

1

a

1

b

1

x

1

a

x a

4

a a

2

a 1

1

x 1

1

x 1

1

x 1

1

x2 1

1

x3 1

2q p

2p2 9pq 5q2

p q

p2 5pq

4

6x2 5x 4

1

3x2 4x

1

2x 1

1

x2 x 20

1

x2 8x 16

3

2x 2

5

x2 1

1

x 1

a2 b2

a2 b2

a

a b

b

b a

x

x a

a

a x

4

x 5

4

5 x

a2 5a 4

a2 16

2a

2a2 8a

ax 1

2ax(x 1)2

3ax

2

(x 1)3

1 1

x

1

x23x

x 2

6

x2 4

4

x 4

4

x 1

1

x 3

3

x 2


28/30

B-28 Appendix B

C

55.

56. Consider the three fractions

(a) If a 1, b 2, and c 3, find the sum of the three

fractions. Also compute their product. What do you

observe?

(b) Show that the sum and the product of the three given

fractions are, in fact, always equal.

b c

1 bc , c a

1 ca , and a b

1 ab

a a 1bba a a 1

bbb

a b 1aba a b 1

abb

46. 47. (x1 2)1 48.

49. 50. (a1 a2)1

51. x(xy)1 y(xy)1

52. x(2x 2y)1 y(2y 2x)1

B

53. 54.1x a1x a

1x a1x a

a b

a b

a b

a b

a b

a b

a b

a b

#ab3 a3b

a2 b2

a 1a1

1

a2b1

(x2 2x)1

x2

xxy

y x

y2

x2 y2 1

Todays familiar plus and minus signswere first used in 15th-century Germany

as warehouse marks. They indicated

when a container held something that

weighed over or under a certain

standard weight. Martin Gardner in

Mathematical Games (Scientific

American, June 1977)

PROPERTIES OF THE REAL NUMBERS

I do not like as a symbol for multiplication, as it is easily confounded with x; . . .

often I simply relate two quantities by an interposed dot. . . . G. W. Leibniz in a letter

dated July 29, 1698

In this section we will first list the basic properties for the real number system. Then

we will use those properties to prove the familiar rules of algebra for working with

signed numbers.

The set of real numbers is closed with respect to the operations of addition

and multiplication. This means that when we add or multiply two real numbers, the

result (that is, the sum or the product) is again a real number. Some of the other most

basic properties and definitions for the real-number system are listed in the following

box. In the box, the lowercase letters a, b, and c denote arbitrary real numbers.

B.6

PROPERTY SUMMARY Some Fundamental Properties of the Real NumbersCommutative properties a b b a ab ba

Associative properties a (b c) (a b) c a(bc) (ab)c

Identity properties 1. There is a unique real number 0 (called zero or the additive identity) such that

2. There is a unique real number 1 (called one or themultiplicative identity) such that

Inverse properties 1. For each real number a there is a real number a (called the additive inverse of a

or the opposite of a) such that

2. For each real number a 0 there is a real number denoted by (or 1a or a1)

and called the multiplicative inverse or reciprocal of a, such that

Distributive properties a(b c) ab ac (b c)a ba ca

a#1

a 1 and 1

a# a 1

1

a

a (a) 0 and (a) a 0a# 1 a and 1# a a

a 0 a and 0 a a


29/30

On reading this list of properties for the first time, many students ask the

question, Why do we even bother to list such obvious properties? One re

that all the other laws of arithmetic and algebra (including the not-so-obviou

can be derived from our rather short list. For example, the rule 0 # a 0

proved by using the distributive property, as can the rule that the product of tw

ative numbers is a positive number. We now state and prove those properties a

eral others.

B.6 Properties of the Real Numbers

Theorem

Let a and b be real numbers. Then

(a) a 0 0 (c) (a) a (e) (a)(b) ab

(b) a (1)a (d) a(b) ab

#

Proof of Part (a)

Now since a 0 is a real number, it has an additive inverse, (a 0). Adding

both sides of the last equation, we obtain

Thus , as we wished to show.

Proof of Part (b)

Now, by adding

a to both sides of this last equation, we obtain

This last equation asserts that a (1)a, as we wished to show.

a (1)a additive identity propertya 0 (1)a additive inverse propertyassociative property of addition

a (a a) (1)a additive identity property anda 0 a [a (1)a]multiplicative identity property a (1)a

distributive property 1# a (1)a

additive inverse property [1 (1)]a

of multiplication

using part (a) and the commutative0 0# a

a# 0 00 a

#0 additive identity prop

0 a# 0 0 additive inverse prope

addition

a# 0 [(a# 0)] a# 0 5a# 0 [(a# 0)]6 associative property oa# 0 [(a# 0)] (a# 0 a# 0) [(a# 0)]

##

a# 0 a# 0 a# 0 distributive property

a# 0 a# (0 0) additive identity property


30/30

B-30 Appendix B

Proof of Part (c)

By adding a to both sides of this last equation, we obtain

This last equation states that (a) a, as we wished to show.

Proof of Part (d)

Thus a(b) ab, as we wished to show.

Proof of Part (e)

Weve now shown that (a)(b) ab, as required.

ab commutative property of multiplication

ba using part (c)

[(ba)] using part (d)

[b(a)] commutative property of multiplication

(a)(b) [(a)b] using part (d)

(ab) using part (b)

(1)(ab) associative property of multiplication

[(1)a]b commutative property of multiplication

[a(1)]b associative property of multiplication

a(b) a[(1)b] using part (b)

(a) a additive identity property(a) 0 a additive inverse property and additive identity property

(a) (a a) a associative property of addition

[(a) (a)] a 0 a(a) (a) 0 additive inverse property

appendix b precalculus

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