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CHAPTER 5

5.1  Positive Integral Exponents.

In Chapter 3 we introduced the notation that

xn=x x x x  (n factors)

where x is any real number and n is a positive integer. The number x is called the base and n the exponent or power. Computations with exponents depend on the following ﬁve basic laws.

If x and y are real numbers and m and n are positive integers, then

E.1  xn xm=xn+m

E.2  (x y)n=xn yn

E.3  (xn)m=xn m

E.4  (xy)n=xnyn

E.5

The justification of these laws involves nothing more than counting the number of factors in a given expression. Law E.1 and the ﬁrst two parts of E.5 were established in Chapter 2. We will establish E.2 and E.3 below and leave E.4 and the rest of E.5 as exercises for the interested reader.
Since (x y)n has n factors of x y, there are n factors of x and n factors of y. Using associativity and commutativity of multiplication we have

(x y)n=(x y) (x y) (x y)

=(x x x) (y y y)

=xn yn

In E.3 the expression (xn)m has m factors of xn. Each of these factors has n factors of x, so that altogether there are n m factors of x. Thus

(xn)m=xn m

Example 1.  We use the laws of exponents to compute each of the following.

(a)  a2 a4=a2+4=a6  (E.1)

(b)  (2 x)4=24 x4=16 x4  (E.2)

(c)  (y2)5=y2·5=y10  (E.3)

(d)  (zw)3=z3w3  (E.4)

(e)  a4a2=a4-2=a2  (E.5)

(f)  a5a5=1  (E.5)

(g)  x2x7=1x7-2=1x5  (E.5)

Example 2.  Simplify (3 a2 b3)3 (a b2)

(3 a2 b3)3 (a b2)

=33 (a2)3 (b3)3 (a b2)

=33 a6 b9 a b2

=27 a6+1 b9+2

=27 a7 b11

Example 3.  Simplify (10 x y4)3(5 x2 y)2.

(10 x y4)3(5 x2 y)2

=103 x2 y4·352 x2·2 y2

=(2·5)3 x3 y1252 x4 y2

=23 53 x3 y1252 x4 y2

=23 53-2 y12-2x4-3

=40 y10x

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5.2  Integral Exponents

In this section we will enlarge our set of exponents to include zero and the negative integers. We want laws E.1 through E.5 to hold for this larger set of exponents. If a0, then in order for a0 to satisfy E.1, we would have

a0 an=a0+n=an

Since 1 is the only real number such that 1 an=an, we deﬁne

a0=1,  a0

Similarly, if n is a positive integer, then in order For a n to satisfy E.1, We would have

a n an=a n+m

=a0

=1

Since 1an is the only real number such that (1an) an=1. we deﬁne

a n=1an,  a0

This deﬁnition gives us the following helpful rule

1a n=11an=an

Note that in the above deﬁnitions, when the exponent is zero or negative, a cannot be zero.
It is straightforward but tedious to show that all the laws of Section 5.1 hold for integral exponents. In fact E.5 may be written in the simpler form

E.5

xnxm=xn-m

In the following examples we will use the new deﬁnitions and laws of exponents to simplify the given expressions. Furthermore, we will rewrite them without zero or negative exponents.

Example 1.  We use the above deﬁnitions to simplify each of the following.

(a)  20=1

(b)  ( 5)0=1

(c)   50= 1

(d)  10 2=1102

(e)  110 3=103

Example 2.  Simplify (u 4 v7) (u2 v 5).

(u 4 v7) (u2 v 5)

=(u 4 u2) (v7 v 5)

=u 4+2 v7-5

=u 2 v2

=v2u2

Example 3.  Simplify (x2 y 2 z)2 (x y4).

(x2 y 2 z)2 (x y4)

=(x2)2 (y 2)2 z2 (x y4)

=x4 y 4 z2 x y4

=(x4 x) (y 4 y4) z2

=x5 y0 z2

=x5 z2

Example 4.  Simplify (a 2 b c3) 2(a2 b 2 c) 3.

(a 2 b c3) 2(a2 b 2 c) 3

=a4 b 2 c 6a 6 b6 c 3

=a4-( 6) b 2-6 c 6-( 3)

=a10 b 8 c 3

=a10b8 c3

Example 5.  Simplify a 3-b 3a 1-b 1.

a 3-b 3a 1-b 1

=1a3-1b31a-1b

=b3-a3a3 b3b-aa b

=(b-a) (b2+b a+a2)a3 b3a bb-a

=b2+b a+a2a2 b2

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Large and small numbers in decimal form occur in scientiﬁc work. Integral exponents can be used to write such numbers in a compact form. For example,

132,000=1.32×100,000

=1.32×105

and

0.0000132=1.32×0.00001

=1.32×10 5

We see that factoring 105 from 132,000 moved the decimal point ﬁve places to the left, while factoring 10 5 from 0.0000132 moved the decimal point ﬁve places to the right. In general if n is a positive integer, then factoring 10n from a decimal number moves the decimal point n places to the left, and factoring 10 n moves the decimal point n places to the right. We can use this idea to write any real number r in decimal form as a number s between 1 and 10, times an integral power of 10; that is,

r=s×10n

where n is an integer. This is called scientiﬁc notation.

Example 6.  Express (a) 312,500, (b) 0.000115, and (c)  0.0927 in scientiﬁc notation.

(a)  312,500=3.125×105

(b)  0.000115=1.15×10 4

(c)   0.0927= 9.27×10 2

Example 7.  Rewrite 3.964×106 and (b) 7.902×10 3in decimal form.

(a)  3.964×106=3,964,000

(b)  7.902×10 3=0.007902

Example 8.  Calculate 2360000.0118.

2360000.0118

=2.36·1051.18·10 2

=2.00×107

=20,000,000

Example 8.  Calculate 132,000,000·0.00056000.

132,000,000·0.00056000

=1.32·108·5·10 46·103

=(1.32)·56×108 x 10 4103

=1.1×108-4-3

=1.1×10

=11

5.3  Rational Exponents

If a and b are real numbers, n is a positive integer, and

bn=a

then b is an nth root of a. When n=2 We say that b is a square root, and when n=3 we say that b is a cube root. For example, 2 and 2 are square roots of 4, since 22=4 and ( 2)2=4. Since  ( 3)3= 27, 3 is a cube root of 27. However, 9 has no square roots since there is no real number b such that b2= 9. The various cases for nth roots of a real number a are as follows.

1.  If n is even and a is positive, then there are two nth roots and one is the negative of the other. The positive nth root is called the principal nth root of a.

2.  If n is even and a is negative, then there are no real nth roots of a.

3.  If n is odd, then for all a, there is exactly one nth root of a, which is called the principal nth root.

4.  If a=0, then for all n,0 is the only nth root of 0, and thus 0 is the principal nth root of 0.

In the last section we enlarged our collection of exponents to include the integers. In this section we extend, by using nth roots, the set of exponents to include all rational numbers in such a way that the laws of exponents E.1-E.5 still hold.

For n, a positive integer, we ﬁrst deﬁne what we mean by a1n. From law E.3, if m=n, then

(a1n)n=a(1n)-n=a1=a

Thus, a1n must be an nth root of a and consequently we deﬁne a1n to
be the principal nth root of a.

Next for m and n positive integers we have, from law E.3,

a(1n)m=a(1n)-m=amn

Therefore, we deﬁne

amn=(a1n)m

When (a1n)m and (am)1nare both deﬁned, then they are equal since each is the principal nth root of am In fact (am)1n is the principal nth root of am by deﬁnition, and (a1n)m can be shown to be the principal nth root of am by examining each of the four cases above.

If a1n is deﬁned, then both(a1n)m and (am)1n exist. In particular, if a is non-negative, then all three exist and we have

amn=(a1n)m=(am)1n

In the remaining sections of the chapter, unless otherwise speciﬁed, we will assume that all the variables in the expressions represent non-negative real numbers.

If m and n are integers, then we deﬁne

a (mn)=1amn

With the above deﬁnitions for rational exponents, it can be shown that the laws E.1-E.5 hold.

Example 1.  Simplify (a) 2723, (b) (116)54, and (c) 8 (23)

(a)  2723=(2713)2=32=9

Note that we could also have written

(b)  (27)23=(272)13=(729)13=9

Often it is simpler to take the root ﬁrst.

(b)  (116)54=[(116)14]5=(12)5=132

(c)  8 (23)=1823=18(13)2=(12)2=14

Example 2.  Simplify (0.0025)32.

(0.0025)32

=(25·10 4)32

=(25)32 (10 4)32

=(2512)3 10 4·32

=53×10 6

=125×10 6

=1.25×10 4

Example 3.  Simplify x14 x23.

x14 x23

=x14+23

=x1112

Example 4. Simplify x (34) (x (14)+x34).

x (34) (x (14)+x34)

=x (34) x (14)+x (34) x34

=x 34-14+x 34+34

=x 1+x0

=1x+1

Example 5. Simplify (x12 y 23x34 y 13) (12).

(x12 y 23x34 y 13) (12)

=(x12-34 y 23+13) (12)

=(x 14 y 13) (12)

=x18 y16

In Section 5.3, we deﬁned x1n to be the principal nth root of x. An alternative notation is

x1n=nx

In the case n=2,2x is simply written as x. There are many cases when this notation will be more convenient to use. The symbol is called the radical, x is called the radicand, n is called the index, and the whole expression is called the radical of x of order n.

Since, from Section 5.3,

xnm=(xn)1m=(x1m)n

for m a natural number and n an integer we have

xnm=mxn=(mx)n

Example 1.  Change (a) 1653 and x 23 to radical form.

(a) 1653=3165=(316)5

(b) x 23=1x32=1x3=1(x)3

Example 2.  Change (a) 653 and (b) 17234to exponential form.

(a) 653=536=512

(b) 17234=12347=23 47

From the deﬁnition of radical and our laws of exponents, the following basic laws for radicals are easily derived:

R.1  nan=a

R.2  na b=na nb

R.3  n(ab)=nanb

For example, R.2 is derived as follows:

na b

=a1n b1n

=na nb

The other two rules are just as easily derived.
A number of operations with radicals involve changes in form, which may be made using R.1, R.2, and R3. We use these rules to simplify the expressions in the following examples.

Example 3.  Simplify 448.

448

=424·3  (R.2)

=424 43  (R.1)

=2 43

Example 4.  Simplify 28 c5 d8.

28 c5 d8

=22·7 c5 d8

=22 (c2)2 (d4)2 7 c  (R.2)

=2 c2 d4 7 c  (R.1)

When the radical has a fractional radicand it is sometimes desirable to remove the denominator from under the radical sign. This is called rationalizing the denominator. This may be accomplished by multiplying both the numerator and denominator by an appropriately chosen factor.

Example 5.  Rationalize the denominator of 532.

532

=525

=5·225·2

=10(23)2

=1023

In the second step, the appropriate factor is 2 since this makes the denominator a perfect square.

Example 6.  Rationalize the denominator of 392 a2 b4.

392 a2 b4

=392 a2 b4·22 a b222 a b2

=39·4 a b223 a3 b6

=336 a b23(2 a b2)3

=336 a b22 a b2

The factor 23 a b2 was chosen in order to make the denominator a perfect cube.

Example 7.  Rationalize the denominator of 2 x y2 z48 x2 y z6.

2 x y2 z48 x2 y z6

=2 x z y2 4(2 x2 y3 z2)423 x2 y z6 42 x2 y3 z2

=2 x z y2 4(2 x2 y3 z2)423 x2 y z6·2 x2 y3 z2

=2 x z y2 4(2 x2 y3 z2)4(2 x y z2)4

=2 x z y2 4(2 x2 y3 z2)2 x y z2

=y·42 x2 y3 z2z

There are also expressions in which the index of the radical can be reduced, as illustrated in the following example.

Example 8.  Simplify 49 u2 v4.

49 u2 v4

=4(3 u v2)2

=[(3 u v2)2]14

=(3 u v2)24

=(3 u v2)12

=(3 u)12 (v2)12

=v3 u

This simpliﬁcation can be made because the exponents of 3,u, and v have the common factor 2, which is a divisor of the index 4.
An expression that involves repeated radicals may be simpliﬁed by the use of the equivalent exponential form.

Example 9.  Simplify 3 3.

3 3

=(3·312)12

=312·314

=427

In general, an expression involving radicals is said to be in standard form if the following three conditions are satisﬁed.
1. The radicand has no factors with exponents as great as, or greater than the index.
2. The index of the radical is as small as possible.
3. The denominator is rationalized.
Note that in Examples 3 through 9 we have simpliﬁed the given expressions by changing them to standard form.

Certain expressions involving radicals can be added and subtracted using the distributive law. For example,

2 5+7 5-3 5

=(2+7-3) 5

=6 5

There are cases where the given terms in an expression are not alike, but when written in standard form common factors appear.

Example 1.  Simplify 50+98.

50+98

=25·2+49·2

=5 2+7 2

=12 2

Example 2.  Simplify 432 c5+72 c3-c4162 c+6 c2 c.

432 c5+72 c3-c4162 c+6 c2 c

=2 c42 c+6 c2 c-3 c42 c+6 c2

= c42 c+12 c2 c

Example 3.  Simplify 8x-18 xx+7 32 x3.

8x-18 xx+7 32 x3

=23 xx2-32·2 xx+7 25 x3

=2 2 xx-3 2 xx+7·22 x2 x

=( 1x+28 x) 2 x

5.6  Multiplication and Division of Radicals

From rules R.2 and R3 of Section 5.4, it is clear that two radicals of the same index may be multiplied or divided by carrying out the operation under the radical sign.

Example 1.

(a)  3 5=3·5=15

(b)  38133=3813=327=3

When the radicals have different indices, change the expressions to exponential form in order to simplify.

Example 2.  simplify 5 a 32 a2.

5 a 32 a2

=(5 a)12 (2 a2)13

512 a12 213 a23

=536 a36 226 a46

=(53 a3 22 a4)16

=653 22 a7

=a653 22 a

In the above example, to write the terms under a common radical it was necessary to express the fractional exponents as equivalent fractions with denominators being their L.C.D. We use the same technique when we divide.

Example 3.  Divide 32 x by 8 x.

3 2 x8 x

=32 x8 x 2 x2 x

=(2 x)13 (2 x)1216 x2

=(2 x)564 x

=632 x54 x

The two expressions ax+by and ax-by; are called conjugates of each other. The product of these expressions is:

(a x+b y) (a x-b y)

=(a x)2-a b x y+a b y x-(b y)2

=a2 x-b2 y

For example, the conjugate of 1+2 3 is 1-2 3, while the conjugate of  2-3 is  2+3. Their respective products are

(1+2 3) (1-2 3)

=1-2 3+2 3-(2 3)2

=1-12

= 11

and

( 2-3) ( 2+3)

=( 2)2-2 3+3 2-(3)2

=2-3

= 1

Conjugates can be used to simplify certain fractions whose denominators are of the form a x+b y, by removing the radicals from the denominators. As in Section 5.4 this also is called rationalizing the denominator.

Example 4.  Simplify 3+53-5by rationalizing the denominator.

3+53-5

=3+53-5·3+53+5

=3+2 3 5+53-5

=8+2 15 2

= 4-15

We recall that in the previous sections we limited our computations with nth roots to non-negative expressions. This allowed us to assume that

amn=nam=(na)m

In considering negative radicands there are certain difﬁculties. In the case that n is odd, the above equations sh']_l hold and the laws of exponents and radicals still apply. In particular, if n is odd,

n a= na

This equality holds since

n a

=n( 1) a

=n 1 na

=( 1) na

= na

For example, 3 8= 38= 2. In the case that n is even, nam may exist while nam may not. For example, ( 3)2=9=3, while ( 3)2 is not deﬁned. This is one of the reasons why it was convenient to consider only non-negative radicands.

We can, however, simplify expressions such as ( 3)2 by using the idea of absolute value. In fact, ( 3)2=| 3|=3. In general, if n is even, then

nxn=|x|

In the following examples the variables are allowed to take on negative values.

Example 1.  Write 416 a6 x4 y2 in standard form.

416 a6 x4 y2

=424 a4 x4 a2 y2

=2 |a| |x| 4a2 y2

=2 |a x| 4a2 y2

=2 |a x| |a y|

Note that in reducing 4a2 y2 we must use |a y| since ay is negative when aa and y have opposite sign.

Example 2.  Simplify 3 3 x416 y2.

3 3 x416 y2

=3 3 x424 y2·22 y22 y

=3 x3 12 x y26 y3

= x (312 x y)4 y

Example 3.  Simplify 36 x2 y4 z.

36 x2 y4 z

=62 x2 y4 z

=6 |x| |y2| z

=6 |x| y2 z

Note that, in the last line, since y2 always non-negative we replace |y2| by y2.

5.8  Complex Numbers.

In our previous discussions of the real number system and radicals, it was pointed out that a is not a real number in the case that n is even and a is negative. For example, 2, 4 8, and 6 3, do not represent real numbers. However, we now introduce new numbers, called complex numbers, which give these expressions meaning. Surprisingly, complex numbers have applications to problems in engineering and physics.

We introduce a new number i whose square is 1. Thus,

1=i

If b is any positive real number, then

b

=b ( 1)

=b 1

=b i

Hence, the square root of any negative real number can be represented as the product of a real number and the number i. For example,

16=16 1=4 i

and

5=5 1=5 i

The number i is called the imaginary unit and any number of the form b i, where b is a real number, is called an imaginary number.
We now consider all expressions of the form a+b i, where a and b are real numbers. These numbers are called complex numbers. The number a is called the real part and b is called the imaginary part of a+b i. For example, 2-3 i is the complex number whose real part is 2 and whose imaginary part is  3. Since

a=a+0 i

every real number a is a complex number. Similarly, every imaginary number b i is a complex number since

b i=0+b i

Two complex numbers a+b i and c+d i are equal if and only if a=c and b=d, that is, if and only if the real parts are equal and the imaginary parts are equal.

The sum and difference of two complex numbers are deﬁned in a manner consistent with how we have performed these operations on other radicals. We have

(a+b i)+(c+d i)

=(a+c)+(b i+d i)

=(a+c)+(b+d) i

and

(a+b i)-(c+d i)

=(a-c)+(b i-d i)

=(a-c)+(b-d) i

For example,

(2+3 i)+( 3+4 i)

=(2-3)+(3 i+4 i)

= 1+7 i

and

(2-3 i)-( 2-i)

=(2+2)+( 3 i+i)

=(4-2 i)

Example 1.  Write each of the following in the form a+b i: (a) 5- 9 (b) 2+ 82

(a) 5- 9

=5-9 1

=5-3 i

(b) 2+ 82

=2+8 12

=2+2 2 i2

=1+2 i

Example 2.  Simplify (2+ 4)+(3+ 25)-( 6- 9).

(2+ 4)+(3+ 25)-( 6- 9)

=( 2+4 1)+(3+25 1)-( 6-9 1)

=2+2 i+3+5 i+6+3 i

=11+10 i

The product of two complex numbers is deﬁned so that it conforms with the product of two binomials. We simply must remember that i2= 1. Thus

(a+b i) (c+d i)

=a (c+d i)+b i (c+d i)

=a c+a d i+b c i+b d i2

=a c+a d i+b c i-b d

=(a c-b d)+(a d+b c) i

Example 3.  Multiply 2+3 i by (3-2 i).

(2+3 i) (3-2 i)

=6-4 i+9 i-6 i2

=6-4 i+9 i+6

=12+5 i

We call the complex number a-b i the complex conjugate of the number a+b i. Therefore, a+b i is the complex conjugate a-b i. If we multiply a complex number by its conjugate we always obtain a real number since

(a+b i) (a-b i)

=a2-a b i+a b i-b2 i2

=a2+b2

We use this fact to develop a method for dividing complex numbers.

For example,

3+2 i1-i

=(3+2 i) (1+i)(1-i) (1+i)

=3+2 i+3 i+2 i21-i+i-i2

=1+5 i2

=12+52 i

The process of division is performed by multiplying numerator and denominator by the complex conjugate of the denominator. In general,

a+b ic+d i

=(a+b i) (c-d i)(c+d i) (c-d i)

=(a+b i) (c-d i)c2+d2

=a c+b dc2+d2+b c-a dc2+d2 i

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