[PDF] Integer Exponents Part 1 - The History Thus Far and the Problem

944_6exponents2.pdf Lecture NotesInteger Exponentspage 1Part 1 - The History Thus Far and the Problem

Recall what we know about exponentiation thus far. Exponential notation expresses repeated multiplication..Denition: We dene27to denote the factor2multiplied by itself repeatedly, such as

2
2
2
2
2
2
2|{z}

7 factors= 2

7When mathematicians agreed to this denition, that was a free choice. They could have gone with other denitions. Once this

denition exists, however, certain properties are automatically true, and we have no other option but to recognize them as true.

They just fell into our laps..Theorem 1.Ifais any number andm,nare any positive integers, thenan
am=an+m

Theorem 2.Ifais any non-zero number andm,nare any positive integers, thenanam=anm Theorem 3.Ifais any number andm,nare any positive integers, then(an)m=anm Theorem 4.Ifa; bare any numbers andnis any positive integer, then(ab)n=anbn Theorem 5.Ifa; bare any numbers,b6= 0, andnis any positive integer, thenab

n=anbnAgain, the denition, immediately followed by the theorems. And then there was a quiet. Another opening for a free choice.

Consider the expression2x. The problem is that the denition of exponentiation only allows for a positive integer value ofx.

The expression2xis meaningful forx= 2or9or100, but it is not meaningful for values ofxsuch as3or35or3:2. In short,

the world of exponents was just the set of all natural numbers. Mathematicians usually don't like that. The best case scenario,

the ultimate hope is that the denition of exponents could be extended to any number forx. That way,2xwould be meaningful,

no matter what the value ofxis.

So, one of the issues was the desire to grow our world of exponents beyond the set of all natural numbers. This will be achieved

in several steps. Today, we are only focusing on enlarging the world of exponents fromNtoZ(i.e. from the set of all natural

numbers to the set of all integers).

The other issue was that as we enlarge our world, we pay especial attention that the new denitions will not conict with the

mathematics we already have. This principle comes up often in our choices, and it is sometimes called theexpansion principle..Denition: Inmanysituations, mathematiciansattempttoincrease, toenlargeourworld. Theexpansion

principleis that when we enlarge our mathematics by adding new denitions, we do so in such a way that the new denitions never create conicts with the mathematics we already have.c Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 2Part 2 - Integer Exponents

Suppose we want to dene20. The repeated multiplication denition can not be applied to zero, so we have complete freedom

to dene20. As it turns out, if we insist on a denition that does not conict with Rule 2,anam=anm, then we do not have all

that many choices for20. Let us think of zero as the result of the subtraction33, and that we would like to dene20so that

Rule 2 is still true.

2

0= 233rule 2=2323=88= 1

This is an expansion principle proof. It did not prove that the value of20is or must be zero. It showed much less; that if we

wanted to dene20without harming Rule 2 in the example given, then the only possible value for20is1. The reader should

imagine a team of mathematicians making rst sure that no part of our good old math is hurt if we dene20= 1. And as it

turned out, this is exactly the case. This computation can be repeated with many different bases. For example, 5

0= 522rule 2=5252=2525= 1or(3)0= (3)22rule 2=(3)2(3)2=99= 1

The only base that is problematic is0. Indeed, division by zero is not allowed and Rule 2,anam=anmdoes not work with

a= 0. If we try to perform the same computation with zero, we ultimately end up in00which is undened..Theorem 6.Ifais any non-zero number, thena0= 1.

0

0is undened.Please note that as we extend our world of exponents, old issues might re-surface. For example,(3)0= 1but30=1is

an important distinction, but not a new one.

Now that we have dened zero exponent, we will similarly try to dene negative integer exponents such as23.

Again, the original denition can not be applied. We cannot write down the factor two negative three times. So we have a

freedom here to dene23in any way we wish. In this decision, we will again use the expansion principle: that we would like

to keep our old rules after having23dened.

We will again use Rule 2,

anam=anmand write3as a subtraction between two positive integers. 2 3= 214Rule 2=2124=216=18=123or, more elegantly,23= 214Rule 2=2124=/2/2
2
2
2=123

When we discovered this rule, we saw that it was true because of cancellation. In case of a negative exponent, we have the same

cancellation, it's just that we run out of factors in the numerator rst. The computation can be repated with any base except for

zero..Theorem 7.Ifais any non-zero number, andnis any positive integer, thenan=1an. 0 nis undened.c Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 3Example 1.Simplify each of the following expressions. Use only positive exponents in your answer.

a)52b)a5c)132d)23 3 e)1x3f)2x3 Solution:a) Recall our new rule,an=1an. We apply this rule:52=152=125. b) We can use the same rule again:a5=1a5. c) In this case, the expression with the negative exponent is in the denominator.

The short story is that

132= 32= 9. The long story is that we apply our new rulean=1anand then we

divide by mutiplying by the reciprocal.

132=1132=11132=11
321=91=9So,1ancan be re-written asan.

d) In this case, the expression with the negative exponent is already a fraction.

The short story is that23

3 =32 3 =278. The long story is that we apply our new rulean=1anand then we divide by mutiplying by the reciprocal.  23
3 =123

3=11827=

11
278=278This computation shows thatab

n=ba n . e) The short story is that

1ancan be re-written asan. The computation below justies this step.

1x3=11x3=111x3=11
x31=x31=x3So,1ancan be re-written asan.

f) It is a common mistake to interpret2x3as(2x)3. Without the parentheses, we perform the exponentiation

before the multiplication. Therefore, the correct computation is

2x3= 2
x3=21
1x3=2x3.c

Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 4.Theorem:The following statements are practical applications of the rulean=1anand frequently

occur in computations.

1an=anandab

n=ba

nProof: As the computation shows, we apply the rulean=1anand then perform the division by multiplying by the reciprocal.

1an=11an=111an=11
an1=an1=anandab

n=1ab n=11anbn=11
bnan=bnan=ba n (end of proof) Example 2.Re-write the expressiona3b5c2d4using only positive exponents. Solution:We re-write the expressions with negative exponents using the rulean=1an. a

3b5c2d4=a3
1b51c2
d4=a

31
1b51c2
d41=a

3b5d4c2=

a3b5
c2d4=a3c2b5d4.

Notice the pattern here. If a factor with a negative exponent is in the numerator, we can re-write it with a positive exponent in

the denominator. Also, if a factor with a negative exponent is in the denominator, we can re-write it with a positive exponent in

the numerator..Theorem:anbmcpdq=bmdqancpwherea;c;dare any non-zero numbers andn,m,p,qare positive integers.The denitions ofa0andanwere developed with the intention that the previous rules (1 through 5) will remain true. Keep

that in mind in case of computations with more complex exponential expressions. Example 3.Simplify each of the given expressions. Present your answer using only positive exponents. a) a2
5b)x2
3x6(x)4c)a3a8d)a2b3a5b3e)2a4b3
5(3a3b2)0

Solution:a) It is much preferred to rst simplify the exponent. Repeated exponentiation means multiplication in the exponent.

a2
5=a2(5)=a10b) Let us re-write the solo negative signs as multiplications by1. Then we will use the rules of exponents to

simplify the exponents. Only after that will we address negative exponents.x2
3x6(x)4=1
x2
3x6(1
x)4=(1)3x2
3x6(1)4x4=(1)3x6x6(1)4x4

Now we get rid of all negative exponents by moving the factors. A factor with exponent5in the numerator can

be re-written as a factor with exponent5in the denominator, and vica versa. (1)3x6x6(1)4x4=(1)4x6x6x4(1)3=1
x161=x16c Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 5c) Solution 1: apply the ruleanam=anm.a3a8=a3(8)=a3+8=a5Solution 2: First we get rid of negative exponents and then apply the ruleanam=anm.

a 3a8=a8a3=a83=a5d) First we get rid of negative exponents. a

2b3a5b3=a5a2b3b3=a5a2b6=a3b6e) We can save a lot of work by noticing that the denominator is just1;because any non-zero quantity raised to the

power zero is1, and so3a3b2
0= 1.

2a4b3
5(3a3b2)0=25a4
5b3
51=25a20b151=a2025b15=a2032b15Part 3 - Scientic Notation Revisited

When we rst saw scientic notation, we learned to use it to handle uncomfortbly large numbers.

Recall the denition of scientic notation:.Denition:We can write numbers in scientic notation. This means to write a number as a product of

two numbers. The rst number is between1and10(can be1but must be less than10), and the second number is a10power. For example, the scientic notation for

428600000000is4:2861011.With negative exponents, we can also use scientic notation to handle extremely small numbers. For example, the mass of an

electron is0:00000000000000000000000000091094grams. Instead of hurds of trailing zeroes, now we are faced with many

zeroes after the decimal point. This number can be re-written as9:1094
1028. Example 4.Re-write the number0:0000000317using scientic notation.

Solution:The rst number in scientic notation needs to be between1and10. In this case, this number is3:17. We just

need to gure out the10power in the second part. We count how many decimal places we move the decimal from

0:0000000317to3:17. We count8decimal places. So the correct answer is3:17
108.

Example 5.Suppose thatA= 3:8
1015andB= 6:5
108. Perform each of the following operations. Present your answer

using scientic notation. a)B2b)AB2c)BA

Solution:a) We will apply rules of exponents.

B

2=6:5
108
2= 6:52
108
2= 42:25
1016

This number is not in scientic notation because42:25is too large for the rst part of scientic notation. Recall

that the rst factor must be between1and10. So we re-write42:25as4:225
10. B

2= 42:25
1016= 4:225
10
1016= 4:225
101+(16)=4:225
1015c

Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 6b) We will apply rules of exponents. AB

2=3:8
1015
6:5
108
2= 3:8
1015
6:52
108
2= 3:8
1015
42:25
1016

= (3:8
42:25)
1015
1016
= 160:55
1015+(16)= 160:55
101= 16:055

This number is not in scientic notation because16:055is too large for the rst part of scientic notation.

AB

2= 16:055 =1:6055
10.

c)

BA=6:5
1083:8
1015=6:53:8

10815=1:7105
1023.Sample Problems

Simplify each of the following. Assume that all variables represent positive numbers. Present your answer without negative

exponents.1.322.1233.m44.1x55.a8
a16.p3p7
p87.x4x98.50a1210a39.t3t410.x011.x012.(x)013.b5
b2
b1
14.1(b5)(b2)(b1)15.m2m516.x3y5z417.18q36q318.23

319.2y320.(2y)321. 35 222.a3b5a2b323.3m3
224.2ab3
325.k3
3(k5)226.2a3b53a3b2 2a3b5
327.2a3
2a2b
428.3p3q5
2(2q0p3)129.2a2b322(a1b)3 230. x3y0x5y3 231. x3y7x5y3

032.x1+y1x2y233.2a2
2b3a0aba2b2
32a2(2a2b)2ab034. a2b1a
5b7(ab2)3!

235.x2
2y3x02yx0y2x2
0yx5(y2x)3(2x1yx3)136.Suppose thatx= 8:5
1012andy= 7:5
107. Perform each of the following operations. Present your answer using

scientic notation. a)xyb)x3c)xy2d)xye)yx5c Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 7Answers

Sample Problems1.192.83.1m44.x55.a76.p47.x58.5a159.1t710.111.112.113.1b414.b415.m316.x3z4y517.3q618.27819.2y320.18y321.25922.a5b823.19m624.b98a325.k26.94a3b27.a58b428.18p9q1029.4a10b1230.x4y631.132.xyyx33.b8234.1b835.2x4y336.a)6:375
104b)6:1413
1034c)4:7813
104d)1:1333
1019e)1:6903
1063c

Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 8

Sample Problems - Solutions

Simplify each of the following. Assume that all variables represent positive numbers. Present your answer without negative

exponents.1.32

Solution: We just apply the rulean=1an.

3 2=132=192.123

Solution: We apply the rulean=1an.

123=1123=118

To divide is to multiply by the reciprocal:118= 1
81= 8

This is true in general:

1an=an

1an=11an= 1
an1=an3.m4

Solution: We apply the rulean=1an.

m 4=1m44.1x5

Solution: We have already proven that

1an=an

1x5=x55.a8
a1

Solution 1: We can apply the rulean
am=an+m

a

8
a1=a8+(1)=a7c

Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 9Solution 2: We can apply the rulean=1anand then the ruleanam=anm.

a

8
a1=a8
1a1=a81
1a=a8a=a8a1=a81=a76.p3p7
p8

Solution 1: We can apply the rulean
am=an+m

p

3p7
p8=p3+(7)+8=p4

Solution 2: We can apply the rulesan=1anandan
am=an+mandanam=anm. p

3p7
p8=p3
1p7
p8=p31
1p7
p81=p3
p8p7=p3+8p7=p11p7=p117=p47.x4x9

Solution 1: We can apply the rule

anam=anm. x 4x9=x4(9)=x4+9=x5 Solution 2: We can apply the rulesan=1anandanam=anm. x 4x9=x9x4=x94=x58.50a1210a3

Solution 1: We can apply the rule

anam=anm.

50a1210a3= 5a12(3)= 5a12+3= 5a15

Solution 2: We can apply the rulesan=1anandanam=anm.

50a1210a3=50a12a310= 5a12+3= 5a159.t3t4

Solution 1: We can apply the rules

anam=anmand thenan=1an. t 3t4=t34=t7=1t7c Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 10Solution 2: We can apply the rulean=1anand thenan
am=an+m. t 3t4=1t4
t3=1t710.x0

Solution: There is a separate rule stating that as long asxis not zero, thenx0= 1. So the answer is1.11.x0

Solution: This is the opposite ofx0and so the answer is1. x0=1
x0=1
1 =112.(x)0 Solution: This is again1because any non-zero riased to the power zero is1.13.b5
b2
b1
Solution 1: We can apply the rulesan
am=an+mand thenan=1an. b5
b2
b1
=b5+2+(1)=b4=1b4 Solution 2: We can apply the rulean=1anand then just cancel. b5
b2
b1
=1b5
b2
1b1=1b5
b21
1b1=b2b6=/b
/b/b
/b
b
b
b
b=1b414.1(b5)(b2)(b1) Solution 1: We can apply the rulesan
am=an+mand thenan=1an.

1(b5)(b2)(b1)=1b5+2+(1)=1b4=11b4= 1
b41=b4

Solution 2: We can apply the rulean=1anand thenanam=anm.

1(b5)(b2)(b1)=b5
b1b2=b6b2=b62=b415.m2m5

Solution 1: We can apply the rules

anam=anmand thenan=1an. m 2m5=m2(5)=m2+5=m3c Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 11Solution 2: We can apply the rulean=1anand thenanam=anm. m 2m5=m5m2=m52=m316.x3y5z4

Solution: Each variable occurs only once and so this problem is just about bringing it to the form required. We can apply

the rulean=1an. We hve alread shown that1an=an. x

3y5z4=x3z4y517.18q36q3

Solution 1: We can apply the rule

anam=anm.

18q36q3=/6
3q3(3)/6
1=3q3+31= 3q6

Solution 2: We can apply the rulesan=1anand thenan
am=an+m.

18q36q3=/6
3q3q3/6
1= 3q618.23

3

Solution: We can apply the rulean=1an.

 23
3 =123

3=123
23
23=

1827= 1
278=278

Note that we basically proved here that

ab n=ba n .19.2y3

Solution: We can apply the rulean=1an. It is important to note that the base of exponentiation isyand not2y.

2y3= 2
1y3=21
1y3=2y3c

Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 1220.(2y)3

Solution: We can apply the rulean=1an. This time the base of exponentiation is2y. So we will apply the rule

(ab)n=anbn. (2y)3=1(2y)3=123y3=18y321. 35 2

Solution 1: We can apply the rulean=1an.

 35 2 =1 35 2=1 35 35 =135
35=

1925= 1
259=259

Solution 2: We proved previously that

ab n=ba n . Using that,  35 2 = 53 2 = 53 53 =25922.a3b5a2b3

Solution 1: We can apply the rule

anam=anmand thenan=1an. a

3b5a2b3=a3(2)b53=a3+2b53=a5b8=a5
1b8=a51
1b8=a5b8

Solution 2: We can apply the rulesan=1anandan
am=an+m. a

3b5a2b3=a3a2b3b5=a5b823.3m3
2

Solution: We can apply the rulean=1anand then(ab)n=anbnand also(an)m=anm.

3m3
2=1(3m3)2=132(m3)2=19m3
2=19m624.2ab3
3

Solution: We can apply the rule(ab)n=anbnand then(an)m=anm. 2ab3
3= (2)3a3b3
3= (2)3a3b3(3)= (2)3a3b9c Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 13We now applyan=1an. (2)3a3b9=1(2)3
1a3
b9=18
1a3
b91=b98a3=b98a325.k3
3(k5)2 Solution: We can apply the rule(an)m=anmand thenanam=anm. k3
3(k5)2=k3(3)k5
2=k9k10=k9(10)=k9+10=k1=k26.2a3b53a3b2 2a3b5
3

Solution:

E=2a3b53a3b2

2a3b5
3=

2a33b5(2)3!

2a3b5
3applyanam=anm = 2a6b5+23 2a3b5
3 = 2a6b73 2a3b5
3applyab n=ba n = 32a6b7

2a3b5
3applyab

n=anbn = (3)2(2a6b7)2a3b5
3apply(ab)n=anbnandan=1an =

922(a6)2(b7)2
1(a3b5)3apply(an)m=anmand(ab)n=anbn

=

94a12b14
1(a3)3(b5)3applyan=1anand(ab)n=anbn

=

9a124b14
1a3
3b(5)3

=

9a124b14
1a9b15applyan=1an

=

9a124b14
b15a9=9a12b154b14a9applyanam=anm

=

9a129b15144=9a3b14=94a3bc

Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 1427.2a3
2a2b
4

Solution:

E=2a3
2a2b
4applyan=1an

= 2
1a3

1(2a2b)4apply(ab)n=anbn

= 21
1a3

1(2)4(a2)4b4apply(an)m=anm

= 2a3
116a8b4applyan=1an = 2a3
a816b4 = 2a8a3
16b4=2a816a3b4applyanam=anm = 1
/2a838
/2b4=a58b428.3p3q5
2(2q0p3)1

Solution:

E=3p3q5
2(2q0p3)1applyq0= 1and1an=an

= 3p3q5
22
1p3
1 = 3p3q5
2
2p3apply(ab)n=anbn = (3)2p3
2q5
2
2p3apply(an)m=anm = 9p3
2q5
2
2p3 = 18p6q10p3applyan
am=an+m = 18p6+3q10= 18p9q1029.2a2b322(a1b)3 2

Solution:

E=2a2b322(a1b)3

2 applyab n=ba n = 22a1b
32a2b3! 2 apply(ab)n=anbn = 4a1
3b32a2b3! 2 apply(an)m=anmc Hidegkuti, Powell, 2008Last revised: October 1, 2018

Lecture NotesInteger Exponentspage 15=

4a1(3)b32a2b3! 2 = 2a3b3a2b3 2 applyanam=anm =  2a3(2)b332 = 2a3+2b33
2 = 2a5b6
2apply(ab)n=anbn = (2)2a5
2b6
2apply(an)m=anm = 4a5
2b6
2= 4a10b12applyan=1an = 4a10
1b12 =

4a101
1b12=4a10b1230.

x3y0x5y3 2

Solution:

E= x3y0x5y3 2 y

0= 1andan
am=an+m

= x3+(5)y3! 2 =1x2y3 2 applyab n=anbn = 1x2
2(y3)2apply(ab)n=anbn = (1)2x2
2(y3)2apply(an)m=anmandan=1an = x2(2)(1)2y3(2)=x41y6=x4y631. x3y7x5y3 0 Solution: Any non-zero quantity raised to the power zero is1:So the answer is1.c Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 1632.x1+y1x2y2

Solution: This problem is very different because there are addition and subtraction involved. Because of that, we can not

simply move the expressions with negative exponents. Instead, this will be a problem involving complex fractions.

E=x1+y1x2y2=1x1+1y11x21y2=1x+1y1x21y2bring fractions to the common denominator =

1
yx
y+1
xy
x1
y2x2
y21
x2y2
x2=yxy+xxyy2x2y2x2x2y2=y+xxyy2x2x2y2to divide is to multiply by the reciprocal

= y+xxy
x2y2y2x2cancel outxy = y+x1
xyy2x2=xy(x+y)y2x2factory2x2via the difference of squares theorem, cancel outx+y = xy(x+y)(yx)(y+x)=xyyx33.2a2
2b3a0aba2b2
32a2(2a2b)2ab0

Solution:

E=2a2
2b3a0aba2b2
32a2(2a2b)2ab0a0=b0= 1andxnxm=xn+m = 2a2
2b3a1+(2)b1+(2)
32a2+1(2a2b)2=2a2
2b31a1b1
32a3(2a2b)2apply(xy)n=xnyn = (2)2a2
2b3(1)3a1
3b1
32a3(2)2(a2)2b2apply(xn)m=xnm =

(2)2a2(2)b3(1)3a1(3)b1(3)2a3(2)2a2(2)b2=(2)2a4b3(1)3a3b32a3(2)2a4b2cancel outa4anda3and(2)2

= b3(1)3b32b2applyxnxm=xn+m = (1)3b3+32b2=(1)3b62b2applyxn=1xn = b6b2(1)32applyxnxm=xn+m = b6+21
2=b82=b82c Hidegkuti, Powell, 2008Last revised: October 1, 2018 Lecture NotesInteger Exponentspage 1734. a2b1a
5b7(ab2)3! 2

Solution:

E= a2b1a
5b7(ab2)3!

2 apply(xy)n=xnyn = a2b1
5a5b7(1)3a3(b2)3! 2 apply(xn)m=xnm = a2b1(5)a5b7(1)3a3b2(3)! 2 =a2b5a5b7(1)3a3b6 2 applyxnxm=xn+m = a2+(5)b5(1)3b7+(6)a3! 2 =1
a3b5(1)3b1a3 2 cancel outa3 = 1b5(1)3b1 2 applyan=1an = 1(1)3b5b1! 2 =1(1)b5b1 2 =1b5b1 2 applyxnxm=xnm = b51
2=b4
2apply(xn)m=xnm =b4(2)=b8=1b835.x2
2y3x02yx0y2x2
0yx5(y2x)3(2x1yx3)1

Solution:

E=x2
2y3x02yx0y2x2
0yx5(y2x)3(2x1yx3)1applya0= 1andanam=an+m = x2
2y3yx5(y2x)3(2x1+3y)1apply(ab)n=anbn = x2
2y3yx5(y2)3x3(2x2y)1apply(ab)n=anbnandanam=an+m = x2
2y3yx5+(3)(y2)3(2)1(x2)1y1apply(an)m=anm = x2(2)y3yx2y2(3)21x2(1)y1=x4y321yx2y6x2y1applyanam=an+m = x4y321y1+6+(1)x2+(2)=x4y321y6x0x0= 1andanam=anm = x4y3621=x4y321applyan=1an =

21x4y3=2x4y3

For more documents like this, visit our page at https://teaching.martahidegkuti.com and click on Lecture Notes. E-mail ques-

tions or comments to mhidegkuti@ccc.edu.c Hidegkuti, Powell, 2008Last revised: October 1, 2018