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Unformatted text preview: A.1 Integers and exgonents Definition. A set of real numbers is called an inductive (a) The number 1 is in the set.
(b) For every x in the set, the number x + 1 is in the set also. The set R+ of positive real numbers is an example of an
inductive set. [The number 1 is in R+ because 1 > 0. And
if x is in R+ (so that x > 0), then x + l is in R+ (since x + 1 > 1 > 0).] Definition. A real number that belongs to £1551
inductive set is called a Rositive integer; such a number is
necessarily positive because R+ is an inductive set. Let P denote the set of positive integers. We prove some basic properties of this set. ~ Theore; ;.. Evgrg element 2: P ;_ greater than 2; egual Egggi, We shall show that the set A of all real
nunbers greater than or equal to l is inductive. It then
follows that every positive integer belongs to this set. The nulber 1 belongs to the set A, since 1 2 1. Suppose. x belongs to the set A. Then x 2 1; it follows
that x + 1 2 l + 1 > 1, so that x + 1 belongs to the set A. Thus A is inductive. a A.2 Theorem 2. 1 is in P. .____——.. Proof. 1 belongs to every inductive set (by definition of "inductive.") Hence 1 belongs to P (by definition of P). 0
Theorem 3. ;_ x ;_ ;_ P, ﬁg 1; x + 1.
Proof. Suppose that x is a given element of P. Let I be an arbitrary inductive set. Then x is in I (by
definition of P). Hence x + 1 is in I (by definition of
"inductive"). Since I is arbitrary. x + l is in I for
every inductive set I. We conclude that x + 1 is in P. (by
definition of P). 0 Theorem 4 (Principle 9: induction). _gt 8 _g a is; 31
positive integers. I: 1 ;_ in S, 23g,;_ £2; giggx x in
S, x + 1 is gig; in S, the; necessarilx S contains giggx
positive integer. giggi. S is inductive, by hypothesis. Therefore every
‘positive integer is in S, by definition of P. in Now we show that P is closed under addition and multiplication.
W. Li’. a m h min. P. agLsgai’b.
' {£221. Let a be a fixed positive integer. Then we letv
S be the set of all positive integers b for which a + b is a positive integer. We shall show that S contains all AI3 positive integers; then the theorem is proved. We use the principle of induction. The number 1 is in S, because a + l is a positive ' integer (by Theorem 3). Given an element b in S, we show that b + l is in S. Now a + b is a positive integer by hypothesis; hence (a+b) + l is a positive integer by Theorem 3. Thus a + (b+1) is'a positive integer, so b + 1 belongs
to S, by definition of 8. Thus 3 is inductive. a
Theorem 6. Li a and b are ;_ P, so i a . b. The proof is left as an exercise. Definition. A number x 'is called an integer if it is O, or is a positive integer, or is the negative of a positive integer. It is easy to see that the negative of any integer is an integer, since —(—a) = a and 0 = 0. Let 2 denote the set of integers. We now show that Z is closed under addition, multiplication, and subtraction.
Closure under multiplication is easy, so we leave the proof as an exercise: difficult: Theorel 8. L_ a and b are in Z, s are a>+ b an a  b. A.4 Step 1. We show that the theorem is true in the case
where a is a positive integer and b = 1. That is, if a is
a positive integer, we show that a + 1 and a  1 are
[email protected] That a + l is an integer (in fact, a positive
integer) has already been proved. We prove that a — l is an
integer, by induction on a. It is true if a = 1, since
a  1 = 0 if a = 1. Supposing it true for a, we prove it true for a + 1. That is, we show (3+1)  1 is an integer. But that is trivial, since (n+1) — l s, which is an integer
by hypothesis (in fact, a positive integer). Step 2. We show the theorem is true if a is any
integer and b = 1. We consider three cases. If a is a positive integer, this result follows from Step 1. If a = O, the result is immediate, since
0 + l = 1 and 0  1 = —1. Finally. suppose a = —c, where c is a positive integer. Then A.5_ Both c  l and c + l are integers, by Step 1; then a + l and a  l are also integers. Step 3. We show the theorem is true if a is any integer and b is a positive integer.v We proceed'by induction on b, holding a fixed. We know the theorem holds if Ah 2 1, by Step 2. Supposing it holds for b, we show it holds for b + 1. That is, we show that a + (b+l) and a  (b+l) are integers. Now a + (b+1) = (a+b) + l, a  (b+l) = (ab) — 1. Both a + b and a  b are integers, by the induction hypothesis; then Step 2 applies to show that (a+b) + l and
(ab)  l are integers.
Step 4. The theoren is true in general. Let a be any integer. The case where b is a positive integer was treated  in Step 3, and the case where b = 0 is trivial. Consider finally the case where b = d, where d is a positive integerl Than Step 3 applies to show that both a  d and a + d are
integers. in Now we prove the ”obvious” fact that if n is an
integer, then n + 1 is the "next" integer after n: Theorem 9. I n i i Z gag n < a < n+1, then a i not i 2. Proof. From the hypothesis of the theorem. it follows that
0 < an < 1. If a were in Z, then a _ n would be an integer, by the
preceding theorem. But 1 is the smallest positive integer, by
Theorem 1. Therefore a is not in Z. u ‘
Now we define integral exponents.l
Definition. Let a be any real nunber. We define 8“,
when. n is a positive integer, by induction, as follows. We .define and supposing an is defined, we define A ._7 Then an is defined for every positive integer n. The number n in this expression is called the exponent, and the number a
is called the £333. pExponents satisfy three basic laws, which are stated in
the following three theorems. They are called the léﬂi g:
exponents. Theorem 10. an  am = an+m‘ Egggi. Let a and n be fixed. We prove the theorem "by induction on n." The theorem is true for n = 1, since an  81 = an . a = an+1 by definition. Suppose it is true for n;' we show it is true for n + 1. .It follows that it holds for all n. We have an  an+1 = on ' (ana) by definition,
= (ans.)  a by associativity of lultiplication,
_ n+l ' . . .
 (a )  a by the induction hypothesis, ; a(n+')*1 by definition, n+(l+1) = s by associativity of addition. Thus the theorel is proved for I + l, as desired. 0 A3 Similar proofs hold for the following two theorems. whose proofs are left as exercises: Theorem 11. (an)m = an”. O
Theorem 12. an bn = (ab)“. U Now we define negative exponents. Definition. Let a be a real number different from
zero. We define zero and negative exponents by the rules:
a0 = l,
~n n . . . . .
a = l/(a ) if n is a positive 1nteger. Theorem 13. 1g; ”lag; g; exponents" gal; gggg n and m
age arbitrary inte e s, pgovided a and b age gag—£353. The proof is left as an exercise. Later on, (in Section G) we shall extend this definition
to define "rational exponents”; that is, we shall define. ar
when .a ‘is positive and r is rational. Still later (in
Section M), we shall extend the definition still further to
define ax when a is positive and x is an arbitrary real nunber.~ In each of these cases. the sane three laws of exponents will hold. Exercises A.9 Prove Theorems 6 and 7. Prove Theorems 11 and 12.
Show that if a set A of integers is bounded abOve,
then A has a largest element. [Hintz Use the
least upper bound axiom.)
Let F be the set of all real numbers of the form
a + bJZ, where a and b are rational. Show that
F is closed under addition, subtraction, multiplica—
tion, and division. Conclude that F is an "ordered
field", that is, that F satisfies Axions 1  9.
Show that F does not contain 43.
Let n and m be positive integers; let a and b
be non—zero real nunbersa Let p be any integer.
Given that the laws of exponents hold for positive
integral exponents, prove them for arbitrary integral
exponents as follows:
(a) Show ans“ = ann' in the three cases n — n > 0 and n  n = 0 and .n  n < 0. n:. o p 3 (b) Show a—na a , and a a (c) Show (an)" an = (sn)'. (d) Shh" (3n) = 8a.. and (8°)p = (ap)o = a°. (e) Show anbn = (ab)n. and sobo = (ab)°. A.10 6. Let a and h be real numbers; let m be a positive integer. Show by induction that if a and
a + h are positive, then (a+h),m 2 aIn + mam—1h [Note: Be explicit about where you use the fact that
a and a + h are positive. Note that h is not
assumed to be positive.) We shall ﬁse this result later on. ...
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 Spring '09
 BRUBAKER

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