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NUMBERS AND SEQUENCES
1. THE NATURAL NUMBERS
The positive integers 1, 2, 3, ... are called natural numbers. Since we intend to do things rigorously, we cannot be satisfied with our everyday familiarity with these numbers, and we should try to axiomatize their properties. Let us first write down five statements concerning the natural numbers that we feel should be true:
(I) 1 is a natural number.
(II) To every natural number n there is associated in a unique way another natural number n' called the successor of n.
(III) 1 is not a successor of any natural number.
(IV) If two natural numbers have the same successor, then they are equal.
(V) Let M be a subset of the natural numbers such that: (i) 1 is in M, and (ii) if a natural number is in M, then its successor also is in M. Then M coincides with the set of all the natural numbers.
From now on we consider the statements (I)–(V) to be axioms. They are called the Peano axioms. The natural numbers will be the objects occurring in the Peano axioms. Axiom (V) is called the principle of mathematical induction.
We denote the successor of 1 by 2, the successor of 2 by 3, and so on. Note that 2 ≠ 1. Indeed, if 2 = 1, then 1 is the successor of 1, thus contradicting (III). Note next that 3 ≠ 2. Indeed, if 3 = 2 then, by (IV), 2 = 1, which is false. In general, one can show that all the numbers obtained by taking the successors of 1 any number of times are all different. The proof of this statement, which we shall not give here, is based on induction, that is, on Axiom (V).
We would like to state Axiom (V) in a form more suitable for application:
(V') Let P(n) be a property regarding the natural number n, for any n. Suppose that (i) P(1) is true, and (ii) if P(n) is true, P(n') also is true. Then P(n) is true for all n.
If we define M to be the set of all natural numbers for which P(n) is true, then (V') follows from (V). If, on the other hand, we define P(n) to be the property that n belongs to M, then (V) follows from (V'). Thus (V) and (V') are equivalent axioms.
The Peano axioms give us objects with which to work. We now proceed to define operations on these objects. There are two operations that we consider: addition (+) and multiplication (·). To any given pair of natural numbers each of these operations corresponds another natural number. The precise definition of this correspondence is given in the following theorem.
THEOREM 1. There exist unique operations "+" and "." with the following properties:
n + 1 = n', n + m' = (n + m)', (1)
n · n, n · m' = n · m + n. (2)
The proof will not be given here. We shall often write mn instead of m · n.
THEOREM 2. The following properties are true for all natural numbers m, n, k:
m + n = n + m, mn = nm (the commutative laws) (3)
(m + n) + k = m + (n + k), (mn)k = m(nk) (the associative laws), (4)
m(n + k) = mn + mk, (n +k)m = nm +km (the distributive laws). (5)
The proof of Theorem 2 can be given by induction; it is based on the properties (1) and (2).
We state, without proof, another theorem, known as the trichotomy law:
THEOREM 3. Given any natural numbers m and n, one and only one of the following possibilities occurs:
(i) m = n.
(ii) m = n + x for some natural number x.
(iii) n = m + y for some natural number y.
If (ii) holds, we write m > n or n < m, and we say that m is larger or greater than n and that n is smaller or less than m. If either (i) or (ii) holds, we write m n or n m, and say that m is larger or equal to n and that n is less than or equal to m.
1. If n > m, then n + k > m + k.
2. If n > m, m > k, then n > k.
3. If n + k m + k, then n m.
4. (n + 1)2 = n2 + 2n + 1, where n2 = n · n.
5. (n + 1)3 = n3 + 3n2 + 3n + 1, where n3 = n2 · n.
6. Prove the associative law (m + n) + k = m + (n + k), by induction on k.
7. Prove the distributive law (m + n)k = mk + nk, by induction on n.
8. If m > n, then mk > nk. Conversely, if mk > nk, then m > n.
2. THE RATIONAL NUMBERS
To every natural number n we correspond a new symbol (-n), or -n, called minus n. We also introduce the symbol 0 called zero. Next we define the operation of addition (-n) + m as follows:
(i) If n = m, then (-n) + m = 0.
(ii) If n = m + x, x a natural number, then (-n) + m = x
(iii) If m = n + y, y a natural number, then (-n) + m = y.
Note, by the trichotomy law, that one and only one of Cases (i), (ii), or (iii) occurs.
Next we define: m + (-n) = (-n) + m, (-n) + (-m) = -(n + m), 0 + m = m + 0 = m, 0 + (-n) = (-n) + 0 = -n, and 0 + 0 = 0.
Multiplication is defined as follows:
m · (-n) = (-n) · m = -(nm), (-m) · (-n) = mn,
0 · (-n) = (-n) · 0 = 0 · m = m · 0 = 0 · 0 = 0
The symbols -n are called the negative integers. The integers consist of the natural numbers (also called the positive integers), the negative integers, and 0. We state without proof:
THEOREM 1. For any integers m, n, k, Rules(3), (4), and (5)of Section 1 are true.
If a is a negative integer -n, then we define a as n. We also define -0 to be 0. Then, for any integer a, a + (-a) = (-a) + a = 0.
Given any two integers a, b, there is a solution x of the equation
a + x = b, x integer. (1)
Indeed, x = (-a) + b is such a solution, since, by Theorem 1, a + [(-a) + b] = [a + (-a)] + b = 0 + b = b. If y is another solution, then a + x = a + y. Hence (-a) + (a + x) = (-a) + (a + y). Using Theorem 1, we get x = y. Thus Equation (1) has a unique solution.
Notation. We write b + (-a) = b - a = -a + b.
If the solution x of (1) is a positive integer, we write b >a or a< b, and we say that b is larger (or greater) than a and that a is smaller (or less) than b. If x is negative, then the equation b + y = a has the positive solution y = -x. Hence a > b.
We now shall introduce fractions. These are symbols that we write in the form a/b or a/b, where a and b are any integers, and b ≠ 0. These symbols are subject to the following definitions:
a/b = c/d if and only if ad = bc (equality), (2)
a/b + c/d = ad + bc/bd (addition), (3)
a/b · c/d = ac/bd (multiplication). (4)
Note that if a/b = c/d and c/d = e/f, then a/b = e/f.
The last two definitions are acceptable only if we can show that b ≠ 0, d ≠ 0 imply that bd ≠ 0. This, however, can be checked by considering the four possibilities: b positive or negative, d positive or negative.
Definitions (3) and (4) would be most unnatural if it turned out that it is possible to have a/b = a'/b', c/d = c'/d' but a/b + c/d is not equal to a'/b' + c'/d' [or (a/b) · (c/d) is not equal to (a'/b') · (c'/d')]. The following theorem shows that this cannot occur.
THEOREM 2. If a'/b' = a/b and c'/d' = c/d, then
a'/b' + c'/d' = a/b + c/d, (5)
a'/b' · c'/d' = a/b · c/d. (6)
Proof. To prove (5) we have to show that
a'd'/b'd' + b'c'/b'd' = ad + bc/bd
a'd'bd + b'c'bd = adb'd' + bcb'd'.
But this follows by multiplying the relation a'b = ab' by dd', the relation c'd = cd' by bb', and adding the resulting equalities. To prove (6) we have to show that
a'c'/b'd' = ac/bd,
a'c'bd = acb'd'.
But this follows from
a'c'bd = (a'b)(c'd) = (ab')(cd') = acb'd'.
A fraction a/b is called negative if either a > 0, b< 0 or a< 0, b > 0. It is called positive if either a > 0, b > 0 or a< 0, b< 0. It is called zero if a = 0. It is easily seen that if a fraction c/d is equal to a fraction a/b, then they are either both positive, or both negative, or both zero.
THEOREM 3. The following properties hold for any fractions a/b, c/d, e/f:
a/b + c/d = c/d + a/b, a/b · c/d = c/d · a/b (the commutative laws), (7)
(a/b + c/d) + e/f = a/b + (c/d + e/f) (the associative laws), (8)
(a/b · c/d) · e/f = a/b · (c/d · e/f) (the distributive laws). (9)
Consider the equation
a/b + x = c/d, x fraction. (10)
It has a solution x = (bc - ad)/bd. If x is positive, then we write a/b< c/d or c/d >a/b, and say that c/d is larger than a/b and that a/b is less than c/d. If x is negative, then the equation
c/d + y = a/b
has the positive solution y = -x, so that a/b >c/d. Note that c/d is positive (negative) if it is larger (smaller) than zero.
The definition of fractions is very intuitive and is, in fact, suggested by our experience with quotients of integers. There is, however, one disturbing feeling about the concept of fractions, due to the fact that fractions having different forms may be equal to each other. This makes it impossible to speak of the zero fraction (since there are many fractions 0/b taking the role of zero). We also cannot assert that Equation (10) has a unique solution. Similarly, the equation
a/b · x = c/d, x fraction (where a ≠ 0) (11)
does not have a unique solution.
To overcome this unpleasant situation, we introduce the concept of a rational number.
DEFINITION. A rational number (a, b) (where a and b are integers, and b ≠ 0) is the class of all the fractions e/f that are equal to a/b.
Addition and multiplication for rational numbers are given by
(a, b) + (c, d) = (ab + cd, bd), (12)
(a, b) · (c, d) = (ac, bd). (13)
Theorem 2 shows that these definitions are meaningful. Theorem 3 implies:
THEOREM 4. Rational numbers satisfy the commutative laws for addition and multiplication, the associative laws for addition and multiplication, and the distributive laws.
Let us write the analog of Equation (11) for rational numbers:
(a, b) · x = (c, d) x rational (where a ≠ 0). (14)
This equation has a unique solution x = (bc, ad). We write this solution also in the form (c, d)/(a, b) or (c, d)(a, b)-1.
Let us write the analog of (10) for rational numbers:
(a, b) + x = (c, d), x rational. (15)
This equation also has a unique solution: x = (bc - ad, bd). We write it also as (c, d) - (a, b).
Note that there is a one-to-one correspondence between the integers a and the rational numbers (a, 1). This correspondence a -> (a, 1) is preserved under addition and multiplication. Indeed, this follows from the relations
(a, 1) + (b, 1) = (a + b, 1). (a, 1) · (b, 1) = (ab, 1).
Hence, if we write an integer a in the form (a, 1), we see that the integers can be identified with a subset of the rational numbers.
In what follows we shall adopt the definition of rational numbers as classes of fractions a/b. However, for brevity, we shall write the rational numbers (a, b) usually in the form a/b. When we write a/b = c/d, we mean that (a, b) = (c, d), that is, ad = bc. The rational numbers b/1 will also be written, briefly, as b. In particular, the rational number zero will be denoted by 0.
Excerpted from ADVANCED CALCULUS by Avner Friedman. Copyright © 1999 Avner Friedman. Excerpted by permission of Dover Publications, Inc..
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