MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS Properties of Real Numbers Some Fundamentals The whole course will be based entirely on the study of sequence of numbers and functions defined on the real line Thus a detail study of the real numbers is important However, such a study will take up quite a bit of time So we shall take a quicker approach by drawing up some essential properties about real numbers, and to deduce some consequences from them (W Rudin, Principles of Mathematical Analysis, or E Landau, Foundations of analysis, 2nd) Proposition (Associative law of addition) If a, b and c are any numbers, then a + (b + c) = (a + b) + c Proposition 2 (Existence of additive identity) There is a number denoted by 0 such that for any number a 0 + a = a + 0 = a Proposition 3 (Existence of additive inverse) For every number a there is a number denoted by a such that a + ( a) = ( a) + a = 0 Proposition 4 (Commutative law for addition) If a, b are any numbers, then a + b = b + a Proposition 5 (Associative law for multiplication) If a, b and c are any numbers, then a (b c) = (a b) c Proposition 6 (Existence of multiplication identity) There is a number denoted by, such that for any number a, a = a = a Moreover, 0 Proposition 7 (Existence of multiplicative inverse) For any number a 0 there is a number denoted by a such that a a = a a = Proposition 8 (Commutative law for multiplication) If a and b are any number, then a b = b a Proposition 9 (Distributive law) If a, b and c are any numbers, then a (b + c) = a b + a c Proposition 0 (Trichotomy law) There is a collection of positive numbers For every number a one and only one of the following follows: (i) a = 0 (ii) a is positive (written a > 0) (iii) a is positive Proposition (Closure under addition) If a > 0 and b > 0, then a + b > 0 Proposition 2 (Closure under multiplication) If a > 0 and b > 0, then a b > 0
2 MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS Definitions a b = a + ( b) a > b means a b > 0 a < b means a b < 0 a b means either a > b or a = b a b means either a < b or a = b From Propositions -2 we may deduce the following consequences: () If x + a = b, then x = b a (x + a) + ( a) = b + ( a) = b a (Definition) x + (a + ( a)) = b a (Proposition ) x + 0 = b a (Proposition 3) x = b a (Proposition 2) (2) If ab = ac and a 0, then b = c a (ab) = a (ac) (Proposition 7) (3) a 0 = 0 for any a (a a)b = (a a)c (Proposition 5) b = c (Proposition 6) b = c a 0 = (0 + 0)a (Proposition 2) = 0a + 0a (Proposition 9) 0a 0a = 0a + (0a 0a) 0 = 0a + 0 0 = 0 a (4) If a b = 0, then either a = 0 or b = 0 If a = 0 and b = 0, then by (3), a b = 0 So there is nothing to prove So we may assume, say, a 0: a (ab) = a 0 (5) ( a)b = (ab) (a a)b = 0 b = 0 b = 0 Therefore ( a)b = (ab) ( a)b + ab = ( a + a)b = 0 b = 0
(6) ( a)( b) = ab MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS 3 ( a)( b) (ab) = ( a)( b) + ( a)b (5) = ( a)( b + b) = a 0 = 0 Therefore ( a)( b) = ab (7) If c < 0, then c > 0 For c < 0 means 0 > c which means 0 c > 0 by definition That is, c > 0 (8) If a < 0 and b < 0, then ab > 0 We have a > 0 and b > 0 from (7) From Proposition 2, ( a)( b) > 0 By (6), ab > 0 Remark a 2 > 0 for any a 0 In particular, = 2 = > 0 If a > 0, then a a > 0 by Proposition 2 If a < 0, then a > 0 and a 2 = ( a)( a) > 0 Definition Let N denotes the set of natural numbers That is, the set {, 2, 3, } Let Z denotes the set of integers That is, the set {, 3, 2,, 0,, 2, 3, } Let Q denotes the set of rational numbers or fractions That is, the set of numbers of the form p q where p and q are integers with no common factors and q 0 Clearly the Z and Q satisfy Proposition -2, and hence their consequences However, we shall show that Q is not complete in the following sense Example If x Q, then x 2 2 Proof Suppose not, let x = m n where m, n Z, n 0, and x2 = 2 Then m 2 n 2 = 2 or m2 = 2n 2 and m is even So we may write m = 2k for some k 0 Hence 2n 2 = m 2 = 4k 2 or n 2 = 2k 2 That is, n is also even This contradicts the fact that m and n have no common factors If we consider a bigger set, R the whole real line then we may say that there exists a number x R such that x 2 = 2 This number is denoted by 2 It is called an irrational number Note that we can merely assume its existence and we are unable to prove it exists at the point Even this pose a problem, since there are many different algebraic equations, so it has many different sorts of irrational numbers Clearly such an approach is unsatisfactory Hence we shall assume ione more property later, and this makes all the differences 2 Inequalities Definition For any number a, we define the absolute value or modulus a of it as follows: { a if a 0 a = a if a 0 Its graph looks like f(x) = x
4 MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS Theorem (I) a b = a b (II) a + b a + b (III) a b a b b a a b Remark We may rewrite (III) as a b a b Proof We distinguish 4 cases: a 0, b 0 2 a 0, b 0 3 a 0, b 0 4 a 0, b 0 Case (I) ab = ab = a b (II) a + b = a + b = a + b Case 4 (I) ab = ab = ( a)( b) = a b (II) a + b = (a + b) = a + ( b) = a + b Case 2 (I) ab = (ab) = a( b) = a b To prove (II) in this case, we must prove that a + b a + b = a b If a + b 0, we need to show a + b = a + b a b That is, b b which is the case If a + b 0, we need to show a + b = a + b a b That is, a b a b That is, a a which is the case again For (III), we need only a = a b + b a b + b That is, a b a b and b = b a + a b a + a That is, b a a b Proposition 3 (Principle of mathematical induction) Let A be any collection (or set) of natural numbers with the properties () A and (2) k + A whenever k A Then A = N Bernoulli s inequality If h + > 0 then ( + h) n + nh for n N Clearly, it is true if n = So suppose it is valid for n That is, suppose ( + h) n + nh ( + h) n+ = ( + h)( + h) n ( + h)( + nh) = + nh + h + nh 2 = + (n + )h + nh 2 > + (n + )h Hence Bernoulli s inequality follows from the principle of induction Theorem 2 Every non-empty set A of natural numbers has a least member Proof Suppose that this is not the case In particular, / A Let B be the set of natural numbers,, n which are not in A Clearly B Suppose the natural number k B, then k + must also belong to B For suppose + k A, then k + would be the least member of A A contradiction Hence k + B Hence B = N from Proposition 3 and A = Remark The empty set is the only set of natural number without a least member We note that Theorem 2 is equivalent to the principle of induction That is, we can deduce Proposition 3 from Theorem 2 We leave this this as an exercise
MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS 5 3 A useful inequality The Greek letter epsilon ɛ is usually need in analysis for a small or arbitrary quantity If a < b + ɛ for any given ɛ (That is, no matter how small) then it follows that a b For suppose it is not true that a b, then a > b must be true Since ɛ is arbitrary, we choose ɛ = a b > 0 Then a < b + ɛ = b + (a b) = a A contradiction Hence a b must be true One can argue similarly that if x a < ɛ is true for any ɛ > 0, where x is a variable and a is fixed Then x = a (Try this!) 4 Least upper bounds Definition A set A of real numbers is said to be bounded above if (there exists) a number a such that x a for all x A Such a number is called an upper bound for A Example If A = {x 0 x } then, 2, 00, 0 are all upper bounds of A Similarly, A = {x 0 x < } has the same upper bounds However, none of the upper bounds actually belongs to A Definition A number a is a least upper bound (or supremum) of A if () a is an upper bound (2) if b is an upper bound then a b Remark Note that if a and b are both least upper bounds for A, then it is easy to see that a = b So we may speak of the least upper bound A set of real numbers is said to be bounded below if there is a number a such that x a for all x a A number a is the greatest lower bound (or infimum) of A if () a is a lower bound (2) if b is a lower bound then a b 5 The least upper bound property of R Proposition 4 If A is a non-empty set of real numbers and A is bounded above, then A has a supremum (sup A) or least upper bound (inf b) We note that Proposition 4 cannot be deduced from Proposition -3 And that it is an assumption only for R For example, this assumption is not true for Q Let A be the set of all real numbers x satisfying x 2 < 2, then there is no rational number q which is an upper bound for A Theorem 3 The set of natural numbers N is not bounded above Proof Suppose that N is bounded above Note that N is not empty Hence there exists a supremum α of N by Proposition 4 That is, n α for all n N That is, n + α for all n N since n + N by Proposition 3 That is, n α for all n N That is, α is an upper bound for N and which is less than α the supremum A contradiction Corollary 4 For any ɛ > 0 there is a natural number n with n < ɛ Proof Suppose this is not true, then there exists an ɛ > 0 such that n ɛ for all n N That is, n ɛ for all n N This contradicts Theorem 3 Theorem 5 There exists a positive real number a such that a 2 = 2 Proof We define S = {s R : s 0 and s 2 < 2} Clearly the set S is non-empty and bounded above Hence it has a supremum α by Proposition 4 We shall show that α 2 < 2 or α 2 > 2 are both impossible
6 MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS First we assume that α 2 < 2 That is, 2 α 2 > 0 Also note that α > Notice that 2 α2 2α+ is a positive number and hence we may find a n N such that This follows from corollary 4 Consider n < 2 α2 2α + (α + n )2 = α 2 + 2 α n + n 2 α2 + (2α + ) n < α 2 + (2 α 2 ) = 2 Hence (α + n ) < 2 and this implies that α + n S, contradicting the fact that alpha is the supremum of S Hence α 2 2 The proof that α 2 > 2 is impossible is left as an exercise Theorem 6 Every interval (0, b) contains a rational numbers Proof Since b a > 0 it follows from Corollary 4 that there exists a rational number k such that k < b a Define A = {n N n k > a} Since A is not empty (why?) and so it must have a least member n 0 say, by Theorem 2 That is, n 0 k > a and n 0 a k Notice that a < n 0 k a + k < b Hence n 0 k is the required rational number Corollary 7 Every interval (a, b) contains an irrational number Proof Simply consider ( a 2, b 2 ) then it contains a rational r by Theorem 6 That is, a 2 < r < b 2 or a < r 2 < b And it is easy to show that r 2 is an irrational 6 Further properties about supremum etc Definition If a set S has a largest element M, we call M the maximum of the set S and write M = max S If S has a smallest element m, we call m the minimum of S and write m = min S Let us recall the example A = {x 0 x } and A = {x 0 x < } Clearly max A = = sup A and min A = 0 = inf A However although min A exists and equals 0, max A doesn t exist But sup A exists and equal Example B = { n : n N} Find max B, min B, sup B, inf B max B = = sup B It is easy to see that there is no min B For suppose min B = n 0 for some n 0 then n 0 + < n 0 We show inf B = 0 First note that inf B 0 For suppose inf B < 0 then 0 > inf B 2 > inf B is a lower bound greater than inf B Hence inf B 0 Suppose inf B > 0 Then by Corollary 4, one can find n N such that < inf B n This contradicts the definition of inf B Hence inf B = 0
MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS 7 Theorem 8 Suppose that S is a non-empty set of real numbers which is bounded above and ξ > 0 Then Proof Let α = sup S So Therefore sup{ξx x S} ξα That is, Let β = supξx then That is, sup ξ x = ξ sup x x α for all x S That is, ξx ξα for all x S sup ξx ξα = ξ sup x ξx β for all x S ξ (ξx) β ξ for all x S(ξ > 0) x ξ β Hence result follows α = sup x ξ β ξ sup x sup ξx We list some simple results (i) ξ > 0, and S is bounded above sup ξs = ξ sup S, ξs = ξ inf S (ii) ξ < 0, sup ξs = ξ inf S, inf ξs = ξ sup S In particular, if S = { x x S} then sup( S) = inf S and inf( S) = sup S (iii) Suppose both A and B are bounded above, then where A + B = {a + b a A, b B} sup A + sup B = sup(a + B), 7 Countable and uncountable sets Some definitions Two sets A and B are said to be equivalent written as A B, if there exists a one-to-one correspondence f : A B (that is, a bijection) A set A R is finite if there is a n N such that A {, 2,, n} A set which is not finite is called infinite A set A R is called countably infinite (or denumerable) if A N A set is called countable if it is either finite or countably infinite A set is called uncountable if it is not countable Theorem 9 If f : A B is an injection and B is countable then A is also countable
8 MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS Proof We may assume that A is infinite, otherwise there is nothing to prove Since A f(a) and hence f(a) B is also infinite This implies that B is countably infinite (that is, not finite) Let ϕ : N B be a bijection and ϕ(n) = b n That is, we put B = {b, b 2, } Now let k be the first integer such that b k f(a); let k 2 k be the next integer so that b k2 f(a), etc Then f(a) = {b k, b k2, b k3, } Now define g : f(a) N by g(b ki ) = i which is clearly a bijection The composed mapping g f : A N shows that A N as required Corollary 0 Every subset of a countable set is countable Theorem If A and B are countably infinite then the Cartesian product A B is countably infinite Proof We may write A = {a, a 2, a 3, } and B = {b, b 2, b 3, } We write A B in the following way (a, b ) (a, b 2 ) (a, b 3 ) (a 2, b ) (a 2, b 2 ) (a 2, b 3 ) (a 3, b ) (a 3, b 2 ) (a 3, b 3 ) Define ϕ : A B N by φ(a, b ) =, φ(a, b 2 ) = 2, φ(a 2, b ) = 3, φ(a, b 3 ) = 4, φ(a 2, b 2 ) = 5, φ(a 3, b ) = 6, φ(a, b 4 ) = 7, Theorem 2 The set Q of all rational numbers is countably infinite Theorem 3 For any a, b R, the interval (a, b) is uncountable Proof It is sufficient to consider the interval (0, ) We assume on the contrary that the set is countable Recall that any number in (0, ) can be written as 0a a 2 a 3 a 4 = a 0 + a 2 0 2 + a 3 0 3 + where a, a 2, a 3, are digits taken from 0 to 9 So we may list (0, ) as {a, x 2, x 3, } such that x = a a 2 a 3 a n x 2 = a 2 a 22 a 23 a 2n x 3 = a 3 a 32 a 33 a 3n x n = a n a n2 a n3 a nn So any number in (0, ) can be found from the above list We now defined a number as follows { 7 a ii 7 x = b b 2 b 3 b n where b i = 8 a ii = 7 Then clearly the x differs from any number of {a, x 2, } by at least one digit That is, x is not on the list A contradiction