POSITIVE INTEGERS, INTEGERS AND RATIONAL NUMBERS OBTAINED FROM THE AXIOMS OF THE REAL NUMBER SYSTEM


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1 MAT 1011 TECHNICAL ENGLISH I Dokuz Eylül University Faculty of Science Department of Mathematics Instructor: Engin Mermut Course assistant: Zübeyir Türkoğlu web: POSITIVE INTEGERS, INTEGERS AND RATIONAL NUMBERS OBTAINED FROM THE AXIOMS OF THE REAL NUMBER SYSTEM Remember that we have assumed a set of axioms for the real number system. Starting from these axioms, one easily obtains the usual algebraic propertiesof R and as we have done in the lectures, we define the set Z + of all positive integers as the smallest inductive subset of R and we obtain the induction principle. Then we define the set of integers and rational numbers and obtain their usual algebraic properties which we are very familiar. Let s see some details for this part following the related parts from the books: Apostol, T. M. Calculus, Volume I: Onevariable calculus, with an introduction to linear algebra. Second edition. John Wiley & Sons, and Stromberg, K. R. Introduction to Classical Real Analysis. Wadsworth, A subset I of R is said to be an inductive subset of R if (i) 1 I, and (ii) For every x R, if x I, then x + 1 I. We define Z + as the intersection of all inductive subsets of R. Then Z + is also an inductive subset of R and it is the smallest inductive subset of R in the sense that Z + I for every inductive subset I of R. This then gives the induction principle because if S is a subset of Z + and S is an inductive subset of R, then S = Z + must hold. Induction principle: Suppose that S is a subset of Z + and it satisfies the following two properties: (i) 1 S, and (ii) for every n Z +, if n S, then n + 1 S. Then S = Z + must hold. This gives us: The method of proof by induction: Suppose that P n is a statement for every n Z +. Suppose that we have shown that: (i) The first statement P 1 is true. (ii) For every n Z +, if the statement P n is true, then the statement P n+1 is true. Then we obtain that the statement P n is true for all n Z +. This is obtained just by applying the above induction principle to the subset S = {n Z + the statement P n is true} of Z +. See the lecture notes for the proofs of the above results. Using the completeness of R (=least upper bound property of R), we also showed in the lectures that Z + is not bounded from above, that is, for every real number x, x cannot an upper bound for Z + and so there exists n Z + such that n > x. This also gives us the important: Archimedean Property of R: For all real numbers x and y, if x > 0, then there exists n Z + such that nx > y. A frequently used corollary of this is that for every real number ɛ > 0, there exits n Z + such that 0 < 1 n < ɛ. A real number x is said to be an integer if x = 0 or x Z + or x Z +. The set of all integers is denoted by Z. By this definition of integers, we have that Z = Z + {0} { n n Z + }. A real number x is said to be a rational number if x = m n = mn 1 for some m, n Z where n 0. The set of all rational numbers is denoted by Q: { m } Q = n m, n Zm, n Z and n 0. A real number that is not a rational number is said to be an irrational number. 1
2 Remember that using the order properties of R, it is shown that x 2 > 0 for every real number x 0 (see Homework 1 for such arguments). This then implies that 1 2 > 0 since 1 0. So 1 = 1 2 > 0. For every n Z +, if n > 0, then n + 1 > = 1 > 0. This argument is the proof by induction that n > 0 for every n Z +. Then by properties of inequalities in R, n < 0 for every n Z +. Thus in the set Z = Z + {0} { n n Z + } of all integers, the positive ones are just the elements of Z + and the negative ones are the elements of Z = { n n Z + } which is called the set of all negative integers. Remember that by the trichotomy law of the order relation in R, we know that for n Z, one and only one of the following is true: n > 0 or n = 0 or n < 0. Theorem. Some properties of the set Z + of positive integers. For all positive integers n, m and for every real number x, we have: (i) n 1. So there exists no integer strictly between 0 and 1. (ii) If n > 1, then n 1 Z +. (iii) If x > 0 and x + 1 Z +, then x Z +. (iv) If x > 0 and x + n Z +, then x Z +. (v) If m > n, then m n Z +. (vi) If n 1 < x < n, then x Z +. Thus there exists no positive integer strictly between n 1 and n. Proof: (i) I = [1, ) = {x R x 1} is an inductive subset of R because 1 I as 1 1 clearly, and for every x R, if x I, then x 1 by definition of I and so x = 2 1 which implies x + 1 I. Since Z + is by its definition the intersection of all inductive subsets of R, we must have Z + I. Thus for every n Z +, n I must hold which implies by the definition of I that n 1. Suppose for the contrary that there exists an integer strictly between 0 and 1, that is, there exists x Z such that 0 < x < 1. Then x is a positive integer that is strictly less than 1. But we have just shown that n 1 for every positive integer n. This contradiction shows that there exists no integer strictly between 0 and 1. (ii) Let S = {1} {n Z + n 1 Z + }. It suffices to prove by induction principle that S = Z +. Firstly 1 S clearly by the definition of S. For every n Z +, if n S, then (n + 1) 1 = n S Z +, that is (n + 1) 1 Z + which then implies by the definition of the set S that the positive integer n + 1 S (note that since Z + is an inductive subset for every n Z +, we must have n + 1 Z + ). Thus by the induction principle S = Z +, that is, S = {1} {n Z + n 1 Z + } = Z +. So if n Z + = S and n > 1, then by the definition of S, since n 1, we must have n 1 Z + as required. (iii) If x R, x > 0 and x + 1 Z +, then for n = x + 1 we have n Z + and n = x + 1 > = 1 as x > 0. Thus by (ii), x = (x + 1) 1 = n 1 Z +. (iv) We shall prove by the method of proof by induction that for every n Z +, the following statement holds: for every real number x, if x > 0 and x + n Z +, then x Z +. For n = 1, this is just (iii). Suppose that this statement is true for n Z + (this is the so called inductive hypothesis). Let s show that it is true for n + 1 also. For that let x > 0 be a real number such that x + (n + 1) Z +. Then for the real number y = x+1, we have y = x+1 > 0+1 = 1 > 0 and y +n = x+1+n = x+(n+1) Z +. That is for the real number y > 0, we have y + n Z + and so by the inductive hypothesis, we obtain y Z +, that is, x + 1 Z +. But then by (iii), x Z + as required. (v) Let m and n be positive integers such that m > n. Let x = m n. Then the real number x = m n > 0, and x + n = (m n) + n = m Z +. By (iv), x Z + must hold, that is, m n Z +. (vi) Suppose for the contrary that there exists n Z + and x Z + such that n 1 < x < n. Then n < x+1 < n+1 and 0 < (x+1) n < 1. Here x+1 Z + since x Z + and n Z +. So for m = x+1, we have that m and n are positive integers satisfying n < m and 0 < m n < 1. By (v), we must have m n Z +. But by (i), there exists no positive integer strictly between 0 and 1. This contradiction shows that for every n Z +, there exists no positive integer strictly between n 1 and n. 2
3 Wellordering principle: Every nonempty subset of Z + has a smallest element. For the proof of this very often used principle, see page 37 from the book Apostol, T. M. Calculus, Volume I: Onevariable calculus, with an introduction to linear algebra. Second edition. John Wiley & Sons, The wellordering principle for Z + means Z + is a wellordered set with respect to its order relation. Remember that a partially ordered set A is said to be ewllordered if every nonempty subset of A has a smallest element. Theorem. The set Z + of positive integers is closed under addition and multiplication: m + n Z + and mn Z + for all m, n Z +. Proof: (i) Fix m Z +. Let S = {n Z + m+n Z + }. Using the induction principle, we shall show that S = Z + which will then imply that m + n Z + for every n Z + = S by the definition of the set S. Firstly 1 S because Z + is an inductive subset of R and so m + 1 Z + since m Z +. Suppose that n S. Then m + n Z + by the definition of the set S. This implies that m + (n + 1) = (m + n) + 1 Z + since Z + is an inductive subset of R. So n + 1 S by the definition of the set S. Thus by the induction principle S = Z +. This means that for every n Z + = S, we have m + n Z +. This holds for every m Z + since we have taken an arbitrary m Z + in this proof. Thus we have shown that Z + is closed under addition. (ii) Fix m Z +. Let S = {n Z + mn Z + }. Using the induction principle, we shall show that S = Z + which will then imply that mn Z + for every n Z + = S by the definition of the set S. Firstly 1 S because m1 = m Z +. Suppose that n S. Then mn Z + by the definition of the set S. This implies that m(n + 1) = mn + m Z + because mn Z + and m Z +, and we have shown in part (i) that the sum of positive integers is a positive integer. Thus n + 1 S by the definition of the set S. Thus by the induction principle S = Z +. This means that for every n Z + = S, we have mn Z +. This holds for every m Z + since we have taken an arbitrary m Z + in this proof. Thus we have shown that Z + is closed under multiplication. Theorem. The set Z of integers is closed under addition, multiplication, additive inverses and subtraction: m + n Z, mn Z, n Z and m n Z, for all m, n Z. Proof: For all m, n Z, we shall prove that (a) m + n Z and mn Z, (b) n Z, (c) m n Z. We shall use the above result that Z + is closed under addition and multiplication and we shall argue by considering the cases according to the signs of m and n. Note that by definition of Z, to prove that a real number x is in Z, we observe that either x = 0 or x Z + or x Z + and so for this last case y = x Z + and x = y. (a) If one of m or n is 0, the result is clear since mn = 0 and m + n equals m or n. Suppose m 0 and n 0. Then there are four cases for their signs: (i) m > 0 and n > 0: In this case, m and n are positive integers and so m+n and mn are also positive integers by the above theorem stating the closure of Z + under addition and multiplication. So it is true that m + n Z and mn Z in this case. (ii) m > 0 and n < 0: In this case, let k = n. Then m and k are positive integers and n = k. So mk Z + by the above theorem stating the closure of Z + under addition and multiplication. Then mn = m( k) = (mn) Z. We have m + n = m k. By the trichotomy property of the order relation in R, there are three cases: m > k or m = k or k < m. If m > k, then m k Z + by part (v) of the theorem above that gives some properties of the set Z + of positive integers. If m = k, then m k = 0. If m < k, then k m Z + by part (v) of the theorem above that gives some properties of the set Z + of positive integers, and so m k = (k m) Z. Thus in any case m + n = m k Z. So it is true that m + n Z and mn Z in this case. (iii) m < 0 and n > 0: This is just the previous case with the roles of m and n interchanged (or by commutativity m + n = n + m and mn = nm and use the previous case). So it is true that m + n Z and mn Z in this case. 3
4 (iv) m < 0 and n < 0: In this case, let t = m and k = n. Then t and k are positive integers, m = t and n = k. So t + k Z + and tk Z + by the above theorem stating the closure of Z + under addition and multiplication. Then mn = ( t)( k) = tk Z + and m + n = t k = (t + k) Z since t + k Z +. So it is true that m + n Z and mn Z in this case. (b) If n Z, then n = 0 or n Z + or n Z +. If n = 0, then n = 0 Z. If n Z +, then n Z by the definition of Z. If n Z +, then n Z by the definition of Z (since Z contains Z + ). Thus in any case n Z for every n Z. (c) Let m, n Z. By (b), n Z. Then m n = m + ( n) Z by (a) since m and n are integers. Theorem. The greatest integer function. Let x be a real number. Then there exists a unique integer n such that n x < n + 1. Proof: Let x R. Let A = {n Z n x}. Firstly A because by the Archimedean property of R, there exists m Z + such that m > x, and so m < x which implies that the integer m A. By the definition of A, x is an upper bound of A. Thus A is a nonempty subset of R that is bounded from above. So by the least upper bound property of R, A has a least upper bound in R, say L = sup(a). Since L = sup(a) is the least upper bound of A and L 1 < L, we must have that L 1 is not an upper bound of A. Thus there exists n 0 A such that n 0 > L 1. Since n 0 A and L = sup(a) is an upper bound of A, we must also have n 0 L. Thus L 1 < n 0 L, and so L < n Our claim is that n n 0 for every n A. Suppose for the conrary that n > n 0 for some n A. Since n A and L = sup(a) is an upper bound of A, we must also have n L. Thus n 0 < n L < n and so n 0 < n < n But then 0 < n n 0 < 1 and n n 0 Z since n and n 0 are in Z and Z is closed under subtraction by the above theorem. Thus n n 0 is an integer strictly between 0 and 1 which is not possible by (i) of the theorem above that gives some properties of the set Z + of positive integers. This contradiction shows that we must have n n 0 for every n A. Thus n 0 is also an upper bound of A. Since L = sup(a) is the least upper bound of A, we must have L n 0. But we have also shown above that n 0 L. Thus we obtain L = n 0 A. So L = sup(a) is in A and thus L is the greatest element of the set A. Since L A, we have L x by the definition of the set A. Since L + 1 is a positive integer that is strictly greater than L and since L is the greatest element of the set A, we must have that L + 1 is not in A. Then by the definition of the set A, L + 1 > x must hold. As a result, we found an integer L such that L x < L + 1. Suppose that n Z also satisfies n x < n + 1. By the trichotomy law for the order relation in R, one and only one of the following holds: n < L or L < n or n = L. If n < L, then n < L x < n + 1 and thus 0 < L n < 1. Since L and n are integers, L n is also an integer. Thus L n is an integer strictly between 0 and 1 which is not possible by (i) of the theorem above that gives some properties of the set Z + of positive integers. This shows that we cannot have n < L. Just by changing the roles of n and L, we see that we cannot have L < n also. Thus the only remaining possibility is n = L. As a result, we have shown that there exists a unique integer n that satisfies n x < n + 1. For every real number x, the unique integer n such that n x < n + 1 is called the greatest integer less than or equal to x and is denoted by x or x or in some books just by [x]. The better notation x is read floor of x, it denotes the greatest integer less than or equal to x. Similarly, we denote by x (read ceiling of x ) to denote the smallest integer greater than or equal to x. So for a real number x, we denote by x and x the unique integers that satisfy the following inequalities: x x < x + 1 and x 1 < x x As a further exercise, prove the existence of the ceiling x, that is, for every real number x, prove that there exists a unique integer n such that n 1 < x n. Theorem. Q is an ordered field but not complete. If x and y are rational numbers, then so are x, x + y and xy. If x 0 is a rational number, then so is x 1. Thus Q is a subfield of R, that is, it also satisfies the field axioms for the operations addition and multiplication that it inherits from R. Q also satisfies the order axioms under the order that it inherites from R. Thus Q satisfies the field and order axioms like R, but unlike R it does not satisfy the completeness property(=least upper bound property). 4
5 Proof: Using the properties of integers, we show that Q is a subfield of R. If x = a b and y = c d are rational numbers, where a, b, c, d Z, b 0 and d 0, then x = a ad + bc, x + y = and xy = ac b bd bd are rational numbers since by the properties of integers a, ad + bc, ac, bd are integers and bd 0. If also x = a b 0, then a 0 and so x 1 = b is also a rational number. a Verify that Q satisfies the field axioms for the operations addition and multiplication that it inherits from R and Q also satisfies the order axioms under the order that it inherites from R. See also Homework 1 to understand more details for these. Remember that it has been shown in the lectures that the subset A = Q {0} {r Q r > 0 and r 2 < 2} has no least upper bound in Q because the set of all its upper bounds in Q is the set B = {b Q b > 0 and b 2 > 2} and B has no smallest element. See Question 21 in Homework 1 and lecture notes to complete the details: (i) If a Q, a > 0 and a 2 < 2, then there exists r Q such that r > a and r 2 < 2. (ii) If b Q, b > 0 and b 2 > 2, then there exists s Q such that s < b and s 2 > 2. (iii) A and B. (iv) Q = A B. (v) a < b for every a A and for every b B. (vi) A B =. (vii) A has no largest element. (viii) B has no smallest element. Note also that the least upper bound of A in R is 2 and as is wellknown 2 is not a rational number so using it we created a gap in Q: A = (, 2) Q and B = ( 2, ) Q. The example given in the proof to show that Q is not complete is not possible in R by the completeness property of R. In these terms, the completeness property of R is formulated as follows: Completeness of R formulated in terms of cuts : Let A and B be subsets of R such that (i) A and B. (ii) A B = R. (iii) a < b for every a A and for every b B. Then there exists exactly one real number L such that for every real number x: (iv) x < L implies x A. (v) x > L implies x B. The real number L is either in A or B but not both since R = A B by property (ii) above and A B = by property (iii) above (if an element a A B, then a A and a B would hold and so a < a would hold by property (iii) above, which is of course not possible). So we obtain: (vi) If L A, then A = (, L] = {x R x L} and B = R \ A = (L, ) = {x R x > L}. (vii) If L B, then A = (, L) = {x R x < L} and B = R \ A = [L, ) = {x R x L}. 5
6 As an exercise to understand the different formulations of the completeness concept, prove the following: (a) Using the least upper bound property of R, that is, the property saying that every nonempty subset of R that is bounded from above has a least upper bound in R, prove the above property of R formulated in terms of cuts. (b) Assuming the above property of R formulated in terms of cuts, prove the least upper bound property of R, that is, prove that every nonempty subset of R that is bounded from above has a least upper bound in R. (c) Using the least upper bound property of R, that is, the property saying that every nonempty subset of R that is bounded from above has a least upper bound in R, prove the greatest lower bound property of R, that is, prove that every nonempty subset of R that is bounded from below has a greatest lower bound in R. (d) Using the greatest lower bound property of R, that is, the property saying that every nonempty subset of R that is bounded from below has a greatest lower bound in R, prove the least upper bound property of R, that is, prove that every nonempty subset of R that is bounded from above has a least upper bound in R. Later we shall also mention two other equivalent formulations of completeness of R in terms of sequences of real numbers, which you shall see in your Analysis courses: (e) Every bounded monotone sequence of real numbers is convergent. (f) Every Cauchy sequence of real numbers is convergent. You will later learn what all these terms mean in the above two formulations means for sequences of real numbers: bounded sequence, monotone sequence, convergent sequence, Cauchy sequence. Theorem. Density of Q and R \ Q in R. (i) Density of Q in R. The set Q of all rational numbers is dense in the set R of all real numbers, that is, between any two distinct real numbers, there exists a rational number. More precisely, for all real numbers a and b, if a < b, then there exits a rational number r such that a < r < b and r Q. (ii) Density of R \ Q in R. The set R \ Q of all irrational numbers is dense in the set R of all real numbers, that is, between any two distinct real numbers, there exists an irrational number. More precisely, for all real numbers a and b, if a < b, then there exits an irrational number s such that a < s < b and s R \ Q. See the Answers to Problems 13 and 20 in Homework 3 for the proofs of these density results. An obvious question to ask: how many irrational numbers are there? Are there too many irrational numbers? More than the rational numbers? Since both rational numbers and irrational numbers are infinite dense sets in R, how can we compare infinities? That is some delicate question which you shall answer later: the concept of cardinalities of sets, a world opened by Cantor for the comparison of different kinds of infinites. How can we count infinity and compare infinities? There are much much more irrational numbers than rational numbers. The set R \ Q is an uncountable set while Q is a countable set, that is, the infinity of Q is like the infinity of Z + but the infinity of R \ Q is strictly more than the infinity of Z +. You shall learn later that Q is a countable set but R is an uncountable set. 6
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