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2.2.2).7. Show that if the product m · n of natural numbers is divisible by a prime p, thatis, m · n = p · k, where k ∈ N, then either m or n is divisible by p.8. It follows from the fundamental theorem of arithmetic that the set of prime numbers is infinite.9. Show that if the natural number n is not of the form k m , where k, m ∈ N, thenthe equation x m = n has no rational roots.10. Show that the expression of a rational number in any q-ary computation systemis periodic, that is, starting from some rank it consists of periodically repeatinggroups of digits.11.
Let us call an irrational number α ∈ R well approximated by rational numbersif for any natural numbers n, N ∈ N there exists a rational number pq such that|α − pq | <1N qn .a) Construct an example of a well-approximated irrational number.b) Prove that a well-approximated irrational number cannot be algebraic, that is,it is transcendental (Liouville’s theorem).9−1 by definition, where m ∈ Z and n ∈ N, derive the12. Knowing that mn := m · n“rules” for addition, multiplication, and division of fractions, and also the conditionfor two fractions to be equal.13. Verify that the rational numbers Q satisfy all the axioms for real numbers exceptthe axiom of completeness.14. Adopting the geometric model of the set of real numbers (the real line), showhow to construct the numbers a + b, a − b, ab, and ab in this model.15.
a) Illustrate the axiom of completeness on the real line.b) Prove that the least-upper-bound principle is equivalent to the axiom of completeness.16. a) If A ⊂ B ⊂ R, then sup A ≤ sup B and inf A ≥ inf B.b) Let R ⊃ X = ∅ and R ⊃ Y = ∅. If x ≤ y for all x ∈ X and all y ∈ Y , thenX is bounded above, Y is bounded below, and sup X ≤ inf Y .c) If the sets X, Y in b) are such that X ∪ Y = R, then sup X = inf Y .d) If X and Y are the sets defined in c), then either X has a maximal elementor Y has a minimal element. (Dedekind’s theorem.)e) (Continuation.) Show that Dedekind’s theorem is equivalent to the axiom ofcompleteness.17. Let A + B be the set of numbers of the form a + b and A · B the set of numbersof the form a · b, where a ∈ A ⊂ R and b ∈ B ⊂ R.
Determine whether it is alwaystrue that9 J. Liouville (1809–1882) – French mathematician, who wrote on complex analysis, geometry,differential equations, number theory, and mechanics.682The Real Numbersa) sup(A + B) = sup A + sup B,b) sup(A · B) = sup A · sup B.18. Let −A be the set of numbers of the form −a, where a ∈ A ⊂ R. Show thatsup(−A) = − inf A.19. a) Showthat for n ∈ N and a > 0 the equation x n = a has a positive root√n(denoted a or a 1/n ).b) Verify that for a > 0, b > 0, and n, m ∈ N√n1ab =√√nna· band√√a = n·m a.n m1c) (a n )m = (a m ) n =: a m/n and a 1/n · a 1/m = a 1/n+1/m .d) (a m/n )−1 = (a −1 )m/n =: a −m/n .e) Show that for all r1 , r2 ∈ Qa r1 · a r2 = a r1 +r2anda r1r2= a r1 r2 .20.
a) Show that the inclusion relation is a partial ordering relation on sets (but nota linear ordering!).b) Let A, B, and C be sets such that A ⊂ C, B ⊂ C, A\B = ∅, and B\A = ∅.We introduce a partial ordering into this triple of sets as in a). Exhibit the maximaland minimal elements of the set {A, B, C}. (Pay attention to the non-uniqueness!)√21. a) Show that,√just like the set Q of rational numbers, the set Q( n) of numbersof the form a + b n, where a, b ∈ Q and n is a fixed natural number that is not thesquare of any integer, is an ordered set satisfying the principle of Archimedes butnot the axiom of completeness.√b) Determine which axioms for the real numbers√ do not hold for Q( n) if thestandard arithmeticoperationsare retained in Q( n) but order is defined√√√ by therule (a + b n ≤ a + b n) := ((b < b ) ∨ ((b = b ) ∧ (a ≤ a ))).
Will Q( n) nowsatisfy the principle of Archimedes?c) Order the set P[x] of polynomials with rational or real coefficients by specifying thatPm (x) = a0 + a1 x + · · · + am x m 0,if am > 0.d) Show that the set Q(x) of rational fractionsRm,n =a0 + a1 x + · · · + am x mb0 + b1 x + · · · + bn x nwith coefficients in Q or R becomes an ordered field, but not an Archimedean ordered field, when the order relation Rm,n 0 is defined to mean am bn > 0 andthe usual arithmetic operations are introduced.
This means that the principle ofArchimedes cannot be deduced from the other axioms for R without using the axiomof completeness.2.2 Classes of Real Numbers and Computations6922. Let n ∈ N and n > 1. In the set En = {0, 1, . . . , n − 1} we define the sum andproduct of two elements as the remainders when the usual sum and product in R aredivided by n. With these operations defined on it, the set En is denoted Zn .a) Show that if n is not a prime number, then there are nonzero numbers m, kin Zn such that m · k = 0.
(Such numbers are called zero divisors.) This means thatin Zn the equation a · b = c · b does not imply that a = c, even when b = 0.b) Show that if p is prime, then there are no zero divisors in Zp and Zp is afield.c) Show that, no matter what the prime p, Zp cannot be ordered in a wayconsistent with the arithmetic operations on it.23.
Show that if R and R are two models of the set of real numbers and f : R → Ris a mapping such that f (x + y) = f (x) + f (y) and f (x · y) = f (x) · f (y) for anyx, y ∈ R, thena) f (0) = 0 ;b) f (1) = 1 if f (x) ≡ 0 , which we shall henceforth assume;c) f (m) = m where m ∈ Z and m ∈ Z , and the mapping f : Z → Z is injective and preserves the order.m d) f ( mn ) = n , where m, n ∈ Z, n = 0, m , n ∈ Z , n = 0 , f (m) = m ,f (n) = n .
Thus f : Q → Q is a bijection that preserves order.e) f : R → R is a bijective mapping that preserves order.24. On the basis of the preceding exercise and the axiom of completeness, showthat the axiom system for the set of real numbers determines it completely up to anisomorphism (method of realizing it), that is, if R and R are two sets satisfying theseaxioms, then there exists a one-to-one correspondence f : R → R that preservesthe arithmetic operations and the order: f (x + y) = f (x) + f (y), f (x · y) = f (x) ·f (y), and (x ≤ y) ⇔ (f (x) ≤ f (y)).25. A number x is represented on a computer asx = ±qpkαnn=1qn,where p is the order of x and M = kn=1 qαnn is the mantissa of the number x ( q1 ≤M < 1).Now a computer works only with a certain range of numbers: for q = 2 usually|p| ≤ 64, and k = 35.
Evaluate this range in the decimal system.26. a) Write out the (6 × 6) multiplication table for multiplication in base 6.b) Using the result of a), multiply “columnwise” in the base-6 system(532)6×(145)6and check your work by repeating the computation in the decimal system.702The Real Numbersc) Perform the “long” division(1301)6 (25)6and check your work by repeating the computation in the decimal system.d) Perform the “columnwise” addition(4052)6×(3125)627. Write (100)10 in the binary and ternary systems.28.
a) Show that along with the unique representation of an integer as(αn αn−1 . . . α0 )3 ,where αi ∈ {0, 1, 2}, it can also be written as(βn βn−1 . . . β0 )3 ,where β ∈ {−1, 0, 1}.b) What is the largest number of coins from which one can detect a counterfeit in three weighings with a pan balance, if it is known in advance only that thecounterfeit coin differs in weight from the other coins?29.
What is the smallest number of questions to be answered “yes” or “no” that onemust pose in order to be sure of determining a 7-digit telephone number?30. a) How many different numbers can one define using 20 decimal digits (forexample, two ranks with 10 possible digits in each)? Answer the same question forthe binary system.
Which system does a comparison of the results favor in terms ofefficiency?b) Evaluate the number of different numbers one can write, having at one’sdisposal n digits of a q-ary system. (Answer: q n/q .)c) Draw the graph of the function f (x) = x n/x over the set of natural-numbervalues of the argument and compare the efficiency of the different systems of computation.2.3 Basic Lemmas Connected with the Completeness of the RealNumbersIn this section we shall establish some simple useful principles, each of which couldhave been used as the axiom of completeness in our construction of the real numbers.1010 SeeProblem 4 at the end of this section.2.3 Basic Lemmas on Completeness71We have called these principles basic lemmas in view of their extensive application in the proofs of a wide variety of theorems in analysis.2.3.1 The Nested Interval Lemma (Cauchy–Cantor Principle)Definition 1 A function f : N → X of a natural-number argument is called a sequence or, more fully, a sequence of elements of X.The value f (n) of the function f corresponding to the number n ∈ N is oftendenoted xn and called the nth term of the sequence.Definition 2 Let X1 , X2 , .
. . , Xn , . . . be a sequence of sets. If X1 ⊃ X2 ⊃ · · · ⊃Xn ⊃ · · · , that is Xn ⊃ Xn+1 for all n ∈ N, we say the sequence is nested.Lemma (Cauchy–Cantor) For any nested sequence I1 ⊃ I2 ⊃ · · · ⊃ In ⊃ · · · ofclosed intervals, there exists a point c ∈ R belonging to all of these intervals.If in addition it is known that for any ε > 0 there is an interval Ik whose length|Ik | is less than ε, then c is the unique point common to all the intervals.Proof We begin by remarking that for any two closed intervals Im = [am , bm ] andIn = [an , bn ] of the sequence we have am ≤ bn . For otherwise we would have an ≤bn < am ≤ bm , that is, the intervals Im and In would be mutually disjoint, while oneof them (the one with the larger index) is contained in the other.Thus the numerical sets A = {am | m ∈ N} and B = {bn | n ∈ N} satisfy the hypotheses of the axiom of completeness, by virtue of which there is a number c ∈ Rsuch that am ≤ c ≤ bn for all am ∈ A and all bn ∈ B.
In particular, an ≤ c ≤ bn forall n ∈ N. But that means that the point c belongs to all the intervals In .Now let c1 and c2 be two points having this property. If they are different, sayc1 < c2 , then for any n ∈ N we have an ≤ c1 < c2 ≤ bn , and therefore 0 < c2 − c1 <bn − an , so that the length of an interval in the sequence cannot be less than c2 − c1 .Hence if there are intervals of arbitrarily small length in the sequence, their commonpoint is unique.2.3.2 The Finite Covering Lemma (Borel–Lebesgue Principle,or Heine–Borel Theorem)Definition 3 A system S = {X} of sets X is said to cover a set Y if Y ⊂ X∈S X,(that is, if every element y ∈ Y belongs to at least one of the sets X in the system S).A subset of a set S = {X} that is a system of sets will be called a subsystem of S.Thus a subsystem of a system of sets is itself a system of sets of the same type.722The Real NumbersLemma (Borel–Lebesgue11 ) Every system of open intervals covering a closed interval contains a finite subsystem that covers the closed interval.Proof Let S = {U } be a system of open intervals U that cover the closed interval[a, b] = I1 .
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