Problem Set 2 Solutions Math 311, Spring 2016

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1 Problem Set 2 Solutions Math 311, Spring 2016 Directions: Name: You must complete (at least) a total of 8 problems, some of which have multiple parts. You may complete more than 8 problems, for possible extra credit. You may replace required problems by optional problems, as long as the total number solved is (at least) 8. Required problems P1. Suppose two integers a,b Z and a positive integer n 1 satisfy a = nq + b, that is, q is the quotient and b is the remainder in dividing a by n. This can be equivalently expressed as n (a b) (read n divides a b ). Given a,b Z, we say a is congruent to b modulo n, and write a b mod n, if n (a b). Based on this, we (informally) define the integers modulo n as the set Z n = {0,1,...,n 1} with modulo n arithmetic operations. For example, Z 2 = {0,1}, since any even integer is equivalent to 0 (mod 2), while any odd number is equivalent to 1 (mod 2). a) Show that any element x Z 3 has an additive inverse; that is, there is a y Z 3 such that x + y 0 mod 3. b) Show that any nonzero element x Z 3 has a multiplicative inverse; that is, there is a w Z 3 such that xw 1 mod 3. c) Now consider the integers modulo 4, Z 4 = {0,1,2,3}. Does every element x Z 4 have an additive inverse? If x 0, does x have a multiplicative inverse? Note that parts a) and b) show that Z 3 is a field, whereas Z 4 turns out not to be a field. For what values of n do you think Z n is a field? Solution 1. Solution 1a) z z z

2 Solution 1b) See table above. Solution 1c) Every element of Z 4 has an additive inverse, but 2z 1 for any z Z 4. Z n is a field for n prime. P2. Let A R be bounded above and let γ R. Consider the following sets γ + A = {γ + a a A}, γa = {γa a A}. a) Show that sup(γ + A) = γ + sup A. b) If γ 0, show that sup(γa) = γ supa. c) If γ < 0, formulate a conjecture for sup(γa) similar to part b). Solution 2. Solution 2a) Suppose M = supa, so a M for all a A. Then γ + a γ + M for all a A, so γ + M is an upper bound for the set γ + A. Since A is bounded, γ + A is bounded, hence sup(γ + A) exists. Assume, for contradiction, that γ + M sup(γ + A). Then, for some ε > 0, (γ + M) ε γ + a for all γ + a γ + A (equivalently, for all a A). Thus, M ε a for all a A, which implies M supa, which is a contradiction. Hence, γ + supa = sup(γ + A). Solution 2b) Again, we will use the fact that M = supa ε > 0, a A such that M ε < a. Let M = supa, so a M for all a A. Since γ > 0, γa γm for all a A, hence γm is an upper bound for γa. Assume, for contradiction, that γm sup(γa). Then, for some ε > 0, γm ε γa for all a A, which implies M γ ε a for all a A. Hence, M supa, since for δ = γ ε > 0, there does not exist a A such that M δ < a. This is a contradiction, so γ sup A = sup(γa) for γ > 0. Solution 2c) If γ < 0, then sup(γa) = γ infa. The proof is similar to b). P3. For each of the following sets, compute the supremum and infimum. a) A 1 = {n N n 2 < 10} b) A 2 = {n/(m + n) m,n N} c) A 3 = {n/(2n + 1) n N} d) A 4 = {n/m m,n N and m + n 10}

3 Solution 3. Solution 3a) infa 1 = 1, supa 1 = 3 Solution 3b) infa 2 = 0, supa 2 = 1 Solution 3c) infa 3 = 1 3, supa 3 = 1 2 Solution 3d) infa 4 = 1 9, supa 4 = 9 P4. Prove that if a A is an upper bound for A, then a = supa. Solution 4. Suppose a A is an upper bound for A. Since A is bounded, supa exists by the Axiom of Completeness. Since a is an upper bound supa a, since the supremum is the least of all upper bounds. Since a is an element of A, a supa. Thus a = supa. P5. Prove the following criterion for the infimum of a set A R. If c R is a lower bound for A, then c = infa if and only if for any ε > 0 there is an a A such that c + ε > a. Solution 5. Suppose c R is an upper bound for A R. For the direction, suppose c = infa. For contradiction, suppose there exists δ > 0 such that c + δ a for all a A. Since c = infa, we have c < c + δ a for all a A, thus c + δ is a lower bound for A that is greater than c. This is a contradiction. For the direction, we will prove the contrapositive. Suppose c infa, so there exists a lower bound m such that c < m a for all a A. Thus, c + δ = m for some δ > 0. In other words, there exists δ > 0 such that c + δ a for all a A. This is the negation of the statement: for all ε > 0 there exists a A such that a < c + ε. Thus, the contrapositive holds. P6. Use the Archimedean property of R to prove that inf{1/n n N} = 0. Solution 6. The set { n 1 n N} is certainly bounded; any number greater than or equal to 1 is an upper bound, while any number less than or equal to 0 is a lower bound. Suppose c = inf{ n 1 n N} and c > 0. By the Archimedean Property, there exists m N such that 0 < m 1 < c, but m 1 { n 1 n N}, so c is not the infimum. This is a contradiction. P7. If x > 0, show that there exists n N such that 1/2 n < x. Solution 7. First, by induction, we show m < 2 m for all m N. If m = 1, then 1 < 2. Assume m < 2 m. Then, 2 m+1 = 22 m > 2m m + 1. Thus, m < 2 m for all m N. Note that the inequality 2m m + 1 for all m N can be proven by induction, as well.

4 Since m < 2 m 1 for all m N, 2 m < m 1 for all m N. By the Archimedean Property, if x > 0, there exists m N such that 0 < m 1 < x. Thus, 1 2 m < m 1 < x. P8. Define intervals I n = (0,1/n) for n N. Prove that n=1 I n =. Why does this not contradict the nested intervals property? Solution 8. Assume, for contradiction, that there exists ε I n. Since ε I n for all n N, we have 0 < ε < n 1 for all n N. But, this contradicts the Archimedean Property, since there exists m N such that 0 < m 1 < ε. n=1

5 Optional problems O1. Assume B A R and A,B are nonempty sets bounded above. Show that supb supa. Solution O1. Suppose, for contradiction, that supa is not an upper bound for the set B. That is, supa < b for some b B. By definition, a supa for all a A. But, b A since B A, so b supa < b, which is a contradiction. Thus, supa is an upper bound for the set B. Since supb is the least upper bound, we have supb supa, by definition. O2. Let A R be nonempty and bounded below. Prove that inf A = sup{ a a A}. Solution O2. Since A is bounded below, the set A = { a a A} is bounded above, hence sup( A) exists. By definition, inf A a for all a A, so a inf A for all a A. Thus, infa is an upper bound for the set A. Since infa is an upper bound, infa = sup( A) if and only if for all ε > 0 infa ε < a for some a A (equivalently, for some a A). To see that this is satisfied, since infa is the infimum of A, for all ε > 0 there exists a A such that a < infa + ε, which implies a > infa ε by negating the inequality. Thus, inf A = sup( A) or, equivalently, inf A = sup( A). O3. Let A be a nonempty set and let f : A R,g : A R be two functions with bounded range. That is, there exist constants M f,m g R such that f (a) M f and g(a) M g for all a A. a) Show that sup{f (a) + g(a) a A} sup{f (a) a A} + sup{g(a) a A}. b) Experiment and conjecture a similar inequality between inf{f (a) + g(a) a A} and inf{f (a) a A} + inf{g(a) a A}. Solution O3. Solution O3a) If s = sup{f (a) a A} and t = sup{g(a) a A}, then f (a) + g(a) s + g(a) s + t for all a A. In other words, s + t is an upper bound for {f (a) + g(a) a A}, hence sup{f (a) + g(a) a A} sup{f (a) a A} + sup{g(a) a A}.

6 Solution O3b) The inequality that holds for infimums is The proof is similar to part a). inf{f (a) a A} + inf{g(a) a A} inf{f (a) + g(a) a A}. O4. Consider the following theorem. Theorem 0.1. If A B, where B is countable, then A is either countable, finite, or empty. Proof. Since B is countable, there exists a bijection f : N B. We want to show that A B is at most countable. To show A is countable, we want to define a bijection g : N A. Let n 1 = min{n N f (n) A} and set g(1) = f (n 1 ). Finish this proof by inductively defining the function g : N A. Solution O4. Continuing the proof started above, we want to completely define g : N A. Let n 2 = min{n N f (n) A\{f (n 1 )}} and set g(2) = f (n 2 ). Inductively, if g(1),...,g(k 1) have been defined, then define n k = min{n N f (n) A\{f (n 1 ),...,f (n k 1 )}} and g(k) = f (n k ). Now that g : N A is defined, we need to show it is a bijection. If m l then n m n l. Since g(m) = f (n m ), g(l) = f (n l ) and f (n m ) f (n l ), we conclude g(m) g(l), since f is a bijection. This shows that g is 1-1. To show that g is onto, we again use the fact that f is a bijection. Since f is onto, for an arbitrary x A there is some ñ N such that f (ñ) = x. By the inductive construction of g, the minimal element is removed at each step. So, ñ must be the minimal element by at least the ñ 1 step. That is, ñ = n k for some k, and hence g(k) = x. Denotes a harder problem, worth 2 regular problems.