Solutions Manual for Homework Sets Math 401. Dr Vignon S. Oussa

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1 1 Solutions Manual for Homework Sets Math 401 Dr Vignon S. Oussa

2 Solutions Homework Set 0 Math 401 Fall (Direct Proof) Assume that x and y are odd integers. Then there exist integers u and v such that Next, Put Then Thus, xy is an odd integer. x = 2u + 1 and y = 2v + 1. xy = (2u + 1) (2v + 1) = 4uv + 2u + 2v + 1 = 2 (2uv + u + v) + 1. w = 2uv + u + v. xy = 2w (Direct Proof) Let a and b be positive real numbers. Assume that Then Now, since a < b then and it follows that a < b. b 2 a 2 = (b a) (b + a). a b < 0 b a > 0. Additionally, since a and b are positive integers then clearly, a + b > 0 (the sum of two positive numbers is positive). Next, since the product of two numbers of the same sign is positive then b 2 a 2 = (b a) (b + a) > 0. This completes the proof. 1

3 3. (Proof by contradiction). Assume that there exists an integer m such that m 2 is even and m is odd. Then m = 2n + 1 for some integer n. Now, squaring each side of the equation above gives m 2 = (2n + 1) 2. Next, foiling (2n + 1) 2, we obtain m 2 = (2n + 1) 2 = 4n 2 + 4n + 1 ( ) = 2 2n 2 + 2n + 1 and it follows that m 2 is odd, and this contradicts our assumption. Thus, it is not possible to find an odd number whose square is even. This completes the proof. 4. (Proof by contradiction) Let us suppose that 2 is a rational number. That is, there exist nonzero integers n, m such that 2 = n m. We shall furthermore assume that the fraction m n is given in its reduced form (n and m have no common divisor). Next, squaring each side of the equation above gives: It follows that 2 = n2 m 2. n 2 = 2m 2. (1) Thus n 2 is an even integer. Now, since n 2 is even then n is also even (see Problem 1). So, there exists an integer k such that n = 2k 2

4 and this implies that n 2 = 4k 2. Next, coming back to (1) we obtain So, 4k 2 = n 2 = 2m 2. m 2 = 2k 2 and this implies that m is also even. Thus, 2 divides both n and m. However this is impossible since the fraction is in its reduced form (the integers n, m have no common factor). 5. (Proof by contradiction) Let us suppose that the set of primes is actually finite. Let us suppose that n m {p 1, p 2,, p k } is the set of all prime numbers. Now, let n = p 1 p 2 p k + 1. Next, let q be a prime divisor of the number n. Then clearly n > 1. Next, since q is a prime number then q must be an element of the set {p 1, p 2,, p k }. Consequently, q must divide the number obtained by taking the product of the elements of {p 1, p 2,, p k } : p 1 p 2 p k. Next, since q also divides n then q must divide However, n p 1 p 2 p k. n p 1 p 2 p k = 1. So q divides 1. This means that q = 1 which contradicts our assumption. We conclude that our assumption that the set of primes is finite is a false assertion. As a consequence, the set of primes must be infinite. 3

5 6. There are two statements to prove here. First, we need to show that if m and n have the same parity then m 2 + n 2 is even. Secondly, we must also show that if m 2 + n 2 is even then m and n have the same parity. For the first statement, let m and n be integers. Assume that m and n have the same parity. There are two cases to consider. First let us assume that m, n are both even. That is, for some integers k, j. Then m = 2k and n = 2j m 2 + n 2 = (2k) 2 + (2j) 2 = 4j 2 + 4k 2 (this is an even number). Secondly, assume that m, n are both odd. That is for some integers k, j. Then m = 2k + 1 and n = 2j + 1 m 2 + n 2 = (2k + 1) 2 + (2j + 1) 2 = 4j 2 + 4j + 4k 2 + 4k + 2 ( ) = 2 2j 2 + 2j + 2k 2 + 2k + 1 which is also clearly even. For the second part of the proof, we want to show that if m 2 + n 2 is even then m and n have the same parity. By ways of contradiction, assume that m 2 + n 2 is even and m and n do not have the same parity. Then, it must be the case that m 2 + n 2 = (2k + 1) 2 + (2j) 2 for some integers k and j. Observe that m 2 + n 2 = (2k + 1) 2 + (2j) 2 = 4j 2 + 4k 2 + 4k + 1 ( ) = 2 2j 2 + 2k 2 + 2k + 1. Thus, m 2 + n 2 is odd and this contradicts our assumption. 4

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40 Homework Set 4 Review 1 Review 2

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52 Homework Set 4 Part 2 Objectives Sharpening skills required to successfully manipulate various inequalities. Mastering techniques used to formally establish the convergence of a sequence. 1. Prove that for any natural number n n n n 7 2n n Prove formally (using the ɛ-n definition) that if lim s n = s then lim ( s n ) = s. 3. Prove formally (using the ɛ-n definition) that lim n n n + 1 = (a) Prove the following identities max {x, y} = 1 (x + y + x y ) 2 (b) Prove that if lim n s n = s and lim r n = r then lim (max {s n, r n }) = max {s, r}. n 1

53 Solutions 1. Given any natural number n, we observe that n n n 7 2n + 1 n n n 5 5n 7 2n n = 5 5n 7 2n n = 5 5n 7 2n n5 5n 7 2n. The first inequality is obtained by using the fact that n 3 is less or equal to n 5 and 1 is also less or equal to the fifth power of any natural number. The third equality is due to the fact that 5n 7 2n + 1 is a positive number for any natural number n. Finally, the last inequality follows from 5n 7 2n + 1 5n 7 2n. Next, we want to show that for any natural number n, there exists a positive number B such that In other words, we want to show that 5n 7 2n Bn 7. (5 B) n 7 2n for all n N. For this to holds, it is enough to pick B = 3. Indeed, since 5n 7 2n 3n 7 for all n N, it follows that n n n 7 2n n5 5n 7 2n 29n5 3n 7 = 29 3n 2 29 n 2 for all natural numbers n. This completes the proof. 2. Assume that lim (s n) = s. n Given any ε > 0 there exists N > 0 such that if n > N then the distance between s n and s is less than ε. That is s n s < ε. Now, let us consider the distance between the opposite of s n and the opposites of s : s n ( s) = s n + s. Since any number is equal to its opposite, and since the opposite of s n + s is s n s then it is clear that s n + s = s n s. Appealing to our stated assumption, for any prescribed positive number ε, there is a positive number N such that as long as n is bigger than N then s n ( s) = s n s < ε. This establishes the desired result lim ( s n ) = s. 2

54 3. We want to show that for any given positive real number ε, there exists a corresponding positive number N such that as long as n > N then n 1 n + 1 < ε. We observe that n 1 n + 1 = n n + 1. n + 1 Multiplying the numerator as well as the denominator of the following fraction n n + 1 n + 1 by the conjugate of n n + 1 yields n n + 1 n + 1 = ( n n + 1 ) ( n + n + 1 ) n + 1 ( n + n + 1 ). Using the difference of perfect square formula, we proceed as follows: n n + 1 n + 1 = = = ( n ) 2 ( n + 1 ) 2 n + 1 ( n + n + 1 ) n (n + 1) n + 1 ( n + n + 1 ) 1 n + 1 n + (n + 1). Therefore the absolute value of the quantity n n+1 1 is given by n 1 n + 1 = 1. n + 1 n + (n + 1) Next, since n + 1 n + (n + 1) is larger or equal to n + 1, we obtain that n 1 n + 1 = 1 1 n + 1 n + (n + 1) n n. Letting N = 1 ε, if n > N then n 1 n n < ε, giving us the desired result. 4. (a) Let x and y be two real numbers. If we assume that x is less or equal to y then x y is less or equal to zero and it follows from the definition of the absolute value that x y = (x y). 3

55 As a result, the quantity 1 2 (x + y + x y ) becomes Since y is either bigger or equal to x then 1 (x + y (x y)) = y 2 1 (x + y (x y)) = y = max {x, y}. 2 Next, let us suppose that x is greater than y. So, the distance between x and y which is given by x y is equal to x y. Next, the quantity 1 2 (x + y + x y ) becomes In summary, we have proved that for any given real numbers. 1 (x + y + x y) = max {x, y}. 2 max {x, y} = 1 (x + y + x y ) 2 (b) Given a positive number ε, we want to establish the existence of a number N such that as long as n > N then max {s n, r n } max {s, r} < ε. In other words, given a positive number ε, we want to establish the existence of a number N such that as long as n > N then 1 2 (s n + r n + s n r n ) 1 (s + r + s r ) 2 < ε. We observe that 1 2 (s n + r n + s n r n ) 1 (s + r + s r ) 2 = s n + r n + s n r n s r s r 2 = (s n s) + (r n r) + s n r n s r. 2 Next, applying the triangle inequality, we obtain (s n s) + (r n r) + s n r n s r ( (s n s) + (r n r) + s n r n s r ) = 1 2 ( (s n + r n ) (s + r) + s n r n s r ). By assumption, we can assert that the limit of the sequence (s n + r n ) n is convergent to the real number s + r. Similarly, the sequence ( s n r n ) n converges to s r. As such, there exists N 1 > 0 such that if n > N 1 then (s n + r n ) (s + r) < ε. 4

56 Additionally, there exists N 2 > 0 such that if n > N 2 then s n r n s r < ε. Put N = max {N 1, N 2 }. As long as n > N, we have (s n s) + (r n r) + s n r n s r 2 1 ε ε {}}{{}}{ (s n + r n ) (s + r) + s n r n s r 2 This completes the proof. = 1 2 (ε + ε) = 1 (2ε) = ε. 2 5

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64 Solutions Homework Set 6 Focus monotone sequences, Cauchy sequences, lim sup, lim inf, Relation between Cauchy sequences and convergent sequences. 1. We consider the quantity s n+1 s n. Computing this s n+1 s n for any natural number n, we obtain that s n+1 s n = Thus, for any natural number n (n + 1) (n + 1) n + 3 5n = 3 5n (n + 1) < 0. s n+1 < s n. It follows that (s n ) is a nonincreasing sequence. 2. First, we observe that for natural numbers n, m n s n s m = cos (t) dt t 2 1 m 1 cos (t) t 2 dt. Without loss of generality, we may assume that n is less than m. So, m s n s m = cos (t) dt n t 2 m cos (t) dt n m n t 2 1 t 2 dt = 1 n 1 m. Now, given any positive ɛ, let N = 2 ɛ. If n, m > N then s n s m 1 n 1 m < ɛ 2 + ɛ 2 = ɛ. We conclude that the sequence is a Cauchy sequence. 3. We complete the table as follows. ( n 1 ) cos (t) dt t 2 n N s n = cos ( 2πn 8 ) lim sup s n lim inf s n 1 1 s n = ( 1 2) n 0 0

65 Justification. For s n = cos ( ) 2πn 8, we observe that ( ) 2πn { cos 8 } { 1 2 : n N =, 0, 1 }, 1, 1 2 It follows that Next, ( lim sup s n = lim sup N ( = lim N { cos ( ) 2πn 8 }) : n > N }) sup { 1 2, 0, 1 2, 1, 1 = lim N 1 since sup = 1. ( lim inf s n = lim N = lim N = lim N ( 1) = 1. { cos { 1 2, 0, 1 2, 1, 1 ( ) 2πn } }) : n > N }) inf 8 ( { 1 2 inf, 0, 1, 1, 1 2 = 1 For s n = ( 1 2) n, and 4. By definition lim sup s n = lim N = lim N = lim = 0 lim inf s n = lim N ( { (1 ) N+1 sup, 2 ( sup 1 N 2 N+1 = lim N 0 = 0. { 1 2 N+1, 1 2 N+2, ( { (1 ) N+1 inf, 2 ( ) }) N+2 1, 2 }) ( ) }) N+2 1, 2 lim sup (x n + y n ) = lim N (sup {x n + y n : n > N}).

66 Since sup {x n + y n : n > N} is the lowest upper bound for the set {x n + y n : n > N} and since sup {x n : n > N} + sup {y n : n > N} is an upperbound for {x n + y n : n > N}, it follows that Thus, sup {x n + y n : n > N} sup {x n : n > N} + sup {y n : n > N}. lim sup (x n + y n ) lim (sup {x n : n > N} + sup {y n : n > N}) N lim (sup {x n : n > N}) + lim (sup {y n : n > N}) N N lim sup (x n ) + lim sup (y n ). No, we do not have equality in general. Here is an example. Let Using the definition of lim sup, we obtain However, Here, x n = ( 1) n and y n = ( 1) n+1. lim sup x n = 1 and lim sup y n = 1. ( { }) lim sup x n + y n = lim sup ( 1) n + ( 1) n+1 : n > N N = lim 0 N = 0. 0 = lim sup x n + y n < lim sup (x n ) + lim sup (y n ) = 2. We have a strict inequality! Now, can you make another example? 5. Let n and m be natural numbers. Then s n s m = 2n + 1 n + 1 2m + 1 m + 1 m n = (m + 1) (n + 1) m + n (m + 1) (n + 1) m + n mn = 1 n + 1 m For any ɛ > 0, let n, m > N = 2 ɛ. Then if Thus, the sequence s n = ( ) 2n+1 n+s n N s n s m 1 n + 1 m < ɛ 2 + ɛ 2 = ɛ. is Cauchy.

67 6. (( s 2n s n = ) ( 1 + n n )) ( n ) n = 1 n n 1 n + n + 1 n + n + 1 ( ) 2n 1 = n = 1 2n 2. Thus, for ɛ = 1 2 and for any given N, we have n > N and s 2n s n 1 2. Clearly, this implies that the sequence (s n ) is not Cauchy. By Theorem 10.11, the sequence is not convergent.

68 Solutions Homework Set 7 1. Let (a n ) n2n be a sequence of real numbers. Let a 2 R: Then there is a positive number, such that for any given real number N; there exists a natural number n such that n > N and Next, let ja n aj : (N 1 ; N 2 ; ; N k ; N k+1 ; ) be a sequence of increasing real numbers. For each natural number k; we choose n k 2 N such that n k > N k : By assumption, it follows that ja nk aj for every natural number k: 2. The statement is true. Indeed, since the sequence (sin (n)) n2n is a bounded sequence, appealing to Bolzano-Weierstrass theorem, it follows that there exists a subsequence (sin n k ) k2n which is convergent. 3. The statement is true. Let (a n ) n2n be a sequence of real numbers. Given a natural number n ja n j < 1 + ja n j : Dividing each side of the inequality above by 1 + ja n j yields ja n j 1 + ja n j < 1 + ja nj 1 + ja n j = 1: Thus the sequence jan j 1 + ja n j n2n is bounded. As such, it admits a convergent subsequence. 4. Let us de ne s n = a ( 1) n + b + 1 n : Next, we verify that lim s 2n = a + b and lim s 2n+1 = a + b: 1

69 Now, solving the system of equations a + b = 1 a + b = 0 we obtain a = 1; b = 1 : The desired result follows then if 2 2 s n = 1 2 ( 1)n n : 5. Let us assume that (a n ) n2n is a bounded sequence. First, we observe that every subsequence of (a n ) n2n is bounded. Next, applying the Bolzano Weierstrass theorem to any xed subsequence of (a n ) n2n, we may safely assert that every subsequence of the given sequence has a convergent subsequence. Secondly, in order to prove the converse of the result, let us suppose that (a n ) n2n is unbounded. Assume that the sequence (a n ) n2n is not bounded above. That is given any real number M > 0 there is a natural number n such that a n > M: Now, choose an increasing sequence 0 < M 1 < M 2 < of real numbers. For a xed natural number k; there is a natural number n k such that a nk > M k : Thus clearly, the subsequence (a n1 ; a n2 ; ) is divergent. Next, assume that the sequence (a n ) n2n is not bounded below. That is given any real number M < 0 there is a natural number n such that a n < M: Now, let us choose a decreasing sequence of real numbers satisfying the conditions 0 > M 1 > M 2 > For a xed natural number k there is a corresponding natural number n k such that a nk < M k : It follows that the subsequence (a n1 ; a n2 ; ) is divergent. This completes the proof. 2

70 Solutions Homework Set 7 1. We consider the subsequence (t k ) given by ( ) 2π (7k + 1) t k = s 7k+1 = cos 7 ( ) 2π (7k) 2π (1) = cos ( = cos 2πk + 2π ) ( ) 2π = cos. 7 7 Clearly, (t k ) is a monotonic sequence, since it is constant. 2. (a) If n = 3k then If n = 3k + 1 then If n = 3k + 2 then ( ) 2πn cos = 1 3 ( ) 2πn cos = cos 3 ( ) 2πn cos = cos 3 ( ) 2π = ( ) 4π = Thus, the set of subsequential limits of (s n ) n N is given by { 1, 1 } 2 (b) { lim sup s n = sup 1, 1 } = 1 2 { lim inf s n = inf 1, 1 } = (c) Since lim sup s n = lim inf s n, the sequence is not convergent. 3. (a) The set of subsequential limits of (s n ) n N is { 5, 1 }. 5 1

71 (b) { lim sup s n = sup 5, 1 } = 5 5 { lim inf s n = inf 5, 1 } = (c) Since lim sup s n = lim inf s n, the sequence is not convergent. 4. lim sup (ks n ) = lim N sup {ks n : n > N} = lim (k sup {s n : n > N}) N ( ) = k lim sup {s n : n > N} N = k lim sup s n 2

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