Solutions for Homework Assignment 2

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1 Solutions for Homework Assignment 2 Problem 1. If a,b R, then a+b a + b. This fact is called the Triangle Inequality. By using the Triangle Inequality, prove that a b a b for all a,b R. Solution. To prove the stated inequality, first note that a = a b)+b, and b = b a)+a. Applying the triangle inequality, we then obtain a a b + b, and b b a + a. Adding b to the first inequality and adding a to the second inequality, and using the fact that b a = b a) = a b, we then obtain a b a b, and b a a b Finally, by definition a b is one of the two numbers a b or a b ). Both numbers are bounded above by a b. Therefore, we have proved that a b a b which is what we wanted to prove. Problem 2. Suppose that a,b R and that a < b. Prove that there exists an irrational number c satisfying a < c < b. Solution. The following result was proved in section 2.5 of our textbook and stated in class. Proposition. Suppose that u,v R and v > u. Then the set Q u,v) is nonempty. Here u,v) = {x x R, u < x < v}. We will use this proposition. However, first we will refine it slightly. According to the proposition, there exists at least one r Q such that 1

2 u < r < v. We can apply the proposition to the interval r,v). It follows that there exists at least one s Q satisfying r < s < v. Note that s r. Note also that s > r > u. Hence we have u < s < v. Therefore, using the above proposition, we have proved that the interval u, v) contains at least two distinct rational numbers. Now consider an interval a,b), where a,b R and a < b. Let u = a/ 2 and let v = b/ 2. Note that u,v R and u < v. Therefore, by the refinement of the above proposition, there exists at least two distinct rational numbers in the interval u, v). At least one of those two rational numbers will be nonzero. Therefore, the interval u, v) contains at least one nonzero rational number. Let r be such a rational number. Hence r 0 and u < r < v. Thus, we have a/ 2 < r < b/ 2 Multiplying by 2 which is a positive real number), we obtain a < r 2 < b Let c = r 2. Then a < c < b. We now prove that c is irrational. The fact that r 0 and r Q will be used. To prove that c is irrational, we give an indirect argument. Suppose that c Q. Since r Q and r 0, we know that r has an inverse in Q under multiplication. We denote that inverse, as usual, by r 1. It is an element of Q. We are using the fact that Q is a field. Thus, since we are assuming that c Q, and we know that r 1 Q, it follows that cr 1 Q. Now cr 1 = r 1 r 2) = 2. Thus, it follows that 2 Q. This is a contradiction because it is known that 2 Q, a fact which was probably proved in your Math 300 course. In the above paragraph, we have used basic facts about Q which are included in the statement that Q is a field. Problem 3. Define a sequence {x n } as follows. Let x 1 = 2. Let x n+1 = 2+x n for all n 1. Use mathematical induction to prove that x n < x n+1 for all n 1. Use mathematical induction to prove that x n < 2 for all n 1. State why you now know that {x n } is a convergent sequence. Determine Lim x n. Solution. First of all, we will prove that x n < x n+1 for all n 1. The Principle of Mathematical Induction tells us that it is sufficient to prove the following two assertions: x 1 < x 2 and x n < x n+1 = x n+1 < x n+2. 2

3 The first statement is true because x 1 = 2, and x 2 = 2+x 1 = 2+ 2 > 2+1 > 2 = x 1 Here we have used the fact that 2 > 1 and also the fact that y = x is an increasing function of x for x > 0. To prove the second assertion, assume that x n < x n+1 for some n 1. It then follows that x n+2 = 2+x n+1 > 2+x n = x n+1. We have again used the fact that y = x is an increasing function of x for x > 0. It follows that x n+1 < x n+2. Therefore, by the Principle of Mathematical Induction, we have proved that x n < x n+1 for all n 1. This means that the sequence {x n } is strictly increasing. It is therefore a nondecreasing sequence. Now we will prove that x n < 2 for all n 1. We will prove this by Mathematical Induction. It suffices to prove x 1 < 2 and x n < 2 = x n+1 < 2 Now x 1 = 2 and so we certainly have x 1 < 2. To prove the above implication, assume that x n < 2. It then follows that x n+1 = 2+x n < 2+2 = 4 = 2 and so it is indeed true that x n < 2 = x n+1 < 2. Therefore, by the Principle of Mathematical Induction, we have proved that x n < 2 for all n 1. The sequence {x n } has been proven to be a nondecreasing sequence which is bounded above. By a theorem proved in class sometimes called the Monotone Convergence Theorem ), we can conclude that {x n } is a convergent sequence. This means that Lim x n = A for some A R. To determine A, we use the equation x n+1 = 2+x n. We have A = Lim x n+1 = Lim 2+xn = 2+ Lim x n = 2+A. 3

4 We have used the fact that the function Fx) = 2+x is continuous at x = A. Note that A 2 and Fx) is actually continuous for all x > 2. ) Therefore, A satisfies the equation A = 2+A. This equation implies that A 2 = 2+A and hence A is a root of the quadratic polynomial x 2 x 2. The roots are x = 2 and x = 1. The x n are positive and so Lim x n 1. Therefore, we have Lim x n = 2. Problem 4. Suppose that a > 0. Define a sequence {x n } as follows. Let x 1 = a. Let x n+1 = a+x n for all n 1. Prove that {x n } is a convergent sequence. Determine Lim x n. Solution. First of all, we will prove that x n < x n+1 for all n 1. The Principle of Mathematical Induction tells us that it is sufficient to prove the following two assertions: x 1 < x 2 and x n < x n+1 = x n+1 < x n+2. The first statement is true because x 1 = a, and x 2 = a+x 1 = a+ a > a+0 = a = x 1 Here we have used the fact that a > 0 and also the fact that y = x is an increasing function of x for x > 0. To prove the second assertion, assume that x n < x n+1 for some value of n 1. It then follows that x n+2 = a+x n+1 > a+x n = x n+1. We have again used the fact that y = x is an increasing function of x for x > 0. It follows that x n+1 < x n+2. Therefore, by the Principle of Mathematical Induction, we have proved that x n < x n+1 for all n 1. This means that the sequence {x n } is strictly increasing. It is therefore a nondecreasing sequence. Now we will prove that the sequence {x n } is bounded above. Let B = Max{2,a}. We will actually prove that x n < B for all n 1. We will use Mathematical Induction. It suffices to prove that x 1 < B and x n < B = x n+1 < B. 4

5 Note that B = 2 if a 2 and B = a if a > 2. Now x 1 = a. If a 2, then we have x 1 = a 2 < 2 = B. and so x 1 < B if a 2. Now consider a > 2. If a > 2, we have x 1 = a < a = B and so we have x 1 < B if a > 2. In all cases, we have proved that x 1 < B. Now we will prove the implication x n < B = x n+1 < B. For this purpose, suppose that x n < B for some n 1. We then have x n+1 = a+x n < a+b B +B = 2B BB = B which implies that x n+1 < B. In the above inequalities, we have used the facts that a B and that 2 B. These inequalities are true because B = Max{a,2}. We have proved the implication x n < B = x n+1 < B. By the Principle of Mathematical Induction, we have proved that x n < B for all n 1. It follows that the sequence {x n } is bounded above. We have also proved that {x n } is a nondecreasing sequence. Therefore, the Monotone Convergence Theorem implies that {x n } is a convergent sequence. Thus, we know that Lim x n = A for some A R. Furthermore, we have A = Lim x n+1 = Lim a+xn = a+a. We have used the fact that Fx) = a+x is a continuous function for all x > 0. Note that A x 1 = a > 0 and so Fx) is continuous at x = A. And so we must have A = a+a. It follows that A 2 = a+a. Hence A is a root of the polynomial x 2 x a. There are two roots of that polynomial, and hence two possibilities for A, namely A = a 2 or A = 1 1+4a 2 We have used the famous formula for the roots of a quadratic polynomial. Now a is positive and therefore we 1+4a > 1 = 1. Therefore, 1 1+4a < 0. The second possibility for A mentioned above is actually negative. But we must have A x 1 = a > 0. Therefore, it is the first possibility which is the correct value of A. We have proved that Lim x n = a 2..

6 Problem 5. Let s n = nn for all n 1. Prove that Lim n!e s n n exists. Do not try to determine this limit. Solution. Note that s n > 0 for all n 1. Thus, the sequence is bounded below. We will prove that {s n } is a strictly decreasing sequence. Therefore, the hypotheses in the Monotone Convergence Theorem more specifically, theorem 2 listed under the completeness properties of R) are satisfied. It will therefore follow that {s n } is a convergent sequence. This means that Lim s n exists. To prove that the sequence {s n } is strictly decreasing, it will be helpful to consider the ratio s n+1 /s n. We have s n+1 /s n = n+1)n+1 n+1)!e n+1 / n n n!e n = Now n+1)! = n+1) n! and therefore We also have n+1) n+1 n n The above remarks imply that = n+1)n n n n+1) = s n+1 /s n = n+1) n+1) n! ) n n n+1)! e n e n+1 ) n! n+1)! = 1 n+1. Also, it is clear that e n e n+1 = 1 e n) n n+1) ) n 1 n+1) n n+1 1 e and therefore we have s n+1 /s n = 1+ 1 ) n / e n This is true for all n 1. Finally, recall a result proved in chapter 1, namely that 1+ 1 n) n < e. for all n 1. It follows that s n+1 /s n < 1. Since s n > 0, it now follows that s n+1 < s n for all n 1. We have proved that {s n } is indeed a strictly decreasing sequence. 6

7 Problem 6. Determine the least upper bound and the greatest lower bound for the following subsets of R. No explanation is necessary. a) S = { a a Q, 2 < a < 100 }. b) S = { a a Q, 2 a < 100 } { a a Z, 50 < a < 105 }. c) S = { 1) n 2 n) 2 n N }. d) S = { 1) n + 1 n N }. n Solutions. The question does not require explanations. Nevertheless we will give explanations below. Solution for part a). It is clear that 100 is an upper bound for S. Suppose that A < 100. Let B = Max{ 2,A}. Then B < 100. According to problem 2, there is an irrational number c such that B < c < 100. Then c 2,100) and c Q. Hence c S. But we also have A < c. Hence A is not an upper bound for S. It follows that every upper bound on S is 100 and hence 100 is the LUB for S. It is clear that 2is alower boundfor S. Wewill showthat 2 istheglbofs. Suppose that A > 2. Let B = Min{100,A}. Then B > 2. According to problem 2, there is an irrational number c such that 2 < c < B. Then c 2,100) and c Q. Hence c S. But we also have c < A. Hence A is not a lower bound for S. It follows that every lower bound for S is 2 and hence 2 is the GLB for S. Solution for part b). The set S is the union of two sets, namely the sets { a a Q, 2 a < 100 } and { a a Z, 50 < a < 105 } The second set is just a finite set. The largest element of that finite set is 104. Note that 104 is an upper bound on both of the above sets. Since 104 S, no upper bound for S can be smaller than 104. Thus, the LUB of S is 104. Note that 2 is a lower bound for S. Just as in part a), one can see that 2 is the GLB of S. Instead of using problem 2, one uses the fact that any interval a,b), where b > a, must contain a rational number. 7

8 Solution for part c). The numbers 1) n 2 n) 2 are zero or negative when n is odd, and hence they are < 2. But if n is even, then n = 2k, where k N, and the number 2 1 is in k the set S. That number is also < 2. It follows that 2 is an upper bound for S. However, Lim 2 1 ) k k and hence it follows that 2 is the LUB for the set S. For even n, the numbers 1) n 2 n) 2 are positive. For n = 1, we obtain the number 0. But for an odd integer n > 1, we can write n = 2k + 1, where k N. The numbers 2 2k+1) 2 are in the set S. All of these numbers are > 2. It follows that 2 is a lower bound for S. Furthermore, Lim 2 2 ) k 2k +1 It follows that 2 is the GLB for the set S. = 2 = 2 Solution for part d). The set S contains 0, 3/2. 2/3, and 5/4. It is clear that 1) n + 1 n 3/2 for all n N and hence 3/2 is an upper bound for S. Since 3/2 S, it follows that 3/2 is the LUB for S. If n is even, then 1) n + 1 = 1+ 1 is positive. But if n is odd and n > 1, n n then n = 2k +1, where k N, and the corresponding number in S is k +1 which is > 1. It follows that 1 is a lower bound for S. Furthermore, and hence the GLB for S is 1. Lim 1+ 1 ) k 2k +1 = 1 7. For each of the sets S in problem 6, determine if the least upper bound is actually in S. Determine if the greatest lower bound is actually in S. No explanation is necessary. 8

9 Solution. For the set in part a), the LUB and the GLB are 100 and 2, respectively. Neither of those numbers is in S. For the set in part b), the LUB and the GLB are 104 and 2, respectively. In fact, 104 is in the set S. But 2 Q and so 2 is not in the set S. For the set in part c), the LUB and the GLB are 2 and 2, respectively. Neither of those numbers is in S. For the set in part a), the LUB is 3/2 and this is in S. However, the GLB is 1 and this is not in S. 9