Solutions Quiz 9 Nov. 8, 2010 1. Prove: If a, b, m are integers such that 2a + 3b 12m + 1, then a 3m + 1 or b 2m + 1. Answer. We prove the contrapositive. Suppose a, b, m are integers such that a < 3m + 1 and b < 2m + 1. Then a 3m and b 2m. Therefore 2a 6m and 3b 6m. Adding these inequalities, 2a + 3b 12m. It follows that 2a + 3b < 12m + 1. We assumed a < 3m + 1 and b < 2m + 1, and deduced 2a + 3b < 12m + 1. This proves the contrapositive: If 2a + 3b 12m + 1, then a 3m + 1 or b 2m + 1. Remarks. This is not true for a, b, m real numbers instead of integers. For eample, take m = 0, a = 1/2, b = 1/3. Then 2a + 3b = 2 and 12m + 1 = 1, so 2a + 3b 12m + 1, yet a < 3m + 1 and b < 2m + 1. So the hypothesis that a, b, m are integers can not be ignored. Also note that the negation of a 3m + 1 or b 2m + 1 is a < 3m + 1 and b < 2m + 1. The original conclusion (a 3m + 1 or b 2m + 1) is true if either condition is true, or both; it is false if both conditions are false, i.e., a < 3m + 1 and b < 2m + 1. Here is a different way to understand it. Assume that it is NOT true that a 3m + 1 or b 2m + 1. In other words we are assuming NOT(a 3m + 1 or b 2m + 1). Then neither one of them can be true, so both a < 3m + 1 and b < 2m + 1. 2. Prove: If is a nonzero real number such that + 1 < 2, then < 0. Answer 1. (contrapositive) We prove the contrapositive: If is a nonzero real number such that 0, then + 1 2. Since 0 and 0, we have > 0. Now, we know that ( 1) 2 0. This holds for any real number; we have not yet used the assumption > 0. (Eercise: Why is ( 1) 2 0?) Multiplying this out, Hence 2 2 + 1 0. 2 + 1 2. Because > 0, we can divide each side of the inequality by without reversing the direction of the inequality, giving us + 1 2, which is what we wanted to show. Answer 2. (contradiction) Let be a nonzero real number such that + 1 < 2. Suppose by way of contradiction 0. Since 0, we have > 0. Then we can multiply both
sides of the inequality by without changing the direction of the inequality, giving us 2 + 1 < 2, hence 2 2 + 1 < 0. Factoring gives us ( 1) 2 < 0. But this is impossible: for any real number y, y 2 0; so for y = 1, it is impossible that y 2 < 0. (If we allowed comple numbers we could have y 2 < 0, but we are only allowing real numbers.) This means our original assumption, 0, is impossible. Therefore < 0. Answer 3. (direct) Let be a nonzero real number such that + 1 < 2. Subtracting 2 from each side of the inequality and putting all terms on a common denominator gives 2 2 + 1 < 0 which we may rewrite as ( 1) 2 < 0. It follows that 1 0: For if 1 = 0, we would have ( 1) 2 / = 0/ = 0, whereas we have ( 1) 2 / < 0. Thus ( 1) 2 > 0. Dividing each side of the inequality by ( 1) 2 gives 1 < 0. Finally, 2 > 0 as well. (For any real number, 2 0, and we were given 0, so 2 0.) Multiplying each side of this inequality by 2 gives as claimed. < 0, 3. Consider the following relation R on the integers Z: Ry if and only if 5 2 + 3y. a. Is 4R4? Eplain. b. Is 6R8? Is 8R6? Eplain. c. Prove or disprove: R is an equivalence relation. Answer. a. We have 2 4 + 3 4 = 8 + 12 = 20, and 5 divides 20. Therefore 4R4. b. We have 26+38 = 12+24 = 36, which is not divisible by 5. Therefore 6 R 8. Similarly, 2 8 + 3 6 = 16 + 18 = 34, which is also not divisible by 5. Therefore 8 R 6. c. We prove that R is an equivalence relation. First, we prove R is refleive. Let be any integer. Then 5 is divisible by 5, by the definition of divisibility. Therefore 5 divides 2 + 3. By the definition of R, we have R. This proves R is refleive.
Net we will prove R is symmetric. Suppose Ry. Then 2 + 3y is divisible by 5, so there is some integer k such that 2 + 3y = 5k. Let us subtract each side of this equation from 5 + 5y: (5 + 5y) (2 + 3y) = (5 + 5y) (5k). Since we are subtracting equal quantities, the results will be equal. This gives us 3 + 2y = 5( + y k). Therefore 2y + 3 is divisible by 5. By the definition of R we have yr. We have shown Ry = yr, so R is symmetric. Finally we show R is transitive. Suppose Ry and yrz. Then 2 + 3y is divisible by 5 and so is 2y + 3z. It follows that their sum, 2 + 5y + 3z, is also divisible by 5. We also have that 5y is divisible by 5. Since 2 + 5y + 3z and 5y are both divisible by 5, so is their difference, 2 + 3z. Therefore Rz. This shows R is transitive. Remark. Here is an alternative approach, which is much trickier. (I wouldn t epect you to come up with this on your own, but it can be helpful to see a different approach.) We begin with two lemmas: Lemma. An integer a is divisible by 5 if and only if 3 is divisible by 5. Proof. If 5 a, say a = 5k, then 3a = 15k = 5 (3k), so 5 3a. Conversely, suppose 5 3a, say 3a = 5k. Then 6a = 10k, so a = 10k 5a = 5(2k a). Since 2k a is an integer, we get 5 a. Lemma. For any integers a and b, 5 a if and only if 5 a 5b. Proof. Eercise. Now, the first lemma shows us that Ry 5 2 + 3y 5 6 + 9y. We can apply the second lemma by subtracting 5 + 10y to get But we can recognize the right-hand side: Ry 5 y. Ry y (mod 5). So this relation R is really just a tricky rewriting of congruence modulo 5. It is proved in the tetbook that congruence modulo 5 is an equivalence relation. 4. The Fibonacci numbers are the sequence defined by F 0 = 1, F 1 = 1, and for n 2, F n = F n 1 + F n 2. Prove that for all n 1, F 2 n F n F n 1 F 2 n 1 = ( 1)n. Answer. We prove this by induction. Our basis step is n = 1. For n = 1 the statement claims that F 2 1 F 1F 0 F 2 0 = ( 1)1.
Since F 1 = F 0 = 1, we have F 2 1 F 1F 0 F 2 0 = 1 2 = 1 and this is indeed equal to ( 1)1. So the statement is true for n = 0. Now, suppose by induction that for some n 1 we have F 2 n F n F n 1 F 2 n 1 = ( 1)n. The Fibonacci number F n+1 is given by F n+1 = F n + F n 1. Therefore F 2 n+1 F n+1f n F 2 n = (F n + F n 1 ) 2 (F n + F n 1 )F n F 2 n. Multiplying this out and collecting like terms gives us F 2 n+1 F n+1f n F 2 n = F 2 n + F n F n 1 + F 2 n 1. Pulling out a factor ( 1) from the right hand side, F 2 n+1 F n+1f n F 2 n = (F 2 n + F n F n 1 + F 2 n 1 ) = ( 1)n = ( 1) n+1, where we used the induction hypothesis for the second equality. This proves that F 2 n+1 F n+1 F n F 2 n = ( 1) n+1, and completes the proof by induction. 5. For all n 1, 1 + 1 4 + 1 9 + + 1 n 2 2 1 n. a. Write out what this statement says for n = 1, 2, 3, 4 and verify that it is true for these values of n. b. Prove the statement for all n 1. (Optional: What happens if you just try to prove 1 + 1 4 +... + 1 2 by induction?) n2 Answer. We have n LHS = value RHS = value 1 1 = 1 2 1 1 = 1 2 1 + 1 4 = 5 4 2 1 = 3 2 2 3 1 + 1 + 1 4 9 = 49 36 2 1 = 5 3 3 4 1 + 1 + 1 + 1 = 205 4 9 16 144 2 1 = 7 4 4 In each case the left hand side (LHS) has a smaller value than the right hand side (RHS). For eample, 205 7 can be checked by cross-multiplication: 205 4 144 7. 144 4 Now, we prove the statement for all n 1. We have already proved it for n = 1. Suppose by induction that we have 1 + 1 4 + 1 9 + + 1 n 2 2 1 n for some n. Adding 1/(n + 1) 2 to each side of this gives us 1 + 1 4 + 1 9 + + 1 n 2 + 1 (n + 1) 2 2 1 n + 1 (n + 1) 2.
The right hand side is 2 1 n + 1 (n + 1) = 2 (n + 1)2 n = 2 n2 + n + 1 2 n(n + 1) 2 (n 2 + n)(n + 1) = 2 n2 + n + 1 1 n 2 + n n + 1. We have n 2 + n + 1 > n 2 + n, so and n 2 + n + 1 n 2 + n n 2 + n + 1 1 n 2 + n n + 1 > 1 n + 1, hence 2 n2 + n + 1 1 n 2 + n n + 1 < 2 1 n + 1. At this point we have shown that 1 + 1 4 + 1 9 + + 1 n + 1 2 (n + 1) 2 1 2 n + 1 (n + 1) = 2 n2 + n + 1 1 2 n 2 + n n + 1 < 2 1 n + 1. This proves that 1 + 1 4 + 1 9 + + 1 n + 1 2 (n + 1) < 2 1 2 n + 1, which completes the proof by induction. > 1 Remark. One upshot of this result is that for all n, 1 + 1 + + 1 < 2. Therefore the infinite 4 n 2 series 1 + 1 + is convergent, and its value is at most 2. This can also be proved using the 4 Integral Test for Convergence, as you may have seen in Math 175 (Calculus 2). You may be interested to know that the value of the infinite seres 1 + 1 + is in fact 4 π 2 /6 = 1.6449.... It was a famous problem to find the value of this series, unsolved for nearly one hundred years, until Leonhard Euler earned his fame by solving it at the age of 28. See http://en.wikipedia.org/wiki/basel_problem for more information. Remark. If we just assume 1 + 1/4 + + 1/n 2 2, then we get 1 + 1 4 + + 1 (n + 1) 2 2 + 1 (n + 1) 2. It s impossible to get this to be 2. The problem is that our hypothesis didn t give us enough to work with. The original statement had a harder conclusion ( 2 1/n instead of easier 2) but also a stronger induction hypothesis, which gave us more to work with. 6. Consider the function of real numbers f : R R given by f () = 3 + 5. Prove f is bijective. Answer. We need to prove f is injective and surjective. First we show f is injective. We must show that if f ( 1 ) = f ( 2 ), then 1 = 2. Thus, suppose 1 and 2 are real numbers such that f ( 1 ) = f ( 2 ). That is, 3 1 + 5 = 3 2 + 5.
Subtracting 5 from each side, Dividing each side by 3 gives 3 1 = 3 2. 1 = 2, and we have shown that f is indeed injective. Net we show f is surjective. We must show that Image( f ) = R. Since the output values y = f () are real numbers, we have Image( f ) R by definition. So we only need to show that R Image( f ). Thus, suppose y R is an arbitrary real number. Let = (y 5)/3. Since y is a real number, y 5 is also a real number; and since 3 0 we can divide by it, so = (y 5)/3 is also a real number. We have ( ) y 5 f () = 3 + 5 = 3 + 5 = (y 5) + 5 = y, 3 which shows that y Image( f ). This completes the proof that f is surjective. Remark. To find the epression for, we could work backward from the conclusion f () = y, i.e., 3 + 5 = y, and solve for. 7. Let f : R R be the function defined by f () = 2 + 3 + 4. a. Show that f is not injective. b. Show that f is not surjective. c. Find the image of f. (Hint: For which r is there a solution 2 + 3 + 4 = r? By the Quadratic Formula, an equation 2 + b + c = 0 has a real-number solution if and only if b 2 4c 0.) Answer. We have f (0) = 4 and f ( 3) = ( 3) 2 + 3( 3) + 4 = 9 9 + 4 = 4. So f (0) = f ( 3), yet 0 3. This shows f is not injective. We also have that f is not surjective; indeed, there is no such that f () = 0. By the Quadratic Formula, 2 + 3 + 4 = 0 for the values = 3 ± 9 16 = 3 ± 7, 2 2 which are not in R. (These are comple values such that f () = 0; but there are no real values that work.) Finally we must find the image of f. A value y is in the image of f if and only if the equation 2 + 3 + 4 = y has a solution with R. By the Quadratic Formula, this equation has a solution in R if and only if 3 2 4(4 y) 0,
in other words, 4y 7 0. So the image of f is the set {y R : y 7/4} or in other words the interval [7/4, ). So, we have found: Let us prove that this is the case. Image( f ) = [7/4, ). Claim. For f : R R given by f () = 2 + 3 + 4, the image is given by Image( f ) = [7/4, ). Proof. First, we show that Image( f ) [7/4, ). Let y Image( f ). By the definition of image, there is an 1 R such that f ( 1 ) = y, i.e., 2 1 + 3 1 + 4 = y. That means the equation 2 + 3 + (4 y) = 0 has a solution ( = 1 is a solution). By the Quadratic Formula, therefore 3 2 4(4 y) 0, therefore 4y 7 0, so y 7/4. Therefore y [7/4, ). This shows Image( f ) [7/4, ). Now we will show that [7/4, ) Image( f ). Let y [7/4, ), i.e., y 7/4. Let 1 = 3 + 4y 7. 2 By the Quadratic Formula, 1 is a solution to the equation 2 +3+(4 y) = 0, i.e., f ( 1 ) = y. (There is another solution with. We only need one solution, so I just chose to use +. The other solution would have worked equally well.) This shows there eists an such that f () = y, i.e., = 1. Therefore y Image( f ).