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CS 173 Homework 6 (due 10/17/08) Fall 2008 CS 173: Discrete Mathematical Structures, Fall 2008 Homework 6 Due at class on Friday, October 17, 2008 1. [10 points] Proof by Induction Use induction to prove 3n < n! for all n > 6 with n ∈ N. Answer. Basis step. The claim is made for n > 6, therefore the base case we need to prove is n = 7. Given that 37 = 2187, and 7! = 5040, the claim does hold for the base case. Induction assumption. Assume the claim holds for n, that is 3 n < n!. Inductive step. The claim we need to prove is: If 3 n < n!, then 3n+1 < (n + 1)!. We can prove this claim directly from the induction assumption: 3n < n! 3 · 3n < 3 · n! 3n+1 < 3 · n! < (n + 1) · n! 3n+1 < (n + 1)! In the second to last step, we used the fact that n + 1 > 3, which is true given that the claim was made for n > 6. 2. [10 points] Proof by Induction Use induction to prove that every positive integer n ∈ Z + can be expressed as the product of an odd number and a power of two. In other words, show n = (2k + 1)2 p where k ∈ N and p ∈ N. Hint: In the inductive step, break the proof down into one case for even numbers and one case for odd numbers. Basis. • For n = 1, choosing k = 0 and p = 0, n = (2 · 0 + 1)2 0 = 1. Induction assumption. Assume the claim holds for n, that is, we can write the number n as the product of an odd number and a power of two. Inductive step. We need to prove: For n ∈ Z + , if for all m ∈ {1, 2, . . . , n} m can be written as the product of an odd number and a power of two, then n + 1 can be written as a product of an odd number and a power of two. We divide the proof in two cases, function of whether n is even. • If n is even, then the result follows easily with a direct proof: If n is even, then (n + 1) is odd. Writing (n + 1)20 completes the proof for this case. Alternatively, using induction, we can write (n+1) = (n−1)+2. Since n is even, then (n−1) is odd, and by the induction assumption can be written as (n−1) = (2k+1)2 p , for some integers k and p. Moreover, since (n − 1) is odd, it does not have 2 as a factor, which means p = 0. We can write (n + 1) = (2k + 1)2 0 + 2 = (2k + 1) + 2 = (2k + 3) = (2(k + 1) + 1) = (2(k + 1) + 1)20 , which completes the proof for this case. 1 CS 173 Homework 6 (due 10/17/08) Fall 2008 • If n is odd, then (n+1) is even. Therefore, (n+1) is divisible by 2. Let m = (n+1)/2. Clearly, m < n. By the induction assumption, m can be written as m = (2k + 1)2 p , for k, p ∈ Z+ . Thus, (n + 1) = 2m = (2k + 1)2p+1 , which completes the proof. 3. [10 points] Proof by Induction If you’ve done much graphics programming, you may have used the following two standard trig identities to rotate objects: cos(x + y) = cos(x) cos(y) − sin(x) sin(y) sin(x + y) = sin(x) cos(y) + cos(x) sin(y) Using these formulas, prove by induction that (cos(x) + i sin(x))n = cos(nx) + i sin(nx) for any natural number n and any real number x (i is the square root of -1.) Do not convert this problem to exponential notation and try to solve it in that form. It doesn’t make things simpler and it will confuse us when we grade. Hint: this is not as bad as it looks. If your equations seem very complicated and won’t simplify, look for a mistake in your algebra. Answer. Basis step. Since n could be any natural number, the base case is n = 0. Given that (cos(x) + i sin(x))0 = 1, and cos(0x) + i sin(0x) = 1 + i0 = 1, the claim is true for the base case. Induction assumption. Assume (cos(x) + i sin(x)) n = cos(nx) + i sin(nx) holds for n. Inductive step. The claim we need to prove is: If (cos(x)+i sin(x)) n = cos(nx)+i sin(nx), then (cos(x)+ i sin(x))n+1 = cos((n + 1)x) + i sin((n + 1)x). Multiplying by cos(x) + i sin(x) both sides of the induction assumption, and then using the identities provided, we have: (cos(x) + i sin(x))n+1 = (cos(x) + i sin(x)) · (cos(nx) + i sin(nx)) = cos(x) cos(nx) + i cos(x) sin(nx) + i sin(x) cos(nx) − sin(x) sin(nx) = (cos(x) cos(nx) − sin(x) sin(nx)) + i(sin(nx) cos(x) + sin(x) cos(nx)) = cos(x + nx) + i sin(x + nx) = cos((n + 1)x) + i sin((n + 1)x) 4. [15 points] Recursive Definitions Reading Section 4.3 of the textbook should help with the following questions. 2 CS 173 Homework 6 (due 10/17/08) Fall 2008 Figure 1. Two intersecting lines decomposing the plane into 4 regions, and three interjecting lines decomposing the plane into 7 regions. Figure 2. Inductive step to count the number of regions of n lines intersecting in the plane. (a) Lines in a Plane Here, we attempt to answer the age-old question: how many slices of pizza can a person obtain by making n straight cuts with a knife? More formally, let L n be the number regions created by drawing n lines in the plane with no lines being parallel and no three lines intersecting at a single point . It is easy to see that L 0 = 1 and L1 = 2 (it’s helpful to draw a picture). i. What are L2 and L3 ? Answer. (see Figure 4(a)i). • L2 = 4. • L3 = 7. ii. Give a recursive definition for L n and briefly explain your answer. Answer. Consider Figure 4(a)ii). Starting with three lines, we find the increase in the number of regions when a fourth line, l, is added. Take l as a ray that progressively intersects the rest of the lines, from left to right (or top to bottom if l is vertical). Every time l meets a line, a region is divided by two, as for regions a, and b. Additionally, when l meets the last line, two regions are divided in two, as for regions c, and d. In general, we find that if there are n lines intersecting according to the conditions given decomposing the plane in L n regions, then adding an extra line increases the number of regions by n + 1. That is, L n+1 = Ln + n + 1. (b) String of 1s and 0s A bitstring is simply a sequence of zeroes and ones. Give a recursive definition for B n the 3 CS 173 Homework 6 (due 10/17/08) Fall 2008 number of bitstrings of length n that contain no consecutive zeroes. Include appropriate bases cases. Briefly explain your answer. Answer. Notation: A bitstring w of length n > 0 is written as w = ax, in which a ∈ {0, 1}, and x is a bitstring of length n − 1. Also, let S n be the set of bitstring of length n that contain no consecutive zeroes. This means Bn = |Sn |. Basis. • There is exactly one bitstring of length 0, the empty string . B 0 = |S0 | = |{}| = 1. • There are exactly two bitstrings of length 1, namely ’0’ and ’1’. B 1 = |S1 | = |{0, 1}| = 2. Recursive step. For n > 2: Consider a bitstring w ∈ S n . For n > 2, it either starts with 0 or with 1. Let Zn ⊂ Sn be the set of bitstrings of length n that start with 0, and let O n ⊂ Sn be the set of bitstrings of length n that start with 1. We can write: Bn = |On | + |Zn |. Consider a bitstring z ∈ On . The bitstring is of the form z = 1x, for some x ∈ S n−1 . Since there are Bn possible choices for x, then |On | = Bn−1 . Similarly, consider y ∈ Zn . The bitstring is of the form y = 0u, for some u ∈ S n−1 . However, in this case we know that u cannot start with 0. That is, u is of the form u = 01t, for some t ∈ S n−2 . Since there are Bn−2 possible choices for t, then |Zn | = Bn2 . Therefore, our recursive step is Bn = Bn−1 + Bn−2 . (Note that Bn looks very much like the Fibonacci numbers F n , but shifted by one. That is, B1 = 2, but F1 = 1 and F2 = 2. Therefore, if we would like to look fancy, then we could say Bn = Fn+1 .) 5. [5 points] An odd sequence The following proof contains a mistake. Specifically and carefully describe what the mistake is. Let a sequence of natural numbers be defined by the recurrence relation a n = an−2 + 2 witha1 = 3 and a2 = 2. We prove by induction on n that for all n ≥ 1 that a n is odd. Basis Step: For n = 1 we have a1 = 3 which is odd Inductive Step: Assume that for some number k, the numbers a 1 , ..., ak are all odd. We know ak+1 = ak−1 + 2. By the inductive hypothesis ak−1 is odd which means ak−1 = 2j + 1 for some natural number j. So ak+1 = (2j + 1) + 2 = 2(j + 1) + 1. So, we know ak+1 is odd. Answer. The base case was not proved for n = 2. The recurrence a n = an−2 + 2 uses the value of an−2 , thus we need to prove two cases in the basis step. The basis of the recurrence for a 2 is an even number, therefore the claim does not hold for the base case. 4