Download U.C. Berkeley CS70 Midterm 1 Solutions: Discrete Math for CS and more Exams Discrete Mathematics in PDF only on Docsity! U.C. Berkeley — CS70 : Discrete Math for CS Midterm 1 Solutions Lecturers: Christos Papadimitriou & Umesh Vazirani February 26, 2004 Midterm 1 1. (10 points) • Is it possible for the propositions P ∨Q and ¬P ∨¬Q to be both false? Justify your answer. Solution. No. If P ∨Q is false, then both P and Q are false, so ¬P ∨ ¬Q is true. • Is it possible for the proposition P ⇒ (¬P ⇒ Q) to be false? Justify your answer. Solution. No. A ⇒ B is true whenever A is false. So for the proposition to be false P must be true. But then ¬P ⇒ Q is also true. Thus the proposition is true. 2. (10 points) Suppose you are proving a proposition P (n) by induction on n. You success- fully prove the induction step, ∀n, P (n) ⇒ P (n + 1). But then you notice that P (2501) is false. Can you conclude anything about P (25)? Justify your answer. Solution. You can conclude that P (25) is false. If P (25) were true, you could use that as the base case of your induction and conclude that ∀n,≥ 25P (n). But since P (2501) is false, this is a contradiction. 3. (20 points) Recall that the Fibonacci numbers F (n) satisfy the recurrence F (n) = F (n− 1) + F (n − 2), with F (0) = F (1) = 1. Prove by induction on n that F (m + n) = F (m)F (n) + F (m− 1)F (n− 1). Solution. As suggested, we will prove the proposition P (n) : F (m + n) = F (m)F (n) + F (m− 1)F (n− 1) by induction on n; the natural number m is fixed, but since we’re not going to assume anything about it in the proof, our argument will establish the claim for all m and n. • Base case: We want to prove P (1), but this just says that F (m + 1) = F (m) + F (m− 1), and this holds by definition. • Inductive Step: We assume that the statement holds for 1, ..., n, and we want to prove it for n + 1: F (m + n + 1) = F (m + n) + F (m + n− 1) = F (m)F (n) + F (m− 1)F (n− 1) + F (m)F (n− 1) + F (m− 1)F (n− 2) = F (m)(F (n) + F (n− 1)) + F (m− 1)(F (n− 1) + F (n− 2)) = F (m)F (n + 1) + F (m− 1)F (n), which is what we wanted to show. 4. (15 points) Evaluate 10050 2510 5 mod 47. Solution. This problem is being graded as an extra-credit problem. The main idea is that if N = p ·q then a(p−1)(q−1) = 1 mod N . So the exponent 502510 5 has to be evaluated mod46 = 2 · 23. So the exponent 25105 has to be evaluated mod22 = 2 · 11. So the exponent 105 has to be evaluated mod10, and is therefore 0. So the exponent in the previous line is 1. So the problem is reduced to evaluating 10050 mod 47 = 64 mod 47 = 48 · 27 mod 47 = 27 mod 47. 1
