CMPSCI 250: Introduction to Computation

Solutions to First Midterm Exam

David Mix Barrington

14 October 2013

Directions:

• Answer the problems on the exam pages.
• There are four problems for 100 total points. Actual scale was A = 90, C = 60.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.

  Q1: 20 points
  Q2: 20 points
  Q3: 30 points
  Q4: 30 points
Total: 100 points

Question text is in black, solutions in blue.

• Question 1 (20): Briefly identify the following terms or concepts (4 points each):
  • (a) DeMorgan's Laws for propositions

    DeMorgan's Laws say that for any propositions p and q, ¬(p ∧ q) is equivalent to ¬p ∨ ¬q and ¬(p ∨ q) is equivalent to ¬p ∧ ¬q.

  • (b) the contrapositive of an implication

    The contrapositive of an implication p → q is the implication ¬q → ¬p. The contrapositive is equivalent to the original implication.

  • (c) the Division Theorem

    The Division Theorem says that if n is any integer and k is any positive integer, then there exist integers q and r such that n = qk + r and 0 ≤ r < k.

  • (d) an odd integer

    An odd integer n is one for which there exists an integer k such that n = 2k + 1.

  • (e) the conclusion of an implication

    The conclusion of an implication p → q is just the statement q.

• Question 2 (20): Let p and q be two propositions. Prove that if the three premises "q → ¬p", "p ∨ q", and "¬(q ∧ ¬p)" are all true, then the statement "p → q" must be false.

  You may use a truth table, apply Proof by Cases, or use other argument patterns that we have learned. (The truth table is probably simplest.)

  
    [(q --> not p) and (p or q) and not(q and not p)] --> not(p --> q)
      0  1   1  0   0   0  0 0   0   1  0  0   1  0    1   0  0  1  0
      1  1   1  0   1   0  1 1   0   0  1  1   1  0    1   0  0  1  1
      0  1   0  1   1   1  1 0   1   1  0  0   0  1    1   1  1  0  0
      1  0   0  1   0   1  1 1   0   1  1  0   0  1    1   0  1  1  1
  

  Here's a case argument. Assume q. Then ¬p follows by Modus Ponens on the first premise, and we have q ∧ ¬p which contradicts the third premise. So this case is impossible. So consider the other case, and assume ¬q. By the second premise, p must be true. But p ∧ ¬q is exactly the negation of p → q.

• Question 3 (30): Let n be any integer. Prove that n³ - n is divisible by 4 if and only if either n is odd or n is divisible by 4. (Hint: Use Proof by Cases based on applying the Division Theorem to n.)

  By the Division Theorem, n is equal to 4q + r where q is an integer and r is 0, 1, 2, or 3. Let A(n) be the predicate "n³ - n is divisible by 4" and let B(n) be the predicate "n is odd or n is divisible by 4". We will prove that in each of the four cases, A(n) ↔ B(n) is true.

  • r = 0, so n = 4q: A(n) is true because both (4q)³ and -4q are divisible by 4. And B(n) is also true because n is divisible by 4. So A(n) ↔ B(n) is true.
  • r = 1, so n = 4q + 1: We compute n³ - n = (4q + 1)³ - (4q - 1). (4q + 1)³ consists of eight terms, seven of which contain 4q as at least one factor and the eighth of which is 1³ = 1. (It multiplies out to 64q³ + 48q² + 12q + 1.) When we subtract 4q + 1, we cancel the only term that was not divisible by 4 and our result is divisible by 4, so A(n) is true. And B(n) is true because n is odd, so A(n) ↔ B(n) is true.
  • r = 2, so n = 4q + 2: This time n³ = (4q + 2)³ has all terms divisible by 4q except for 2³ = 8, so it is itself divisibe by 4. That means that since n = 4q + 2 is not divisible by 4, n³ - n is not divisible by 4 and A(n) is false. But B(n) is also false because n is even, but not divisible by 4. So A(n) ↔ B(n) is true in this case as well.
  • r = 3, so n = 4q + 3: Now n³ = (4q + 3)³ has seven terms divisible by 4q and one term that is 3³ = 27. When we subtract n = 4q + 3, we get terms divisible by 4q, then 27 - 3 - 24 which is divisible by 4, so n³ - n is divisible by 4 and A(n) is true. B(n) is true because n is odd, so A(n) ↔ B(n) is true.

• Question 4 (30): We define a sequence of numbers (a₁, a₂, a₃,...) by the following recurrence. We define a₁ to be 1 and for any k with k ≥ 2, we define a_k to be 2a_k-1 + k.

  We would like to prove by induction that for any positive integer n, this sequence obeys the closed formula a_n = 2ⁿ⁺¹ - n - 2.

  • (a, 10) What is the base case of the induction? Prove the base case.

    The statement P(n) is "a_n = 2ⁿ⁺¹ - n - 2". The base case P(1) is "a₁ = 2¹⁺¹ - 1 - 2". The base case is true because a₁ = 1 by the definition, and 2² - 1 - 2 = 4 - 1 - 2 = 1.

  • (b, 10) In the inductive step of the proof, what do we assume and what do we intend to prove? State both the hypothesis and the conclusion carefully.

    We assume P(n), which says that a_n = 2ⁿ⁺¹ - n - 2. We want to prove P(n+1), which says that a_n+1 = 2^(n+1)+1 - (n+1) - 2 = 2ⁿ⁺² - n - 3.

  • (c, 10) Complete the proof of the inductive step.

    We begin with a_n+1 and transform it into the desired expression. We first use the definition with k = n + 1 to say that a_n+1 = 2a_n + (n + 1). We then use the assumption to replace a_n, getting a_n+1 = 2(2ⁿ⁺¹ - n - 2) + (n + 1) = 2ⁿ⁺² - 2n - 4 + n + 1 = 2ⁿ⁺² - n - 3. This proves that P(n) → P(n+1) and completes the induction.

  Last modified 8 October 2014