CMPSCI 250: Introduction to Computation

Solutions to Practice First Midterm Exam

David Mix Barrington

2 October 2013

Directions:

• Answer the problems on the exam pages.
• There are four problems for 100 total points. Scale will be determined after the exam but a good guess is 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

Exam questions are in black, solutions in blue.

• Question 1 (20): Briefly identify the following terms or concepts (4 points each):
  • (a) a predicate

    A predicate is a statement with some variables whose value is not given. It becomes a proposition when the value of those variables is known.

  • (b) a counterexample

    A counterexample to a universal statement is an example of a value for which it is false. For example, a counterexample to the claim "all unicorns are green" would be any unicorn that was not green.

  • (c) two logically equivalent statements

    Two statements are logically equivalent if they have the same truth value in all possible situations, such as values for their free variables.

  • (d) a rational number

    A rational number is any number that can be written in the form a/b, where a is an integer and b is a nonzero integer.

  • (e) the principle of mathematical induction

    The principle of mathematical induction is a way to prove a statement P(n) is true for all possible positive integers n. We first prove P(1), then prove that if all the values P(1), P(2),..., P(n) are true, then P(n+1) is true. (We usually use just the fact that P(n) is true to prove P(n+1).)

• Question 2 (20): Let p, q, and r be three propositional variables. Use a truth table to show that if p ∨ ¬q and q ∧ ¬r are both true, then p → r must be false.

     (p or not q) and (q and not r) --> not (p --> r)
      0  1  1  0   0   0  0   1  0   1   0   0  1  0
      0  1  1  0   0   0  0   0  1   1   0   0  1  1
      0  0  0  1   0   1  1   1  0   1   0   0  1  0
      0  0  0  1   0   1  0   0  1   1   0   0  1  1
      1  1  1  0   0   0  0   1  0   1   1   1  0  0
      1  1  1  0   0   0  0   0  1   1   0   1  1  1
      1  1  0  1   1   1  1   1  0   1   1   1  0  1
      1  1  0  1   0   1  0   0  1   1   0   1  1  1
  

  Since the implication is true in all eight cases, the two premises imply the conclusion.

• Question 3 (30): Prove by cases that if n is any integer, (n+1)(n+2)(n+3) is divisible by 6. (Hint: You don't necessarily have to multiply out the polynomial, though you may.)

  The number n must have of the form 6q + r by the Division Theorem, with q an integer and r in the range from 0 through 5.

  • If r = 0, (n+1)(n+2)(n+3) = (6q+1)(6q+2)(6q+3) = (terms with 6q's in them) + 1*2*3, divisible by 6
  • If r = 1, (n+1)(n+2)(n+3) = (6q+2)(6q+3)(6q+4) = (terms with 6q's in them) + 2*3*4, divisible by 6 as last term is 24
  • If r = 2, (n+1)(n+2)(n+3) = (6q+3)(6q+4)(6q+5), divisible by 6 as the last term is 3*4*5 = 60
  • If r = 3, r = 4 or r = 5, (n+1)(n+2)(n+3) includes a multiplication by 6q + 6 and so is divisible by 6

• Question 4 (30): Suppose we want to use mathematical induction to prove that for every positive integer n, (n+1)(n+2)(n+3) is divisible by 6.
  • (a, 10) What is the base case?

    We must prove the n = 1 case, that (1+1)(2+1)(3+1) is divisible by 6, which is true since this product is 2*3*4 = 24.

  • (b, 10) What are the hypothesis and conclusion of the inductive case?

    The hypothesis is the n case, that (n+1)(n+2)(n+3) is divisible by 6. (We will not need to use the cases for smaller n.) The conclusion is the n + 1 case, that (n+2)(n+3)(n+4) is divisible by n.

  • (c, 10) Prove the inductive case.

    We use the fact that (n+1)(n+2)(n+3) and (n+2)(n+3)(n+4) have a common factor of (n+2)(n+3), so we can say that (n+2)(n+3)(n+4) = (n+1)(n+2)(n+3) + 3(n+2)(n+3). The first term is divisible by 6 by our hypothesis. The number (n+2)(n+3) must be even because one of the numbers is even and the other is odd. So (n+2)(n+3) may be written as 2k for some integer k, and the second term is thus of the form 3*3k or 6k, and it is divisible by 6. The sum of two numbers each divisible by 6 is itself divisible by 6.

  Last modified 14 October 2013