CMPSCI 190DM: A Mathematical Foundation for Informatics

Solutions to Practice Second Exam

David Mix Barrington

20 October 2015

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.

Question text is in black, solutions in blue.

  Q1: 15 points
  Q2: 15 points
  Q3: 20 points
  Q4: 20 points
  Q5: 30 points
Total: 100 points

Question 1 (15): Briefly identify the following terms or concepts (3 points each):

(a) for x to be a multiple of y
An integer x is a multiple of y if it equals yz for some integer z.
(b) a proof by contradiction
A proof by contradiction is where you prove a proposition p by first assuming "not p", then using that and other true facts and valid proof methods to derive a contradiction. You have then proved "not p → 0", which is equivalent to p.
(c) a rational number
A rational number is a real number that can be written in the form x/y, where x and y are integers and y ≠ 0.
(d) the base case of a proof by induction
When we use induction to prove a statement P(n) for all positive integers n, the base case is to prove the proposition P(1).
(e) a predicate on the positive integers
A predicate on the positive integers is some statement that contains one free variable of type "positive integer". If P(x) is such a statement, then P(1), P(2), P(3), etc. are propositions obtained by substituting each number for n in the definition of P(n).

Question 2 (15): Determine whether each of these five statements is true or false. No explanation is needed or wanted, and there is no penalty for guessing.

(a) 91 is not divisible by 7.
FALSE. Since 91 = 7 times 13, 91 is divisible by 7.
(b) (31 mod 6) = (25 mod 4) (Remember that E&C's "mod" is the same as Java's "%" on non-negative integers.)
TRUE. Both divisions have remainder 1.
(c) The binary representation of the decimal number "19" is "10101".
FALSE. 10101 is the binary for 16 + 4 + 1 = 21.
(d) Every perfect square can be written as either "4k" or "4k+1" for some integer k.
TRUE. If n = 2m, then n² = 4m² which is divisible by 4. If n = 2m + 1, then n² = 4m² + 4m + 1 which is a multiple of 4, plus 1. By the Division Theorem these are the only two possible cases.
(e) The square of every irrational number is irrational.
FALSE. We proved in lecture that the square root of 2 is irrational. So it is an irrational number whose square is rational.

Question 3 (20): Prove that if k is any integer, the number 7k + 5 is not a perfect square. (Hint: Argue the contrapositive, and apply the Division Theorem with m = 7 to the number you are squaring.)

We will show that if n is any integer, then n² is not of the form 7k + 5 because it has a remainder mod 7 of other than 5. Any integer n can be written as 7q + r where r is 0, 1, 2, 3, 4, 5, or 6. If r = 0, n² = 49q² which is a multiple of 7 and hence not of the form 7k + 5. If r = 1, m² = 49q² + 14q + 1, which has remainder 1, not 5, when divided by 7. In each of the other five cases, n² = 49q² + 14qr + r². This number, when divided by 7, has the same remainder as does r² because the other two terms are divisible by 7. With r = 2 this remainder is 4, with r = 3 it is 2 because (9 mod 7) = 2, with r = 4 it is (16 mod 7) = 2, with r = 5 it is (25 mod 7) = 4, and with r = 6 it is (36 mod 7) = 1. So it is never 5 for any perfect square.

Question 4 (20): Define a number sequence so that a₁ = 3 and for all k with k > 1, a_k = a_k-1 + 4k - 7. Prove by induction that for any positive integer n, a_n = 2n² - 5n + 6.

For our base case, we must prove P(1), which says that a₁ = 2(1)² - 5(1) + 6. Since 2 - 5 + 6 = 3, this matches the given value of a₁.

For the inductive case, we assume that a_k-1 = 2(n-1)² -5(n-1) + 6, which is 2n² - 4n + 2 - 5n + 5 + 6 = 2n² -9n + 13. Then, by the rule, a_n = a_n-1 + 4n - 7 = 2n² -9n + 13 + 4n - 7 = 2n² -5n + 6. This is exactly what the statement P(n) says that it should be.

Question 5 (30): Define a number sequence so that b₁ = 2, b₂ = 3, and for all k with k > 2, b_k = b_k-1 - b_k-2 + 1.

(a, 5) Calculate b_i for all integers i from i through 10.
b₃ = 3 - 2 = 1, b₄ = 1 - 3 = -2, b₅ = -2 - 1 = -3, b₆ = -3 - (-2) = -1, b₇ = -1 - (-3) = 2, b₈ = 2 - (-1) = 3, b₉ = 3 - 2 = 1, and b₁₀ = 1 = 3 = -2.
(b, 5) Determine the value of b₁₀₀. (Hint: I don't expect you to determine all the intermediate values -- find a pattern in part (a) or use the result of part (c).)
By the result of part (c), b₁₀₀ = b₉₄ = b₈₈, and so on removing 6 at a time until we reach b₄ = -2.
(c, 20) Prove by induction that for any n with n > 6, b_n = b_n-6. (Hint: Prove the base cases of n = 7 and n = 8, then use the sequence definition in your inductive case.)
First base case: We observe that b₇ and b₁ are both 2.
Second base case: We observe that b₈ and b2 are both 3.
Inductive case: Let n be any number with n > 8. We assume that b_n-1 = b_n-7, and that b_n-2 = b_n-8. We can do this because n-1 and n-2 are each greater than 6. Now the rule says that b_n = b_n-1 - b_n-2, and by our two assumptions this equals b_n-7 - b_n-8. But by the rule again, this latter expression equals b_n-6, so we have proved that b_n = b_n-6.

Last modified 20 October 2015