Math 215: Final Exam (Solutions)

Prof. S. Smith  1–3 pm, LC D1  Tuesday 30 April 2002

Each of the 4 problems is worth 50 points. (Some with parts indicate subdivision of points). Notation: N denotes the natural numbers 1, 2, 3, …; Z is the integers; R is the real numbers.

Problem 1: Define a function f : R^{2} → R^{2} by f(x_{1}, x_{2}) = (x_{2}, x_{1} + x_{2}).

(a) (25 points) Is f injective ? You must PROVE your answer.

Yes: For general (a_{1}, a_{2}) and (b_{1}, b_{2}), we assume f(a_{1}, a_{2}) = f(b_{1}, b_{2}), and we must show this forces (a_{1}, a_{2}) = (b_{1}, b_{2}). PROOF: We are assuming (a_{2}, a_{1} + a_{2}) = (b_{2}, b_{1} + b_{2}); so a_{2} = b_{2} and a_{1} + a_{2} = b_{1} + b_{2}. Substituting the first equation into the second, we get a_{1} = b_{1}. So our assumption implies (a_{1}, a_{2}) = (b_{1}, b_{2}), and hence f is injective.

(b) (25 points) Is f surjective ? You must PROVE your answer.

Yes: For general (y_{1}, y_{2}), we assume there is some (x_{1}, x_{2}) with f(x_{1}, x_{2}) = (y_{1}, y_{2}), and we must solve for (x_{1}, x_{2}) in terms of (y_{1}, y_{2}). PROOF: We are assuming (x_{2}, x_{1} + x_{2}) = (y_{1}, y_{2}), so x_{2} = y_{1} and x_{1} + x_{2} = y_{2}. Substituting the first equation into the second, we get x_{1} = −y_{1} + y_{2}. So we do get a solution (x_{1}, x_{2}) = (−y_{1} + y_{2}, y_{1}), and hence f is surjective.

Problem 2: Let S denote the set of 3 × 3 “skew-symmetric” matrices over the integers Z:

These are the matrices of form ( \begin{matrix}0 & a & b \\ −a & 0 & c \\ −b & −c & 0 \end{matrix} ) where a, b, c ∈ Z.

Show that S is denumerable (i.e., countably infinite). You may quote standard theorems. Define a map f : S → Z^{3} taking the above matrix to (a, b, c). Then f is easily checked to be a bijection. So by standard theorems: First Z is denumberable, then the Cartesian product Z^{3} is denumerable, and so we conclude S is denumerable.

Problem 3: Three customers each purchase one television from 4 possible models.

(a) (25 points) How many possible purchases are there ? (i.e., keeping track of which customer purchased what)

The set C of customers has size |C| = 3,

and the set T of choices for a model of a television has size |T | = 4. The purchases are functions F un(C, T), or more simply just lists, i.e. triples in T^{3}. Thus the number of purchases is |T |^{|C|} or |T |^{3}, namely 4^{3} = 64.

(b) (25 points) How many possible purchases are there, if we only keep track of how many of each model was purchased ? (that is, NOT keeping track of which customer ordered which model) Show work.

There are 3 cases: all 3 customers order the same model; or 2 customers order one model while the third customer orders a different model; or all 3 customers order different models:

Same: There are \binom{4}{1} = 4 choices of one model.

2 and 1: There are \binom{4}{1} = 4 choices of a model ordered by 2 customers;

then \binom{3}{1} = 3 choices for the model ordered by the remaining customer, hence 4 · 3 = 12 choices in this case.

All different: There are \binom{4}{3} = 4 choices of 3 different models. Hence there are 4 + 12 + 4 = 20 different possible purchases.

Problem 4:

(a) (25 points) Show that if a positive integer n is a perfect square (that is, n = a^{2} for a ∈ Z), then n has remainder 0, 1, or 4 when divided by 5. Conclude 111111113 is not a perfect square.

Since a has remainder 0,1,2,3, or 4, and these numbers square to 0,1,4,9,16,

we see that n = a^{2} has remainder 0,1,4,4,1; that is, 0,1, or 4. So since 111111113 has remainder 3, it is not a perfrect square.

(b) (25 points) Use the Euclidean algorithm to find the greatest common divisor d = gcd(210, 112). Then for that number d, find m, n ∈ Z such that d = 210m + 112n.

Euclidean algorithm: 210 = 112 · 1 + 98 112 = 98 · 1 + 14 98 = 14 · 7 + 0, so that 14 = gcd(210, 112). Then by the “starting at the bottom” method: 14 = 112 − 98 = 112 · 1 − (210 · 1 − 112 · 1) = 210 · (−1) + 112 · 2

so one possibility is m = −1, n = 2.