Math 215: Hour Exam 2 (Solutions) Prof. S. Smith Friday 5 April 2002

Each problem is worth 25 points. (Some problems 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:

(a) (10 points) As usual π means 3.1415 …. Let P denote the statement:

(∀x ∈ Z) (∃y ∈ Z) , x + y = π.

(i) Use the rules to write out not-P (so quantifiers come first, with “not” later). (i) not-P becomes : (∃x ∈ Z) (∀y ∈ Z) , x + y ≠ π.

(ii) Which is true, P or not-P ? (Why?)

(ii) not-P is true; PROOF: Since x, y ∈ Z, also x + y ∈ Z; so x + y ≠ π as π ∉ Z.

(b) (15 points: New problem, forget about P above) Prove the new statement:

(∀x ∈ Z) , [ (∃y ∈ Z, x = 2y) ⇒ not-( (∃z ∈ Z), x^{2} = 2z + 1 ) ] .

What does this statement say in the usual language of even and odd integers ?

Probably easiest by contradiction: Given: [(∃y), x = 2y] and [(∃z), x^{2} = 2z + 1]; Goal: contradiction. PROOF: 4y^{2} = x^{2} = 2z + 1, so 1 = 4y^{2} − 2z = 2(2y^{2} − z) is even, contradicting that 1 is odd. In other words, again: “The square of an even integer is even.”

Problem 2: Define f : R → R by the formula f(x) = 5x − 7.

(a) (10 points) Is f injective ? (Why/why not)

Yes. PROOF: If 5x − 7 = 5y − 7, then 5x = 5y, and so x = y.

(b) (10 points) Is f surjective ? (Why/why not)

Yes. PROOF: Solve y = 5x − 7 for x; that is, 5x = y + 7 so x = \frac{y+7}{5}.

(c) (5 points) Is f invertible ? (If not, why not; if so, give f^{−1})

Yes. PROOF: By (a) and (b), f is bijective. By (b), f^{−1}(x) = \frac{x+7}{5}.

Problem 3: For sets A, B, and C:

(a) (10 points) Assume that |A| = 3 and |B| = 4. What is the number |F un(A, B)| of functions from A to B ?

A function f : A → B can be given as a list of values (f(a_{1}), f(a_{2}), f(a_{3})) where each f(a_{i}) ∈ B; so the number of choices is 4^{3} = 64.

(b) (15 points) Assume that:

|A| = 5, |B| = 6, |C| = 7; |A ∩ B| = 4, |A ∩ C| = 3, |B ∩ C| = 2; and |A ∩ B ∩ C| = 1. What is |A ∪ B ∪ C| ? (Why?)

By inclusion/exclusion, |A ∪ B ∪ C| = (5 + 6 + 7) − (4 + 3 + 2) + 1 = 18 − 9 + 1 = 10. My numbers were inconsistent: reverse 4,3,2 to 2,3,4 in order to make the Venn diagram all

nonnegative.

Problem 4: Let M denote the set of 2 × 2 matrices ( \begin{matrix}a & b \\ c & d \end{matrix} ) with coefficients a, b, c, d ∈ Z. Show that M is a denumerable set (that is, countably infinite).

(Hint: Give a bijection of M with Z^{m} for a suitable m > 1. Then apply known theorems.) PROOF: The map taking ( \begin{matrix}a & b \\ c & d \end{matrix} ) to (a, b, c, d) is a bijection (easy to prove) from M to Z^{4}. By standard theorems: Z is denumerable, so the Cartesian product Z^{4} is denumerable;

so M is denumerable.