Download Solutions for Exam 1 - University of Hawaii Mathematics

ICS 141 Section 2
Exam 1
Name:
Thursday, September 16, 2004
Student ID:
Instructions. Write your name and your student id number in the spaces provided above. Each
problem is worth 12 points. Write your answers in the spaces provided on this exam. Do not use
your own paper. If you need scratch paper, use the back pages of the exam. You must justify your
answers to receive full credit. You may use a calculator. This exam is closed book and closed notes.
1. Complete the following sentences so that each one becomes an informal definition of the word
in italics:
(a) A proposition is a declaritive sentense that is either true or false, but not both.
(b) A predicate is a declaritive sentense that describes properties of its subject or subjects.
2. Write the following compound statements in symbols. Use the following letters to represent the
statements:
c: it is cold d: it is dry r: it is rainy w: it is warm
(a) It is neither cold nor dry
¬c ∧ ¬d ≡ ¬(c ∨ d)
(b) It is rainy if it is not dry
¬d → r
(c) To be warm it is necessary that it be dry
w→d
(d) It is cold or dry, but not both.
c⊕d
3. Let P (m, n) be the statement m ≤ n, where the universe of discourse for m and n is the set of
nonnegative integers. Determine the truth value of
(a) ∀n P (0, n). P (0, n) says 0 ≤ n, which is true for all nonnegative integers.
(b) ∃n∀m P (m, n). This proposition says that there is a nonnegative integer, n, that is greater
than or equal to every nonnegative integer. This is false since for any nonnegavite integer
n we may select the nonnegative integer m = n + 1 to see that m > n. Said another way,
∀n∃m ¬P (m, n) ≡ ¬(∃n∀m P (m, n)) is true.
(c) ∀m∃n P (m, n). For every nonnegative integer, m, we may select the nonnegative integer
n = m to see that m ≤ n. This proposition is true.
4. Let f : Z → Z be a function from the integers to the integers. In the notation of predicate logic,
to say that the function f is increasing means
∀x∀y
x < y → f (x) < f (y),
where the universe of discourse for x and y is the set of integers, Z. Using the notation of
predicate logic, state what it means to say that the function f is not increasing. (Hint: negate
the above proposition.)
¬ ∀x∀y
x < y → f (x) < f (y) ≡ ∃x∃y
x < y ∧ f (x) ≥ f (y)
2
5. Let A and B be subsets of a given universal set, U . Prove the following identities:
(a) The first absorption law: A = A ∪ (A ∩ B). Prove by showing mutual containment.
(i) Let x ∈ A. Then (x ∈ A) ∨ (x ∈ (A ∩ B)) (by addition) and therefore x ∈ A ∪ (A ∩ B).
Since every element of A is an element of A ∪ (A ∩ B), A ⊆ A ∪ (A ∩ B).
(ii) Let x ∈ A ∪ (A ∩ B), then (x ∈ A) ∨ (x ∈ (A ∩ B)) ≡ (x ∈ A) ∨ ((x ∈ A) ∧ (x ∈ B)).
The absorption law for propositions shows this is equivalent to x ∈ A. Since every
element of A ∪ (A ∩ B) is an element of A, A ∪ (A ∩ B) ⊆ A.
(iii) Since A ⊆ A ∪ (A ∩ B) and A ∪ (A ∩ B) ⊆ A, A = A ∪ (A ∩ B).
(b) One of De Morgan’s laws: A ∩ B = A ∪ B. Prove using a membership table.
A B A B A∩B A∩B A∪B
1 1 0 0
1
0
0
1 0 0 1
0
1
1
0
1
1
0 1 1 0
0 0 1 1
0
1
1
Since the last two columns are identical, A ∩ B = A ∪ B.
6. Determine whether the following argument is valid. Name the rule of inference or the fallacy.
If n is a real number such that n > 2, then n2 > 4. Suppose that n ≤ 2. Then n2 ≤ 4.
This is the fallacy of denying the hypothesis, so the argument is invalid.
7. Recall that if A is a set, then we denote the cardinality of A by |A|, and we denote the power set
of A by P (A). Suppose we are given two finite sets A and B, and suppose further that |A| = n,
|B| = m.
(a) What is |A × B|?
|A × B| = |A| · |B| = n · m.
(b) What is |P (A)|?
|P (A)| = 2|A| = 2n .
(c) Given |A ∩ B| = p, what is |A ∪ B|?
|A ∪ B| = |A| + |B| − |A ∩ B| = n + m − p.
8. Prove or disprove: if A, B, and C are sets, then A − (B ∩ C) = (A − B) ∩ (A − C).
Let U = {a, b, c} be the universal set, and let A = {a, b, c}, B = {b}, C = {c}. For this
example, A − (B ∩ C) = {a, b, c} and (A − B) ∩ (A − C) = {a}. So, there exist sets A, B and C
such that A − (B ∩ C) 6= (A − B) ∩ (A − C), which disproves the claim. (Note: I also accepted
solutions using Venn diagrams.)
3
9. For each set S, either find a set A for which S = P (A) or explain why A does not exist. P (A)
is the power set of A.
(a) S = {∅, {a}}.
A = {a}.
(b) S = {∅, {a}, {∅, a}}.
|S| = 3 and |P (A)| must be a power of 2, so there cannot be a set A such that S = P (A).
(c)
S = {∅, {∅}, {a}, {{a}}, {{{a}}}, {∅, a}, {∅, {a}}, {∅, {{a}}},
{a, {a}}, {a, {{a}}}, {{a}, {{a}}}, {∅, a, {a}},
{∅, a, {{a}}}, {∅, {a}, {{a}}}, {a, {a}, {{a}}}, {∅, a, {a}, {{a}}}}.
A = {∅, a, {a}, {{a}}}
10. (a) Which rule of inference justifies both
∃x∀y P (x, y)
∃y∀x ¬P (x, y)
and
?
∴ ∀y P (c, y), for some c
∴ ∀x ¬P (x, d), for some d
Existential Instantiation justifies both inferences.
(b) Use a proof by contradiction to prove that ∃x∀y P (x, y) → ∀y∃xP (x, y). Hint: the above
inferences will help.
Assume ∃x∀y P (x, y) ∧ ∃y∀x¬P (x, y). Then
(i) ∃x∀y P (x, y) by simplification,
(ii) and existential instantiation gives ∀y P (c, y), for some c.
(iii) A nearly identical argument gives ∀x ¬P (x, d), for some d.
Using the constant d from (iii) and universal instantiation, ∀y P (c, y) → P (c, d). Using
the constant c from (ii) and universal instantiation again, ∀x ¬P (x, d) → ¬P (c, d). The
conjuction inference yields P (c, d)∧¬P (c, d), a contradiction. Hence the original assumption
is false, and this proves that ∃x∀y P (x, y) → ∀y∃xP (x, y).