Problem Set Week 1
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Southern New Hampshire University *
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Course
230
Subject
Mathematics
Date
Apr 3, 2024
Type
docx
Pages
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Uploaded by ColonelJaguar2786
MODULE ONE PROBLEM SET
This document is proprietary to Southern New Hampshire
University. It and the problems within may not be posted on
any non-SNHU website.
Rafael V. Canseco
1
Directions: Type your solutions into this document and be
sure to show all steps for arriving at your solution. Just giving a
final number may not receive full credit. section*Problem 1
In the following question, the domain of discourse is a set of
male patients in a clinical study. Define the following
predicates:
•
P
(
x
) : x was given the placebo
•
D
(
x
) : x was given the medication
•
M
(
x
) : x had migraines
Translate each of the following statements into a logical
expression. Then negate the expression by adding a negation
operation to the beginning of the expression. Apply De
Morgan’s law until each negation operation applies directly to a
predicate and then translate the logical expression back into
English.
Sample question: Some patient was given the placebo and the medication.
•
∃
x (
P
(
x
) ∧
D
(
x
))
•
Negation: ¬
∃
x (
P
(
x
) ∧
D
(
x
))
•
Applying De Morgan’s law: ∀
x (¬
P
(
x
) ¬
∨
D
(
x
))
•
English: Every patient was either not given the placebo or not
given the medication (or both).
(a)
Every patient was given the medication or the placebo or both.
(b)
Every patient who took the placebo had migraines. (Hint: you
will need toapply the conditional identity, p → q ≡¬
p ∨
q
.)
(c)
There is a patient who had migraines and was given the
placebo.
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Problem 1
In the following question, the domain of discourse is a set of
male patients in a
clinical study. Define the following predicates:
•
P
(
x
) : x was given the placebo
•
D
(
x
) : x was given the medication
•
M
(
x
) : x had migraines
Translate each of the following statements into a logical
expression. Then negate the expression by adding a negation
operation to the beginning of the expression. Apply De
Morgan’s law until each negation operation applies directly to a
predicate and then translate the logical expression back into
English.
Sample question: Some patient was given the placebo and the medication.
•
∃
x (
P
(
x
) ∧
D
(
x
))
•
Negation: ¬
∃
x (
P
(
x
) ∧
D
(
x
))
•
Applying De Morgan’s law: ∀
x (¬
P
(
x
) ¬
∨
D
(
x
))
•
English: Every patient was either not given the placebo or not
given the medication (or both).
(a)
Every patient was given the medication or the placebo or both.
Logical expression: x(D(x) P(x))
∀
∨
Negation: ¬(
x(D(x) P(x)))
∀
∨
Applying De Morgan’s law: x(¬D(x) ¬P(x))
∃
∧
English Translation: There exists at least one patient
who was not given the medication and not given the
placebo.
(b)
Every patient who took the placebo had migraines. (Hint: you
will need toapply the conditional identity, p → q ≡¬
p ∨
q
.)
Logical expression: x(D(x) P(x))
∀
∨
Negation: ¬(
x(D(x) P(x)))
∀
∨
Applying De Morgan’s law: x(¬D(x) ¬P(x))
∃
∧
English Translation: There exists at least one patient
who was not given the medication and not given the
placebo.
(c)
There is a patient who had migraines and was given the
placebo.
Logical expression: x(D(x) P(x))
∀
∨
Negation: ¬(
x(D(x) P(x)))
∀
∨
Applying De Morgan’s law: x(¬D(x) ¬P(x))
∃
∧
English Translation: There exists at least one patient
who was not given the medication and not given the
placebo.
(d)
Problem 2
Use De Morgan’s law for quantified statements and the laws
of propositional logic to show the following equivalences:
(a)
¬
∀
x (
P
(
x
) ¬
∧
Q
(
x
)) ≡ ∃
x (¬
P
(
x
) ∨
Q
(
x
))
Proof:
o
Apply De Morgan's Law: ¬
x (P(x) ¬Q(x)) ≡ x
∀
∧
∃
¬(P(x) ¬Q(x))
∧
o
Apply De Morgan's Law again: ≡ x (¬P(x) ∃
∨
Q(x))
(b)
¬
∀
x (¬
P
(
x
) → Q
(
x
)) ≡ ∃
x (¬
P
(
x
) ¬
∧
Q
(
x
))
Proof: o
Apply the contrapositive: ¬
x (P(x) Q(x))
∀
∨
o
Apply De Morgan's Law: ≡ x ¬(P(x) Q(x))
∃
∨
o
Apply De Morgan's Law again: ≡ x (¬P(x) ∃
∧
¬Q(x))
Proof: o
Apply De Morgan's Law: ¬
x(¬P(x)
∃
(Q(x) ¬R(x))) ≡ x ¬(¬P(x) ∨
∧
∀
∨
(Q(x) ¬R(x)))
∧
o
Apply De Morgan's Law again: ≡ x
∀
(P(x) (¬Q(x) R(x)))
∧
∨
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Problem 3
The domain of discourse for this problem is a group of three
people who are working on a project. To make notation easier,
the people are numbered 1
, 2
, 3. The predicate M
(
x, y
) indicates
whether x has sent an email to y
, so M
(2
, 3) is read “Person 2
has sent an email to person 3.” The table below shows the value
of the predicate M
(
x, y
) for each (
x, y
) pair. The truth value in
row x and column y gives the truth value for M
(
x, y
).
M
1
2
3
1
T
T
T
2
T
F
T
3
T
T
F
Determine if the quantified statement is true or false. Justify your answer.
(a)
∀
x
∀
y (
x 6= y
) → M
(
x, y
))
All pairs of distinct people have received emails, which matches the predicted M(x,y) for all x≠y.
(b)
∀
x
∃
y ¬
M
(
x, y
)
For each person, there exists at least one person to whom they have not sent email.
(c)
∃
x
∀
y M
(
x, y
)
There does not exist a single person who has sent emails to everyone else. Problem 4
Translate each of the following English statements into
logical expressions. The domain of discourse is the set of all
real numbers.
(a)
The reciprocal of every positive number less than one is
greater than one.
(a) Translation: x (x > 0 x < 1) → (1/x > 1)
∀
∧
o
Explanation: For every real number x greater
than 0 and less than 1, the reciprocal of x is
greater than 1.
(b)
There is no smallest number.
(b) Translation: ¬
x (x is the smallest number
∃
o
Explanation: There does not exist a smallest
number.
(c)
Every number other than 0 has a multiplicative inverse.
(c) Translation: x ≠ 0, (
y, y ≠ 0, xy = 1)
∀
∃
i
Explanation: For every real number x not equal to 0,
there exists a real number y not equal to 0 such that
their product equals 1.
Problem 5 The sets A
, B
, and C are defined as follows:
A
= tall,grande,venti
B
= foam,no – foam
C = non −
fat,whole
Use the definitions for A
, B
, and C to answer the questions. Express
the elements using n
-tuple notation, not string notation.
(a) Write an element from the set A × B × C
.
(a) An element from A × B × C: (tall,
foam, non-fat)
i
Explanation: This represents an
element
where
the
first
component is from set A (tall),
the second component is from
set B (foam), and the third
component is from set C (non-
fat).
(b) Write an element from the set B × A × C
.
(b) An element from B × A × C:
(foam, tall, non-fat)
i
Explanation: This represents an
element
where
the
first
component is from set B (foam),
the second component is from
set A (tall), and the third
component is from set C (non-
fat).
(c) Write the set B × C using roster notation.
(c) Set B × C using roster notation:
{(foam, non-fat), (foam, whole), (no-
foam, non-fat), (no-foam, whole)}
i
Explanation: This set contains
tuples
where
the
first
component is from set B and the
second component is from set C.
It
includes
all
possible
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combinations of elements from
sets B and C.