Math 280
Strategies of Proof
Homework Solutions
2.1 Propositional Functions and Quantified Statements
Spring 2019
4. For each of the following sentences:
(i) Rewrite the sentence formally using the variable n, with domain Z, the set of all integers.
(ii) Translate each to symbolic form using the following propositional functions with domain Z.
p(n) : n is even.
r(n) : n is divisible by 3.
q(n) : n is a perfect square.
s(n) : n is divisible by 4.
(iii) Determine the truth value of each statement. Justify your conclusions.
(iv) Find the negation of each statement in both symbolic form and in English.
(a) All integers are either perfect squares or are divisible by 4.
This is equivalent to
For all integers n, n is a perfect square or n is divisible by 4.
This has symbolic form
⇥
⇤
8n q(n) _ s(n) .
This is true provided that the propositional function q(n) _ s(n) is true for all integers n. Let
n = 2. Then
q(2) _ s(2) = F _ F = F.
Therefore the quantified statement
⇥
⇤
8n q(n) _ s(n) = F
with counterexample n = 2.
The negation is
⇥
⇤
⇥
⇤
⇥
⇤
⇠ 8n q(n) _ s(n) ⌘ 9n ⇠ q(n) _ s(n) ⌘ 9n ⇠ q(n)^ ⇠ s(n)
In English, this is
There exists an integer n such that n is not a perfect square and n is not divisible by 4.
(b) All perfect squares are either divisible by 3 or divisible by 4.
This is a universal statement so should be interpreted as a conditional statement, which is equivalent to
For all integers n, if n is a perfect square, then n is divisible by 3 or divisible by 4.
This has symbolic form
⇥
⇤
8n q(n) ! r(n) _ s(n) .
This is true provided that the propositional function q(n) ! r(n)_s(n) is true for all integers n.
Let n = 25. Then
q(25) ! r(25) _ s(25) = T ! (F _ F) = F.
Therefore the quantified statement
⇥
⇤
8n q(n) ! r(n) _ s(n) = F
with counterexample n = 25.
The negation is
⇥
⇤
⇥
⇤
⇠ 8n q(n) ! r(n) _ s(n) ⌘ 9n ⇠ q(n) ! r(n) _ s(n)
⇥
⇤
⌘ 9n q(n)^ ⇠ r(n) _ s(n)
⇥
⇤
⌘ 9n q(n)^ ⇠ r(n)^ ⇠ s(n)
In English, this is
There exists an integer n such that n is a perfect square but n is not divisible by 3 and n is
not divisible by 4.
24
Math 280 Homework Solutions page 25
2.1
Propositional Functions and Quantified Statements
4. (continued)
(c) Each even integer is a perfect square.
This is equivalent to
For all integers n, if n is even, then n is a perfect square.
This has symbolic form
⇥
⇤
8n p(n) ! q(n) .
This is true provided that the propositional function p(n) ! q(n)) is true for all integers n. Let
n = 6. Then
p(6) ! q(6) = T ! F = F.
Therefore the quantified statement
⇥
⇤
8n p(n) ! q(n) = F
with counterexample n = 6.
The negation is
⇥
⇤
⇥
⇤
⇥
⇤
⇠ 8n p(n) ! q(n) ⌘ 9n ⇠ p(n) ! q(n) ⌘ 9n p(n)^ ⇠ q(n)
In English, this is
There exists an integer n such that n is even but n is not a perfect square.
(d) Some even integers are perfect squares or divisible by 3.
Since this is an existential statement, this should be interpreted as an “and” statement, which
is equivalent to
There exists an integer n such that n is even and either n is a perfect square or n is divisible
by 3.
This has symbolic form
⇥
⇤
9n p(n) ^ q(n) _ r(n) .
This is true provided that there exists at least one integer n for which the propositional function
p(n) ^ q(n) _ r(n) is true. Let n = 36. Then
p(36) ^ q(36) _ r(36) = T ^ (T ^ T) = T ^ T = T.
Therefore the quantified statement
with example n = 36.
The negation is
⇥
⇤
9n p(n) ^ q(n) _ r(n) = T
⇥
⇤
⇥
⇤
⇠ 9n p(n) ^ q(n) _ r(n) ⌘ 8n ⇠ p(n) ^ q(n) _ r(n)
⇥
⇤
⌘ 8n ⇠ p(n) _ ⇠ q(n)^ ⇠ r(n)
In English, this is
For all integers n, either n is not even or n is neither a perfect square nor is divisible by 3.
2.1
Propositional Functions and Quantified Statements
Math 280 Homework Solutions page 26
4. (continued)
(e) Some perfect squares are neither even nor divisible by 3.
Since this is an existential statement, this should be interpreted as an “and” statement, which
is equivalent to
There exists an integer n such that n is a perfect square but n is neither even nor divisible
by 3.
This has symbolic form
⇥
⇤
9n q(n) ^ ⇠ p(n)^ ⇠ r(n) .
This is true provided that there exists at least one integer n for which the propositional function
q(n) ^ ⇠ p(n)^ ⇠ r(n) is true. Let n = 25. Then
q(25) ^ ⇠ p(25)^ ⇠ r(25) = T ^ (⇠ F^ ⇠ F) = T ^ T = T.
Therefore the quantified statement
⇥
⇤
9n q(n) ^ ⇠ p(n)^ ⇠ r(n) = T
with example n = 25.
The negation is
⇥
⇤
⇥
⇤
⇠ 9n q(n) ^ ⇠ p(n)^ ⇠ r(n)
⌘ 8n ⇠ q(n) ^ ⇠ p(n)^ ⇠ r(n)
⇥
⇤
⌘ 8n ⇠ q(n) _ p(n) _ r(n)
In English, this is
For all integers n, n is not a perfect square or n is even or n is divisible by 3.
(f) All even perfect squares are divisible by 3.
This is equivalent to
For all integers n, if n is even and n is a perfect square, then n is divisible by 3.
This has symbolic form
⇥
⇤
8n p(n) ^ q(n) ! r(n) .
This is true provided that the propositional function p(n) ^ q(n) ! r(n) is true for all integers
n. Let n = 4. Then p(4) ^ q(4) ! r(4) = (T ^ T) ! F = F. Therefore the quantified statement
⇥
⇤
8n p(n) ^ q(n) ! r(n) = F
with counterexample n = 4.
The negation is
⇥
⇤
⇥
⇤
⇥
⇤
⇠ 8n p(n) ^ q(n) ! r(n) ⌘ 9n ⇠ p(n) ^ q(n) ! r(n) ⌘ 9n p(n) ^ q(n) ^ ⇠ r(n)
In English, this is
There exists an integer n such that n is even and n is a perfect square but n is not divisible
by 3.
Math 280 Homework Solutions page 27
2.1
Propositional Functions and Quantified Statements
4. (continued)
(g) Some even perfect squares are not divisible by 3.
This is equivalent to
There exists an integer n such that n is even and n is a perfect square but n is not divisible
by 3.
This has symbolic form
⇥
⇤
9n p(n) ^ q(n)^ ⇠ r(n) .
This is true provided that there exists at least one integer n for which the propositional function
p(n) ^ q(n)^ ⇠ r(n) is true. Let n = 4. Then
p(4) ^ q(4)^ ⇠ r(4) = T ^ T^ ⇠ F = T.
Therefore the quantified statement
with example n = 4.
The negation is
⇥
⇤
9n p(n) ^ q(n)^ ⇠ r(n) = T
⇥
⇤
⇥
⇤
⇠ 9n p(n) ^ q(n)^ ⇠ r(n) ⌘ 8n ⇠ p(n) ^ q(n)^ ⇠ r(n)
⇥
⇤
⌘ 8n ⇠ p(n)_ ⇠ q(n) _ r(n))
In English, this is
For all integers n, n is not even or n is not a perfect square or n is divisible by 3.
(h) Any integer is a perfect square if and only if it is divisible by 3 or by 4.
This is equivalent to
For all integers n, n is a perfect square if and only if n is divisible by 3 or n is divisible by 4.
This has symbolic form
⇥
⇤
8n q(n) $ r(n) _ s(n) .
This is true provided that the propositional function q(n) $ r(n) _ s(n) is true for all integers
n, and is false provided that the propositional function is false for at least one integer x. Let
n = 25. Then
q(25) $ r(25) _ s(25) = T $ (F _ F) = T $ F = F.
Therefore the quantified statement
⇥
⇤
8n q(n) $ r(n) _ s(n) = F
with counterexample n = 25.
The negation is
⇥
⇤
⇥
⇤
⇠ 8n q(n) $ r(n) _ s(n) ⌘ 9n ⇠ q(n) $ r(n) _ s(n)
⇥
⇤
⌘ 9n q(n) ^ ⇠ r(n)^ ⇠ s(n) _ ⇠ q(n) ^ r(n) _ s(n)
In English, this is
There exists an integer n such that either n is perfect square but is neither divisible by 3 nor
by 4 or n is not a perfect square but is divisible by either 3 or 4.
2.1
Math 280 Homework Solutions page 28
Propositional Functions and Quantified Statements
5. For each of the following mathematical definitions, transform the condition (the part following the
phrase “provided that”) in the definition to symbolic form. Then find the negation of the condition
in symbolic form and carefully and precisely write out in English what it means for the definition not
to be satisfied.
(a) Two sets A and B are equal provided that for every x, x 2 A if and only if x 2 B.
In symbolic form, A and B are equal provided that
Alternatively, we can let
8x [(x 2 A) $ (x 2 B)].
p(x) : x 2 A
q(x) : x 2 B.
Then the symbolic form can be written as
8x [p(x) $ q(x)].
Taking the negation, we obtain that A and B are not equal provided that
⇠ 8x [(x 2 A) $ (x 2 B)] ⌘ 9x ⇠ [(x 2 A) $ (x 2 B)]
⇥
⇤
⌘ 9x (x 2 A) ^ (x 2
/ B) _ (x 2
/ A) ^ (x 2 B)
Therefore two sets A and B are not equal provided that there exists at least one element x such
that either x 2 A but x 2
/ B or x 2 B but x 2
/ A.
(b) An integer n is divisible by an integer d provided that there exists an integer k such that n = kd.
In symbolic form, n is divisible by d provided that
9k [n = kd],
where the domain is the set of integers Z. Alternatively, we can let
p(k) : n = kd.
Then the symbolic form can be written as
9k p(k).
Taking the negation, we have n is not divisible by d provided that
⇠ 9k [n = kd] ⌘ 8k ⇠ [n = kd] ⌘ 8k [n 6= kd].
Therefore n is not divisible by d provided that for all integers k, n 6= kd.
(c) An element m of a set S of real numbers is a minimal element of S provided that for any x 2 S,
if x  m, then x = m.
In symbolic form, m is a minimal element of S provided that
8x [(x  m) ! (x = m)],
where the domain for x is the set S. Alternatively, we can let
p(x) : x  m
q(x) : x = m.
Then the symbolic form can be written as
8x [p(x) ! q(x)].
Taking the negation, m is not a minimal element of S provided that
⇠ 8x [(x  m) ! (x = m)] ⌘ 9x ⇠ [(x  m) ! (x = m)]
⌘ 9x [(x  m)^ ⇠ (x = m)]
⌘ 9x [(x  m) ^ (x 6= m)]
Therefore m is not a minimal element of S provided that there exists an element x in S such that
x  m but x 6= m, which is equivalent to x < m.