Download MATH 2320. Problem set 1.

MATH 2320. Problem set 1.
Instructions. Answer each question completely; justify your answers. This assignment
is due at 17:00 on Tuesday January 17 in Assignment Box #45.
1. The contrapositive and converse of an implication.
(a) (Contrapositive) Write down the truth tables of p → q and its contrapositive
¬q → ¬p and compare them.
Solution:
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
p
T
T
F
F
q
T
F
T
F
¬q → ¬p
T
F
T
T
An implication and its contrapositive are equivalent statements.
(b) (Converse) Write down the truth tables of p → q and its converse q → p and
compare them.
Solution:
p
T
T
F
F
q
T
F
T
F
p→q
T
F
T
T
p
T
T
F
F
q
T
F
T
F
q→p
T
T
F
T
An implication and its converse are in general non equivalent statements.
2. Each of the following statements is an implication. In each case identify the hypothesis
and the conclusion; and write down the converse, the contrapositive, and the negation.
(a) If n is an odd integer, then n2 + n − 2 is an even integer.
Solution:
•
•
•
•
•
(hypothesis) n is an odd integer.
(conclusion) n2 + n − 2 is an even integer
(converse) If n2 + n − 2 is an even integer, then n is an odd integer.
(contrapositive) If n2 + n − 2 is an odd integer, then n is a even integer.
(negation) n and n2 + n − 2 are odd integers.
(b) The product of rational numbers is rational.
Solution:
First we write the statement in the ”if-then” form; the trick is to give names to
the objects that the statement is refering to. In this case two arbitrary numbers
x and y
if x and y are rational numbers then x + y is rational.
•
•
•
•
(hypothesis) x and y are rational numbers.
(conclusion) x + y is a rational number.
(converse) If x + y is a rational number, then x and y are rational numbers.
(contrapositive) If x + y is an irrational number, then either x is irrational
or y is irrational.
• (negation) x and y are rational numbers, and x + y is an irrational number.
(c) If p(x) is a polynomial of odd degree, then p(x) has at least one real root.
Solution: The objects that statement is refering to are polynomials:
• (hypothesis) p(x) is a polynomial of odd degree.
• (conclusion) p(x) has at least one real root.
• (converse) If p(x) is a polynomial with at least one root, then p(x) has odd
degree.
• (contrapositive) If p(x) is a polynomial with no roots, then p(x) has even
degree.
• (negation) p(x) is a polynomial of odd degree and p(x) has no roots.
3. Write down the negation of each of the following statements.
(a) sin(x) is a periodic function and cos(x) is an even function.
Solution: sin(x) is not a periodic function or cos(x) is not an even function.
(b) Every integer is divisible by a prime.
Solution: There is an integer that is not divisible by a prime.
(c) There exists an infinite set whose proper subsets are all finite.
Solution: Every infinite set has a proper infinite subset.
4. Exhibit a counterexample to each of the following statements.
(a) If a and b are integers and ab = 1 then a = b = 1.
Solution: Let a = −1 and b = −1 and observe that ab = 1 but a 6= 1.
(b) If the average of four different integers is 10, at least one of the integers must be
greater than 12.
Solution: Consider the collection of integers 8, 9, 11, 12; they are all different,
their average is 10, and all them are less than or equal to 12.
5. Rewrite each of the following statements using the quantifiers ”for all” (or ”for every”)
and ”there exists” as appropiate.
(a) All positive real numbers have real square roots.
Solution: Any of the following answers is correct:
• For every positive real number x, there is a real number y such that y 2 = x.
• For every real number x, if x is positive then there exists y such that y 2 = x.
• ∀x ∈ R(x > 0 → ∃y(y 2 = x)).
(b) Not every polynomial has a real root.
Solution: Any of the following answers is a correct answer. The second answer
captures the fact the verb tohave is an instance of a quantifier.
• There exists a polynomial p(x) with no real roots.
• There exists a polynomial p(x) such that for every real number α, p(α) 6= 0.
(c) There is no smallest positive real number.
Solution: Any of the following answers is a correct answer.
• For every positive real number x there is a positive real number y such that
y < x.
• For every real number x, if x is positive then there is a real number y such
that 0 < y and y < x.
• ∀x ∈ R(x > 0 → ∃y(0 < y ∧ y < x)).
6. Construct a truth table for the statement (p ∧ q) ∨ ((¬p) ∨ q).
Solution: Before constructing the truth table, we can simplify the statement using
the basic logical equivalences. For example, the following is a sequence of logical
equivalences:
1. (p ∧ q) ∨ ((¬p) ∨ q)
2. ((p ∧ q) ∨ (¬p)) ∨ q
3. ((p ∨ ¬p) ∧ (q ∨ ¬p)) ∨ q
4. (q ∨ ¬p) ∨ q
5. (q ∨ ¬p) ∨ q
6. q ∨ ¬p
Therefore (p ∧ q) ∨ ((¬p) ∨ q) and q ∨ ¬p are logically equivalent, hence they have the
same truth table:
p
T
T
F
F
q
T
F
T
F
¬p ∧ q
F
F
T
F
(p ∧ q) ∨ ((¬p) ∨ q)
F
F
T
F
7. Find a compound statement in p, q, r with the following truth table.
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
r
T
F
T
F
T
F
T
F
?
T
T
F
F
F
F
F
F
Solution: We discussed an algorithm that solves this problem; the solution is
(p ∧ q ∧ r) ∨ (p ∧ q ∧ ¬r).