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Chapter 1
Logic
Mathematicians make explicit our assumptions about mathematical structures.
These assumptions are call axioms. From axioms we deduce, logically, other properties and connections. Hence, a requirement for being able to write a mathematical
proof is an understanding of logic. While we will not delve into the foundations
of logic here, we will clean up and make more precise our everyday experience of
logical notions (without ever saying what logic is).
If we examine many theorems in mathematics, we notice a few properties. First
theorems are (hopefully) true statements. They often contain connective words like
if, then, such that, exists, for all, or, and, ... interspersed with mathematical symbols. Moreover, the proofs of theorems do not involve using the English language
to argue their validity. To begin, we examine the words if, then, such that, exists,
for all, or, and, in the context of logic, more carefully.
First, as always, we carefully define a new idea or concept.
Definition 1.1. A statement is a sentence that is either true or false.
Example 1.2. Consider the following sentences.
(1) 1=0.
(2) There exists a θ such that cos θ = θ.
(3) This statement is false.
(4) x2 − 1 = 0
The first two are statements. The last two are not. In (3) the sentence is neither
true or false. In the last, (4), we are not told what x is, and so we are unable to
check if it is true or false.
We can combine two statements to form a new statement. We do this with
logical connectives. Table 4 below summarizes some of the possible connectives we
will encounter. Here they are in detail.
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1. Logic
Conjunction: If P and Q are two statements, then P and Q, denoted P ∧ Q is
called the conjunction. The new connective statement is true only when both P
and Q are true and is false otherwise. One way summarize the truth value of P ∧ Q
is with a truth table:
P
T
F
T
F
Q
T
T
F
F
P ∧Q
T
F
F
F
Table 1. Truth table for a Conjunction.
Disjunction: If P and Q are two statements, then P or Q, denoted P ∨ Q is called
the disjunction of P and Q. In mathematics or is always inclusive. That is, P ∨ Q
is true if either P is true, Q is true or both are true. In the English language or is
almost always exclusive. For example, if someone in a restaurant said
you can have French fries or a salad
you would probably interpret or to mean French fries or a salad, but not both. If
you wanted to convey an inclusive or, you would have to say “you can have French
fries or a salad or both.”
There is no ambiguity in the use of or in mathematics - it is always inclusive.
Here is a truth table for a disjunction:
P
T
F
T
F
Q
T
T
F
F
P ∨Q
T
T
T
F
Table 2. Truth table for a disjunction.
Negation: If P is a statement, then ¬P is call the negation of P . When P is true,
¬P is false, and when P is false, ¬P is true.
Example 1.3. Find the negation of the following statements.
(a) Today is Friday
(b) There is a broken chair in this room.
(c) x is less than 2.
(d) Seven is an even number
(e) All the chairs are broken in the room.
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1. Logic
Solution.
(a) It is false today is Friday. Or, today is not Friday.
(b) All of the chairs are not broken in this room.
(c) It is false x is less than 2. Or, x is greater or equal to 2.
(d) Seven is not an even number. Or, seven is an odd number.
(e) At least one chair is not broken in the room.
Conditional Statements The conditional connective is probably the most frequent connective we encounter. It is denoted P → Q and read “if P , then Q.”
There are many equivalent ways of expressing “P implies Q.” Indeed, the following
are all equivalent.
- If P , then Q
- Q if P
- P only if Q
- P is a sufficient condition for Q
- Q is a necessary condition for P
- Q whenever P
Always P is called the hypothesis and Q is called the conclusion. The following
truth table summarizes the true value for P → Q
P
T
F
T
F
Q
T
T
F
F
P →Q
T
T
F
T
Table 3. Truth table for a conditional connective.
A summary of the most common used connectives is given in Table 4.
Name
Symbol
Meaning
Negation
¬
not P
Conjunction
∧
P and Q
Disjunction
∨
P or Q
Conditional P → Q
P implies Q
Biconditional P ↔ Q P if and only if Q
Condition for Truth
P false
both true
at least one true
Q true whenever P true
P , Q same truth value
Table 4. List of connective symbols and their meaning.
Definition 1.4. The converse of P implies Q is Q implies P . That is, the converse
of P → Q is Q → P .
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Some examples follow.
Example 1.5. Suppose you are playing music in a band and your part says
Play only if four trombones
and you notice there are four trombone players present. Do you play the part?
Solution. We let P be the statement “we play the part”, and Q be the statement
“there are four trombones.” Then P only if Q is equivalent to if P , then Q. That
is, if we play, P is true, then Q must be true. However, if Q is true, then the truth
value of P does not matter, the connective statement will always be true. Thus if
four trombones are present, we can play the part or not. However, if there are only
three trombones, then we cannot play the part.
Example 1.6. Suppose the following conditional statement is true.
If Jack is younger than his father, then Jack will not lose the contest.
Does Jack win the contest?
Solution. Let P be the statement “Jack is younger than his father” and Q be
”Jack will not lose the contest.” Then P → Q is a true statement. Since P is true,
it must be that Q is true. Jack will win.
Example 1.7. How would you prove P → Q is false? Use it to prove “If x > 0,
then log7 x > 0” is false.
Solution. We need Q to be false with P true to show P → Q is false. Recall
log7 x = y if and only if 7y = x. If we set x = 1/7, then x > 0 and log7 x = −1.
Then P is true while Q is false.
Example 1.8. Write the following statements using the logical connectives in Table
4.
(a) an exclusive or
(b) e ≥ 2
(c) 2 ≤ e < 3
√
(d) 7 ≤ 2
Solution.
(a) An exclusive “P or Q” means P or Q but not both. That is, (P ∨Q)∧¬(P ∧Q).
(b) Let P be the statement e is larger than 2, and Q be the statement e equals 2.
Then e ≥ 2 is equivalent to P ∨ Q.
(c) Similar to Part (b) we see 2 ≤ e ≤ 3 is the same as [(2 < e) ∧ (e = 2)] ∧ (e < 3).
√
√
(d) This is equivalent to ¬[( 7 < 2) ∨ ( 7 = 2)].
Example 1.9. State the hypothesis and conclusion in the following conditional
statements.
(a) The sum of the first n natural numbers is
n(n+1)
.
2
(b) r is irrational if r is real and satisfies r2 = 2.
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1.1. Algebra of Connectives
(c) When x is a real number, the minimum value of x(x − 1) is at least 1/4.
Solution.
n
i, then M =
(a) We could rewrite the statement as “If M =
n(n+1)
.”
2
Thus the
i=1
hypothesis is M is the sum of the first n natural numbers, and the conclusion
is M = n(n+1)
.
2
(b) The hypothesis is r is real and r2 = 2. The conclusion is r is irrational.
(c) We could rewrite the statement “If m is the minimum of x(x − 1) for x ∈ R,
then m ≥ 1/4.” The hypothesis is m minimizes x(x − 1), and the conclusion
is m ≥ 1/4.
1.1. Algebra of Connectives
We will have on occasion to manipulate a statement into an equivalent statement.
Definition 1.10. Let P1 and P2 be two statements. If P1 implies P2 and P2 implies
P1 , we say P1 and P2 are equivalent.
As a consequence of our definition, we see two statements are equivalent iff
they have the same truth table.
Example 1.11. Show ¬(P ∧ Q) is equivalent to ¬P ∨ ¬Q.
Solution. We can either apply the definition directly or construct a truth table.
We will do both. We apply the Definition directly 1.10 first.
Suppose ¬(P ∧ Q) is true. Then P ∧ Q is false. Thus at least on of P or Q is
false. That is, ¬P ∨ ¬Q is true. Conversely, suppose ¬P ∨ ¬Q is true. Then one or
both of ¬P or ¬Q is true. This mean that one of P or Q is false, and so P ∧ Q is
false. If follows ¬(P ∧ Q) is true.
Alternatively we can simply construct a truth table. We have
P
T
F
T
F
Q
T
T
F
F
P ∧Q
T
F
F
F
¬(P ∧ Q)
F
T
T
T
P
T
F
T
F
Q ¬P
T F
T T
F F
F T
¬Q
F
F
T
T
¬P ∨ ¬Q
F
T
T
T
Table 5. Left: truth table for ¬(P ∨ Q). Right: truth table for ¬P ∨ ¬Q.
Since the first two and last columns agree, the two statements have the same
truth table.
Similarly we have
Example 1.12. Show ¬(P ∨ Q) is equivalent to ¬P ∧ ¬Q.
Solution. The truth table for both sides is
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1. Logic
P
T
F
T
F
Q
T
T
F
F
P ∨Q
T
T
T
F
¬(P ∨ Q)
F
F
F
T
P
T
F
T
F
Q ¬P
T F
T T
F F
F T
¬Q
F
F
T
T
¬P ∧ ¬Q
F
F
F
T
Table 6. Left: truth table for ¬(P ∨ Q). Right: truth table for ¬P ∧ ¬Q.
Since the first two and last columns agree, the two statements have the same
truth table.
The importance of the next theorem cannot be understated.
Theorem 1.13. The following are equivalent:
(i) P → Q
(ii) ¬(P ∧ ¬Q)
(iii) ¬P ∨ Q
(iv) ¬Q → ¬P
Part (iv) is called the contrapositive of P → Q.
Proof. We construct truth tables for all four.
P →Q
T
T
F
T
P
T
F
T
F
Q
T
T
F
F
P
T
F
T
F
Q ¬P
T F
T T
F F
F T
P
T
F
T
F
Q
T
T
F
F
¬P ∨ Q
T
T
F
T
¬Q
F
F
T
T
P
T
F
T
F
P ∧ ¬Q
F
F
T
F
Q ¬P
T F
T T
F F
F T
¬Q
F
F
T
T
¬(P ∧ ¬Q)
T
T
F
T
¬Q → ¬P
T
T
F
T
This shows all four are equivalent.
Corollary 1.14. The negation of P → Q is P ∧ ¬Q.
Sometimes the equivalent statements in Theorem 1.13 lead to strange sentences.
Example 1.15. Let P be the statement “I will buy the boat” and Q be “I will
buy the trailer.” Apply Theorem 1.13 to if P , then Q. Would they make sense to
you if someone said them to you?
If I buy the boat, I’ll buy the trailer.
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1.1. Algebra of Connectives
I won’t buy the boat, and not the trailer.
I won’t buy the boat or I’ll buy the trailer.
If I don’t buy the boat, I won’t buy the trailer.
The following table will be useful for manipulating logical connectives.
DeMorgann’s Laws
¬(P ∧ Q) ≡ ¬P ∨ ¬Q
¬(P ∨ Q) ≡ ¬P ∧ ¬Q
Commutative Laws
P ∧Q≡Q∧P
P ∨Q≡Q∨P
Associative Laws
P ∧ (Q ∧ R) ≡ (P ∧ Q) ∧ R
P ∨ (Q ∨ R) ≡ (P ∨ Q) ∨ R
Idempotent Laws
P ∧P ≡ P
P ∨P ≡ P
Distributive Laws
P ∧ (Q ∨ R) ≡ (P ∧ Q) ∨ (P ∧ R)
P ∨ (Q ∧ R) ≡ (P ∨ Q) ∧ (P ∨ R)
Absorption Laws
P ∧ (P ∨ Q) ≡ P
P ∨ (P ∧ Q) ≡ P
Double Negation Laws
¬¬P ≡ P
Example 1.16. Prove P → (Q → R) is equivalent to (P ∧ Q) → R three different
ways.
Proof. One way is to use the logic laws and Theorem 1.13. We find
P → (Q → R) ≡
P → (¬Q ∨ R)
≡
≡
(¬P ∨ ¬Q) ∨ R
¬(P ∧ Q) ∨ R
≡
≡
¬P ∨ (¬Q ∨ R)
(P ∧ Q) → R.
Alternatively we could have used truth tables or Definition 1.10.
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Example 1.17. Prove (P1 ∧ P2 ) → Q is equivalent to (¬Q ∧ P1 ) → ¬P2 . This is
a form of the contrapositive.
Proof. We use logic laws and Theorem 1.13. We find
(P1 ∧ P2 ) → Q
≡ ¬(P1 ∧ P2 ) ∨ Q
≡ (¬P2 ∨ ¬P1 ) ∨ Q
≡ ¬P2 ∨ (¬P1 ∨ Q)
≡ (¬P1 ∨ Q) ∨ ¬P2
≡ ¬(P1 ∧ ¬Q) ∨ ¬P2
≡ (P1 ∧ ¬Q) → ¬P2 .
Alternatively we could have used truth tables or Definition 1.10. We will find this
relation useful later on.
About Definitions. Mathematicians define carefully new concepts and quantities.
The form of the definition (most authors) is “Name iff given conditions.” For
example we might say a function if differentiable iff (conditions follow). Or we
might say an integer z is even iff z = 2k for some k ∈ Z. You might wonder why
we put “iff” in the definition.
Let’s suppose a definition has the form N if C, where N means a name (a
function is continuous, differentiable, integrable..) and C stands for the conditions.
We might want to know what ¬N means. It should be “if ¬C, then ¬N .” If we
made definitions of the form “if C, then N ,” then we would want the contrapositive
of the converse to be true (this is called the inverse). In order for the converse to
be true, we need the definition to be biconditional. Thus we require definitions to
be in the form N iff C and this is equivalent to ¬N iff ¬C.
1.2. Homework
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1.2. Homework
Exercise 1.1. Write the negation of each statement.
(a) The functions f and g are differentiable.
(b) Eight is prime or nine is odd.
(c) If x ∈ D, then f (x) < 7.
(d) The relation R is reflexive or transitive.
(e) x ∈ A or x ∈ B.
(f) If x < 3, then f is not differentiable.
(g) If |x − 3| < δ, then |f (x) − f (3)| < ǫ.
(h) four and nine are relatively prime.
Exercise 1.2. More connectives. The symbol ↓ means nor. That is, P ↓ Q means
neither P or Q.
(a) Make a truth table for P ↓ Q.
(b) Find a formula for P ↓ Q using only ∧, ∨, and ¬.
(c) Find formals for ¬P , P ∧ Q, P ∨ Q using only ↓.
Exercise 1.3. More connectives. The symbol | means nand. That is, P |Q means
P and Q are not both true.
(a) Make a truth table for P |Q.
(b) Find a formula for P |Q using only ∧, ∨, and ¬.
(c) Find formulas for ¬P , P ∧ Q, P ∨ Q using only |.
Exercise 1.4. More connectives. The symbol + means an exclusive or. That is,
P + Q means P or Q but not both.
(a) Make a truth table for P + Q.
(b) Find a formula for P + Q using only ∧, ∨, and ¬.
Exercise 1.5. There does not seem to be an established symbol for the connective
unless. However, Q unless P means if not P , then Q. Write the following statements
in the form if P then Q.
(a) I will go shopping, unless it rains.
(b) The function f is continuous, unless x < 2.
Exercise 1.6. Use the laws of logic to prove the following.
(a) (P ∨ Q) → R ≡ (P → R) ∧ (Q → R).
(b) P → (Q ∨ R) ≡ (P → Q) ∨ (P → R).
Exercise 1.7. Find a formula for P ∧ Q using only the connectives ¬ and →.
Exercise 1.8. Find a formula for P ↔ Q using only the connectives ¬, ∧, and ∨.
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1. Logic
1.3. Quantifiers
Sometimes a statement, or sentence, depends on a parameter. For example, let
P (x) be x > 2. We should not call P (x) a statement since we cannot determine if
it is true or false - until x is quantified.
The allowed values of x are called the universe of discourse. To become a
statement we must be more specific about x to make P (x) a statement. We will
precede the sentence P (x) with two types of quantifiers - for all and there exists.
If the quantified statement depends on more than one parameter, each parameter
must be quantified using “for all,” a universal quantifier, and “there exists,” a
existential quantifier. The following table will help.
Universal (∀)
helpers Existential (∃)
for all, for every
for some
if
then
there exists
whenever, for, given
at least one
every, any
Satisfies some
a, arbitrary
must, is has a
let
be
helpers
such that
for which
satisfies
such that
Example 1.17. Write the statement “the square of a real number is nonnegative”
using quantifier notation.
Solution. The P (x) in this case is x2 ≥ 0. To make a statement out of P (x) we
need to quantify x. In this case we need all real x. Thus the given statement can
be expressed (∀x ∈ R)(x2 ≥ 0).
Example 1.18. Write the statement
If n is even, n is the sum of two odd numbers
using quantifier notation.
Solution. Let P (n, x, y) be the sentence, if n is even, then n = x + y. We need to
quantify x and y. Let E be the set of all even integers and O be the set of all odd
integers. The given statement is equivalent to
(∃x ∈ O)(∃y ∈ O)(n ∈ E → n = x + y).
Alternatively, let P (n, x, y) be the sentence n = x + y. All three variables must
be quantified to make the sentence P (n, x, y) a statement. The given statement is
equivalent to
(∀n ∈ E)(∃x ∈ O)(∃y ∈ O)(n = x + y).
Example 1.19. Write the statement
Let a, b be real numbers. Then ax2 + bx = a has a real solution.
using quantifier notation.
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1.3. Quantifiers
Solution. Let P (a, b, x) be ax2 + bx = a. We must quantify a, b, and x to make a
statement. Thus
(∀a ∈ R)(∀b ∈ R)(∃x ∈ R)P (a, b, x),
or
(∀a ∈ R)(∀b ∈ R)(∃x ∈ R)(ax2 + bx = a).
Note that the order we put the quantifier matters! Consider
(∀x ∈ A)(∃y ∈ B)P (x, y)
and
(∃y ∈ B)(∀x ∈ A)P (x, y).
The first means for each x ∈ A there is a y ∈ B such that P (x, y). The y depends
on the x. In the second y comes first and does not depend on x. There is a y which
works for all x.
Example 1.20. Consider the statement attributed to A. Lincoln
You can fool some of the people all of the time and all of the people some
of the time, but you cannot fool all of the people all of the time.
Write this in quantifier notation.
Solution. Let P (x, t) be the statement x can be fooled at time t. Then “you
can fool some of the people all of the time” could be (∃x)(∀t)P (x, t). This would
mean that there exists a person who can be fooled all of the time. Or perhaps Mr.
Lincoln meant (∀t)(∃x)P (x, t). This would mean that for each time, a person can
be found who is fooled. Since Mr. Lincoln is unavailable for questions, let’s assume
he meant the second version.
The second part is “(you can fool) all of the people some of the time.” This
could be written (∀x)(∃t)P (x, t) or (∃t)(∀x)P (x, t). We see a similar dilemma as
before. We will take this to mean (∀x)(∃t)P (x, t) - each person has a time in which
they can be fooled. The last statement is unambiguous. Thus the original quote
can be written
(1.1)
(∀t)(∃x)P (x, t) ∧ (∀x)(∃t)P (x, t) ∧ ¬(∀x)(∀t)P (x, t).
Example 1.21. Consider the statement “Susan likes everyone who dislikes Joes.”
Write this in quantifier notation.
Solution. Let L(x, y) mean “x likes y.” The original statement can be changed
to “if x dislikes Joe, then Susan likes x. Thus we may write the original statement
(∀x)(¬L(x, j) → L(s, x)).
1.3.1. Negation of Quantified Statements. Consider a statement of the form
(∀x)P (x). For example, P (x) could be x did the homework. Then (∀x)P (x) means
everyone did the homework. The negation of this statements would be “it is false
everyone did the homework” or “at least one person did not do the homework.” So
apparently
¬(∀x)P (x) ≡ (∃x)¬P (x).
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1. Logic
If the statement is of the form (∃x)P (x), a similar results holds. For example, if
P (x) is x is perfect, then (∃x)P (x) would mean “someone is perfect.” The negation
would be “it is false someone is perfect” or ”everyone is not perfect. That is,
¬(∃x)P (x) ≡ (∀x)¬P (x).
Suppose now the statement is in form (∀x)(∃y)P (x, y). Set P̃ (x) = (∃y)P (x, y).
Then
¬ (∀x)(∃y)P (x, y)
≡
¬ (∀x)P̃ (x)
≡
≡
(∃x)¬P̃ (x)
(∃x)¬(∃y)P (x, y)
≡
(∃x)(∀y)¬P (x, y).
We see negating quantified statements is easy - we simply change ∀ to ∃ and ∃ to
∀ negate P .
Example 1.22. Negate
You can fool some of the people all of the time and all of the people some
of the time, but you cannot fool all of the people all of the time.
Solution. We need to negate Equation (1.1). We find
(∃t)(∀x)¬P (x, t) ∨ (∃x)(∀t)¬P (x, t) ∨ (∀x)(∀t)P (x, t).
In English this would be
There exists a time such that all people cannot be fooled or there exists
people such that for all time they cannot be fooled or you can fool all of
the people all of the time.
Example 1.23. Consider the definition
A function f : D → R is bounded iff there exists M ≥ 0 such that
|f (x)| ≤ M for all x ∈ D.
What does is mean for f to be unbounded?
Solution. We first have to express the definition in logical notation. A function is
bounded iff
(∃M ≥ 0)(∀x ∈ D)(|f (x)| ≤ M ).
Now its negation is easy. A function is not bounded if
(∀M ≥ 0)(∃x ∈ D)(|f (x)| > M ).
That is, a function is not bounded if for all M ≥ 0 there exists x ∈ D such that
|f (x)| > M .
Example 1.24. Consider the definition
A function f : R → R is even iff for all x ∈ R, f (−x) = f (x).
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1.3. Quantifiers
What does is mean for f not to be even?
Solution. We first have to express the definition in logical notation. A function is
even iff
(∀x ∈ R)(f (−x) = f (x)).
Now its negation is easy. A function is not even if
(∃x ∈ R)(f (−x) = f (x)).
That is, a function is not even if x ∈ R exists so that f (−x) = f (x).
1.3.2. Some Notation about Sets. A set is a collection of objects. The objects
are called elements of the set. We denote a set by listing the elements and putting
brackets around the list. For example, the set containing the elements 1, 2, 3 is
denoted {1, 2, 3}. When there are too many elements to list we typically provide
a description of the elements in the set. For example, {x | x is even} would be the
collection of all even integers. We use the symbol ∈ to denote an element is in a
set. So y ∈ {x | x is even} would mean y is an even number.
We can combine two sets to form another set. Let A and B be two sets. Then
(i) A ∪ B = {x | x ∈ A ∨ x ∈ B},
(ii) A ∩ B = {x | x ∈ A ∧ x ∈ B},
(iii) A \ B = {x | x ∈ A ∧ x ∈ B}.
The statement A ⊆ B means for all x ∈ A, x ∈ B. In quantifier notation A ⊆
B ≡ (∀x ∈ A)(x ∈ B) or A ⊆ B ≡ (∀x)(x ∈ A → x ∈ B). The set containing no
elements is called the empty set and denoted ∅. If A ∩ B = ∅, A and B are called
disjoint.
For future reference
N = {1, 2, 3, 4, . . .},
Z = {0, ±1, ±2, ±3, ±4, . . .},
Q = {n/m | n ∈ Z ∧ m ∈ N},
R is the set of real numbers,
where N is called the natural numbers, Z is called the integers, and Q is called the
rational numbers.
Example 1.25. Write A ∩ B ⊆ C in logical notation.
Solution. The original statement is the same as “it is false that A ∩ B ⊆ C.” This
is equivalent to ¬[(∀x)((x ∈ A∧x ∈ B) → x ∈ C)] or (∃x)((x ∈ A∧x ∈ B)∧x ∈ C)].
You may have to review Theorem 1.13 to see this.
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1. Logic
1.4. Homework
Exercise 1.1. Express the following in an equivalent form using only the symbols
∀, ∃, →, ∈, ∈, ∧, ∨
a) A ⊆ B
b) A \ B = ∅
c) x ∈ A \ B
d) A \ B = ∅
e) (A \ B) ⊆ C
f ) (A ∪ B) ⊆ C
Exercise 1.2. The universe of discourse for the following quantified statements is
the set of integers. Find the truth values of the quantified statements. No formal
proof necessary, but explain your reasoning.
a) (∀x)(∃y)(x + 2y = 2). (Recall x, y ∈ Z)
b) (∀y)(∃x)(x + 2y = 2)
c) (∀x)(∀y)(x + y = 1 → x · y ≤ 1)
d) (∀x)[x > 2 → (∃y)(y − x = 2)]
e) Is d) equivalent to (∀x > 2)(∃y)(y − x = 2)?
f ) Negate d).
Exercise 1.3. Is the statement (∃x)(∃y)(∀z)(∃u)(xz > y u ) true if the universe of
discourse is the natural numbers?
Exercise 1.4. Is the statement (∀x)(∃y)(∀u)(∃v)(x + y = uv) true if the universe
of discourse is the natural numbers?
Exercise 1.5. True or False?
a) (∀x ∈ R)(∀y ∈ R)(x > y)
b) (∀x ∈ R)(∃y ∈ R)(x > y)
c) (∀y ∈ R)(∃x ∈ R)(x > y)
d) (∃x ∈ R)(∀y ∈ R)(x > y)
e) (∃x ∈ R)(∃y ∈ R)(x > y)
f ) Using the appropriate quantifiers, write down the negation of the assertion in
part (c) of this question.
Exercise 1.6. We say a function defined on the real line is periodic iff a k > 0
exists so that f (x + k) = f (x) for all x ∈ R. What does it mean for a function not
to be periodic?
Exercise 1.7. We say a function defined on the real line is odd iff a f (−x) = −f (x)
for all x ∈ R. What does it mean for a function not to be odd?
Exercise 1.8. We say a function f : A → B is injective (one-to-one) iff for all
x1 ∈ A, x2 ∈ A, f (x1 ) = f (x2 ) implies x1 = x2 . What does it mean for a function
not to be injective?
Exercise 1.9. We say a function f : A → B is surjective (onto) iff for all y ∈ B,
x ∈ A exists so that f (x) = y. What does it mean for a function not to be
surjective?
1.4. Homework
15
Exercise 1.10. Consider the statement “For every integer n > 0 there is some real
number x > 0 such that x < 1/n.” Negate, without using words of negation (it is
false that ...). Which statement is true?
Exercise 1.11. We say f : R → R is increasing if and only if, for all x, y ∈ R with
x ≤ y, we have f (x) ≤ f (y). What does it mean for a function not to be increasing.
Exercise 1.12. Consider a fixed function f : R → R. Let S be the set of function
defined the following way. We say g ∈ S iff there exists positive constants a, c ∈ R
such that |g(x)| ≤ c|f (x)| for all x > a. (The functions g are in some sense bounded
by f .) What does it mean for g ∈
/ S?
Exercise 1.13. Describe the meaning of the following two statements and their
negations. Which one implies the other? Here E ⊂ R.
a) There is a real number M such that, for every x in the set E, |x| ≤ M .
b) For every x in E, there is a real number M such that |x| ≤ M .
Exercise 1.14. For a ∈ R and f : R → R, what is the meaning of the following two
statements (state in sentences). What is the difference between the two? Negate
them.
a) (∀ǫ > 0)(∃δ > 0)(∀x ∈ R)[(|x − a| < δ) → (|f (x) − f (a)| < ǫ)]
b) (∃δ > 0)(∀ǫ > 0)(∀x ∈ R)[(|x − a| < δ) → (|f (x) − f (a)| < ǫ)]
Exercise 1.15. Let f : R → R. What is the difference between the following two
statements. Does one imply the other? Negate them.
a) For every ǫ > 0 and every a ∈ R, there is a δ > 0 such that for all x ∈ R,
|f (x) − f (a)| < ǫ whenever |x − a| < δ.
b) For every ǫ > 0 there is a δ > 0 such that for all x ∈ R, |f (x) − f (a)| < ǫ
whenever |x − a| < δ and a is a real number.
Exercise 1.16. Let f : R → R and c ∈ R. Write each of the following statements
using logic notation. Negate both versions.
a) For every x ∈ R and all δ > 0, there exists a ǫ > 0 such that |x| < δ implies
|f (x) − c| < ǫ.
b) For all x ∈ R there is a δ > 0, such that for all ǫ > 0 we have |x| < δ implies
|f (x) − c| < ǫ.
Exercise 1.17. A set K in a linear space is called convex iff for any two points x
and y in K, each point tx + (1 − t)y, with 0 < t < 1, also belongs to K. What does
it mean for a set not to be convex?
Exercise 1.18. A second-order linear differential operator D is called hypoelliptic
iff every function u ∈ C 2 (Ω) that satisfies Du = 0 is in C ∞ (Ω). What does it mean
for a second-order linear differential operator not to be hypoelliptic?