Download Chapter 1 LOGIC AND PROOF

Chapter 1
LOGIC AND PROOF
To be able to understand mathematics and mathematical arguments, it is necessary
to have a solid understanding of logic and the way in which known facts can be
combined to prove new facts. In this chapter we take a careful look at the rules of
logic and the way in which mathematical arguments are constructed.
1.1 Logical statements
The study of logic is concerned with the truth of falsity of statements.
Definition 1 A statement ∗
is a sentence which can be classified as true or false
without ambiguity. The truth or falsity of the statement is known as the
truth value.
For a sentence to be a statement, it is not necessary that we actually know
whether it is true or false, but it must be clear that it is one or the other.
Example 2
Consider the following sentences:
1. “Every continuous function is differentiable” is a statement with truth value
“false”.
2.
”x < 2” is true for some x and false for some others.
If we have a particular
context in mind, then it could be a statement. Otherwise it is not.
3. ”Every even number greater than 2 is the sum of two primes” is a statement,
whose truth value is not known...yet.
Exercise 3
Which of these sentences are statements?
1. If x is a real number, then
x2 ≥ 0.
2. Seven is a prime number
3. This sentence is false
∗ Some
authors also call it proposition.
8
Logic and Proof
1.2 Logical Connectives
In studying mathematical logic we shall not be concerned with the truth value of
any particular simple statement. What will be important is how the truth value of
a
compound
statement is determined from the truth values of its simpler parts.
To obtain such compound statements it is necessary to introduce the concept of a
connective
Definition 4 A sentential connective is a logic symbol representing an operator
that combines statements into a new statement.
Statements with connectives are called compound statements. Statements
without connectives are known as atomic statements. The sentential connectives
are ”not”, ”and”, ”or”, ”if ...then”, and ”if and only if”. The respective operators
for these connectives are negation, conjunction, disjunction, implication and
equivalence respectively. We shall introduce each of them now
•
Negation : Let p stand for a given statement. Then ˜p ( read not p ) represents
the logical opposite of p. When p is true, then ˜p is false and viceversa. This can
be summarized in a “truth table”, which gives the mapping from the truth
value of the individual statements to the truth value of the resulting compound
statement. Some authors also use for the negation connective. In this case
p
T
F
where
T
stands for true and
p
F .
T
˜
F
for false.
Note that the truth table of a
connective is an alternative way of defining a connective, since these are defined in
terms of the truth value of the resulting compound statement, given the truth value
of its components.
•
Conjunction:
q)
If
p and q
is true only when both
are statements, then the statement
p
and
p
T
T
F
F
q
p∧q
are true, and is false otherwise.
q
T
F
T
F
p∧q
T
F .
F
F
( read
p and
Logical Connectives
9
For example, given p : ”2 > 0” and q : ”0 > 2” then the compound statement
p ∧ q : ”2 > 0 and 0 > 2” is false.
•
(Inclusive)Disjunction
( read p
or q )
: If p and q are statements, then the statement p ∨ q
is true when at least one of the two statements is true, and is
false when both are false.
p
T
T
F
F
p∨q
T
T .
T
F
q
T
F
T
F
Note that the inclusive disjunction doesn’t complete the list of disjunctions
used in everyday life.
when either
p
the connective
q : ”0 > 2”
•
q
or
or
In fact, we also have the exclusive disjunction, which is true
is true, but not when both are true.
is for the inclusive meaning.
then the compound statement
Implication
p
: ”2
>0
or 0
> 2”
p
: ”2
>
0” and
is true.
: A statement of the form
If
is called an
p∨q
In logic the only use for
For example, given
implication
or a
p,
then
q
conditional statement.
is called the antecedent and the then-statement
q
The if-statement
is called the consequent.
The
convention adopted for the truth value of the implication is that it will be called
false only when the antecedent is true and the consequent is false. If we denote the
implication ”if
p,
then
q”
by
p ⇒ q,
p
T
T
F
F
we obtain the following truth table
q
T
F
T
F
p⇒q
T
F .
T
T
The last row of the table may seem counterintuitive. However, the usage of ’if
... then” as a connective or as a mathematical language is quite different from that
of daily language.
The reason for giving truth value
T
⇒ −
to the last case may be not
understandable in the following mathematical expression: “3 + 1 = 7
6
1 = 2”
Logic and Proof
10
−
−
Nevertheless, we can easily prove the implication is indeed true: 3 + 1
so 1 = 4. But then 6
p
⇒ q.
1 = 6
4 = 2.
−
3 = 7
−
3 = 4
In English there are several ways to express the same mathematical condition
These are
p, then q ”
p implies q ”
” p only if q ”
”q if p”
q
”if
”
”
”
p”
p”
provided that
”
q
whenever
q is a necessary condition for p”
p is a sufficient condition for q”
”
• Equivalence: The statement ”p if and only if q” is defined as the conjunction
of the two implications p ⇒ q and q ⇒ p. A statement of this form is called an
equivalence and is denoted by p ⇔ q. In written form, the abbreviation ”iff” is
used instead of ”if and only if”. The truth table for equivalence can be obtained
by computing the truth table for the compund statement (p ⇒ q) ∧ (q ⇒ p)
p
T
T
F
F
q
T
F
T
F
p
⇒ q q ⇒ p p ⇐⇒ q
T
F
T
T
T
T
F
T
T
F
F
T
Thus we see that p ⇔ q is true precisely when p and q have the same truth
values. Since p ⇔ q is equivalent to (p ⇒ q) ∧ (q ⇒ p) we can use the terminology
seen above for the implication connective and say
”q if p” and ”q only if p”
”p is a sufficient condition for q” and ”p is a necessary condition for q”
Two compound statements p and q are said to be logically equivalent if one
is true if and only if the other is true. In other words, two compound propositions
are logically equivalent whenever they display the same truth table. In this case we
write p ≡ q.
Example 5 (˜(˜p)) is logically equivalent to p
p
T
F
˜p
F
T
˜ (˜p)
T
F
Logical Connectives
Example 6
11
˜(p ∧ q)
p
T
T
F
F
q
T
F
T
F
Exercise 7 (Implication)
[(˜p) ∨ q] .
connector.
to
[(˜p) ∨ (˜q)]
˜(p ∧ q) ˜p ˜q [(˜p) ∨ (˜q)]
is logically equivalent to
F
T
T
T
F
F
T
T
F
T
F
T
F
T
T
T
Show that the implication symbol is not a ”primitive”
In other words, show that the implication
p
⇒
q
is logically equivalent
This is actually another convenient way to interpret the implication
connector.
When two compound statements p and q are logically equivalent, their equiv-
alence is a new statement with a truth value ”true” in all cases. Such a statement is
known as a tautology, being its definition
Definition 8 A compound proposition is said to be a tautology if it is always true
regardless of the truth value of the simple propositions from which it is constructed.
It is a contradiction if it is always false. Hence a contradiction and a tautology are
a negation of each other.
Example 9 p ∨ (˜p) is a tautology, while p ∧ (˜p) is a contradiction.
p
T
F
(˜p) p ∨ (˜p) p ∧ (˜p)
F
T
T
T
F
F
Example 10 [p ∧ (p =⇒ q)] =⇒ q is a tautology
p
T
T
F
F
q
T
F
T
F
p⇒q
T
F
T
T
p ∧ (p ⇒ q)
T
F
F
F
[p ∧ (p ⇒ q )] ⇒ q
T
T
T
T
Lets now prove the equivalence of a compound statement that will be useful
when studying methods of proof
Example 11 (Contrapositive)
p
T
T
F
F
q
T
F
T
F
p
⇒q
T
F
T
T
p
F
F
T
T
˜
q
F
T
F
T
˜
p
⇒q
q
(˜ )
q
is logically equivalent to [(˜ )
⇒
T
F
T
T
p
(˜ )
p
(
⇒q ⇔
)
q
[(˜ )
T
T
T
T
⇒
⇒
p
p
(˜ )]
(˜ )]
Logic and Proof
12
Exercise 12 (De Morgan’s Law) Prove that ˜(p∨q) is logically equivalent to [(˜p) ∧ (˜q)]
Exercise 13
1.
2.
Establish the logical equivalence of these compound statements
⇒ q ) ⇔ [p ∧ ˜q]
(p ⇒ q ) ⇔ (q ⇒ p)
˜ (p
Remark 14
converse q
p
⇒
q
⇒ p. Claiming the opposite is a very common mistake you should not
As you have already proven
is not logically equivalent to its
make in the future.
Exercise 15 Write the negation of each statement
a) If K is closed and bounded, then K is compact.
b) If K is compact, then K is closed and bounded.
c) A continuous function is differentiable.
Exercise 16 Construct a truth table for each statement
⇒ (q ∧ ˜q)] ⇔ ˜p
⇒ ˜p) ⇔ (p ∧ q)
∧ q) ⇔ (q ∧ p) (commutative property)
∨ q) ⇔ (q ∨ p)
∧ (q ∧ r)] ⇔ [(p ∧ q) ∧ r] (associative property)
∨ (q ∨ r)] ⇔ [(p ∨ q) ∨ r]
∧ (q ∨ r)] ⇔ [(p ∧ q) ∨ (p ∧ r)] (distributive property)
∨ (q ∧ r)] ⇔ [(p ∨ q) ∧ (p ∨ r)]
a) [ p
b) (q
c) (p
d) (p
e) [p
f) [p
g) [p
h)[p
1.3 Quantifiers
In Section 1 we saw that a mathematical sentence that involves a variable, for example ”x
> 3”,
needs to be considered in a particular context in order to become a
statement. The quantifiers help create statements by limiting the role of variables in
a mathematical sentence.
is a symbol that can assume several specifications. A
propositional function is a sentence which becomes a statement once we replace a
variable by one of its specifications.
Definition 17
A
variable
Let p(x) denote a propositional function, p being the sentence and x being the
variable. Thus p (x) : x > 3 is a propositional function.
Let S denote a given set.† There are two quantifiers:
†
You may think of a set S as a collection of objects, who are called the elements of S . When x
is an element of S we write x ∈ S and we read “x is an element of S ”.
Quantifiers
•
13
Universal quantifier
(“for all”), denoted by
∀,
with the following interpreta-
tion:
∀x ∈ S, p(x)]
[
an example, if
means that
S
=
{4, 5},
p(x)
x in S . As
> 3) ∧ (5 > 3)].
is true provided it is true for each
∀x ∈ S, p(x)
then [
:
x>
3]
≡
[(4
S is implicit in the context, and can be omitted. The notation
therefore becomes more compact, leading to ∀x, p(x).
Sometimes the set
•
Existential quantifier
∃x ∈ S p(x)]
[
meaning that
p(x)
(“there exists”), denoted by
∃:
reads ” There exists an x belonging to S such that
p(x)
is true provided there exists at least an
is true. As an example, if
[(1 > 3) ∨ (2 > 3) ∨ (4 > 3)].
S
=
x
in
S
p (x) ”‡
for which
{1, 2, 4}, then [∃x ∈ S p(x) : x > 3] ≡
Exercise 18 Let S be a finite set (i.e. {1, 2}). Use de De Morgan’s law to show that
˜[∃x ∈ S p(x)] ⇔ [∀x ∈ S, ˜p(x)],
tautologies.
and
˜ [∀x ∈ S , p(x)] ⇔ [∃x ∈ S, ˜p(x)]
are
∃! x to denote the case when a unique value
exists for the variable x that makes p(x) true. The universal and the existential
quantifiers are thus seen as extensions of the logical connectives ∧ and ∨, to deal with
infinitely many assertions, or assertions about infinitely many things, x. Moreover,
one can combine the existential quantifier with negation: ˜∃ means “there exists no”
(often written as ∃
/).
Sometimes
we use the symbol
Example 19 ∃/ x > 0 (x + 1) = 0 is logically equivalent to ∀x > 0, x + 1 = 0.
Exercise 20
1.
2.
3.
Write the negation of the following statements
∀x ∈ A, f (x) > 5
∃ y > 0 0 ≤ g (y ) ≤ 3
∀ε > 0 ∃ N ∀n, if n > N,
convergence)
then
∀x ∈ S, | fn (x) − f (x) | < ε (uniform
As we saw in the last exercise, we can use both ∀ and ∃ in one statement. It
is important to clarify the following point about the order in which quantifiers are
used. While [∀x, ∀y, p(x, y)] ≡ [∀y, ∀x, p(x, y)] ≡ [∀x, y, p(x, y)], the propositions
[∃ y ∀x p(x, y)] and [∀x, ∃ y p(x, y)] are not logically equivalent. In this case, in
fact, the order in which quantifiers appear affects the meaning — and the truth value
— of the statement. The first statement says that for at least one y, p(x, y) is true for
all x. In other words, the choice of y is independent of x. On the other hand, the
second statement establishes that for all x there exist at least one y such that p(x, y)
is true. This means that the choice of y is allowed to depend on x.
‡
stands for “such that”. Another ways of saying “such that” are given by the symbol “:” or
by writing ”s.t.”.
14
Logic and Proof
As an example consider p(x, y) being given by x + 1 = y. The statement
[∃ y ∀x p(x, y)] establishes that there exists an y such that for all x, x +1 = y. This
is clearly a false statement. The second proposition, on the other hand, states that
for all x there exists an y such that x + 1 = y. Unlike the former, this statement is
true.
Exercise 21 Find a p (x, y) such that [∃ y ∀x p(x, y)] and [∀x, ∃ y
the same truth value.
p(x, y )] have
Exercise 22 The following statements give properties of functions that we shall encounter later in the course.
You have to do two things.
a) rewrite the defining
conditions in logical symbolism and b) write the negation of part a) using the same
symbolism.
1. A function f is
odd iff
2. A function f is
even
3. A function f is
f (x).
≤ f (y) .
5. A function f is
f (x) > f (y ) .
6. A function
then
iff for every x,
increasing
iff for every x and for every y, if
strictly decreasing iff for every x
f :A
x = y.
−f (x) = f (−x).
f (x) = f (−x).
periodic iff there exists a k > 0 such that for every x, f (x + k ) =
4. A function f is
f (x)
for every x,
−→ B
is
and every y, if
injective iff for every x and y in A, if
≤
x
x
y,
then
≤ y,
then
f (x) = f (y) ,
f : A −→ B is surjective iff for every y in B there exists an x in
A such that f (x) = y.
7. A function
f : D −→ R is continuous at c ∈ D iff for every ε > 0 there exists
a δ > 0 such that |f (x) − f (c)| < ε whenever |x − c| < δ and x ∈ D.
8. A function
uniformly continuous on a set S iff for every ε > 0 there is
δ > 0 such that |f (x) − f (y)| < ε whenever x and y are in S and x y < δ.
9. A function f is
a
|
ε > 0 there
0 < |x − c| < δ.
is a
δ>0
such that
|
f : D −→ R at the point c iff
|f (x) − L| < ε whenever x ∈ D and
10. The real number L is the limit of the function
for each
−
Techniques of Proof
15
1.4 Techniques of Proof
A proof is a method of establishing the truthfulness of an implication. Typically, one
has to prove proposition of the sort if H1,H2,...,Hn , then C. Propositions H1,...,Hn are
often referred to as the
conclusion,
hypotheses of the proof, whereas proposition C
is referred
or thesis. A formal proof of such a proposition consists of a
sequence of valid propositions ending with conclusion C. To be valid, a proposition in
the sequence must be either one of the hypotheses H1 , ...Hn , or an axiom, a definition,
a tautology or a previously proved proposition, or it must be derived from previous
propositions using either substitution or logical implications.
As an example, if you have to prove that given a set S , [∀x ∈ S, p(x)], then
you need to prove that [x ∈ S =⇒ p(x)]. Similarly, in case you have to prove that
[∃x ∈ S : p(x)], you “just” need to find an x in S such that p(x) is true.
There are several ways of proving a proposition, all of them being useful in
different situations. These methods are enumerated below.
to as the
Direct To show that
q is true.
Example 23
p =⇒ q is true, we first assume that p is true and conclude that
If
x > 1,
then
x2 > x.
Here
proof goes as follows:
p:x>1
p:x>1
•
Let
•
from the axiom ”If
•
But by definition
•
substituting in the above expression leads to
•
which is the conclusion we wanted
and
q : x2 > x.The
direct
(Hypothesis);
x x > 1 x;
a>b
and
c > 0,
x x = x2
and
1 x = x;
then
a c > b c”
we have
x>
1
⇒
x x > 1 x ⇒ x2 > x
x > 1 ⇒ x2 > x.
Contrapositive Associated with the implication
p ⇒ q there is a logically equivalent statement
˜q
˜p, called the contrapositive (see example 11). Thus one way to prove
an implication is to give a direct proof of its contrapositive. In other words,
assume that ˜q is true and conclude that ˜p is true. Then we can conclude that
(p
q ) is true.
⇒
⇒
Example 24 Consider the statement for m ∈ N , ”If 7m is an odd number,
then m is an odd number”. The contrapositive of the statement is ”If m is not
an odd number, then 7m is not an odd number”, or equivalently ”If m is an
even number, then 7m is an even number”. This statement is much easier to
prove.
Logic and Proof
16
Proof. Let ∼ q
k
: ”m is an even number”. Then, by definition m = 2k for some
∈ N. But then substituting into m we have 7m = 7
(2 k ) = 2 (7k ) = 2
k
which
is also even.
Exercise 25 Try the direct proof of the above statement.
Decomposition Suppose we want to prove that p ⇒ q, and that p can be decomposed into
two disjoint propositions p1, p2 such that p1 ∧ p2 is a contradiction. Then p ≡
(p1
∨ p2) ∧ ˜(p1 ∧ p2 ) ≡ (p1 ∨ p2 ) .
Given this choice of
(p
p1
and
⇒ q) ⇔
⇔
⇔
p2
we have
∨ q) ⇔ [˜ (p1 ∨ p2 ) ∨ q ]
[(˜p1 ∧ ˜p2 ) ∨ q ] ⇔ [(˜p1 ∨ q) ∧ (˜p2 ∨ q)]
[(p1 ⇒ q) ∧ (p2 ⇒ q )]
(˜p
meaning that you only need to show that
method works also if you decompose
p
p1
⇒q
and
p2
⇒ q.
Note that this
into a number of propositions bigger
than 2 as far as these propositions are mutually exclusive (which means that
every pair of them is a contradiction).
Example 26 Take the contrapositive of the example given for the direct proof
method. The contrapositive is ”x2 ≤ x ⇒ x ≤ 1”.
Proof. By definition x2 ≥ 0. Then p : 0 ≤ x2 ≤ x and q : x ≤ 1. Decompose
p into p1 : x > 0 and p2 : x = 0. Then p1 ∧ p2 is always false and p ≡
(p1 ∨ p2 ) ∧ ˜(p1 ∧ p2 ). To show that p2 ⇒ q is trivial since 0 ≤ 1. To prove
p1 ⇒ q, we only need to use the implication ”∀ c > 0, x ≤ y ⇒ xc ≤ yc ”. Given
that x > 0 by assumption, x2 ≤ x ⇒ xx ≤ xx ⇒ x ≤ 1
2
Exercise 27
Show the following statement ”Let x ∈ R. If |x| ≥ 1 ⇒ x2 ≥ x”.
Construction This approach is used when the statement includes an existential quantifier. i.e.,
the conclusion is of the form: ∃x, p(x). To prove this, simply find (“construct”)
a value of x such that p(x) is true when H is true. For example, if A =
{1, 2, 3, 4, 5} and the proposition is: ∃x ∈ A x > 2, observe that for x = 3 we
have x > 2 and x ∈ A.
Exercise 28
Show that f (x) = x is a continuous function at every x0 ∈ R . You
have to prove that ”∀ε > 0 , ∃ δ (x0, ε) > 0 |x − x0| < δ (x0 , ε) ⇒ |f (x) − f (x0)| <
ε”.
Techniques of Proof
17
Contradiction Suppose we want to prove p ⇒ q. This method of proof — also known as ”reductio ad absurdum”— consists of assuming that [p ∧ (˜q)] is true and deriving
a contradiction c. To show that both statements are logically equivalent, we
compare their truth tables
p q p ⇒ q ˜q (p ∧ ˜q ) c (p ∧ ˜q) ⇒ c
T
T
F
F
T
F
T
F
T
F
T
T
F
T
F
T
F
T
F
F
F
F
F
F
T
F
T
T
.
An equivalent way of making a proof by contradiction is given by the following tautology: (p =⇒ q) ⇔ [(p ∧ (˜q)) =⇒ (˜p)]. This is a particular case of the
general definition, since [(p ∧ (˜q)) =⇒ (˜p)] is equivalent to [(p ∧ (˜q)) =⇒ ((˜p) ∧ p)]
(since p was already an assumption), and given that ((˜p) ∧ p) ⇔ c then
[(p ∧ (˜q)) =⇒ c] .The proof of their equivalence is left as an exercise.
Exercise 29
⇒q ⇔ p∧
Show that (p =
)
[(
⇒
(˜q )) =
(˜p)] .
Example 30 We will show again that ”If 7m is odd, then m is odd”.
Proof. Here the statements are p ” 7m is odd ” and q ”m is odd ”.
To prove the result by contradiction we will use the equivalence p ⇒ q ⇔
p∧ q
⇒ p , so we have to assume first that p ∧ q , that is ” 7m
is odd and m is even ”. Since m is even by definition we have m = 2k for some
k. Then 7m= 7(2k) = 2 (7k) = 2 j for j=7k. But this means that 7m is even
p , contradicting the assumption.
:
:
(
[(
(˜ )) =
(˜ )]
(
=
)
(˜ ))
(˜ )
Remark 31 Be very careful when writing proofs by contradiction ! A very strong
word of caution can be found in Royden§ ”All students are enjoined in the strongest
possible terms to eschew proofs by contradiction! There are two reasons for the prohibition: First such proofs are very often fallacious, the contradiction on the final
page arising from an erroneous deduction on an earlier page, rather than from the
incompatibility of p with ˜q. Second, even when correct, such a proof gives little insight into the connection between p and q whereas both the direct proof and the proof
by contraposition construct a chain of argument connecting p and q.One reason why
mistakes are so much more likely in proofs by contradiction than in direct proofs is
that in a direct proof ( assuming the hypothesis is not always false) all deduction
from the hypothesis are true in those cases where the hypothesis holds. One is dealing
with true statements, and ones’s intuition and knowledge about what it is true help to
keep one from making erroneous statements. In proofs by contradiction, however, you
are (assuming the theorem is true) in the unreal world where any statement can be
derived, and so the falsity of a statement is no indication of an erroneous deduction.
§
H.L. Royden ”Real Analysis” pg3.
Logic and Proof
18
Exercise 32 Let x be a real number. Prove that if x>0 then 1/x>0.
Induction It is used for statements of the form
p(n), ∀n ∈ {m, m + 1, m + 2, ...}.
The
proof consists of two steps:
Basis of Induction Prove that for the first element in the set of interest — in this case
p(m)
p(k ) ⇒ p(k + 1), for k m.
prove that p(k + 1) is also true.
Inductive step Show
m
—,
is true,
In other words, assume
p(k )
is true and
Exercise 33 The following tautologies are widely used in the methods of proof. Some
of them have already been seen before. a) Verify that they are indeed tautologies, b)
Interpret them and associate them with the different methods of proof
1. (p ⇔ q) ⇔ [(p ⇒ q) ∧ (q ⇒ p)]
2. (p ⇔ q) ⇔ [(p ⇒ q) ∧ (˜p ⇒ ˜q)]
3. (p ⇒ q) ⇔ (˜q ⇒ ˜p)
4. p ∨ ˜p
5. (p ∧ ˜p) ⇔ c
6. (˜p ⇒ c) ⇔ p
7. (p ⇒ q) ⇔ [(p ∧ ˜ q) ⇒ c]
8. [ p ∧ (p ⇒ q)] ⇒ q
9. [˜q ∧ (p ⇒ q)] ⇒ ˜p
10. [˜p ∧ (p ∨ q)] ⇒ q
11. (p ∧ q) ⇒ p
12. [(p ⇒ q) ∧ (q ⇒ r)] ⇒ (p ⇒ r)
13. [(p1 ⇒ p2) ∧ (p2 ⇒ p3) ∧ ... ∧ (pn−1 ⇒ pn )] ⇒ (p1 ⇒ pn)
14. [(p ∧ q) ⇒ r] ⇔ [p ⇒ (q ⇒ r)]
15. [(p ⇒ q) ∧ (r ⇒ s) ∧ (p ∨ r)] ⇒ (q ∨ s)
16. [p ⇒ (q ∨ r)] ⇔ [(p ∧ ˜q) ⇒ r]
17. [(p ∨ q) ⇒ r] ⇔ [(p ⇒ r) ∧ (q ⇒ r)]
1.5 References
S.R Lay , Analysis with an Introduction to Proof . Chapter 1. Third Edition. Prentice Hall.
A. Matozzi, Lecture Notes Econ 897 University of Pennsylvania Summer
2001.
H.L. Royden, Real Analysis . Third Edition. Prentice Hall.