Download (a) (p q)

Discrete Mathematical
Structures
离散数学结构
Bernard Kolman
Robert C. Busby
Sharon Cutler Ross
《离散数学》教学组
中山大学软件学院
Chapter 2 Logic
2.1 Propositions and Logical Operations
2.2 Conditional Statements
2.3 Methods of Proof
2.4 Mathematical Induction
2
推理的符号化（形式化）
推理规则举例：
假言推理（modus ponens）
如果张三是软院2011级学生，那么离散数学是张三的必修课
张三是软院2011级学生
所以，离散数学是张三的必修课
符号化：
若p则q
p
-----------q
3
p => q
p
-----------q
推理规则举例：
三段论（Syllogism）
所有软院2011级学生必修离散数学
张三是软院2011级学生
所以，张三必修离散数学
符号化：
x(A(x) -> B(x))
A(a)
-----------B(a)
4
Reasoning in Mathematics
Gottfried Wilhelm Leibniz’s (July 1, 1646 – November
14, 1716) Symbolic thought: The only way to rectify our reasonings is to
make them as tangible as those of the Mathematicians, so that we can find our error at a
glance, and when there are disputes among persons, we can simply say: Let us calculate
[calculemus], without further ado, to see who is right.
It is obvious that if we could find characters or signs suited for expressing all our
thoughts as clearly and as exactly as arithmetic expresses numbers or geometry
expresses lines, we could do in all matters insofar as they are subject to reasoning all that
we can do in arithmetic and geometry. For all investigations which depend on reasoning
would be carried out by transposing these characters and by a species of calculus.
Formal Logic
George Boole (2 November 1815 -8 December 1864 )
invented Boolean Logic.
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中山大学软件学院
2.1 Propositions and Logical Operations
1) Statement or Proposition
A statement or proposition (命题) is a declarative
sentence that is either true or false, but not both.
Ex. Which of the following are statements?
(b) 2 + 3 = 5.
Yes
Yes
(c) Do you speak English?
No, it is a question.
(a) The earth is round.
(d) 3 – x = 5
No, it is a declaration sentence, not a statement.
(e) Take two aspirins.
No, it is a command.
(f) The temperature on the surface of the planet Venus (金星) is
800F.
Yes, it is a statement.
(g) The sun will come out tomorrow. Yes, it is a statement.
6
Proposition examples (compound propositions):
2 is an even number and 3 is an odd number.
2 is an even number and 3 is an even number.
2 is an odd number and 3 is an odd number.
2 is an odd number and 3 is an even number.
2 is an odd number and I am a student.
Pattern of the proposition: with connective “and”
Symbol: /\ (conjunction)
Truth of the pattern, or the truth table of “/\”
7
Compound proposition examples:
2 is an even number or 3 is an odd number.
2 is an even number or 3 is an even number.
2 is an odd number or 3 is an odd number.
2 is an odd number or 3 is an even number.
2 is an odd number or 2 is an even number.
Pattern of the proposition: with connective “or”
Symbol: \/ (disjunction)
Truth of the pattern, or the truth table of “\/”
8
Compound proposition examples:
If I am hungry, then I’ll eat.
If it is raining, then 2+3 = 5.
If it is raining, then 2+3 = 4.
If 2 is an even number, then 3+3 = 4.
If 2 is an odd number, then 3 +3 = 4.
If 2 is an odd number, then the sun rise on the west.
Pattern of the proposition: with connective “if … then
…”
Symbol: => (implication, antecedent, consequent)
Truth of the pattern, or the truth table of “=>”
9
Compound proposition examples
2 is not an odd number.
2 is not an even number.
Pattern of the proposition: with connective “not”
Symbol: ~
Truth of the pattern, or the truth table of “~”
10
Equivalence of propositions
p => q and ~p \/ q
Bidirectional
p<=>q
Tautologies
Absurdity
Videos:Inventing on Principle
http://worrydream.com/ABriefRantOnTheFutureOfInte
ractionDesign/
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中山大学软件学院
2.1 Propositions and Logical Operations
2) Proposition variable and Compound statements
In Logic, the letters p, q, r, … denote proposition variable
(命题变量), they are replaced by statements.
p: The sun is shining today.
q: It is cold.
Statement or proposition variables can be combined by
logical connectives (逻辑联结词) to obtain compound
statements(复合命题).
Ex. and
p and q: The sun is shining today and it is cold.
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中山大学软件学院
2.1 Propositions and Logical Operations
3) Negation of statement – 否定命题
If p is a statement, the negation (非) of p is the statement
not p, denote by ~p.
~p is the statement “it is not the case that p”.
If p is true, ~p is false. If p is false, ~p is true.
The truth table(真值表) is a table that shows the truth
valuesof a compound statement in terms of its component
parts.
The truth table
p
~p
13
T
F
F
T
中山大学软件学院
2.1 Propositions and Logical Operations
3) Negation of statement
not is not a connective, it does not join two statements.
not p is not compound statement.
not is a unary operation for collection of statements.
~p is statement if p is.
Ex. Give the negation of the following statements:
(1) p: 2 + 3 > 1.
(2) q: it is cold.
Solution:
(1) ~p: 2 + 3  1.
(2) ~q: it is not cold.
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中山大学软件学院
2.1 Propositions and Logical Operations
4) Conjunction of statements – 合取式
If p and q are statements, the conjunction of p and q is
“p and q”, denote by pq.
 stands for connective and.
The compound pq is true if both p and q are true,
otherwise, it is false.
The truth table of pq
p q pq
T
T
F
F
15
T
F
T
F
T
F
F
F
中山大学软件学院
2.1 Propositions and Logical Operations
4) Conjunction of statements
Ex. From the conjunction of p and q for each of the
following.
(a) p: It is snowing, q: I am cold.
(b) p: 2 < 3, q: -5 > -8.
(c) p: It is snowing, q: 3 < 5.
Solution:
(a) pq: it is snowing and I am cold.
(b) pq: 2 < 3 and -5 > -8.
(c) pq: It is snowing and 3 < 5.
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中山大学软件学院
2.1 Propositions and Logical Operations
5) Disjunction of statements – 析取式
17
If p and q are statements, the disjunction of p and q is “p
or q”, denote by pq.
 denotes the connective or.
The compound pq is true if at least one of p or q is true,
it is false when both p and q are false.
The truth table of pq
p q pq
T T
T
T F
T
F T
T
F F
F
中山大学软件学院
2.1 Propositions and Logical Operations
5) Disjunction of statements
Ex. From the disjunction of p and q for each of the
following.
(a) p: 2 is a positive integer q: 21/2 is a rational number.
(b) p: 2 + 3  5
q: London is the capital of France.
Solution:
(a) pq: 2 is a positive integer or 21/2 is a rational number.
p is true, q is false, so pq is true.
(b) pq: 2 + 3  5 or London is the capitcal of France.
p and q are false, so pq is false.
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中山大学软件学院
2.1 Propositions and Logical Operations
6) Compound statement – 复合命题
The compound statement has many component parts,
both of component parts are connected by connective.
pq(pr) is compound statement and has three
compound parts (propositions): p, q and r.
p q r pr q(pr) pq(pr)
pq(pr) = pqr
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T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
T
T
T
F
T
F
T
T
F
F
T
F
F
F
T
T
T
T
T
F
F
F
中山大学软件学院
2.1 Propositions and Logical Operations
7) The truth table – 真值表
20
If a compound statement s has n component statements,
there will need to be 2n rows in the truth table for s. its truth
table may be systematically constructed in the following way.
(1). The first n columns of the table are labeled by
statements, further columns are included for all intermediate
expression, the last column is for the full statement.
(2). Under each of the first n headings, we list the 2n
possible n-tuples of truth values for the n component
statements.
(3). For each of the remaining columns, we compute the
remaining truth values in sequence.
中山大学软件学院
2.2 Conditional Statements
21
If p and q are statements, the compound statement “if p
then q” denoted by p  q, is called a conditional statement
(条件命题), or implication (蕴涵).
To p  q, p is called the antecedent (前件) or hypothesis
(假设), and q is called the consequent (后件) or conclusion(结
论).
Ex. Write the implication p  q for each of the following.
(a) p: I am hungry.
q: I will eat.
(b) p: It is snowing.
q: 3 + 5 = 8
Solution:
(a) If I am hungry, then I will eat.
(b) If it is snowing, then 3 + 5 = 8.
中山大学软件学院
2.2 Conditional Statements
p  q: p implies q; q if p; p only if q.
p is a sufficient condition for q, q is necessary condition
for p.
If p  q is a implication, the converse (逆命题) of p  q
is the implication q  p, and the contrapositive (逆否命题) of
p  q is the implication ~q  ~p.
22
Ex. Give the converse and contrapositive of the implication “If it is
raining, then I get wet.”
Solution:
p: It is raining; q: I get wet.
the converse is q  p: If I get wet, then It is raining.
the contrapositive is ~q  ~p: If I don’t get wet, then it is not
raining.
中山大学软件学院
2.2 Conditional Statements
23
If p and q are statements, the compound statement p if
and only if q, denoted by p  q, is called an equivalence (等
价) or biconditional (双条件).
The connective if and only if is denoted by the symbol .
conditional statement (条件命题), or implication (蕴涵).
The truth table of p  q is following.
p  q is true only when both The truth table of p  q
p q pq
p and q are true or when both p
T T
T
and q are false.
T F
F
The equivalence p  q can
F T
F
also be stated as p is a necessary
F F
T
and sufficient condition for q.
中山大学软件学院
2.2 Conditional Statements
Ex. Compute the truth table of the statement (p  q) 
(~q  ~p).
Solution:
p q p  q ~q ~p ~q  ~p (p  q)  (~q  ~p)
T
T
F
F
24
T
F
T
F
T
F
T
T
F
T
F
T
F
F
T
T
T
F
T
T
T
T
T
T
中山大学软件学院
2.2 Conditional Statements
A statement that is true for all possible values of its
propositional variables is called a tautology(重言式/永真式).
A statement that is always false is called a contradiction
(矛盾/永假式) or an absurdity (谬论).
A statement that can be either true or false, depending
on the truth values of its propositional variables, is called a
contingency (不定式).
Ex.
(a) (p  q)  (~q  ~p) is a tautology.
(b) p  ~p is an absurdity.
(c) (p  q)  (p  q) is a contingency.
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中山大学软件学院
2.2 Conditional Statements
If p  q is a tautology, p and q are logically equivalent
(逻辑等价), or simply equivalent (等价).
“p is equivalent to q” is denoted by p  q.
Ex. p  q  q  p.
p q p  q q  p (p  q)  (q  p)
T T
T
T
T
T F
T
T
T
F T
T
T
T
F
F F
F
T
The truth table for (p  q)  (q  p) shows the statement
is a tautology.
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中山大学软件学院
2.2 Conditional Statements
Ex. p  q is equivalent to (~p  q).
p q p  q ~p ~p  q
T
T
F
F
T
F
T
F
T
F
T
T
F
F
T
T
T
F
T
T
Column 1 and 3 in the above table show that for any
truth values of p and q, p  q and (~p  q) have the same
truth values.
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中山大学软件学院
2.2 Conditional Statements
Theorem 1 The operations for propositions have the
following properties.
(1) Commutative properties
pqqp
(2) Associative properties
p  (q  r)  (p  q)  r
(3) Distributive properties
p  (q  r)  (p  q)  (p  r)
(4) Idempotent properties
ppp
(5) Properties of Negation
~(~p)  p
~(p  q)  (~p)  (~q)
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pqqp
p  (q  r)  (p  q)  r
p  (q  r)  (p  q)  (p  r)
ppp
~(p  q)  (~p)  (~q)
中山大学软件学院
2.2 Conditional Statements
Theorem 2
(a) (p  q)  ((~p)  q)
(b) (p  q)  (~q  ~p)
(c) (p  q)  ((p  q)  (q  p))
(d) ~(p  q)  (p  ~q)
(e) ~(p  q)  ((p  ~q)  (q  ~p))
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中山大学软件学院
2.2 Conditional Statements
Theorem 4 Each of the following is a tautology.
(a) (p  q)  p
(b) (p  q)  q
(c) p  (p  q)
(d) q  (p  q)
(e) ~p  (p  q)
(f) ~(p  q)  p
(g) (p  (p  q))  q
(h) (~p  (p  q))  q
(i) (~q  (p  q))  ~p
(j) ((p  q)  (q  r))  (p  r)
30
Formalization of math statements
Theorem If A and B are finite sets, then |A∪B|=|A|+|B||A∩B|.
Implicitly, we are saying that for all sets A and B,
if A and B are finite, then |A∪B|=|A|+|B|-|A∩B|.
Here we have a predicate P(x): x is a finite set.
31
Theorem Let a, b and c be integers.
If a | b and a | c, then a | (b+c).
Implicitly, we are saying that for all integers a, b and c If
a | b and a | c, then a | (b+c).
Here we have a binary predicate (or a relation) R(x,y): x
| y.
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中山大学软件学院
2.1 Propositions and Logical Operations
8) Quantifiers – 量词
A = { x | P(x) }
The set A is the collection that an object t for which P(t)
is true.
P(x) is called predicate(谓词) or propositional function
(命题函数).
To each choice of x, P(x) is a proposition that is either
true ot false.
Ex. A = { x | x is an integer less than 8 }.
P(x) is the sentence “x is an integer less than 8”.
P(1) and P(2) are true, P(9) is false.
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中山大学软件学院
2.1 Propositions and Logical Operations
9) Universal quantification – 全称量化
The universal quantification of a predicate P(x) is the
statement “For all values of x, P(x) is true”, it is denoted by
xP(x).
The symbol  is called universal quantifier(全称量词).
Ex. P(x): -(-x) = x, x is a real number.
xP(x) is true because for all real numbers, -(-x) = x.
Q(x): x + 1 = 4.
xQ(x) is false because Q(1) is false.
34
判定xP(x)真假的一般方法:
(1) 当要确定xP(x)为“真”时, 则要确定所有P(x)都为“真”;
(2) 当要确定xP(x)为“假”时, 只要找一个y, 使得: P(y)为“假”.
中山大学软件学院
2.1 Propositions and Logical Operations
9) Universal quantification
A predicate may contain many variables. The universal
quantification may be applied to each of the variables.
if P(x, y) is a predicate, its universal quantification is
xyP(x, y).
For example,
P(x, y): x + y = y + x
xyP(x, y) is true for every x and y.
Q(x, y): x ∪ y = y ∪ x
xyQ(x, y) is true for any set x and y.
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中山大学软件学院
2.1 Propositions and Logical Operations
10) Existential quantification – 存在量化
The existential quantification of a predicate P(x) is the
statement “There exists a value of x for which P(x) is true”, it
is denoted by xP(x).
The symbol  is called existential quantifier(存在量词).
Ex. P(x): x + 1 < 4, x is a real number.
xP(x) is true because P(2) is true.
Q(y): y + 2 = y.
yQ(y) is false because Q(y) is false for any y.
36
判定xP(x)真假的一般方法:
(1) 当要确定xP(x)为“真”时, 只要找到一个x, 使得: P(x)为“真”;
(2) 当要确定xP(x)为“假”时, 则要确定所有的x, 都有: P(x)为“假”.
中山大学软件学院
2.1 Propositions and Logical Operations
10) Existential quantification
A predicate may contain many variables. The existential
quantification may be applied to each of the variables.
if P(x, y) is a predicate, its existential quantification is
xyP(x, y).
Ex.
P(x, y): x * y = y + x
xyP(x, y) is true because x = 2 and y = 2.
Q(x, y): x∩y = x - y
xyQ(x, y) is false for any set x and y.
x=?
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中山大学软件学院
2.1 Propositions and Logical Operations
10) The order of quantifiers
The order of same quantifiers is not important, it does
n’t affect the truth value. The order of different quantifiers is
important, it affects the truth value.
Ex.
P(x, y): x + y = 0
xyP(x, y) is true, xyP(x, y) is false.
Q(x, y): x * y = 0
v2
xyP(x, y) is true, xyP(x, y) is true too.
v1
v3
R(x, y): <x, y>  E(G), x, y  V(G)
v4
xyR(x, y) is true, xyR(x, y) is false.
图G
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中山大学软件学院
2.1 Propositions and Logical Operations
11) Negation of quantifiers
Let p: xP(x), what is the negation of p? (~p)
~p: ~xP(x) = x~P(x).
Let q: xQ(x), what is the negation of q? (~q)
~q: ~xQ(x) = x~Q(x).
39
3
中山大学软件学院
2.2 Conditional Statements
Theorem 3
(a) ~(xP(x))  x~P(x)
(b) ~(xP(x))  x(~P(x))
(c) x(P(x)  Q(x))  xP(x)  xQ(x)
(d) x(P(x)  Q(x))  xP(x)  xQ(x)
(e) x(P(x)  Q(x))  xP(x)  xQ(x)
(f) xP(x)  xQ(x)  x(P(x)  Q(x)) is a tautology
(g) x(P(x)  Q(x))  xP(x)  xQ(x) is a tautology
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中山大学软件学院
2.3 Methods of Proof
If an implication p  q is a tautology, where p and q may
be compound statements involving any number of proposition
variables, we say that q logically follows from p.
Suppose that an implication of the form (p1  p2  …  pn)
 q is a tautology. We say that q logically follows from p1, p2,
…, pn, and write
p1
p2
┆
pn
If p1, p2, …, pn are true, we know
 q
q is true.
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中山大学软件学院
2.3 Methods of Proof
(p1  p2  …  pn)  q
The pi’s are called the hypotheses (假设) or premises (前
提), and q is called the conclusion(结论).
To “prove the theorem” means to show that the implication is a tautology. If p1, p2,… and pn are true, q will be true.
For this reason, mathematical proofs often begin with the
statement “suppose that p1, p2,… and pn are true” and
conclude with the statement “therefore, q is true.”
The proof does not show that that q is true, but simply
shows if the pi are all true, then q has to be true.
42
Methods of Proofs
Arguments based on tautologies represent universally
correct methods of reasoning. The validity only depends on
the form of the statements involved and not on the truth
values of the variables they contain. Such arguments are
called rules of inference.
A mathematical proof of a theorem must start with the
hypotheses, proceed through various steps, each justified
by some rule of reference, and arrive at the conclusion.
(see page 63).
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中山大学软件学院
2.3 Methods of Proof
Ex. ((p  q)  (q  r))  (p  r)
The argument
pq
qr
 pr
is valid, and so is a rule of inference.
Ex. Is the following argument valid?
If you invest in the stock market, then you will get rich.
If you get rich, then you are happy.
 If you invest in the stock market, then you are happy.
The argument is valid, although the conclusion may be
false.
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中山大学软件学院
2.3 Methods of Proof
A important rule of inference is
p
pq
 q
That is, p is true and p  q is true, so q is true.
The rule is called modus ponens (假言推理), or the
method of asserting (论断方法).
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中山大学软件学院
2.3 Methods of Proof
Ex. Is the following argument valid?
Smoke is healthy.
If Smoke is healthy, then cigarettes are prescribed by physicians.
 Cigarettes are prescribed by physicians.
The argument is valid. However, the conclusion is false.
Ex. Is the following argument valid?
If taxes are lowered, then income rises.
Income rises.
 Taxes are lowered.
46
Solution:
pq
p: taxes are lowered, q: income rises.
q
p  q may be true with p being false. Then the
p

conclusion p is false. The argument is not valid.
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2.3 Methods of Proof
Ex. Let n be an integer. Prove that if n2 is odd, then n is
odd.
Proof:
p: n2 is odd.
q: n is odd.
We have to prove that p  q is true.
Suppose that ~q is true, i.e. n is not odd.
Then n = 2k, where k is an integer.
n2 = (2k)2 = 4k2 = 2(2k2), so n2 is even.
So we have that if n is even, then n2 is even, which is the
contrapositive of the given statement.
Hence the given statement has been proved.
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中山大学软件学院
2.3 Methods of Proof
An important indirect proof technique is proof by
contradiction(反证法). This method is based on the tautology
((p  q)  (~q))  ~p
Thus the rule of inference is valid.
pq
~q
 ~p
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Suppose we prove that a statement q logically follows from
statements p1, p2, …, pn.
If p1  p2  …  pn  (~q) implies a contradiction, then at least
one of the statements p1, p2, …, pn, (~q) must be false.
If all the pi’s are true, then ~q must be false, so q is true.
This is proof by contradiction.
中山大学软件学院
2.3 Methods of Proof
Ex. Prove there is no rational number p/q whose square
is 2. In other words, show 21/2 is irrational.
Proof:
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Assume (p/q)2 = 2 for some integers p and q, which have no common
factors.
Then p2 = 2q2, so p2 is even.
Since the square of an odd number is odd, so p is even.
Suppose that p = 2k for some integer k.
2q2 = (2k)2 = 4k2
q2 = 2k2
Thus, q2 is even, and q is even.
Now p and q are even, and have a common factor 2.
This is a contradiction to the assumption. Thus the assumption is
false.
中山大学软件学院
2.3 Methods of Proof
In order to prove a theorem of the form (p1  p2  … 
pn)  q, we begin with the hypothesis p1, p2, …, pn and show
that some result r1 logically follows. Then, using p1, p2, …, pn,
r1, we show that some other statement r2 logically follows.
We continue this process, producing intermediate
statements r1, r2, …, rk, called steps in the proof, until we can
finally show that the conclusion q logically follows from p1,
p2, …, pn, r1, r2, …, rk.
Each logical step must be justified by some valid proof
technique, based on the rules of inference, or on some other
rules that come from tautological implications.
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中山大学软件学院
2.3 Methods of Proof
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Ex. Prove or disprove the statement that if x and y are
real number, (x2 = y2)  (x = y).
Proof:
The statement is restated in the following form
xy((x2 = y2)  (x = y))
To prove it, we would need to prove that for all x and y it
is true.
To disprove it, we need only find one example for which
the implication is false.
Let x = -3, y = 3.
Since (-3)2 = 32, but -3  3, i.e. x  y.
Thus the statement is false.
The example of x and y is called a counterexample(反例).
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2.4 Mathematical Induction
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We have the statement nn0P(n), where n0 is some
fixed integer.
If we prove P(n) is true for all integers n  n0, the
following steps shall be done.
Step 1. P(n0) is true;
Step 2. If P(k) is true for some k  n0, then P(k+1) must
be true.
So P(n) is true for n  n0.
The method of proof is called the principle of
mathematical induction (数学归纳法).
Step 1 is the basis step (基础步), step 2 is called the
induction step (归纳步).
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Ex. Show, by mathematical induction, that for all n  1,
1 + 2 + 3 + … + n = n(n+1)/2
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Proof:
Let P(n) be the statement 1+2+3+…+n = n(n+1)/2.
(1). Basis Step
P(1): 1 = 1(1+1)/2, so P(1) is true.
(2). Induction Step
Assume that for some fixed k  1, then P(k) is true.
P(k): 1+2+3+…+k = k(k+1)/2
Then P(k+1) is the following statement.
P(k+1): 1+2+…+k+(k+1)= (k+1)((k+1)+1)/2
1+2+…+k + (k+1) = k(k+1)/2 + (k+1)
= (k+1)(k/2 + 1)
= (k+1)(k+2)/2
= (k+1)((k+1)+1)/2
So P(k+1) is true.
Thus by the mathematical induction, P(n) is true for all n  1.
中山大学软件学院
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Ex. Let A1, A2, …, An be any n stes, we show by mathematical
induction that ∪i=1..nAi = ∩i=1..nAi.
(it is an extended version of one of De Morgan’s laws)
Proof:
Let P(n) be the predicate that the equality holds for any n sets. We
prove by mathematical induction that for all n  1, P(n) is true.
(1). Basis Step
P(1): A1 = A1, so P(1) is true.
(2). Induction Step
Assume that for some fixed k  1, then P(k) is true.
P(k+1): ∪i=1..(k+1)Ai = ∩i=1..(k+1)Ai
The left-hand sideof P(k+1) is
∪i=1..(k+1)Ai = A1∪A2 … ∪Ak∪Ak+1
= (A1∪A2 … ∪Ak)∪Ak+1
= (A1∪A2 … ∪Ak)∩Ak+1
= (∩i=1..kAi)∩Ak+1
So P(k+1) is true.
= ∩i=1..(k+1)Ai
Thus by the mathematical
induction, P(n) is true for all n  1.
中山大学软件学院
2.4 Mathematical Induction
Strong Induction - 强归纳法
In strong form of mathematical induction, or strong
induction, the induction step is to show that
P(n0)P(n0+1)…P(k-1)P(k)  P(k+1)
is a tautology for k  n0.
We must prove: if each P(j) is true (j=n0..k), then P(k+1)
is true.
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中山大学软件学院
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Ex. Prove that every positive integer n > 1 can be written uniquely
as p1a1p2a2…psas, where the pi are primes and p1<p2<…<ps.
Proof:
(1). Basis Step
Here n0 is 2, P(2) is clearly true, since 2 = 21.
(2). Induction Step
Assume that P(2), P(3), …, P(k) are true.
Now we need to prove that P(k+1) is true.
(2.1). (k+1) is a prime
P(k+1) = (k+1)1, so P(k+1) is true.
(2.2). (k+1) is not prime
Let (k+1) = mn, where 2  m, n  k.
P(m) = q1b1q2b2…qubu, P(n) = r1c1r2c2…cvcv.
k+1 = mn = q1b1q2b2…qubu  r1c1r2c2…cvcv = p1a1p2a2…psas
if pi = qj = rl, then ai = bj+cl, otherwise pi = qj and ai = bj, or pi =
rl and ai = cl.
Since the factorization of m and n are unique, so is the factorization
of (k+1).
Inductive Definitions, Induction and
Recursion
Definition of naturals
Definition of well formed formulas
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Summary
 Important concepts: propositions, connectives,
quantifiers, truth value, truth table, tautology, absurdity,
contingency, logically equivalent, reference rules, induction
 To be able to construct the truth table of a proposition,
decide if a proposition is a tautology and if two
propositions are equivalent.
 To be to do mathematical induction proofs.
 Understand mathematical proofs formally.
 Understand inductive definitions and recursion.
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