Chapter 2. Logic

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Chapter 2. Logic

• Weiqi Luo (???)
• School of Software
• Sun Yat-Sen University
• Emailweiqi.luo_at_yahoo.com OfficeA309

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Chapter two Logic

• 2.1. Propositions and Logical Operation
• 2.2. Conditional Statements
• 2.3. Methods of Proof
• 2.4. Mathematical Induction
• 2.5. Mathematical Statements
• 2.6. Logic and Problem Solving

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2.1. Propositions and Logical Operation

• Statement (Proposition)
• A statement or proposition is a declarative
  sentence that is either true or false but not
  both
• Example 1
• (a) The earth is round.
• (b) 235
• (c) Do you speak English?
• (d) 3-x5
• (e) Take two aspirins.
  
• (f) The temperature on the surface of the
  planet Venus is 800 F
• 
  
• (g) The sun will come out tomorrow

Yes.
Yes.
No. This is a question.
No. May true or false
No. This is a command.
Yes.
yes
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2.1. Propositions and Logical Operation

• Paradox
• A male barber shaves all and only those men
  who do not shave themselves
• Paradox A paradox is a seemingly true
  statement or group of statements that lead to a
  contradiction or a situation which seems to defy
  logic or intuition . Paradox is not a statement.
  
• Refer to
• http//en.wikipedia.org/wiki/List_of_paradoxe
  s
• Others
• I am lying, this sentence is wrong, so
  on and so forth

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2.1. Propositions and Logical Operation

• Propositional variables
• In logic, the letters p, q, r denote
  propositional variables, which are replaced by
  statements
• p 12 5
• q It is raining.
• Compound statements
• Propositional variables can be combined by
  logical connectives to obtain compound
  statements. E.g.
• p and q 12 5 and it
  is raining.

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2.1. Propositions and Logical Operation

• Negation (a unary operation)
• If p is a statement, the negation of p is
  the statement not p, denoted by p, meaning it
  is not the case that p.
• if p is true, then p is false, and if p is
  false, then p is true.
• Truth Table List the truth value of a
  compound statement in terms of its component
  parts.

p q
T F
F T
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2.1. Propositions and Logical Operation

• Example
• Give the negation of the following statements
• (a) p 23 gt1
• (b) q It is snowing.
• Solution
• (a) p 23 is not greater than 1, namely,
  23 lt1
• (b) q It is not the case that it is
  snowing. More simply, q It is not snowing.

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2.1. Propositions and Logical Operation

• Conjunction
• If p and q are statements, the conjunction
  of p and q is the compound statement p and q
  denoted by p? q .
• Truth Table
• Note p? q is T if and only if p is T and q
  is T.

p q p? q
T T T
T F F
F T F
F F F
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2.1. Propositions and Logical Operation

• Example
• Form the conjunction of p and q
• p 1gt3 q It
  is raining.
• Solution
• p? q 1 gt 3 and It is raining.
  
• Note
• In logic, unlike in everyday English, we may
  join two totally unrelated statements by logical
  connectives.

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2.1. Propositions and Logical Operation

• Disjunction
• If p and q are statements, the disjunction of
  p and q is the compound statement p or q,
  denoted by p V q
• Truth Table
• Note p V q is F is and only if q is F and q
  is F.

p q p V q
T T T
T F T
F T T
F F F
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2.1. Propositions and Logical Operation

• The connective or
• (a) I left for Spain on Monday or I left for
  Spain on Friday.
• (b) I passed math or I failed French
• Note
• Case (a) Both could not have occurred.
  or is an excusive sense in this case.
• Case (b) Both could have occurred. or is
  an inclusive sense in this case.
• In mathematics and computer science. We
  agree to use the connective or in inclusive
  manner.

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2.1. Propositions and Logical Operation

• Example
• Form the disjunction of p and q
• p 2 is a positive integer q sqrt(2)
  is a rational number
• Solution
• p V q 2 is a positive integer or sqrt(2)
  is a rational number. Since p is true, p V q is
  true, even through q is false.

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2.1. Propositions and Logical Operation

• Algorithm for making Truth Table
• Step 1 The first n columns of the table are
  labeled by the component propositional variables.
  Further columns are included for all intermediate
  combinations of the variables, culminating in a
  column for the full statement.
• Step 2 Under each of the first n headings, we
  list the 2n possible n-tuples of truth values for
  the n component statements.
• Step 3 For each of the remaining columns, we
  compute, in sequence, the remaining truth values.
  

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2.1. Propositions and Logical Operation

• Example 5
• Make a truth table for the statement (p ? q)
  V ( p)
• Truth Table

p q p? q p V
T T T F T
T F F F F
F T F T T
F F F T T
(1)
(2)
(3)
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2.1. Propositions and Logical Operation

• Propositional function (predicate)
• An element of a set x P(x) is an object
  t for which the statement P(t) is true. P(x) is
  called a propositional function (or predicate) ,
  because each choice of x produces a proposition
  P(x) that is either true or false (well-defined)
• E.g. Let A x x is an integer less than
  8.
• Here P(x) is the sentence x is an integer
  less than 8
• P(1) denotes the statement 1 is an integer
  less than 8 (true)
• P(8) denotes the statement 8 is an integer
  less than 8 (false)

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2.1. Propositions and Logical Operation

• Universal Quantifiers (?)
• The Universal Quantifiers of a predicate
  P(x) is the statement for all values of x, P(x)
  is true , denoted by ? x P(x)
• Example 8
• (a) P(x) -(-x) x is a predicate that
  makes sense for all real number x.
• then ? x P(x) is true statement.
  Since ? x ? R, -(-x) x
• (b) Q(x) x1lt4.
• then ? x Q(x) is a false statement,
  since Q(5) is false

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2.1. Propositions and Logical Operation

• Existential Quantifiers (?)
• The Existential Quantifiers of a predicate
  P(x) is the statement there exists a value of x
  for which P(x) is true, denoted by
• ? x
  P(x)
• Example 9
• (a) Let Q(x) x1lt4. then the existential
  quantification of Q(x), ? x Q(x), is a true
  statement, since Q(1) is a true statement
• (b) The statement ? y y2y is false since
  there is no value of y for which the
  propositional function y2y produces a true
  statement.

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2.1. Propositions and Logical Operation

• The order of the Quantifiers ? ?
• The order does not affect the output for the
  same quantifiers, while it may produce different
  results for different quantifiers.
• E.g. P(x, y) x y 1
• ? x ? y P(x) is true, ? y ?
  x is false.
• P(x, y) x y 0
• ? x ? y P(x) is true, ? y ?
  x is true too.

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2.1. Propositions and Logical Operation

• The negation of Quantifiers ? ?
• (a) let p ?x P(x),
• then p there must be at least one
  value of x for which P(x) is false, namely,
• ?x P(x) ?x P(x)
• (b) let p ?x P(x),
• then p for all x, P(x) is false,
  namely,
• ?x P(x) ?x P(x)

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2.1. Propositions and Logical Operation

• Homework
• Ex. 2, Ex. 4, Ex. 16, Ex. 28

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2.2. Conditional Statements

• Conditional statement
• If p and q are statements, the compound
  statement if p then q, denoted pgtq, is called
  a conditional statement or implication.
• p antecedent or hypothesis q
  consequent or conclusion
• if then gt
• Truth Table
• Note when p is F, then pgtq is T.

p q p gt q
T T T
T F F
F T T
F F T
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2.2. Conditional Statements

• Example
• Form the implication pgtq for each the
  following
• (a) p I am hungry. q I will eat.
• (b) p 225 q I am the king
  of England.
• Solution
• (a) If I am hungry, then I will eat
• (b) If 225, then I am the king of English
  
• Note There is no cause-and effect relationship
  between p and q in case (b). And (b) is true,
  since 225 is false.

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2.2. Conditional Statements

• Converse and Contrapositive
• If pgtq is an implication, then its converse
  is the implication q gt p
• and its contrapositive is the implication
  q gt p
• E.g. Give the converse and the contrapositive
  of the implication If it is raining, then I get
  wet
• Converse If I get wet, then It is
  raining.
• Contrapositive If I do not get wet, then It is
  not raining.

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2.2. Conditional Statements

• Equivalence (biconditional)
• If p and q are statements, the compound
  statement p if and only if q, denoted by p ? q,
  is called an equivalence or biconditional.
• Truth Table
• Note p ltgt q is T when p and q are both T or
  both F.

p q p ? q
T T T
T F F
F T T
F F T
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2.2. Conditional Statements

• Example 3
• Is the following equivalence a true
  statement?
• 3gt2 if and only if 0lt 3
  2
• Solution
• Let p 3gt2 and q 0lt 3 2,
• since p and q are both true, we then
  conclude that
• p ? q is true
  statement.

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2.2. Conditional Statements

• Example 4.
• Compute the truth table of the statement
• (pgtq) ? (q gt
  p)
• Truth Table

p q pgtq q p qgtp (pgtq) ? (q gt p)
T T T F F T T
T F F T F F T
F T T F T T T
F F T T T T T
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2.2. Conditional Statements

• Tautology
• A statement that is true for all possible
  values of its propositional variables called
  Tautology. (e.g. Example 4)
• Contradiction (Absurdity)
• A statement that is false for all possible
  values of its propositional variables called
  Contradiction or Absurdity. (e.g. p ? p)
• Contingency
• A statement that can be either true or false,
  depending on the truth values of its
  propositional variables, is called a contingency.
  
• (e.g. p gt q)

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2.2. Conditional Statements

• Logically Equivalent
• If two statements p and q are always either
  both true or both false, for any values of the
  propositional variables, namely
• p ? q is a tautology
• Then we call p and q are logically
  equivalent.
• Denoted by
• p q

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2.2. Conditional Statements

• Example 6
• Show that p V q and q V p are logically
  equivalent
• The truth table of (p V q ) ? (q V p ) are
  shown as follows
• Truth Table

p q p V q q V p p V q ?q V p
T T T T T
T F T T T
F T T T T
F F F F T
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2.2. Conditional Statements

• Two Structures with similar properties
• (Theorem 1)
• Structure 1 (logic operations)
• (propositions, ?, V, )
• Structure 2 (sets operations)
• (sets, U, n , - )

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2.2. Conditional Statements

• Theorem 1
• Commutative properties
• p ? q q ?p, p? q q
  ? p
• Associative Properties
• p ? (q ? r) (p ? q) ?r , p? (q ? r)
  (p ? q) ? r
• Distributive Properties
• p ? (q ? r) (p ? q) ?
  (p ? r)
• p ? (q ? r) (p ? q) ?
  (p ? r)

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2.2. Conditional Statements

• Idempotent Properties
• p ? p p , p? p
  p
• Properties of Negation
• (p) p
• (p ? q) (p) ? (q)
• (p ? q) (p) ? (q) De
  Morgans laws

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2.2. Conditional Statements

• Theorem 2
• (p ? q) ((p) ? q)
• (p ? q) (q ? p)
• (p ? q) ((p ? q) ? (q ? p))
• (p ? q) (p ? q)
• (p ? q) ((p ? q) ? (q ? p))

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2.2. Conditional Statements

• Theorem 3
• (?xP(x)) ?xP(x)
• (?xP(x)) ?x(P(x))
• ?x(P(x) ? Q(x)) ?xP(x) ? ?xQ(x)
• ?x(P(x) ? Q(x)) ?xP(x) ? ?xQ(x)
• ?x(P(x) ? Q(x)) ?xP(x) ? ?xQ(x)
• ?xP(x) ? ?xQ(x) ? ?x(P(x) ? Q(x)) is a tautology
• ?x(P(x) ? Q(x)) ? ?xP(x) ? ?xQ(x) is a tautology

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2.2. Conditional Statements

• Theorem 4 Each of the following is a tautology
• (p ? q) ? p , (p ? q) ? q
• p ? (p ? q) , q ? (p ? q)
• p ? (p ? q) , (p ? q) ? p
• (p ? (p ? q)) ? q , (p ? (p ? q)) ? q
• (q ? (p ? q)) ? p
• ((p ? q) ? (q ? r)) ? (p ? r)

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2.2. Conditional Statements

• Properties of Quantifiers ? and ?
• (a) ? x (P(x) V Q(x)) ?x P(x) V ?x (Q(x))
• (b) ? x (P(x) ? Q(x)) ?x P(x) ?x Q(x)

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2.2. Conditional Statements

• Homework
• ex.2, ex.7, ex.12, ex.15, ex.21 ex.34

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2.3. Method of Proof

• Logically Follow
• 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, denoted by

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2.3. Method of Proof

• (p1 ? p2 ? ? pn) ? q
• The pis are called the hypotheses or
  premises, and q is called the conclusion.
• Note we are not trying to show that q is
  true, but only that q will be true if all the pi
  are true.
• ? denotes therefore

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2.3. Method of Proof

• Rules of inference
• Arguments based on tautologies represent
  universally correct methods of reasoning. Their
  validity depends only on the form of the
  statements involved and not on the truth values
  of the variables they contain.
• Example 1
• ((p ? q) ? (q ? r)) ? (p ? r) is tautology,
  then the argument
• is universally valid, and so is a rule of
  inference.

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2.3. Method of Proof

• Example 2 Is the following argument valid?
• If you invest in the stock market,
  then you will get rich.
• If you get rich, then you will be
  happy.
• ? If you invest in the stock market, then
  you will be happy.
• let p you invest in the stock market, q
  you will get rich
• r you will be happy
• The above argument is of the form given in
  Example 1, hence the argument is valid!

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2.3. Method of Proof

• Example 3
• The tautology
• (p ? q) ?((p ? q) ? (q ? p))
• means that the following two arguments are
  valid
• 
  p ? q
• p ? q
  q ? p
• ? (p ? q) ? (q ? p) ? p ? q

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2.3. Method of Proof

• Equivalence (p ? q)
• They are usually stated p if and only if q.
  We need to prove both pgtq and qgtp by the
  tautology mentioned in example 3.
• Algorithm
• Step one Assuming p is true, show q must be
  true.
• Step two Assuming q is true, show p must be
  true.

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2.3. Method of Proof

• Modus Ponens
• p is true, and pgtq is true, so q is true
• (Theorem 4(g) in Section 2.2.)
• p
• pgtq
• ? q

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2.3. Method of Proof

• Example 4 Is the following argument valid?
• Smoking is healthy.
• If smoking is healthy, then cigarettes are
  prescribed by physicians.
• ? Cigarettes are prescribed by physicians
• p Smoking is healthy.
• q cigarettes are prescribed by physicians
• The argument is valid since it is of form modus
  ponens.

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2.3. Method of Proof

• Example 5 Is the following argument valid?
• If taxes are lowered, then income
  rises
• Income rises
• ? taxes are lowered
• Solution
• Let p taxes are lowed q income rise
• pgtq
• q
• ? p
• Then argument is not valid , since pgtq and q
  can be both true with p being false.

(or show the truth table of (pgtq) ? q gt p ,
and determine whether or not it is a tautology)
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2.3. Method of Proof

• Indirect Method of Proof
• The tautology
• (pgtq ) ? (q) gt (p )
• (An implication is equivalent to its
  contrapositive)
• Note
• To proof pgtq indirectly, we assume q is false
  (q) and show that p is then false (p)

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2.3. Method of Proof

• Example 6
• Let n be an integer. Prove that if n2 is odd,
  then n is odd.
• Solution
• Let p n2 is odd , q n is odd.
• To prove that (pgtq)
• We try to prove its contrapositive qgtp
• Assuming that n is even (q), let n2k, k is
  an integer, then we have n2 (2k) 2 4k2 is
  even (p).
• Hence, the given statement has been proved.

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2.3. Method of Proof

• The tautology
• ((p ? q) ? (q)) ? p
• If a statement p implies a false statement q,
  then p must be false.
• Proof by contradiction
• To prove (p1 ? p2 ? ? pn) ? q ,
• We add (q) into hypothesis p1 ? p2 ? ? pn,
  if the enlarged hypothesis p1 ? p2 ? ? pn ? (q)
  implies a contradiction, then we can conclude
  that q follows from p1 ,p2 , and pn .

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2.3. Method of Proof

• Example 7
• Prove there is no rational number a/b whose
  square is 2, namely, sqrt(2) is irrational.
• Solution
• Let p a, b are integers and no common
  factors, and b is not 0
• q (a/b)2 is not 2
• In order to prove p gt q ,
• We try to find the contradiction from p ? q
  
• Refer to Example 7 for more details.

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2.3. Method of Proof

• Example 9
• Prove or disprove the statement that if x and
  y are real numbers, (x2y2) ? (xy)
• Solution
• Since (1)2(-1)2, but -1 ? 1, the result is
  false. Our example is called counterexample, and
  any other counterexample would do just as well.
• Note
• If a statement claims that a property holds
  for all objects of a certain type, then to prove
  it, we must use steps that are valid for all
  objects of that type and that do not make
  references to any particular object. To disprove
  such a statement, we need only show one
  counterexample.

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2.3. Method of Proof

• Homework
• Ex. 6, Ex. 8, Ex. 9
• Ex. 19, Ex. 20
• Ex. 31, Ex. 34

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2.4 Mathematical Induction

• To Prove
• ?ngtn0 P(n)
• where n0 is some fixed integer
• Two Steps
• 1) Basis Step
• Prove that P(n0) is true
• 2) Induction Step
• Prove that P(k) gt P(k1) is a
  tautology
• (if P(k) is true, then P(k1) must be true)

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2.4 Mathematical Induction

• Example 1 Prove that for all ngt0
• 123n n(n1)/2
• Solution let P(n)n(n1)/2
• Basis step P(1) 1 12/21 is true.
• Induction Step assuming P(k) is true, then
• P(k1) 12 3 k (k1)
• P(k) (k1)
• k (k1)/2 (k1)
  P(k) is true
• (k1)(k2)/2
• P(k1)

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2.4 Mathematical Induction

• Example 3 Prove any finite, nonempty set is
  countable (the elements can be arranged in a
  list)
• Solution Let P(n) be the predicate that if A is
  any set with Angt0, then A is countable
• Basis Step P(1) is true ( A x )
• Induction Step assuming P(k) is true
• Let B denotes any finite, nonempty set with
  k1 elements. We first pick any element x in B,
  then B-x is a set with k elements, and it is
  countable (P(k) is true), and listed by x1,x2,xk
  . Then we can also list the elements of B as
  x1,x2,xk , x (P(k1) is true)

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2.4 Mathematical Induction

• Example 5 Consider the following function given
  in pseudocode
• Function SQ(A)
• 1. C lt- 0
• 2. D lt- 0
• 3. WHILE (D is not A)
• a. C lt- CA
• b. D lt- D1
• 4. RETURN (C)
• End of Function SQ

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2.4 Mathematical Induction

• Strong Induction

Weak Induction Strong Induction
Basis Step P(no) is true (or the first several statements are true) P(no) is true (or the first several statements are true)
Induction Step P(k) ? P(k1) is a tautology P(n0) ?P(n1) ? P(k) ? P(k1) is a tautology
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2.4 Mathematical Induction

• Example 7 Prove that every positive integer ngt1
  can be written uniquely as p1a1p2a2psas, where
  pi are primes and p1ltp2lt..Ps
• Basis Step P(2) is true, since 2 is prime
  and 2 21 (unique )
• Induction Step Assuming P(2), P(3), P(k)
  are true
• if k1 is prime, then k1 (k1)1
• if k1 is not prime, then we let k1Lm,
  where L, m are positive integers less than k1.
  Using P(L) and P(m) are true, we have k1Lm
  q1b1q2b2qsbs (unique form)
• Why? Proof by contradiction

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2.4 Mathematical Induction

• Homework
• Ex. 2, Ex. 14, Ex. 17, Ex. 22, Ex. 34