Review I

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Review I
Rosen 1.1-1.5, 3.1
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Know your definitions!
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Definition 1. Negation of p

• Let p be a proposition. The statement It is
  not the case that p is also a proposition,
  called the negation of p or p (read not p)

p The sky is blue. ?p It is not the case that
the sky is blue. ?p The sky is not blue.
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Definition 2. Conjunction of p and q

• Let p and q be propositions. The proposition p
  and q, denoted by p?q is true when both p and q
  are true and is false otherwise. This is called
  the conjunction of p and q.

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Definition 3. Disjunction of p and q

• Let p and q be propositions. The proposition p
  or q, denoted by p?q, is the proposition that is
  false when p and q are both false and true
  otherwise.

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Definition 4. Exclusive or of p and q

• Let p and q be propositions. The exclusive or
  of p and q, denoted by p?q, is the proposition
  that is true when exactly one of p and q is true
  and is false otherwise.

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Definition 5. Implication p?q

• Let p and q be propositions. The implication p?q
  is the proposition that is false when p is true
  and q is false, and true otherwise. In this
  implication p is called the hypothesis (or
  antecedent or premise) and q is called the
  conclusion (or consequence).

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Implications

• If p, then q
• p implies q
• if p,q
• p only if q
• p is sufficient for q
• q if p
• q whenever p
• q is necessary for p

• Not the same as the if-then construct used in
  programming languages such as If p then S

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Implications

• How can both p and q be false, and p?q be true?
• Think of p as a contract and q as its
  obligation that is only carried out if the
  contract is valid.
• Example If you make more than 25,000, then you
  must file a tax return. This says nothing about
  someone who makes less than 25,000. So the
  implication is true no matter what someone making
  less than 25,000 does.
• Another example
• p Bill Gates is poor.
• q Pigs can fly.
• p?q is always true because Bill Gates is not
  poor. Another way of saying the implication is
• Pigs can fly whenever Bill Gates is poor which
  is true since neither p nor q is true.

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Related Implications

• Converse of p ? q is q ? p

• Contrapositive of p ? q is
  the proposition ?q ? ?p

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Definition 6. Biconditional

• Let p and q be propositions. The biconditional
  p?q is the proposition that is true when p and q
  have the same truth values and is false
  otherwise. p if and only if q, p is necessary
  and sufficient for q

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Logical Equivalence

• An important technique in proofs is to replace a
  statement with another statement that is
  logically equivalent.
• Tautology compound proposition that is always
  true regardless of the truth values of the
  propositions in it.
• Contradiction Compound proposition that is
  always false regardless of the truth values of
  the propositions in it.

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Logically Equivalent

• Compound propositions P and Q are logically
  equivalent if P?Q is a tautology. In other
  words, P and Q have the same truth values for all
  combinations of truth values of simple
  propositions.
• This is denoted P?Q (or by P Q)

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Example DeMorgans

• Prove that ?(p?q) ? (?p ? ?q)
• p q (p?q) ?(p?q) ?p ?q (?p ? ?q)

T T T F F T F F
T F F F F
T F F T F
T F T F F

F T T T T

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List of Logical Equivalences
p?T ? p p?F ? p Identity Laws p?T ? T
p?F ? F Domination Laws p?p ? p p?p ? p
Idempotent Laws ?(?p) ? p Double Negation
Law p?q ? q?p p?q ? q?p Commutative
Laws (p?q)? r ? p? (q?r) (p?q) ? r ? p ? (q?r)

Associative Laws
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List of Equivalences
p?(q?r) ? (p?q)?(p?r) Distribution Laws p?(q?r)
? (p?q)?(p?r) ?(p?q)?(?p ? ?q) De Morgans
Laws ?(p?q)?(?p ? ?q) Misc. , Table 6 p ? ?p
? T Or Tautology p ? ?p ? F And
Contradiction (p?q) ? (?p ? q) Implication
Equivalence p?q?(p?q) ? (q?p) Biconditional
Equivalence
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Prove (p??q) ? q ? p?q

• (p??q) ? q Left-Hand Statement
• ? q ? (p??q) Commutative
• ? (q?p) ? (q ??q) Distributive
• ? (q?p) ? T Or Tautology (Misc. T6)
• ? q?p Identity
• ? p?q Commutative

Begin with exactly the left-hand side
statement End with exactly what is on the
right Justify EVERY step with a logical
equivalence
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Prove (p??q) ? q ? p?q

• (p??q) ? q Left-Hand Statement
• ? q ? (p??q) Commutative
• (q?p) ? (q ??q) Distributive
• Why did we need this step?

Our logical equivalence specified that ? is
distributive on the right. This does not
guarantee distribution on the left! Ex. Matrix
multiplication is not always commutative (Note
that whether or not ? is distributive on the left
is not the point here.)
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Prove or Disprove

• p ? q ? p ? ?q ???
• To prove that something is not true it is enough
  to provide one counter-example. (Something that
  is true must be true in every case.)
• p q p?q p??q
• F T T F
• The statements are not logically equivalent

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Method to construct DNF

• Construct a truth table for the proposition.
• Use the rows of the truth table where the
  proposition is True to construct minterms
• If the variable is true, use the propositional
  variable in the minterm
• If a variable is false, use the negation of the
  variable in the minterm
• Connect the minterms with ?s.

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How to find the DNF of (p Ú q)Ør

• p q r (p Ú q) Ør (p Ú q)Ør
• T T T T F F
• T T F T T T
• T F T T F F
• T F F T T T
• F T T T F F
• F T F T T T
• F F T F F T
• F F F F T T
• There are five sets of input that make the
  statement true. Therefore there are five
  minterms.

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• p q r (p Ú q) Ør (p Ú q)Ør
• T T T T F F
• T T F T T T
• T F T T F F
• T F F T T T
• F T T T F F
• F T F T T T
• F F T F F T
• F F F F T T
• From the truth table we can set up the DNF
• (p Ú q)Ør ? (p?q??r) ? (p??q??r) ? (?p?q??r) ?
  (?p??q?r) ? (?p??q??r)

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Quantifiers
Universe of Discourse, U The domain of a
variable in a propositional function.
Universal Quantification of P(x) is the
propositionP(x) is true for all values of x in
U. Existential Quantification of P(x) is the
proposition There exists an element, x, in U
such that P(x) is true.
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Universal Quantification of P(x)
?xP(x) for all x P(x) for every x
P(x) Defined as P(x0) ? P(x1) ? P(x2) ? P(x3) ?
. . . for all xi in U Example Let P(x) denote
x2 ? x If U is x such that 0 lt x lt 1 then ?xP(x)
is false. If U is x such that 1 lt x then ?xP(x)
is true.
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Existential Quantification of P(x)
?xP(x) there is an x such that P(x) there is
at least one x such that P(x) there exists at
least one x such that P(x) Defined as P(x0) ?
P(x1) ? P(x2) ? P(x3) ? . . . for all xi in
U Example Let P(x) denote x2 ? x If U is x such
that 0 lt x ? 1 then ?xP(x) is true. If U is x
such that x lt 1 then ?xP(x) is true.
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Quantifiers

• ?xP(x)
• True when P(x) is true for every x.
• False if there is an x for which P(x) is false.
• ?xP(x)
• True if there exists an x for which P(x) is true.
• False if P(x) is false for every x.

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Negation (it is not the case)

• ??xP(x) equivalent to ?x?P(x)
• True when P(x) is false for every x
• False if there is an x for which P(x) is true.
• ? ?xP(x) is equivalent to ?x?P(x)
• True if there exists an x for which P(x) is
  false.
• False if P(x) is true for every x.

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Quantification of Two Variables(read left to
right)

• ?x?yP(x,y) or ?y?xP(x,y)
• True when P(x,y) is true for every pair x,y.
• False if there is a pair x,y for which P(x,y) is
  false.
• ?x?yP(x,y) or ?y?xP(x,y)
• True if there is a pair x,y for which P(x,y) is
  true.
• False if P(x,y) is false for every pair x,y.

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Quantification of Two Variables

• ?x?yP(x,y)
• True when for every x there is a y for which
  P(x,y) is true.
• (in this case y can depend on x)
• False if there is an x such that P(x,y) is false
  for every y.
• ?y?xP(x,y)
• True if there is a y for which P(x,y) is true for
  every x.
• (i.e., true for a particular y regardless (or
  independent) of x)
• False if for every y there is an x for which
  P(x,y) is false.
• Note that order matters here
• In particular, if ?y?xP(x,y) is true, then
  ?x?yP(x,y) is true.
• However, if ?x?yP(x,y) is true, it is not
  necessary that ?y?xP(x,y) is true.

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Basic Number Theory Definitionsfrom Chapters
1.6, 2

• Z Set of all Integers
• Z Set of all Positive Integers
• N Set of Natural Numbers (Z and Zero)
• R Set of Real Numbers
• Addition and multiplication on integers produce
  integers. (a,b ? Z) ? (ab) ? Z ? (ab) ? Z

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Number Theory Defs (cont.)
? such that

• n is even is defined as ?k ? Z ? n 2k
• n is odd is defined as ?k ? Z ? n 2k1
• x is rational is defined as ?a,b ? Z ? x a/b,
  b?0
• x is irrational is defined as ??a,b ? Z ? x
  a/b, b?0 or ?a,b ? Z, x ? a/b, b?0
• p ? Z is prime means that the only positive
  factors of p are p and 1. If p is not prime we
  say it is composite.

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Methods of Proof

• p? q (Example if n is even, then n2 is even)
• Direct proof Assume p is true and use a series
  of previously proven statements to show that q is
  true.
• Indirect proof Show ?q ??p is true
  (contrapositive), using any proof technique
  (usually direct proof).
• Proof by contradiction Assume negation of what
  you are trying to prove (p??q). Show that this
  leads to a contradiction.

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Direct Proof

• Prove ?n?Z, if n is even, then n2 is even.
• Tabular-style proof
• n is even hypothesis
• n2k for some k?Z definition of even
• n2 4k2 algebra
• n2 2(2k2) which is algebra and mult of
• 2(an integer) integers gives integers
• n2 is even definition of even

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Same Direct Proof

• Prove ?n?Z, if n is even, then n2 is even.
• Sentence-style proof
• Assume that n is even. Thus, we know that n 2k
  for some integer k. It follows that n2 4k2
  2(2k2). Therefore n2 is even since it is 2 times
  2k2, which is an integer.

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Structure of a Direct Proof

• Prove ?n?Z, if n is even, then n2 is even.
• Proof
• Assume that n is even. Thus, we know that n 2k
  for some integer k. It follows that n2 4k2
  2(2k2). Therefore n2 is even since it is 2 times
  2k2 which is an integer.

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Example of an Indirect Proof

• Prove If n3 is even, then n is even.
• Proof The contrapositive of If n3 is even,
  then n is even is If n is odd, then n3 is odd.
  If the contrapositive is true then the original
  statement must be true.
• Assume n is odd. Then ?k?Z ? n 2k1. It
  follows that n3 (2k1)3 8k38k24k1
  2(4k34k22k)1. (4k34k22k) is an integer.
  Therefore n3 is 1 plus an even integer.
  Therefore n3 is odd.

Assumption, Definition, Arithmetic, Conclusion
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Discussion of Indirect Proof

• Could we do a direct proof of If n3 is even, then
  n is even?
• Assume n3 is even . . . then what?

We dont have a rule about how to take n3 apart!
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Example Proof by Contradiction

• Prove The sum of an irrational number and a
  rational number is irrational.
• Proof Let q be an irrational number and r be a
  rational number. Assume that their sum is
  rational, i.e., qrs where s is a rational
  number. Then q s-r. But by our previous proof
  the sum of two rational numbers must be rational,
  so we have an irrational number on the left equal
  to a rational number on the right. This is a
  contradiction. Therefore qr cant be rational
  and must be irrational.

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Structure of Proof by Contradiction

• Basic idea is to assume that the opposite of what
  you are trying to prove is true and show that it
  results in a violation of one of your initial
  assumptions.
• In the previous proof we showed that assuming
  that the sum of a rational number and an
  irrational number is rational and showed that it
  resulted in the impossible conclusion that a
  number could be rational and irrational at the
  same time. (It can be put in a form that implies
  n ? ?n is true, which is a contradiction.)

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Using Cases
Prove ?n ?Z, n3 n is even. Separate into cases
based on whether n is even or odd. Prove each
separately using direct proof. Proof We can
divide this problem into two cases. n can be even
or n can be odd. Case 1 n is even. Then ?k?Z ?
n 2k. n3n 8k3 2k 2(4k3k) which is even
since 4k3k must be an integer.
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Cases (cont.)

• Case 2 n is odd. Then ?k?Z ? n 2k1.
• n3 n (8k3 12k2 6k 1) (2k 1) 2(4k3
  6k2 4k 1) which is even since 4k3 6k2
  4k 1 must be an integer.
• Therefore ?n ?Z, n3 n is even

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Proof?

• Prove if n3 is even then n is even.
• Proof Assume n3 is even.
• Then ?k?Z ? n3 8k3 for some integer k. It
  follows that n 3?8k3 2k. Therefore n is
  even.
• Statement is true but argument is false.
• Argument assumes that n is even in making the
  claim n38k3, rather than n3 2k. This is
  circular reasoning.

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Prove or Disprove

• If m and n are even integers, then mn is
  divisible by 4.
• The sum of two odd integers is odd.
• The sum of two odd integers is even.
• If n is a positive integer, then n is even iff
  3n28 is even.
• n2 n 1 is a prime number whenever n is a
  positive integer.
• n2 n 1 is a prime number whenever n is a
  prime number.
• x y ? x y when x,y ? R.
• ?3 is irrational.