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Week 4 - Friday
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What did we talk about last time?
Floor and ceiling
Proof by contradiction
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By their nature, manticores lie every Monday,
Tuesday and Wednesday and the other days
speak the truth
However, unicorns lie on Thursdays, Fridays and
Saturdays and speak the truth the other days of
the week
A manticore and a unicorn meet and have the
following conversation:
Manticore: Yesterday I was lying.
Unicorn:
So was I.
On which day did they meet?
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The most common form of indirect proof is a
proof by contradiction
In such a proof, you begin by assuming the
negation of the conclusion
Then, you show that doing so leads to a
logical impossibility
Thus, the assumption must be false and the
conclusion true
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Theorem: x, y  Z+, x2 – y2  1
Proof by contradiction: Assume there is
such a pair of integers
Theorem: 2 is irrational
Proof by contradiction:
3.
4.
5.
6.
Suppose 2 is rational
2 = m/n, where m,n Z, n  0 and m
and n have no common factors
2 = m2/n2
2n2 = m2
2k = m2, k Z
m = 2a, a Z
7.
8.
9.
10.
2n2 = (2a)2 = 4a2
n2 = 2a2
n = 2b, b Z
2|m and 2|n
1.
2.
11.
QED
2 is irrational
1.
2.
3.
4.
5.
6.
Negation of conclusion
Definition of rational
Squaring both sides
Transitivity
Square2 of integer is integer
Even x implies even x
(Proof on p. 202)
7.
Substitution
8.
Transitivity
2
9.
Even x implies even x
10. Conjunction of 6 and 9,
contradiction
11. By contradiction in 10,
supposition is false
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Claim: ∀𝑎, 𝑝 ∈ ℤ 𝑝 is prime ˄ 𝑝 ∣ 𝑎 → 𝑝 ∤ 𝑎 + 1
Proof by contradiction:
1. Suppose ∃𝑎, 𝑝 ∈ ℤ such that
𝑝 is prime ˄ 𝑝 ∣ 𝑎 ˄ 𝑝 ∣ 𝑎 + 1
2. 𝑎 = 𝑝 ∙ 𝑟, 𝑟 ∈ ℤ
3. 𝑎 + 1 = 𝑝 ∙ 𝑠, 𝑠 ∈ ℤ
4. 𝑎 + 1 − 𝑎 = 1
5. 𝑝 ∙ 𝑠 − 𝑝 ∙ 𝑟 = 1
6. 𝑝(𝑠 − 𝑟) = 1
7. 𝑝 ∣ 1
8. 𝑝 ≤ 1
9. 𝑝 > 1
10. Contradiction
11. ∀𝑎, 𝑝 ∈ ℤ 𝑝 is prime ˄ 𝑝 ∣ 𝑎 → 𝑝 ∤
𝑎+1
QED
1. Negation of conclusion
2.
3.
4.
5.
6.
7.
8.
Definition of divides
Definition of divides
Subtraction
Substitution
Distributive law
Definition of divides
Since 1 and -1 are the only integers
that divide 1
9. Definition of prime
10. Statement 8 and statement 9 are
negations of each other
11. By contradiction at statement 10
Theorem: There are an infinite number of primes
Proof by contradiction:
1.
2.
Suppose there is a finite list of all
primes: p1, p2, p3, …, pn
Let N = p1p2p3…pn + 1, N  Z
pk | N where pk is a prime
pk | p1p2p3…pn + 1
p1p2p3…pn = pk(p1p2p3…pk-1pk+1…pn)
p1p2p3…pn = pkP, P  Z
pk | p1p2p3…pn
pk does not divide p1p2p3…pn + 1
pk does and does not divide
p1p2p3…pn + 1
10. There are an infinite number of
primes
3.
4.
5.
6.
7.
8.
9.
QED
1.
2.
Negation of conclusion
Product and sum of integers
is an integer
3.
Theorem 4.3.4, p. 174
4.
Substitution
5.
Commutativity
6.
Product of integers is integer
7.
Definition of divides
8.
Proposition from last slide
9.
Conjunction of 4 and 8,
contradiction
10. By contradiction in 9,
supposition is false
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Don't combine direct proofs and indirect
proofs
You're either looking for a contradiction or
not
Proving the contrapositive directly is
equivalent to a proof by contradiction
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Statements
AND, OR, NOT, IMPLIES
Truth tables
Logical equivalence
De Morgan's laws
Tautologies and contradictions
Name
Law
Dual
pqqp
pqqp
Associative
(p  q)  r  p  (q  r)
(p  q)  r  p  (q  r)
Distributive
p  (q  r)  (p  q)  (p  r)
p  (q  r)  (p  q)  (p  r)
Identity
ptp
pcp
Negation
p  ~p  t
p  ~p  c
Double Negative
~(~p)  p
Idempotent
ppp
ppp
Universal Bound
ptt
pcc
De Morgan’s
~(p  q)  ~p  ~q
~(p  q)  ~p  ~q
Absorption
p  (p  q)  p
p  (p  q)  p
~t  c
~c  t
Commutative
Negations of t and c
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Can be used to write an if-then statement
Contrapositive is logically equivalent
Inverse and converse are not (though they
are logically equivalent to each other)
Biconditional:
 pqqp
A series of premises and a conclusion
Using the premises and rules of inference, an argument is
valid if and only if you can show the conclusion
 Rules of inference:
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Modus Ponens
Modus Tollens
Generalization
Specialization
Conjunction
Elimination
Transitivity
Division into cases
Contradiction rule
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The following gates have the same function as
the logical operators with the same names:
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NOT gate:
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AND gate:
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OR gate:
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A predicate is a sentence with a fixed number
of variables that becomes a statement when
specific values are substituted for to the
variables
The domain gives all the possible values that
can be substituted
The truth set of a predicate P(x) are those
elements of the domain that make P(x) true
when they are substituted
We will frequently be referring to various sets of numbers
in this class
 Some typical notation used for these sets:
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Symbol Set
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Examples
R
Real numbers
Virtually everything
Z
Integers
{…, -2, -1, 0, 1, 2,…}
Z-
Negative integers
{-1, -2, -3, …}
Z+
Positive integers
{1, 2, 3, …}
N
Natural numbers
{1, 2, 3, …}
Q
Rational numbers
a/b where a,b  Z and b  0
Some authors use Z+ to refer to non-negative integers
and only N for the natural numbers
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The universal quantifier  means “for all”
The statement “All DJ’s are mad ill” can be
written more formally as:
x  D, M(x)
 Where D is the set of DJ’s and M(x) denotes that x is
mad ill
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The existential quantifier  means “there exists”
The statement “Some emcee can bust a rhyme”
can be written more formally as:
y  E, B(y)
 Where E is the set of emcees and B(y) denotes that y
can bust a rhyme
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When doing a negation, negate the predicate
and change the universal quantifier to
existential or vice versa
Formally:
 ~(x, P(x))  x, ~P(x)
 ~(x, P(x))  x, ~P(x)
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Thus, the negation of "Every dragon breathes
fire" is "There is one dragon that does not
breathe fire"
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Any statement with a universal quantifier
whose domain is the empty set is vacuously
true
When we talk about "all things" and there's
nothing there, we can say anything we want
"All mythological creatures are real."
Every single one of the (of which there are
none) is real
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Recall:
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pq
~q  ~p
qp
~p  ~q
Statement:
Contrapositive:
Converse:
Inverse:
x, P(x)  Q(x)
x, ~Q(x)  ~P(x)
x, Q(x)  P(x)
x, ~P(x)  ~Q(x)
These can be extended to universal statements:
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Statement:
Contrapositive:
Converse:
Inverse:
Similar properties relating a statement equating a
statement to its contrapositive (but not to its converse and
inverse) apply
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p is a sufficient condition for q means p  q
p is a necessary condition for q means q  p
These come over into universal conditional
statements as well:
x, P(x) is a sufficient condition for Q(x)
means x, P(x)  Q(x)
x, P(x) is a necessary condition for Q(x)
means x, Q(x)  P(x)
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With multiple quantifiers, we imagine that
corresponding “actions” happen in the same
order as the quantifiers
The action for x  A is something like, “pick
any x from A you want”
Since a “for all” must work on everything, it
doesn’t matter which you pick
The action for y  B is something like, “find
some y from B”
Since a “there exists” only needs one to work,
you should try to find the one that matches
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For negation,
 Simply switch every  to  and every  to 
 Then negate the predicate
Changing the order of quantifiers can change the
truth of the whole statement but does not always
 Furthermore, quantifiers of the same type are
commutative:
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 You can reorder a sequence of  quantifiers however you
want
 The same goes for 
 Once they start overlapping, however, you can’t be sure
anymore
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Universal instantiation: If a property is true for
everything in a domain (universal quantifier), it is
true for any specific thing in the domain
Universal modus ponens:
 x, P(x)  Q(x)
 P(a) for some particular a
 Q(a)
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Universal modus tollens:
 x, P(x)  Q(x)
 ~Q(a) for some particular a
 ~P(a)
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To prove x  D  P(x) we need to find at
least one element of D that makes P(x) true
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To disprove x  D, P(x)  Q(x), we need to
find an x that makes P(x) true and Q(x) false
If the domain is finite, we can use the method of
exhaustion, by simply trying every element
 Otherwise, we can use a direct proof
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1.
2.
3.
Express the statement to be proved in the form x  D, if P(x)
then Q(x)
Suppose that x is some specific (but arbitrarily chosen)
element of D for which P(x) is true
Show that the conclusion Q(x) is true by using definitions,
other theorems, and the rules for logical inference
Direct proofs should start with the word Proof, end with
the word QED, and have a justification next to every
step in the argument
 For proofs with cases, number each case clearly and show
that you have proved the conclusion for all possible cases
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If n is an integer, then:
 n is even   k  Z  n = 2k
 n is odd   k  Z  n = 2k + 1
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If n is an integer where n > 1, then:
 n is prime   r  Z+,  s  Z+, if n = rs, then r = 1 or s = 1
 n is composite   r  Z+,  s  Z+  n = rs and r  1 and s  1
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r is rational   a, b Z  r = a/b and b  0
For n, d  Z,
 n is divisible by d  k Z  n = dk
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For any real number x, the floor of x, written x, is defined as
follows:
 x = the unique integer n such that n ≤ x < n + 1
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For any real number x, the ceiling of x, written x, is defined as
follows:
 x = the unique integer n such that n – 1 < x ≤ n
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Unique factorization theorem: For any integer
n > 1, there exist a positive integer k, distinct
prime numbers p1, p2, …, pk, and positive
integers e1, e2, …, ek such that
e1 e2 e3
ek
1
2
3
k
Quotient remainder theorem: For any integer n
and any positive integer d, there exist unique
integers q and r such that
n  p p p ...p
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 n = dq + r and 0 ≤ r < d
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Theorem: for all odd integers n,
𝑛2
4
=
𝑛2 +3
4
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Exam 1!
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Exam 1 is Monday in class!