Chapter 3, Number Theory
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Introductory number theory
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A good proof should have:
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a statement of what is to be proven
"Proof:" to indicate where the proof starts
a clear indication of flow
a clear indication of the reason for each step
careful notation, completeness and order
a clear indication of the conclusion
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Some definitions
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Z is the integers
Q is the rational numbers (quotients of integers)
r  Q  (a,b  Z) [(r = a / b)  (b  0)]
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Irrational numbers are those which are not rational
R is the real numbers
A superscript of + indicates the positive portion only of
one of these sets of numbers
A superscript of – indicates the negative portion only of
one of these sets of numbers
Other superscripts can be used, such as Zeven, Zodd , Q>5
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We can define the closure of these sets for an operation
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Z is closed under what operations?
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Integer definitions
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An even integer
n  Zeven  (k  Z) [n = 2k]
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An odd integer
n  Zodd  (k  Z) [n = 2k + 1]
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A prime integer (Z>1)
n  Zprime  (r,s  Z+) [ (n = r  s)  (r = 1)  (s = 1)]
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A composite integer (Z>1)
n  Zcomposite  (r,s  Z+) [(n = r  s)  (r  1)  (s  1)]
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Constructive proof of existence
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How can we prove the following?
(n  Z, p, q, r, s  Z )
[p  q  r  s  p  r  p  s  q  r  q  s 
n  p 3  q 3  n  r 3  s3 ]
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Discrete Structures
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Lecture 13
February 25, 2008
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Methods of proving
universally quantified statements
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Method of exhaustion
– prove the statement is true for each and every member of the
domain
– (r  Z+)[23 < r < 29  ( p,q  Z+) [(r = p  q)  (p  q)]]
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Generalizing from the generic particular
– suppose that x is a particular but arbitrarily-chosen element of
the domain
– show that x satisfies the property
– then the result holds for every element of the domain
– example: (r  Z) [r  Zeven  r2  Zeven ]
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Examples of generalizing from the generic
particular
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The product of any two odd integers is also odd.
– (m,n  Z) [(m  Zodd  n  Zodd)  m  n  Zodd ]
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The product of any two rationals is also rational.
– (m,n  Q) [m  n  Q]
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Discrete Structures
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Lecture 14
February 27, 2008
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Disproof by counterexample
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( r  Z) [r2  Z+  r  Z+]
– counterexample: r2 = 9  r = –3
• r2  Z+ since 9  Z+ so the antecedent is true
• but r  Z+ since –3  Z+ so the consequent is false
• this means the implication is false for r = –3, so this is a valid
counterexample
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When you give a counterexample you must justify that it
is a valid counterexample, by showing the algebra (or
other interpretation needed) to support your claim
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Divisibility
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Definition: d | n  (k  Z)[n = d × k]
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n is divisible by d
n is a multiple of d
d is a divisor of n
d divides n
Results involving divisibility:
(a, b  Z>1)[a | b→ a | (b + 1)]
(x  Z>1)(p  Zprime)[p | x]
– note another proof method would be used here, proof by
division into cases
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Proof by contrapositive
For all positive integers n, if n does not divide a number
of which d is a factor, then n can not divide d.
(n,d,c  Z+) [n | dc  n | d]
(n,d,c  Z+) [n | d  n | dc]
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Discrete Structures
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Lecture 15
February 29, 2008
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Proof by contradiction
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The number of primes is infinite
Assume this is false, that there is some largest prime
number, so there are only n different prime numbers
Consider the number
k  (p1  p2  ...  pn 1  pn )  1
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Prime factored form
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The Unique Factorization Theorem (Theorem 3.3.3)
Given any integer n > 1,
(kZ)(p1,p2,…pkZprime)(e1,e2,…ekZ+)[n = p1e1 × p2e2 × p3e3 × …× pkek]
where the p’s are distinct and any other expression of n is identical to
this except maybe in the order of the factors.
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Standard factored form
n = p1e1 × p2e2 × p3e3 × … × pkek
pi < pi+1
(mZ)[8×7×6×5×4×3×2×m = 17×16×15×14×13×12×11×10]
– does 17 | m ??
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More integer definitions
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div and mod operators
n
n div d- integer quotient for d
n
d
n mod d- integer remainder for
(n div d = q) ^ (n mod d = r)  n = d × q + r
where n  Z0, d  Z+, r  Z, q  Z, 0  r < d
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Relating “mod” to “divides”:
d|n
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0 = n mod d
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0 d n
Definition of equivalence in a mod:
x d y  d | (x–y)
(note: their remainders are equal)
sometimes written as x  y mod d, meaning (x  y) mod d
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Discrete Structures
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Lecture 16
March 3, 2008
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Modular arithmetic (Theorem 10.4.3)
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Let a, b, c, d, n  Z, and n > 1. Suppose a  c (mod
n) and b  d (mod n). Then:
1. (a + b)  (c + d) (mod n)
2. (a – b)  (c – d) (mod n)
3. ab  cd (mod n)
4. am  cm (mod n) for all integers m
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Proof using this definition:
(m  Z+)(a,b  Z)[a m b  (k  Z) [a = b + km]]
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Discrete Structures
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Lecture 18
March 7, 2008
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Floor and ceiling
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Definitions:
– n is the floor of x where x  R  n  Z
x = n  n  x < n+1
– n is the ceiling of x where x  R  n  Z
x = n  n–1 < x  n
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Proofs using floor and ceiling:
(x,y  R) [ x+y = x + y ]
(x  R)(y  Z)[ x+y = x + y ]
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More proof by division into cases
The floor of (n/2) is either
a)
n/2 when n is even
or b)
(n–1)/2 when n is odd
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Discrete Structures
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Lecture 19
March 10, 2008
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The quotient remainder theorem
(n  Z)(d  Z+)(q,r  Z)[(n = dq + r)  (0  r < d)]
Proving definition of equivalence in a mod using the quotient
remainder theorem
This means
prove that if [m d n], then [d | (n-m)]
where m,n  Z and d  Z+
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Proof by division into cases again
(n  Z) [3 | n  n2 3 1]
or, alternatively, (n  Z) [3 | n  n2  1 (mod 3)]
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Steps toward proving the unique
factorization theorem
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Every integer greater than or equal to 2 has at least
one prime that divides it
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For all integers greater than 1,
if a | b, then a | (b+1)
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There are an infinite number of primes
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Discrete Structures
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Lecture 20
March 12, 2008
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Using the unique factorization theorem
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Prove that (a  Z+)(q  Zprime) [q | a2  q | a]
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Prove that
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Discrete Structures
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Lecture 21
March 14, 2008
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Summary of proof methods
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Constructive proof of existence
Proof by exhaustion
Proof by generalizing from the generic particular
Proof by contraposition
Proof by contradiction
Proof by division into cases
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Errors in proofs
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Arguing from example for universal proof
Misuse of variables
Jumping to the conclusion (missing steps)
Begging the question
Using "if" about something that is known
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