Download Lecture 13 - Direct Proof and Counterexample III

Direct Proof and
Counterexample III
Lecture 13
Section 3.3
Fri, Feb 10, 2006
Divisibility
Definition: An integer a divides an integer
b if a  0 and there exists an integer c
such that
ac = b.
 Write a | b to indicate that a divides b.
 Divisibility is a “positive” property.

Prime Numbers
Definition: An integer p is prime if p  2
and the only positive divisors of p are 1
and p.
 A prime number factors only in a trivial
way: p = 1  p.
 Prime numbers: 2, 3, 5, 7, 11, …
 Is this a positive property?
 Is there a positive characterization of
primes?

Composite Numbers
Definition: An integer n is composite if
there exist integers a and b such that a > 1
and b > 1 and n = ab.
 A composite number factors in a non-trivial
way.
 Composite numbers: 4, 6, 8, 9, 10, 12, …
 Is this a positive property?

Units and Zero

Definition: An integer u is a unit if u | 1.


The only units are 1 and –1.
Definition: 0 is zero.
Example: Direct Proof
Theorem: If a | b and b | c, then a | c.
 Proof:

Let a, b, c be integers and assume a | b
and b | c.
 Since a | b, there exists an integer r such
that ar = b.
 Since b | c, there exists an integer s such
that bs = c.
 Therefore, a(rs) = (ar)s = bs = c.
 So a | c.

Example: Direct Proof
Theorem: Let a and b be integers. If a | b
and b | a, then a = b.
 Proof:

Let a and b be integers.
 Suppose a | b and b | a.
 There exist integers c and d such that ac =
b and bd = a.
 Therefore, acd = bd = a.

Example: Direct Proof
Divide by a to get cd = 1.
 Thus, c = d = 1 or c = d = -1.
 Then a = b or a = -b.

Example: Direct Proof
Corollary: If a, b  N and a | b and b | a,
then a = b.
 This is analogous to the set-theoretic
statement that if A  B and B  A, then
A = B.
 Preview: This property is called
antisymmetry.


If a ~ b and b ~ a, then a = b.
Example: Direct Proof
Theorem: Let a, b, c be integers. If a | b
and b | a + c, then a | c.
 Proof:

Let a, b, and c be integers.
 Suppose a | b and b | a + c.
 There exist integers r and s such that
ar = b and bs = a + c.

Example: Direct Proof
Substitute ar for b in the 2nd equation to get
(ar)s = a + c.
 Rearrange the terms and factor to get
a(rs – 1) = c.
 Therefore, a | c.

Example: Direct Proof
Theorem: If n is odd, then 8 | (n2 – 1).
 Proof:

Let n be an odd integer.
 Then n = 2k + 1 for some integer k.
 So n2 – 1 = (2k + 1)2 – 1 = 4k2 + 4k
= 4k(k + 1).

Example: Direct Proof
Either k or k + 1 is even.
 Therefore, k(k + 1) is a multiple of 2.
 Therefore, n2 – 1 is a multiple of 8.


Can you think of an alternate, simpler
proof, based on the factorization of n2 – 1?
Example: Direct Proof
Theorem: If n is odd, then 24 | (n3 – n).
 Proof:


?
Proving Biconditionals

To prove a statement
x  D, P(x)  Q(x),
we must prove both
x  D, P(x)  Q(x)
and
x  D, Q(x)  P(x).
Proving Biconditionals

Or we could prove both
x  D, P(x)  Q(x)
and
x  D, P(x)  Q(x).
Proving Biconditionals
A half-integer is a number of the form
n + ½, for some integer n.
 Theorem: Let a and b be real numbers.
Then a + b and a – b are integers if and
only if a and b are both integers or both
half-integers.

Proving Biconditionals

Proof ():
Let a and b be real numbers and suppose
that a + b and a – b are integers.
 Let a + b = m and a – b = n for some
integers m, n.
 Add to get 2a = m + n and subtract to get
2b = m – n.
 Divide by 2 to get a = (m + n)/2 and b = (m
– n)/2.

Proving Biconditionals

Case I: Suppose m and n are both even or
both odd.
• Then m + n and m – n are both even.
• Let m + n = 2r and m – n = 2s for some
integers r and s.
• Then a = r and b = s, so a and b are integers.
Proving Biconditionals

Case II: Suppose one of m and n is even
and the other is odd.
• Then m + n and m – n are odd.
• Let m + n = 2u + 1 and m – n = 2v + 1.
• Then a = u + ½ and b = v + ½, so a and b are
half-integers.
Proving Biconditionals

Proof ():
Let a and b be real numbers and suppose
that a and b are both integers or both halfintegers.
 Case I: Suppose that a and b are both
integers.

• Then a + b and a – b are integers.
Proving Biconditionals

Case II: Suppose a and b are both halfintegers.
• Let a = p + ½ and b = q + ½.
• Then a + b = p + q + 1 and a – b = p – q.
• So a + b and a – b are both integers.
The Fundamental Theorem
of Arithmetic

Theorem: Let a be a positive integer.
Then a = p1a p2a …pka , where each pi is a
prime and each ai is a nonnegative integer.
Furthermore, this representation is unique
except for the order of factors.
1
2
k
Application of the Fundamental
Theorem of Arithmetic
Let a and b be positive integers.
 Then
a = p1a p2a …pka
and
b = p1b p2b …pkb .
 Then the g.c.d. of a and b is
gcd(a, b) = p1min(a , b )p2min(a , b )…pkmin(a , b )

1
2
1
1
k
2
1
k
2
2
k
k
Application of the Fundamental
Theorem of Arithmetic
and the l.c.m. of a and b is
lcm(a, b) = p1max(a , b )p2max(a , b )…pkmax(a , b ).
1
1
2
2
k
k
Example: gcd’s and lcm’s
Let a = 4200 and b = 1080.
 Then a = 23315271 and b = 23325170.
 Then
gcd(a, b) = 23315170 = 120
and
lcm(a, b) = 23325271 = 13600.

Example: gcd’s and lcm’s

Corollary: Let a and b be positive integers.
Then gcd(a, b)lcm(a, b) = ab.