Transcript Document 7511371

Direct Proof and
Counterexample III
Part 2
Lecture 16
Section 3.3
Tue, Feb 13, 2007
Example: Direct Proof
Theorem: Let a and b be integers. If a | b
and b | a, then a = b.
 Proof:

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: A property ~ is called
antisymmetry if


a ~ b and b ~ a implies that a = b.
Example: Direct Proof
Theorem: Let a, b, c be integers. If a | b
and b | a + c, then a | c.
 Proof:

Example: Direct Proof
Theorem: If n is odd, then 8 | (n2 – 1).
 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.
…

Proving Biconditionals

Case I: Suppose m and n are both even or
both odd.
• Then …
Proving Biconditionals

Case II: Suppose one of m and n is even
and the other is odd.
• Then …
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.

•…
Proving Biconditionals

Case II: Suppose a and b are both halfintegers.
•…
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.
 Comment: The Euclidean algorithm is a lot
faster.
