1. Direct Proof and Counterexample III
Lecture 13
Section 3.3
Fri, Feb 10, 2006

2. 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.

3. 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?

4. 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?

5. Units and Zero Definition: An integer u is a unit if u | 1.
The only units are 1 and –1.
Definition: 0 is zero.

6. 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.

7. 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.

8. Example: Direct Proof Divide by a to get cd = 1.
Thus, c = d = 1 or c = d = -1.
Then a = b or a = -b.

9. 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.

10. 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.

11. 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.

12. 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).

13. 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?

14. Example: Direct Proof Theorem: If n is odd, then 24 | (n3 – n). Proof:?

15. 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).

16. Proving Biconditionals
Or we could prove both
x  D, P(x)  Q(x)
and
x  D, P(x)  Q(x).

17. 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.

18. 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.

19. 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.

20. 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.

21. Proving Biconditionals
Proof ():
Let a and b be real numbers and suppose that a and b are both integers or both half-integers.
Case I: Suppose that a and b are both integers.
Then a + b and a – b are integers.

22. Proving Biconditionals
Case II: Suppose a and b are both half-integers.
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.

23. The Fundamental Theorem of Arithmetic
Theorem: Let a be a positive integer. Then a = p1a1p2a2…pkak, where each pi is a prime and each ai is a nonnegative integer. Furthermore, this representation is unique except for the order of factors.

24. Application of the Fundamental Theorem of Arithmetic
Let a and b be positive integers.
Then
a = p1a1p2a2…pkak
and
b = p1b1p2b2…pkbk.
Then the g.c.d. of a and b is
gcd(a, b) = p1min(a1, b1)p2min(a2, b2)…pkmin(ak, bk)

25. Application of the Fundamental Theorem of Arithmetic
and the l.c.m. of a and b is
lcm(a, b) = p1max(a1, b1)p2max(a2, b2)…pkmax(ak, bk).

26. 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 =

27. Example: gcd’s and lcm’s
Corollary: Let a and b be positive integers. Then gcd(a, b)lcm(a, b) = ab.