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 half-integers. 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 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.

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.

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)

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).

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.