# Chapter 3 Elementary Number Theory and Methods of Proof.

## Presentation on theme: "Chapter 3 Elementary Number Theory and Methods of Proof."— Presentation transcript:

Chapter 3 Elementary Number Theory and Methods of Proof

3.3 Direct Proof and Counterexample 3 Divisibility

Definition – If n and d are integers, then n is divisible by d if, and only if, n = dk for some integers k. – The notation d|n is read “d divides n.” – Symbolically, if n and d are integers, – d|n ⇔ ∃ an integer k such that n = dk.

Examples Divisibility – Is 21 divisible by 3? Yes, 21 = 3 * 7 – Does 5 divide 40? Yes, 40 = 5 * 8 – Does 7|42? Yes, 42 = 7 * 6 – Is 32 a multiple of -16? Yes, 32 = (-16) * (-2) – Is 7 a factor of -7? Yes, -7 = 7 * (-1)

Divisors Divisors of zero – If d is any integer, does d divide 0? – Recall: d|n ⇔ ∃ an integer k such that n = dk – n = 0, is there an integer k such that dk = 0 (n) – Yes, 0 = d*0 (0 is an integer) Divisors of one – Which integers divide 1? – 1 = dk, 1*1 or (-1) * (-1)

Divisors Positive Divisors of a Positive Number – Suppose a and b are positive integers and a|b. – Is a≤b? Yes. a|b means that b = ka for some integer k. k must be positive because b and a are positive. 1 ≤ k a ≤ k * a = b, hence, a ≤ b

Divisibility Algebraic expressions – 3a + 3b divisible by 3 (a & b are integers)? – Yes. By distributive law 3a + 3b = 3(a + b), a & b are integers so the sum of a, b are integers. – 10km divisible by 5 (k & m are integers)? 10km = 5 * (2km)

Prime Numbers An alternative way to define a prime number is to say that an integer n > 1 is prime if, and only if, its only positive integer divisors are 1 and itself.

Transitivity of Divisibility Prove that for all integers a, b, and c, if a|b and b|c, then a|c. – Starting point: Suppose a, b, and c are particular but arbitrarily chosen integers such that a|b and b|c. – To show: a|c a|c, c = a*(some integers) since a|b, b = ar for some integer r And since b|c, c = bs c = (ar)s (substitue for b) c = a(rs) (assoc law) c = ak (such that rs is integer due to close property of int) Theorem 3.3.1 Transitivity of Divisibility

Divisibility by a Prime Any integer n > 1 is divisible by a prime number. – Suppose n is a integer that is greater than 1. – If n is prime, then n is divisible by a prime number (namely itself). If n is not prime, then n = r 0 s 0 where r 0 and s 0 are integers and 1 < r 0 < n and 1 < s 0 < n. it follows by definition of divisibility that r 0 |n. if r 0 is prime, then r 0 is a prime number that divides n, and we are done. If r 0 is not prime, then r 0 = r 1 s 1 where r 1 and s 1 are integers and 1 < r 1 < r 0 etc.

Counterexamples and Divisibility Is it true or false that for all integers a and b, if a|b and b|a then a = b? – Starting Point: Suppose a and b are integers such that a|b and b|a. b = ka & a =lb (for some integers k and l) b = ka = k(lb) = (kl)b factor b assuming b≠0 1 = kl Thus, k = l = 1 or k = l = -1

Unique Factorization Theorem Theorem 3.3.3 Fundamental Theorem of Arithmetic – Given any integer n > 1, there exist a positive integer k, distinct prime numbers p 1, p 2, …, p k, and positive integers e 1, e 2, …, e k such that – n = p 1 e1 p 2 e2 p 3 e3 p 4 e4 … p k ek, – and any other expression of n as a product of prime numbers is identical to this except, perhaps, for the order in which the factors are written.

Standard Factored Form Given any integer n > 1, the standard factored form of n is an expression of the form – n = p 1 e1 p 2 e2 p 3 e3 p 4 e4 … p k ek, – where k is a positive integer; p1, p2, …,pk are prime numbers; e1, e2, …, ek are positive integers; and p1 < p2 < … < pk.

Example Write 3,300 in standard factored form. – 3,300 = 100 * 33 = 4 * 25 * 3 * 11 – = 2 * 2 * 5 * 5 * 3 * 11 – = 2 2 * 5 2 * 3 * 11