Transcript MTH 231
MTH 231 Section 4.1 Divisibility of Natural Numbers Overview: NCTM Says… • Throughout their study of numbers, students in grades 3 – 5 should identify classes of numbers and examine their properties • Tasks involving factors, multiples, prime numbers, and divisibility can afford opportunities for problem solving and reasoning. Factors, Divisors, Multiples, Divides • Suppose a and b are whole numbers with b not equal to 0. If there exists a whole number q such that a = bq, then: 1. b divides a 2. b is a factor (or divisor) of a 3. a is a multiple of b. It follows that everything that is true for b is also true for q. Even and Odd Numbers • A whole number a is even if a is divisible by 2. • A whole number that is not even is odd. Finding the Factors of a Number • Use the array model “in reverse”: given a certain number of objects, try to arrange them into as many different rectangles, or arrays, as possible. • The dimensions of the array, or rectangle, will be the factors of the numbers. An Example Prime and Composite Numbers • A prime number is a natural number that has exactly two factors. • A composite number is a natural number that has more than two factors. • The number 1 is neither prime nor composite. • 2 is the only even prime number. • Every natural number other than 1 is either prime, or a product of primes. The Sieve of Eratosthenes • A simple algorithm used to find all the prime numbers on a specified interval. A Slightly Longer List… Two Questions About Primes 1. Is there a largest prime number, or is the set of primes infinite? 2. How do you determine if a given number is prime? The First Question • Suppose that 7 was the largest prime. Then every number starting with 8 would be composite, which means every number starting with 8 would be the product of some combination of primes. • Now, consider 2 x 3 x 5 x 7 (the product of 7 and all the primes less than 7). This product, 210, certainly fits the requirement listed above. Continued… • • • • • • Now consider 211, which is 210 +1. 211 divided by 2 is 105 with a remainder of 1. 211 divided by 3 is 70 with a remainder of 1. 211 divided by 5 is 42 with a remainder of 1. 211 divided by 7 is 30 with a remainder of 1. So 211 is not divisible by 2, 3, 5, or 7, which are the only primes. Continued… • But 211 is also not divisible by 8 or anything larger, because every composite number is the product of primes (and nothing 8 or larger is prime). • So, 211 must be prime. But that can’t be because 7 is the largest prime. • Therefore, the original supposition, that 7 is the largest prime, must be wrong. The Second Question • Use a calculator to find the square root of the number in question (if the square root is a natural number, the number in question is composite). • Divide the number by all of the primes up to but less than that square root. • If none of them are factors, the original number is prime. • Keep in mind that every even number other than 2 is composite, and every number ending in 5 is divisible by 5 (more on this in Section 4.2).