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MTH 231 Section 4.1 Divisibility of Natural Numbers.

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1 MTH 231 Section 4.1 Divisibility of Natural Numbers

2 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.

3 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.

4 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.

5 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.

6 An Example

7 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.

8 The Sieve of Eratosthenes A simple algorithm used to find all the prime numbers on a specified interval.

9 A Slightly Longer List…

10 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?

11 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.

12 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.

13 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.

14 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).


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