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3.3 Divisibility Definition If n and d are integers, then n is divisible by d if, and only if, n = dk for some integer k. d | n There exists an integer k such that n = dk Example 10 divided by 2 = 5 2 times 5 = 10

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Proving Properties of Divisibility 1 st number divides 2 nd number, 2 nd number divides 3 rd number, therefore 1 st number divides 3 rd number Example 10 divides 20 20 divides 40, therefore 10 divides 40

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Theorem 3.3.1 Transitivity of Divisibility For all integers a, b, and c, if a divides b and b divides c, then a divides c.

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Proof: Suppose a, b, and c are [ particular but arbitrarily chosen ] integers such that a divides b and b divides c. [ We must show that a divides c.] By definition of divisibility, b = ar and c = bs for some integers r and s.

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Proof continued : By substitution, c = bs = (ar)s = a(rs) by basic algebra Let k = rs. Then k is an integer since it is a product of integers, and therefore c = ak where k is an integer. Thus a divides c by definition of divisibility. [This is what we needed to show]

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Exam Question Determine whether the statement is true or false. Prove the statement directly from the definitions if it is true, and give a counterexample if it is false. 1. A necessary condition for an integer to be divisible by 9 is that it be divisible by 3.

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Related Homework Problems Section 3.3: 15, 16, 19 – 28, 33.

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Prime Numbers and Divisibility An integer n > 1 is a prime number if, and only if, its only positive integer divisors are 1 and itself. n is prime for all positive integers r and s, if n = r * s then r = 1 or s = 1 Example 5 is greater than 1 try divisors2,3,4,5 Its divisors are 1 and 5

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Theorem 3.3.2 Divisibility by a Prime Any integer n > 1 is divisible by a prime number

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Proof: Suppose n is a [ particular but arbitrarily chosen ] integer that is greater than 1. [ We must show that there is a prime number that divides n.] If n is prime, then n is divisible by a prime number (namely itself), and we are done. 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.

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Proof: continued 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 and 1 < s 1 < r 0. It follows by the definition of divisibility that r 1 | r 0. But we already know that r 0 | n. Consequently, by transitivity of divisibility, r 1 | n. If r 1 is prime, then r 1 is a prime number that divides n, and we are done. If r 1 is not prime, then r 1 = r 2 s 2 where r 2 and s 2 are integers and 1 < r 2 < r 1 and 1 < s 1 < r 1. It follows by definition of divisibility that r 2 | r 1. But we already know that r 1 | n. Consequently, by transitivity of divisibility, r 2 | n.

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Proof: continued If r 2 is prime, then r 2 is a prime number that divides n, and we are done. If r 2 is not prime, then we may repeat the process by factoring r 2 as r 3 s 3. We may continue this way. factoring successive factors of n until we find a prime factor. We must succeed in a finite number of steps because each new factor is both less than the previous one (which is less than n) and greater than 1,and there are fewer than n integers strictly between 1 and n.* Thus we obtain a sequence r 0, r 1, r 2, …, r K, where k >= 0, 1 < r K < r K -1 < … < r 2 < r 1 < r 0 < n, and r i | n for each i = 0, 1, 2,…, k. The condition for termination is that r K should be prime. Hence r K is a prime number that divides n. [This is what we needed to show]

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Exam Question Is 273 divisible by a prime number?

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Related Homework Problems Section 3.3: 35, 36.

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Direct Proof and Counterexample III

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