# Four Basic Sets. Divisors

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1 Four Basic Sets Z = the integers Q = the rationals R = the real numbers C = the complex numbers Divisors Definition. Suppose a 0 and b = ax, where a, b, and x are integers. Then we say a divides b (or a is a divisor of b or b is a multiple of a). Notation: Write a b for a divides b and a b for a does not divides b. Example: 2 2, Divisor Facts For each of the following divisibility statements, () give an example and (2) give an argument explaining why the statement is true. Here a, b, c, r, and s are integers. () a b implies a rb (2) a b and b c implies a c (3) a b and a c implies a rb + sc (3 ) [Special case of (3)] a b and a c implies a b + c and a b c (4) a b and b a implies a = ±b (5) a b, a > 0, b > 0 implies a b (6) What can you say about a b (i) if b = 0; (ii) if a =?

2 2 The Division Algorithm The Division Algorithm. Suppose a and b are integers and a > 0. Then there is a unique pair of integers q and r such that b = aq + r where 0 r < a. Examples. (a) 23 and 5; (b) 7 and 3; (c) 666 and 2. Well Ordering Well Ordering Principle. A nonempty set S of nonnegative integers always contains a smallest element m, that is, m satisfies the following two conditions: (i) m is in S and (ii) m n for every number n in S. Example. There is a smallest prime. Example. All positive integers are interesting. Proof of the Division Algorithm. Use Well-Ordering Principle. Use the set S consisting of all numbers of the form b n a, where (i) n ranges over all integers in Z and (ii) b n a 0. Induction Principle of Induction P (n) is a statement about the positive integer n. In order to show that P (n) is true for all positive integers n, it suffices to show that (i) First Case: P () is true; (ii) Next Case: If P (n) is true for the integer n, then statement P is true for the next integer n +. Example: Find the sum of consecutive odd numbers. Let s experiment: = + 3 = = = 6

3 3 Question: What is the pattern, 4, 9, 6? Answer: They are all squares. Question: Does this pattern continue for the next sum when n = 5? Answer: Let s check: = = 25, another square. Conjectured Formula: (2n ) = n 2 Geometric Proof: For example, when n = 5, we can decompose a 5 square into five L-shaped pieces whose area are, 3, 5, 7, and Induction Proof: P (n) : (2n ) = n 2 First Step: P () is true: = Next Step: Assume P (n) is true and show that P (n + ) is also true. The idea is to start with the formula P (n). Add the next odd number 2(n + ) to both sides, and hopefully transform the equation into P (n + ), like this: (2n ) = n 2 [P (n)] (2n ) + (2(n + ) ) = n 2 + 2(n + ) + [add 2(n + ) + ] (2n ) + (2(n + ) ) = n 2 + 2n + [algebra] (2n ) + (2(n + ) ) = (n + ) 2 [P (n + )]

4 4 Proof 2 of the Division Algorithm. Induction. We can verify the division algorithm by induction on the variable b. It simplifies the discussion to assume that the divisor a is greater than. First Step: b =. Here = 0 a + When b =, the quotient is q = 0 and the remainder is r =. Next step: Assume the division algorithm holds for the positive integer b, that is, assume b = q a + r Can we figure out the values of q and r for b +? That s easy. b + = q a + (r + ) Use the old value of q and just increase r by. Wait a minute. What if r = a? When we add to r, the new value will not lie between 0 and a. For example, if a = 5 and b = 22, then from 22 = we get the next statement 23 = But if we start with 24 = , the next statement becomes 25 = , which is true, but the remainder 5 does not lie in the required range. How can be handle this case? What about the case where b is negative? What about the case where a =? The Floor Function For any real number x, the floor of x, written x is defined to be the largest integer n such that n x. x is also called the greatest integer function. For example, π = 3, 2 =, 6.57 = 7. Note that if n = x, then n x < n +. Proof 3 of the Division Algorithm. Use the floor function.

5 5 Given integers a > 0 and b, let q = b a. Then q b a < q + = qa b < (q + )a = 0 b qa < a Let r = b qa. It follows that the only choices for the integer r are 0,, 2,..., a. What about uniqueness? Suppose Tom divides b by a and gets b = q a + r and Becky divides b by a and gets b = q 2 a + r 2. Is it possible that both Tom and Becky could have correct, but different, answers? Calculus Example: Find an antiderivative of (x + ) 2 Tom: 3 (x + )3 Becky: 3 x3 + x 2 + x Proof of uniquesness of q and r: We have q a + r = q 2 a + r 2 = (q q 2 )a = r 2 r = a r 2 r We know 0 r a and 0 r 2 a = 0 r a and (a ) r 2 0 = (a ) r r 2 a = r r 2 a. Fact (5) about division says that if a r r 2, a > 0, and r r 2 0, then a r r 2. Assuming r r 2 0 puts us in a bind: a r r 2 and r r 2 < a. The only way out of this bind is to have r r 2 = 0, or r = r 2. Why does this imply q = q 2?

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