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Unformatted text preview: CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1 . Given two integers a and b we say a divides b if there is an integer c such that b = ac . If a divides b , we write a  b . If a does not divide b , we write a 6  b . Discussion Example 1.1.1 . The number 6 is divisible by 3, 3  6 , since 6 = 3 · 2 . Exercise 1.1.1 . Let a , b , and c be integers with a 6 = 0 . Prove that if ab  ac , then b  c . Using this definition, we may define an integer to be even if it is divisible by 2 and odd if it is not divisible by 2. This concept is one of the simplest of properties of numbers to define, yet it is among the most complicated of all mathematical ideas. Keep in mind that we are talking about a very restricted notion of what it means for one number to “divide” another: we can certainly divide 7 by 3 and get the rational number 7 3 = 2 . 3333 ··· , but, since the result is not an integer, we say that 3 does not divide 7, or 3 6  7. For this reason, you should avoid using fractions in any discussion of integers and integer arithmetic. 1.2. Basic Properties of Divisibility. Theorem 1.2.1 . For all integers a , b , and c , 1. If a  b and a  c , then a  ( b + c ) . 2. If a  b , then a  ( bc ) . 3. If a  b and b  c , then a  c . Discussion 146 1. INTEGERS AND DIVISION 147 Theorem 1.2.1 states the most basic properties of division. Here is the proof of part 3: Proof of part 3. Assume a , b , and c are integers such that a  b and b  c . Then by definition, there must be integers m and n such that b = am and c = bn . Thus c = bn = ( am ) n = a ( mn ) . Since the product of two integers is again an integer, we have a  c . Exercise 1.2.1 . Prove part 1 of Theorem 1.2.1. Exercise 1.2.2 . Prove part 2 of Theorem 1.2.1. 1.3. Theorem 1.3.1  The Division Algorithm. Theorem 1.3.1 . (Division Algorithm) Given integers a and d , with d > , there exists unique integers q and r , with ≤ r < d , such that a = qd + r . Notation 1.3.1 . We call a the dividend , d the divisor , q the quotient , and r the remainder . Discussion The division algorithm is probably one of the first concepts you learned relative to the operation of division. It is not actually an algorithm, but this is this theorem’s traditional name. For example, if we divide 26 by 3, then we get a quotient of 8 and remainder or 2. This can be expressed 26 = 3 · 8 + 2. It is a little trickier to see what q and r should be if a < 0. For example, if we divide 26 is by 3, then the remainder is not 2. We can, however, use the equation 26 = 3 · 8 + 2 to our advantage: 26 = 3 · ( 8) 2 = [3 · ( 8) 3] 2 + 3 = 3( 9) + 1 So dividing 26 by 3 gives a quotient of 9 and remainder 1. The condition 0 ≤ r < d makes r and q unique for any given a and d ....
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This note was uploaded on 11/27/2011 for the course MCH 108 taught by Professor Penelopekirby during the Fall '08 term at FSU.
 Fall '08
 PenelopeKirby

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