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Unformatted text preview: Lecture Notes, Concepts of Mathematics (21127) Lecture 1, Recitation AD, Spring 2008 5 Rational and Real Numbers 5.1 Rational Numbers READ: Textbook pages 111120. EXERCISES: Pages 121123, #152. Proposition 5.1 The relation defined on Z ( Z \{ } ) by ( a, b ) ( c, d ) iff ad = bc is an equivalence relation. Proof: ( is reflexive) Let ( a, b ) Z ( Z \{ } ). Then, since multiplication is commutative, we have that ab = ba . Thus ( a, b ) ( a, b ), i.e. is reflexive. ( is symmetric) Let ( a, b ) , ( c, d ) Z ( Z \ { } ) with ( a, b ) ( c, d ). Then ad = bc . It is easy to see that we also have cb = da , or ( c, d ) ( a, b ). Thus is symmetric. ( is transitive) Let ( a, b ) , ( c, d ) , ( e, f ) Z ( Z \{ } ) with ( a, b ) ( c, d ) and ( c, d ) ( e, f ). Then we have ad = bc and cf = de . h We want to show that ( a, b ) ( e, f ), or af = be . i Now cf = de implies c = de/f . Then ad = bc implies ad = b de f , or, multiplying both sides by f , afd = bed , or af = be . Hence ( a, b ) ( e, f ), and thus is transitive. Since is reflexive, symmetric, and transitive, is an equivalence relation on Z ( Z \{ } ). The set of equivalence classes of this relation is the set of rational numbers Q , and the equivalence class containing ( a, b ) is denoted by a b . We will introduce the definition of the greatest common divisor , which we will study in detail later: Definition 5.1 The greatest common divisor of two integers a and b , not both zero, is the largest positive integer k dividing both of a and b and is denoted by k = gcd( a, b ) . Example 5.1 gcd(12 , 4) = 4 , gcd(7 , 5) = 1 , gcd( a, 0) =  a  , gcd(0 , 0) = 0 , gcd(1 , a ) = 1 . Proposition 5.2 (Prop. 5.11, pg. 112) Any nonzero rational number r can be expressed in a unique way as a fraction r = a/b with b > and gcd( a, b ) = 1 ....
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 Spring '08
 howard
 Math, Real Numbers

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