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Unformatted text preview: Applied Algebra Lecture 1 Audrey Terras Math. Dept., U.C.S.D., San Diego, Ca 92093-0112 email: firstname.lastname@example.org web: http://math.ucsd.edu/~aterras/ March 28, 2010 The Integers The set of integers is Z = f ; & 1 ; & 2 ; & 3 ;::: g : It is de&ned by various axioms. First there are the algebraic axioms. Algebra Axioms 0) uniqueness: a = a ;b = b in Z implies a + b = a + b 1) closure: a;b 2 Z implies a + b; a b; a b 2 Z 2) commutative laws: a + b = b + a; a b = b a 3) associative laws: a + ( b + c ) = ( a + b ) + c ; a ( b c ) = ( a b ) c 4) distributive law: a ( b + c ) = a b + a c 5) zero: 0 + a = a 6) one: 1 a = a 7) additive inverses exist: 8 ( for any) integer a, 9 ( there exists) an integer a so that a + ( a ) = 0 : 8) cancellation: c 6 = 0 ;ca = cb implies a = b . These axioms hold for all a;b;c 2 Z : Later (next week probably) we will say Z is a "group" under addition (and next quarter we will say that Z is a "ring" - actually an "integral domain"). This quarter our main subject is groups. Next quarter it will be rings. There is one more important property of the integers. They are ordered. You view them as equally spaced points on the real line. j 4 j 3 j 2 j 1 j j 1 j 2 j 3 j 4 We say a < b if a is to the left of b . There are many properties of this ordering, all of which can be deduced from the following de&nition. Defn . a < b means b a 2 Z + = the positive integers = f 1 ; 2 ; 3 ; 4 ;::: g : The set of positive integers Z + satis&es the following axioms. Ordering Axioms 1) addition: a;b 2 Z + implies a + b 2 Z + 2) multiplication: a;b 2 Z + implies a b 2 Z + 3) trichotomy: 8 a 2 Z one and only one of the following alternatives holds: i) a 2 Z + ii) a = 0 iii) a 2 Z + : Every other fact about inequalities can be deduced from these 3 axioms and the de&nition of a < b: 1 Example 1. Transitive Law. a < b;b < c = ) a < c: Proof. c & a = c & b + b & a which is in Z + by ordering axiom 1 since c & b 2 Z + and b & a 2 Z + : Example 2. Multiplying an inequality by a negative number changes < to > . a < b; c < = ) ac > bc: Proof. ac & bc = ( a & b ) c = ( b & a )( & c ) 2 Z + by ordering axiom 2 since c < says & c 2 Z + : Example 3. a < b = ) a + c < b + c: Proof....
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