{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Math 300 HW1

# Math 300 HW1 - Math 300 Homework 1 Alex Lemon In this...

This preview shows pages 1–4. Sign up to view the full content.

Math 300: Homework 1 Alex Lemon January 25, 2008 In this problem set, the set of real numbers, R , is taken as a generic set, with two closed operations (+ and · ) and an order operation ( < ), satisfying certain axioms. For the sake of completeness, these axioms are presented below. The axioms will often be referenced according to the enumeration scheme below. Let a , b and c denote arbitrary elements of R . A1: Associativity of Addition ( a + b ) + c = a + ( b + c ) A2: Commutativity of Addition a + b = b + a A3: Existence of an Additive Identity 0 R : a + 0 = a, a R A4: Existence of Additive Inverses a R , ( - a ) R : a + ( - a ) = 0 M1: Associativity of Multiplication ( a · b ) · c = a · ( b · c ) M2: Commutativity of Multiplication a · b = b · a M3: Existence of a Multiplicative Identity 1 R , 1 6 = 0 : a · 1 = a, a R M4: Existence of Multiplicative Inverses a R , a 6 = 0 , a - 1 R : a · a - 1 = 1 D: Distributivity ( a + b ) · c = a · c + b · c 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
O1: Trichotomy Law a, b R , exactly one of the following is true 1. a < b 2. b < a 3. a = b O2: Transitivity If a < b and b < c , then a < c O3: Inequality and Addition If a < b , then a + c < b + c, c R O4: Inequality and Multiplication If a < b and 0 < c , then a · c < b · c . O5: Archimedean Ordering Principle If 0 < a and 0 < b , then n N : b < n · a = a + a + · · · + a ( n summands) 1 Problem # 1 Suppose a, b R . Using the axioms for R from the notes, show that a = b if and only if a + c = b + c and a · c = b · c for all c R . 1.1 a = b a + c = b + c The first statement is a = b if and only if a + c = b + c . This requires that two things being shown: 1. a = b a + c = b + c 2. a + c = b + c a = b 1.1.1 Argue that a = b a + c = b + c Suppose that a, b R and a = b . Addition is implicitly defined to be a function, + : R 2 7→ R , where R 2 is the Cartesian product of R with itself and the Cartesian product of two sets, S and T , is defined by S × T = { ( s, t ) | s S and t T } . Thus, addition can be formally written as a + b +( a, b ). Since addition is a function, we know that + associates a unique value, x R , to each ( x 1 , x 2 ) R 2 . Suppose we are able to identify two real numbers, a, b R ; that is, we can write a = b . Then, the definition of equality in R 2 allows us to identify ( a, c ) = 2
( b, c ) , c R . One property of functions is that they map every element in their domains to one and only one element in their ranges (though multiple elements of the domain can map to a single element of the range). Using the implicit assumption that + is a function, we can therefore conclude that ( a, c ) = ( b, c ) +( a, c ) = +( b, c ) . This conclusion can be converted to more familiar notation. Given a, b R : a = b a + c = b + c, c R 1.1.2 Show that a + c = b + c a = b Suppose that a + c = b + c . Property (A4) guarantees the existence of additive inverses: a R , ( - a ) R : a + ( - a ) = 0 (A4) Since we have just argued that a = b a + c = b + c and we have assumed that a + c = b + c : ( a + c ) + ( - c ) = ( b + c ) + ( - c ) a + ( c + ( - c )) = b + ( c + ( - c )) (by A1) a + 0 = b + 0 (by A4) a = b (by A3) a + c = b + c a = b 1.1.3 Conclude that a = b a + c = b + c It has been shown that, a, b, c R : 1. a = b a + c = b + c (i.e. a = b only if a + c = b + c ).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern