Math 301 Problem Set 1 Solutions

Math 301 Problem Set 1 Solutions - Math 301/ENAS 513,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 301/ENAS 513, Homework 1 Solution Triet Le September 17, 2007 1. Recall: Z = { m- n : m, n N } . Each n N is identified with the expression ( n + m )- m for any m N . Addition on Z is defined as: ( m- n )+( m- n ) = ( m + m )- ( n + n ). Let k, m, n N such that k = m + n . Show that m- k is the additive inverse of n and it is unique. Solution 1. Note that if m- k and m- k are two additive inverses of n ( n + p )- p , p N , then (( n + p )- p ) + ( m- k ) = 0 = (( n + p )- p + ( m- k ) ( n + p + m )- ( k + p ) = ( n + p + m )- ( k + p ) ( n + p + m ) + ( k + p ) = ( n + p + m ) + ( k + p ) ( n + p + p ) + ( m + k ) = ( n + p + p ) + ( m + k ) m + k = m + k m + k m + k . Therefore, we have uniqueness. Let p N , we have (( n + p )- p )+( m- k ) = ( n + p + m )- ( p + k ) = (( m + n )+ p )- ( k + p ) = ( k + p )- ( k + p ) = 0 . Therefore, ( m- k ) is the additive inverse of ( n + p )- p n . 2. Recall: A set S Q is called a cut if S satisfies C1: S is not empty and is not all of Q . C2: If r S , s Q and s < r , then s S ....
View Full Document

This note was uploaded on 07/18/2008 for the course MATH 301 taught by Professor Trietle during the Fall '07 term at Yale.

Page1 / 3

Math 301 Problem Set 1 Solutions - Math 301/ENAS 513,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online