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Math 301 Problem Set 1 Solutions

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

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

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