problems03-sol - MATH S-104 Lecture 3 In-class Problem...

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June 30, 2009 Problem 1 : Determine an appropriate set of cases for a proof by cases of the following: if x is a real number, then ± x - 1 ± + ± x + 5 ± 6. Case i. x < - 5: x - 1 and x + 5 are both negative. x - 1 < - 6 and x + 5 < 0 ± x - 1 ± > 6 and ± x + 5 ± > 0 ± x - 1 ± + ± x + 5 ± > 6 Case ii. - 5 x < 1: x - 1 is negative and x + 5 is non-negative. ± x - 1 ± + ± x + 5 ± = - ( x - 1 ) + ( x + 5 ) = 6 Case iii. x 1: x - 1 and x + 5 are both non-negative. x - 1 0 and x + 5 6 ± x - 1 ± 0 and ± x + 5 ± 6 ± x - 1 ± + ± x + 5 ± 6 Problem 2 * : Show that if n is an odd integer, then there is an integer k such that n is the sum of k - 2 and k + 3. n = ( k - 2 ) + ( k + 3 ) means that k = n - 1 2 . Because n is odd, we know that n - 1 is even, so n - 1 2 is an integer. Problem 3 : Show that the integer k from the problem 2 is unique. Suppose for integers s and t , n = ( s - 2 ) + ( s + 3 ) and n = ( t - 2 ) + ( t + 3 ) . Then ( s - 2 ) + ( s + 3 ) = ( t - 2 ) + ( t + 3 ) 2 s
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problems03-sol - MATH S-104 Lecture 3 In-class Problem...

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