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Unformatted text preview: CS103X: Discrete Structures Homework Assignment 1: Solutions Due January 25, 2008 Exercise 1 (10 Points). Prove or give a counterexample for each of the following: (a) If A ⊆ B and B ⊆ C , then A ⊆ C . (b) If A ∈ B and B ∈ C , then A ∈ C . Solution: (a) Consider any element a ∈ A . Since A ⊆ B every element of A is also an element of B , so a ∈ B . By same reasoning, a ∈ C since B ⊆ C . Thus every element of A is an element of C , so A ⊆ C . (b) Let A = { 1 } ,B = { 1 , { 1 }} , and C = {{ 1 , { 1 }} , 3 , { 4 }} . These sets satisfy A ∈ B and B ∈ C , but A / ∈ C . Exercise 2 (10 Points). If a ( t ), b ( t ), and c ( t ) are the lengths of the three sides of a triangle t in nondecreasing order (i.e. a ( t ) ≤ b ( t ) ≤ c ( t )), we define the sets: • X := { Triangle t : a ( t ) = b ( t ) } • Y := { Triangle t : b ( t ) = c ( t ) } • T := the set of all triangles Using only set operations on these three sets, define: (a) The set of all equilateral triangles (all sides equal) (b) The set of all isosceles triangles (at least two sides equal) (c) The set of all scalene triangles (no two sides equal) Solution: (a) We require a ( t ) = b ( t ) and b ( t ) = c ( t ) (this obiviously implies a ( t ) = c ( t )), so the set is X ∩ Y (b) An isoceles triangle t can have 1. a ( t ) = b ( t ), or 2. b ( t ) = c ( t ), or 3. a ( t ) = c ( t ). 1 Now we have assumed that a ( t ) ,b ( t ), and c ( t ) are in nondecreasing order, so the last condition holds if and only if both the first two do. So the required set is X ∪ Y . (c) A scalene triangle has its two smaller sides a ( t ) and b ( t ) unequal (set T \ X ) and its two larger sides b ( t ) and c ( t ) unequal (set T \ Y ). Since the sides are listed in the nondecreasing order, either of the above conditions guarantess a ( t ) 6 = c ( t ). So the required set is ( T \ X ) ∩ ( T \ Y ). An alternative argument is: A triangle is scalene if and only if it is not isosceles. So using the result of the previous part, the set of scalene triangles is T \ ( X ∪ Y ). It’s easy to confirm that the answers given by the two arguments are actually the same  this is an instance of a general rule called De Morgan’s Law. Exercise 3 (10 Points) The Fibonacci sequence is defined as follows: a 1 = 1 a 2 = 1 a n = a n 1 + a n 2 for all n ≥ 3 Prove that a n = ( 1+ √ 5 2 ) n ( 1 √ 5 2 ) n √ 5 Solution: The proof proceeds by strong induction. Let P ( n ) denote the hypothesis that a n = ( 1+ √ 5 2 ) n ( 1 √ 5 2 ) n √ 5 . For n = 1 , a 1 = ( 1+ √ 5 2 ) ( 1 √ 5 2 ) √ 5 = √ 5 √ 5 = 1. Thus, P (1) is true. Similarily, P (2) is true since a 2 = 1. Now we assume that the claim holds for all integers between 1 and n ( n ≥ 3). We need to show that P ( n + 1) is true to complete the proof....
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 Spring '10
 DavidG.

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