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Unformatted text preview: CS 2800 – Fall 2008 Homework #3 Solution Section 2.4 Problem 20 [4pts] Using the given hint, we get: k = n X k =1 1 k ( k + 1) = k = n X k =1 1 k 1 k + 1 = k = n X k =1 1 k + 1 1 k = 1 n + 1 1 1 (from telescoping sum formula) = 1 1 n + 1 = n n + 1 Problem 22 [6pts] First, we observe that k 3 ( k 1) 3 = 3 k 2 3 k + 1. Summing both sides, we have: k = n X k =1 ( k 3 ( k 1) 3 ) = k = n X k =1 (3 k 2 3 k + 1) i.e. n 3 = 3 k = n X k =1 k 2 3 k = n X k =1 k + k = n X k =1 1 (simplified LHS using telescoping sum) = 3 k = n X k =1 k 2 3 n ( n + 1) 2 + n Rearranging terms, we have: 1 3 k = n X k =1 k 2 = n 3 + 3 n ( n + 1) 2 n = 2 n 3 + 3 n 2 + n 2 = n ( n + 1)(2 n + 1) 2 i.e. k = n X k =1 k 2 = n ( n + 1)(2 n + 1) 6 Problem 28 [2pts] n ! = i = n Y i =1 i Problem 34 [7pts] (a) The set is countable. The integers in this set are ± 1 , ± 2 , ± 4 , ± 5 , ± 7 . . . . We can list these numbers in the order 1 , 1 , 2 , 2 , 4 , 4 , 5 , 5 , 7 , 7 , . . . to build the desired corre spondence i.e. the correspondence is 1 ↔ 1 , 2 ↔  1 , 3 ↔ 2 , 4 ↔  2 , . . . . Note that in general, the subset of a countable set is countable. (b) The set is countable. We can build the correspondence in a similar fashion to part (a), by listing the integers in increasing order of magnitude, with the positive integers occurring before the negative ones: 5 , 5 , 10 , 10 , 15 , 15 , 20 , 20 , 25 , 25 , 30 , 30 , 40 , 40 , 45 , 45 . . . . (c) The set is countable. We can arrange the numbers in a table as follows: . 1 . 11 . 111 . 1111 . 11111 . . . 1 . 1 1 . 11 1 . 111 1 . 1111 1 . 11111 . . . 11 . 1 11 . 11 11 . 111 11 . 1111 11 . 11111 . . . 111 . 1 111 . 11 111 . 111 111 . 1111 111 . 11111 . . . . . . . . . . . . . . . . . . Each row in the table is a countable set and we have a countable number of rows. We have thus shown that the set of interest is the union of a countable number of countable sets — therefore, by exercise 41, this set is countable as well. 2 (d) The set is uncountable. This can be proven using a diagonalization argument, similar to the one used for proving that the set R is uncountable (example 21 in the text). We make a simple modification to the proof presented in the text — choose d i = 1 when d ii = 9 and d i = 9 when d ii = 1 or blank (i.e. the decimal expansion is finite)....
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This note was uploaded on 12/04/2008 for the course CS 2800 taught by Professor Selman during the Fall '07 term at Cornell.
 Fall '07
 SELMAN

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