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CS 201-31, Discrete Structures, Spring 2001, Assignment 3 Solution
Section 2.1, page 70:
4d) {..., -5, -3, -1, 1, 3, 5, 7, ...}. 4e) { x : x = 10k, k
N }. 4f) { } or
, but NOT {
}
Section 2.7, page 96: Prove the following statement: If x is odd, then x2 is odd. Proof:
(1) Assume x is odd.
(2)
x = 2k + 1, for some k
Z.
(3)
x2 = (2k + 1) 2 .
(4)
x2 = 4k2 + 4k + 1.
(5)
x2 = 2(2k2 + 2k) + 1.
(6) Since m = 2k2 + 2k
Z , x2 can be written in the form x2 = 2m + 1, for some m
Z.
(7)
x2 is odd. PROOF COMPLETE.
Section 2.2, page 73:
4) B = {a, b, c, d, e}. List all subsets of B having exactly 2 elements
{a, b}, {a, c}, {a, d}, {a, e}, {b, c}, {b, d}, {b, e}, {c, d}, {c, e}, {d, e} 10 of them.
6a) Is every subset of a finite set finite? This is true. Need to show:
If A
B and B is finite then A is finite.
Proof:
(1) A
B and B is finite.
(2) B is finite means we can write B in list format without ellipses.
(2) A
B , so therefore all elements in A are also in B.
(4) This means that A can be written in list format without ellipses—simply take the representation
of set B in step (2), and cross out any elements in B that are not in A.
(5) Thus A is finite. PROOF COMPLETE.
6b) Is every subset of an infinite set infinite? This question translates to determining whether the following
statement is True or False:
If A
B and B is infinite then A is infinite.
We prove this is FALSE. An if-then is false exactly when the premise is TRUE and the conclusion is FALSE.
A counterexample suffices: Let B be the natural numbers and let A be the set {1,2,3}.
Section 2.4, pages 81, 82:
2) S = {s : s = r2, r <= 10, r
N } = {s: s = r2 , r
{1,2,3,4,5,6,7,8,9,10}}
= {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
T = {2, 4, 6, 8, ..., 38}
X = {2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83,
86, 89, 92, 95, 98, 101, 104, ... }
2c) S
T = {1, 2, 4, 6, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 25, 26, 28, 30, 32, 34, 36, 38, 49, 64, 81, 100}
2d) S
X’ = S = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
4) A = {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, ...}
B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, ... }
C = {4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70,
73, 76, 79, 82, 85, 88, 91, 94, 97, 100, 103, ...}
4c) A
(B
C) = set of elements in A and (B or C), so the first such elements are: 7, 28, 49, 56, 70, 84
4e) (A
B) ’ = set of elements neither in A nor B (assuming N is the universe): 1, 2, 3, 5, 6, 9 are first six.