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CITS2211 Discrete Structures Week 6 Tutorial 1. Which of the following functions are injective, which surjective, which bijective? (a) f1 : N → N where f1 (n) = 2n. (b) f2 : N → E where f2 (n) = 2n (recall E is the set of natural even numbers). (c) f3 : N → N where f3 (n) = n/2 if n is even and f3 (n) = (n + 1)/2 otherwise. (d) f4 : R → R where f4 (x) = 2x. (e) f5 : R → R>0 where f5 (x) = x2 . Solution: (a) f1 is injective (if f1 (m) = f1 (n) then 2m = 2n, so m = n) but not surjective (1 is not f1 (n) for any n ∈ N). (b) f2 is injective (if f2 (m) = f2 (n) then 2m = 2n, so m = n) and surjective (every even number n is the image of an integer, namely n/2), hence bijective. (c) f3 is not injective (f3 (2) = 1 = f3 (1) but 2 6= 1) but it is surjective (every even number n is the image of an integer, namely 2n). (d) f4 is injective (if f4 (x) = f4 (y) then 2x = 2y, so x = y) and surjective (every real number x is the image of a real number, namely x/2), hence bijective. (e) f5 is not injective (f5 (1) = 1 = f5 (−1) but 1 6= −1) but it is surjective (every non-negative real number x is the image of a √ real number, namely x). 1 2. Let A = {1, 2, 3} and B = {a, b}. (a) How many functions f : A → B are there? (b) How many functions f : A → B are injective, how many surjective? (c) How many functions f : A → B are bijective? Solution: (a) There are 2 choices for f (1), 2 choices for f (2) and 2 choices for f (3), so there are 2 ∗ 2 ∗ 2 = 8 such functions. (b) Since A has size 3 and B has size 2, there are no such injections. To count surjections, it is easier to count the “non-surjections”. For f not to be a surjection, it has to map every element of A to the same element of B, either a or b. Hence there are 2 non-surjections. The total number of functions, as seen in Part (a), is 8. Thus there are 6 surjections. (c) If there existed a bijection from A to B, we would have |A| = |B|. So there are no bijections. 3. Show that |A ∪ B| = |A| + |B| − |A ∩ B|, when A and B are finite sets. Solution: We have that A ∪ B is the disjoint union of A − B, B − A and A ∩ B, so |A ∪ B| = |A − B| + |B − A| + |A ∩ B|. Now A is the disjoint union of A − B and A ∩ B, so |A| = |A − B| + |A ∩ B|. Similarly |B| = |B − A| + |A ∩ B|. Thus |A − B| = (|A| − |A ∩ B|) + (|B| − |A ∩ B|) + |A ∩ B| = |A| + |B| − |A ∩ B|. Solution: (alternative) Divide A ∪ B into two disjoint parts: A ∪ B = A ∪ (B − A). We know they’re disjoint, so the cardinality is |A ∪ B| = |A| + |B − A|. Now let’s work out the cardinality of B − A. B − A = B − (A ∩ B), 2 and we know they are disjoint, too, so |B − A| = |B| − |A ∩ B|. So the cardinality of the whole expression is |A ∪ B| = |A| + |B| − |A ∩ B. 4. Of the integers x, 100 ≤ x ≤ 999: how many are divisible by 5? How many are not divisible by 5? How many are divisible by 4? How many are divisible by 4 and 5? How many are divisible by 4 or 5? How many are divisible by neither 4 nor 5? Solution: There are 900 numbers altogether. Let A denote those divisible by 5 and B denote those divisible by 4. Then A = {100 + 5k | k ∈ {0, 1, . . . 179}} and so |A| = 180. So |A| = 900 − 180 = 720, that is the number of numbers not divisible by 5. Then B = {100 + 4k | k ∈ {0, 1, . . . 224}} and so |B| = 225. Then A ∩ B contains all the numbers divisible by 20, so A ∩ B = {100 + 20k | k ∈ {0, 1, . . . , 44}} and so |A ∩ B| = 45. Divisible by 4 or 5 is the set A ∪ B which has 180 + 225 − 45 = 360 elements. This leaves |A ∪ B| = 900 − 360 = 540 divisible by neither. 5. Show that the union of two countable sets is countable. 3 Solution: If the two sets are finite, it is obvious. If one is finite and the other infinite countable, make a list by first putting the finite set in any order then the infinite set, which we know can be ordered in a list since it is countable. Now assume both sets A and B are infinite countable. We know we can order both sets A = {a1 , a2 , a3 , . . .} and B = {b1 , b2 , b3 , . . .}. Now we can order A ∪ B the following way: a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 . . . b1 b2 b3 b4 b5 b6 b7 b8 b9 b10 . . . so A ∪ B is infinite countable. 6. Which of the following sets are countable? Justify your answer. (a) The set P(N>0 ), the power set of the positive natural numbers. (b) The set S of all functions f : {0, 1} → N>0 . (c) The set T of all functions f : N>0 → {0, 1}. (d) The set A × A, where A is an infinite countable set. Solution: (a) The set P(N>0 ) is uncountable because the power set of a set always has greater cardinality than the set. (b) The set S of all functions f : {0, 1} → N>0 is countable because it has the same cardinality as N>0 × N>0 , which we have seen in lectures is countable. To prove that we will exhibit a bijection between S and N>0 × N>0 : g : S → N>0 × N>0 defined by g(f ) = (f (0), f (1)). • g is injective since if g(f1 ) = g(f2 ) then (f1 (0), f1 (1)) = (f2 (0), f2 (1)) so f1 (0) = f2 (0) and f1 (1) = f2 (1), which means that f1 and f2 are the same function. 4 • g is surjective since (a, b) ∈ N>0 × N>0 is g(f ) for the function f mapping 0 to a and 1 to b. (c) The set T of all functions f : N>0 → {0, 1} is uncountable because it is the same size as P(N>0 ). To prove that we will exhibit a bijection between T and P(N>0 ): g : T → P(N>0 ) defined by g(f ) = {n ∈ N>0 |f (n) = 1}. • g is injective since if g(f1 ) = g(f2 ) then {n ∈ N>0 |f1 (n) = 1} = {n ∈ N>0 |f2 (n) = 1} so f1 (n) = 1 exactly whenever f2 (n) = 1. It follows f1 (n) 6= 1 (so f1 (n) = 0) exactly (You can refer to question 6(c).)whenever f2 (n) 6= 1 (so f2 (n) = 0), which means that f1 and f2 are the same function. • g is surjective since C ∈ P(N>0 ) is g(f ) for the function f mapping every element in C to 1 and everything else to 0. (d) The set A × A is countable, by a zig-zag argument similar to the one we saw in lectures for N>0 × N>0 . Solution: (alternative solution for (b)): Any function f from the set {0, 1} to the positive natural numbers will be of the form: f (n) = when n is 0, f (n) = x when n is 1, f (n) = y where x and y are two positive natural numbers. Let us give these functions names - fx,y will be the name of the function for which f (0) = x and f (1) = y. We can now put the names of this functions in a grid, as we have done in lectures with pairs of integers: f1,1 f1,2 f1,3 ... f2,1 f2,2 f2,3 ... f3,1 f3,2 f3,3 ... ... We can then exhibit a bijection between these and the natural num- 5 bers, by enumerating them in a “zig-zag” fashion (refer to to lecture notes for the function). Therefore, the number of functions is countable. (Trivia: a path which goes left and then right alternately is a boustrophedon path - from the Greek for “ox turning”, from the pattern an ox makes when plowing a field. So we could call this “zig-zag” a diagonal boustrophedon path.) 7. (challenge problem) There is a smallest legal Java program. It looks like: public class A{public static void main(String[] a){}} What is the cardinality of the set of all legal Java programs, J? There is also a set of functions from N to {T, F } - call it P . What is the cardinality of this set? (You can refer to question 6(c).) How do the cardinalities of J and P compare - is one larger than the other? If so, what do you deduce from this? 6