Foundations of Computing Discrete Mathematics Solutions to exercises for week 2


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1 Foundations of Computing Discrete Mathematics Solutions to exercises for week 2 Agata Murawska September 16, 2013 Note. The solutions presented here are usually one of many possiblities. The fact that your solution is different does not make it bad. Note. The level of detail in the solutions presented here is similar to the level we expect you to present for the exam. Exercise (2.1.2). Use set builder notation to give a description of each of these sets. a) {0, 3, 6, 9, 12} = {3n n N n 4} b) { 3, 2, 1, 0, 1, 2, 3} = {x Z x 3} c) {m, n, o, p} = {c c is a letter m c p} Exercise (2.1.12). Suppose that A, B and C are sets such that A B, B C. Show that A C. Let us first expand the definitions of subsets. We know: A 1 : A B = x, x A x B We want to show: A 2 : B C = x, x B x C S : A C = x, x A x C Let us fix some x. We assume that this x is an element of A and we want to show that it is also an element of C. This changes S to S : x C and introduces another assumption, A 3 : x A. To show that x 1
2 is in C we will want to use the assumption A 2 instantiated with x. This creates another assumption, A 2 : x B x C. Now to conclude x C we need to know that x B. In order to show this, we can again instantiate one of our assumptions, this time A 1. Assumption created this way is A 1 : x A x B. These are the assumptions we have at this point of the proof: A 1 : x, x A x B A 2 : x, x B x C A 3 : x A A 1 : x A x B A 2 : x B x C The goal we want to achieve is showing S : x C Now what we can do is use rule of inference: modus ponens 1 two times. First we use it on A 1 and A 3. This produces another assumption: A 4 : x B Finally, to finish the proof we use modus ponens again, this time on A 2 and A 4. The conclusion is x C and that is exactly what we were supposed to show. Exercise (2.1.16). Can you conclude that A = B if A and B are two sets with the same power set? Yes, we can. First note the following: P(A) = P(B) means that X, X P(A) X P(B), so (by the definition of powerset) also X, X A X B. We know that A A, therefore also A B, similarly we know B B and so B A. From the two: A B and B A we can conclude A = B. Exercise (2.2.2). Let A = {a, b, c, d, e} and B = {a, b, c, d, e, f, g, h}. Find a) A B = {a, b, c, d, e, f, g, h} b) A B = {a, b, c, d, e} c) A B = 1 From A B and A we can conclude B 2
3 d) B A = {f, g, h} Exercise (2.2.8). Let A and B be sets. Show that 2 : a) A B A A B = {x x A x B}, A B = x, x A x B So A B A means x, x {x x A x B} x A; we can simplify this definition into: x, x A x B x A. To prove this, we take an arbitrary x. We want to show that x A x B x A. Let us assume that x A x B. From that we want to conclude that x A. But if x A x B then surely x A. d) A (B A) = This translates to x, x A (B A) x. Equivalently, we can write it as x, x A x (B A) x x, x A (x B x / A) x. Let us prove this by taking an arbitrary x and showing that: x A (x B x / A) x. Now, by the definition of empty set ( ), we know that no element is in it: x, x /, so in particular x /. This means that x is always false. By the definition of biconditional, this means that x A (x B x / A) must also be false. This is in fact the case, as by regrouping we get x A (x B x / A) x A x B x / A x A x / A x B (x A x / A) x B false x B false Exercise (2.3.4). Find the domain and range of these functions. Note that in each case (...) a) the function that assigns to each nonnegative integer its last digit domain: (nonnegative integers) Z (or: Z + {0}, or: N) range: (the set of all digits) {n Z 0 n 9} b) the function that assigns the next largest integer to a positive integer domain: Z + range: {n N 2 n} 2 We only show two examples here, the rest can be solved using similar methods 3
4 c) the function that assigns to a bit string the number of one bits in the string domain: (the set of all bit strings) {0, 1} range: N d) the function that asigns to a bit string the number of bits in the string domain: (the set of all bit strings) {0, 1} range: N Exercise (2.3.8). Determine whether each of these functions from Z to Z is onetoone. a) f(n) = n 1 it is onetoone: there are no two different elements x 1, x 2 such that x 1 1 = x 2 1, as this would imply x 1 = x 2 b) f(n) = n is not onetoone: for example take 1 and 1, for both of them the image is 2 c) f(n) = n 3 is onetoone; to see this informally notice that sign of f(n) is the same as the sign of n and the values are increasing for n 0 and decreasing for n < 0. d) f(n) = n/2 is not onetoone; for example take 1 and 2, then f(1) = f(2) = 1. Exercise (2.4.2). What are the terms a 0, a 1, a 2 and a 3 of the sequence {a n }, where a n equals a) 2 n + 1 : 1, 3, 5, 9 b) (n + 1) n+1 : 1, 4, 81, 256 c) n/2 : 0, 0, 1, 1 d) n/2 + n/2 : 0, 1, 2, 3 Exercise (2.4.20). Compute each of these double sums. a) (i + j) = j=1 ((i + 1) + (i + 2) + (i + 3)) = (3i + 6) = (3 + 6) + (6 + 6) = = 21 4
5 b) (2i + 3j) = ((2i + 0) + (2i + 3) + (2i + 6) + (2i + 9)) = j=0 (8i + 18) = (0 + 18) + (8 + 18) + ( ) = 78 c) i = j=0 (i + i + i) = (3i) = = 18 d) (ij) = j=1 (i + i 2 + i 3) = (6i) = = 18 5
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