Tutorial 3 Q&A. En la pregunta 7 de la sección 2.2 el cual dice: 7. Prove the domination laws in Table 1 by showing that: a)a U = U b)a =

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1 Tutorial 3 Q&A Question 1: 1) Can the range be considered a subset of a function's codomain? No, not always. There are cases that it is like that, but there are many that not. 2) Why is it that if the symmetrical difference between sets A and B equals A, B has to be an empty set? Why doesn't the (intersection of A and B) = ø work? Because then B would be a subset of A, and the empty set is a subset of every set. 3) What is the difference between (A x B) x C and A x B x C if A, B, C are all sets? (For the benefit of the class, the professor already answered this one.) Let A = {1,2}, B = {3,4}, C = {5,6}. Then, A x B = {(1,3),(1,4),(2,3),(2,4)}. Therefore (A x B) x C = {((1,2),5),((1,2),6),((1,3),5), ((1,3),6)...}. While A x B x C = {(1,3,5),(1,3,6),(1,4,5),(1,4,6),(2,3,5),(2,3,6),(2,4,5),(2,4,6)...} 4) What is the difference between a finite sequence and a tuple? Since they are both ordered collections of items, what is the fundamental difference? There's no difference. Strictly speaking in terms of set theory, a tuple is a sequence in itself. En la pregunta 7 de la sección 2.2 el cual dice: 7. Prove the domination laws in Table 1 by showing that: a)a U = U b)a = La a) la entendí, pero en la b) cuando empezé a resolverlo me quedé en esta parte: A = {x x A x } = { x x A F} Por qué en la contestación del libro sale AND en vez de OR? The reason is because the definition of intersection is that A B={x ( x A ) ( x B ) } En la pregunta nueve de la misma sección 2.2 el cual dice: 9. Prove the complement laws in Table 1 by showing that a) A A(Complements of A) = U. b) A A =. Mi contestación fue: A A(Complements of A) = {x x A x A(Complement)} = {x x A x (No Pertenece) A} = {x x A (x A) } = {x T (T) } = {x T } = {x U} Entonces la contestación del libro fue: A A={x x A x A}=U

2 Eso viene siendo equivalente a lo que yo tengo? It is. Question 3 1. The first one is in exercise 30 (pg 169). I am not pretty sure if I am understanding this notation since it is completely new to me, for example when we try to solve (C), what this notation suggests is that I add up 1/j but replacing j with the first element of the set S and then keep like that or is my approach incorrect? Your approach is correct. In the summation notation you must add iteratively. The number of iterations performed is equal to the number of the elements in the set given or in the range (for example when it says i=1 below the sum and n up of the sum symbol means n iterations). If there s some function that depends of the value of the iteration, then it must be replaced by that value. 2. The second one is the 25 of this same section (pg 169). Even though I have found some formulas I have been having some problems with most of the exercises, for example (e), is there some type of technique that could be use to find this formulas or is it just intuitive? Some basic notions to obtain a closed formula for a sequence: 1. Does the same value appear many times? 2. Are some terms obtained from previous terms by adding a fixed amount? 3. Are some terms obtained from previous terms by multiplying by a fixed amount? 4. Are some terms obtained from previous terms by other means? 5. Are there any cycle among the terms? I suggest that the problem you mention may fall within the last question, as it follows some kind of cycle. Question 4: For each of these lists of integers, provide a simple formula or rule that generates the terms of an integer sequence that begins with the given list. Assuming that your formula or rule is correct, determine the next three terms of the sequence. e) 0, 2, 8, 26, 80, 242, 728, 2186, 6560, 19682,... I found that from n=2 it substracts [(Xn-1 - Xn-2) *3 + Xn-1] = 4(Xn-1)-3(Xn-2) but I cant find the sequence for the n=0 and n=2 numbers. Then this sequence is like the Fibonacci sequence, where the first two terms are specifically stated.

3 Question 5 How can I make proof like the problems below?: 1. Let A be a set. Show that ø X A = A X ø= ø Since ordered pairs are built using two elements, when we try to create the each ordered pair we notice that we have an element from A, but not one from the empty set. Since we can't make an ordered pair with one element, we obtain as a result a set that contains no ordered pairs, or an empty set. 2. The defining property of an ordered pair is that two ordered pairs are equal if and only if their first elements are equal and their second elements are equal. Surprisingly, instead of taking the ordered pair as a primitive concept, we can consruct ordered pairs using basic notions from set theory. Show that if we define the ordered pair (a,b) to be {{a}, {a,b}}, then (a,b)=(c,d) if and onlyif a=c and b=d. [Hint: First show that {{a}, {a,b}}={{c}, {c,d}} if and only if a=c and b=d]. Taking the given hint, we know that in set theory, the order of the set elements doesn't matters. Therefore, a set A is only equal to a set B if and only if they have the same elements. So, let's begin with the subsets of A = {{a}, {a,b}} and B = {{c}, {c,d}}. For A and B to be equal, we know that {a} = {c} so that a must be c for those two sets. In the same way, {a,b} = {c,d}. Question Data are transmitted over a particular Ethernet network in blocks of 1500 octets (blocks of 8 bits). How many blocks are required to transmit the following amounts of data over this Ethernet network? (Note that a byte is a synonym for an octet, a kilobyte is 1000 bytes, and a megabyte is 1,000,000 bytes.) a) 150 kilobytes of data b) 384 kilobytes of data c) megabytes of data d) 45.3 megabytes of data In this exercise you must use the fact that 1,500 octets are actually 1,500 bytes. Therefore, the first one, is 150 KB = 150,000 bytes, and this quantity divided by 1,500 yields 100 blocks of 8 bit or 100 blocks of bytes. The rest are done the same way. 20. Assume that the population of the world in 2010 was 6.9 billion and is growing at the rate of 1.1% a year. b) Find an explicit formula for the population of the

4 world n years after c) What will the population of the world be in 2030? b. P(n) = 6,900,000,000 + (0.011*n*6,900,000,000) c. P(20) = 8,418,000,000 Question 7: Exercise 46 from section 2.3 states "Show that floor function(x+.5) is the closest integer to the number x, except when x is midway between two integers, when it is the larger of these two integers." I understand that since it is a floor function in the end x will always be the largest integer that is less than or equal to x. In this case if x ends with a.5 of higher it will end up becoming x+1 in the end because of the 0.5 in the floor function. But how can I show or prove this? Let x = n + e, where n is an integer part, and e is a fractional part. Then, performing a proof by cases: Case 1: e = 0 floor(x+0.5) = floor(n + 0.5) = n Case 2: 0 < e < 1 floor (x+0.5) = floor(n e) where 0 < e < 1. Therefore, we have two additional cases: Case 3: e < 0.5 If e < 0.5, then we can see that e < 1. Let's call m = e < 1, therefore m < 1. floor(x + 0.5) = floor ( n + m ) where n is an integer number and 0 < m < 1. Therefore, floor (n+m) = n. Case 4: 0.5 < e < 1 If e > 0.5 and e < 1, we can see that e > 1. Therefore, we know that floor(e + 0.5) = 1. Therefore, we know: floor(x + 0.5) = floor (n + e + 0.5), where e+0.5 > 1. It is safe to round up e+0.5 = 1, since for the purposes of the floor function, it will act the same. So, floor (n + 1) = n + 1. Question 8: In example 2 of page 172 I don't get how you can give a room in a fully booked hotel by moving everyone to the next room. How can you create an empty room at the end of the sequence? How is that different than just giving the next room to the new guest? Remember that a sequence may be infinite. There's no way to move guests to a new room except

5 if we accept the facts that the rooms are infinite. Question 9: 28. Show that the function f(x)= ex from the set of real numbers to the set of real numbers is not invertible, but if the codomain is restricted to the set of positive real numbers, the resulting function is invertible.. BY assuming contradiction I e^x it is invertible, so f=ex and if 1/f or f-1(x) x<0 then it does not exist therefore it's not invertible...is it true? It is true. Question 10: 13. Show that the hexadecimal expansion of a positive integer can be obtained from its binary expansion by grouping together blocks of four binary digits, adding initial zeros if necessary, and translating each block off our binary digits into a single hexadecimal digit. Assume a decimal positive integer number whose binary expansion is: a n a n 1...a 2 a 1 a 0 We take into account that hexadecimal numbers go from 0 to F, where F = 15 (F has a value of 15 in decimal number). Also, we know that with n bits, we can represent 2 n different numbers. Therefore, with 4 bits, we can represents 16 different numbers, which are exactly the same quantity of numbers represented in hexadecimal system. Therefore, we can take the first four binary numbers, a 3 a 2 a 1 a 0 and this will be a number that is less or equal to F (or 15). Therefore, we can always group block of 4 binary numbers and exchange them for an hexadecimal number. Notice that if there are not more than 4 bits, this doesn't affect, with 1 bit set to 1 we are saying also 1 hexadecimal; with the combination 0011 we are saying 3 hexadecimal, and so on. Question 11: My questions for the topics of the next tutorials are; How the operations of sets apply to sets that is not finite or that is define by predicate logic statements? Especially those that contain a universal quantifier. There's no difference. Most probably the set operations will be performed over a subset of that infinite subset.

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7 Question 12 On slide 16 of the power point the element 0 is described as always being part of a set. however i did not understood the reason for it. Because a is always false, S, for every set S. The empty set is always part of any other set because it s empty. It s a property. Question 13 Buenas, estoy tratando de hacer el ejercicio 6 de la seccion 2.2 el cual dice: Prove the identity laws: a) A union (null set) = A b)a interseccion U=A En la parte a) hice lo siguiente: Let x exista en A union (null set) x existe en A Luego de eso me quede trancado. En la parte b) hice lo siguiente: Let x exista en A Therefore x existe en A implica que x existe en U No estoy muy seguro de este ultimo paso A = A Let a A. a is always false. The union acts like an or, then, therefore (a A ) ( a ) (a A ) or the set A. A U niv Let a A. a U niv is always true. The intersection acts like an and, then, therefore (a A ) (a U niv ) (a A ) or the set A. Question 14 1.The Cartesian product A B is A B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} The Cartesian product B A is B A = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)} A x B is not equal to B x A Porqué no son iguales si la definición de un SET dice que es una colección de elementos sin orden y la única diferencia que veo es el orden de los elementos? Because the result of a cartesian product is a set contianing ordered pairs. Ordered pairs are tuples of 2 elements in which the order of the occurrence of its elements matters.

8 Question 15 On question 34, section 2.3. Notice that it asks the following: 34. If "f" and "f o g" are one-to-one, does it follow that "g" is one-to-one? Justify your answer. The justifying part is naturally the difficult area of the exercise. What approach would you follow? I can start by assuming that f: B --> C and g: A --> B, and thus f o g: A --> C. We need to prove that if "f" is one-to-one, then "g" is one-to-one. Or do I suppose that "g" is not one-to-one and go from there (like a sort of proof by contradiction)? It would be of great help if we do this exercise on the tutorials. As always, thank you. Suppose that g(x) is not surjective, that is, there is a real number in g that is not mapped onto (e.g. if g(x) = x, -3 would be a counterexample). Then f(g(x)) would not be defined everywhere because g(x) has an incomplete domain so f(g(x)) would not be surjective over all real numbers (another way to phrase it -- the cardinality of the range of f(g(x)) is less than or equal to the cardinality of the domain of g(x), so there must be some f(g(x)) value that is not mapped onto). This contradicts our claim so g(x) must be surjective. (reference: