Discrete mathematics. Elementary and beyond
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Numerous examples, figures, and exercises are spread throughout the book. Levertijd We doen er alles aan om dit artikel op tijd te bezorgen.
Het is echter in een enkel geval mogelijk dat door omstandigheden de bezorging vertraagd is. Bezorgopties We bieden verschillende opties aan voor het bezorgen of ophalen van je bestelling. Welke opties voor jouw bestelling beschikbaar zijn, zie je bij het afronden van de bestelling. Schrijf een review. E-mail deze pagina. Co-auteur: Katalin L. Samenvatting Aimed at undergraduate mathematics and computer science students, this book is an excellent introduction to a lot of problems of discrete mathematics. Recensie s From the reviews: The goal of this book is to use the introduction to discrete mathematics Consequently, the authors The language is very informal and easy to read.
The level is always introductory which makes it possible to give a taste of a wide range of topics There are a lot of exercises Eisenkoelbl, Monatshefte fur Mathematik, Vol. The authors discuss a number of selected results and methods of discrete mathematics Wherever possible, the authors use proofs and problem solving to help students understand the solutions to problems. In addition, there are numerous examples, figures, and exercises spread throughout the book. Zentralblatt fur Didaktik der Mathematik, January, The title of this book is quite apposite It follows that if a1 ,.
Discrete Mathematics: Elementary and Beyond - Lovasz Et. Al, László Lovász - Google книги
We have given two proofs of Theorem 1. You may wonder why we needed two proofs. Unlike in a legal procedure, a mathematical proof either gives absolute certainty or else it is useless. For that matter, we could ask you to take our word for it, and not give any proof. Later, in some cases this will be necessary, when we will state theorems whose proofs are too long or too involved to be included in this introductory book.
So why did we bother to give any proof, let alone two proofs of the same statement? The answer is that every proof reveals much more than just the bare fact stated in the theorem, and this revelation may be more valuable than the theorem itself. The second proof introduced the idea of enumerating these subsets la- beling them with integers 0, 1, 2,.
We also saw an important method of counting: We established a correspondence between the objects we wanted to count the subsets and some other kinds of objects that we can count easily the numbers 0, 1,. In this correspondence: — for every subset, we had exactly one corresponding number, and — for every number, we had exactly one corresponding subset. A correspondence with these properties is called a one-to-one correspon- dence or bijection. If we can make a one-to-one correspondence between the elements of two sets, then they have the same number of elements.
This is a large number, but how large? It would be good to know, at least, how many digits it will have in the usual decimal form. Can we at least estimate the order of magnitude of it? This gives us a reasonably good idea of the size of With a little more high-school math, we can get the number of digits exactly. What does it mean that a number has exactly k digits? We can also say that we obtain k by rounding x down to the next integer. Thus has 31 digits. Proceeding in a similar manner, we get that the number of ways to choose the whole string is 3n.
In fact, the number 3 has no special role here; the same argument proves the following theorem: Theorem 1. The following problem leads to a generalization of this question. The number will certainly be large. We already know from theorem 1. Now how do we determine the number of ways these can be combined? We can formulate the following generalization of Theorem 1.
As another special case, consider the following problem: how many non- negative integers have exactly n digits in decimal? Exercise 1. It is not required that every child get something; it could even happen that we give all the presents to the same child. In how many ways can we distribute the presents? We want to give presents to 12 children.
Again, it is not required that every child gets something; but no child can get two copies of the same present. In how many ways can we give presents? Yet another way of saying this is to order the people in a single line, or write down an ordered list of their names. We also call the rearrangement that does not change anything a permutation somewhat in the spirit of calling the empty set a set. So the question is to determine the number of ways n objects can be ordered, i. This product is so important that we have a notation for it: n! In other words, n! With this notation, we can state our second theorem.
Again, we can illustrate the argument above graphically Figure 1.
Making a decision, we can follow one of the arrows down to the next node. This carries the next decision problem: whom do we put in the second chair? The two arrows out of the node represent the two possible choices. If we make a decision and follow the corresponding arrow to the next node, we know who sits in the third chair. The node carries the whole seating order.
It is clear that for a set with n elements, n arrows leave the top node, and hence there are n nodes on the next level. The bottom level has n! This shows that there are exactly n! In how many ways can they all dance simultaneously?
We assume that only couples of mixed gender dance with each other. This question can be answered along the lines of the arguments we have seen. To give a more mathematical form to the result, we can replace the athletes by any set of size n. We may also view this as selecting a subset of the athletes containing k elements, and then ordering them. Thus we have the following theorem. Check that this is the same number as given in Theorem 1. Of course, if we want to know the number of unordered subsets, then we have overcounted; every subset was counted exactly k!
So we have to divide this number by k! In the next theo- rem we collect some of these; some other identities will occur in the exercises and in the next chapter.
We prove 1. We have an n-element set, say S. After substitution, the identity becomes n!
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Finally, we prove 1. Again, let S be an n-element set. In the whole sum, every subset of S is counted exactly once. We know that 2n the right-hand side is the number of all subsets of S. This proves 1. Review Exercises 1. List all subsets of A whose intersection with B has 1 element. How many elements can their union and their intersection have? What set do you get?