Kenneth H. Rosen. Rosen. SEVENTH EDITION. VENTH. ITION. Discrete. Mathematics and Its. Applications. Disc rete Ma th e m atic s a n d Its. Ap plica tio ns. Discrete Mathematics And Its Applications [ 7th Edition] Kenneth H. Rosen. Identifier-arkark://t58d42v5r. OcrABBYY FineReader DISCRETE MATHEMATICS AND ITS APPLICATIONS, SIXTH EDITION .. Kenneth H. Rosen has had a long career as a Distinguished Member of the Technical.

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Discrete. Mathematics and Its. Applications. Seventh Edition. Kenneth H. Rosen. Monmouth University. (and formerly AT&T Laboratories). Discrete Mathematics and Its Applications: With Combinatorics and Graph Theory . Front Cover. Kenneth H. Rosen, Rosen. McGraw-Hill Offices, Discrete Mathematics and its Applications is a focused introduction to the primary themes in a discrete mathematics course, as introduced through extensive.

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I think I will keep learning from the book and I would say all in all it's a good textbook. I really want to know more about discrete probability sooner rather than later. Jun 12, Pedro Montoto rated it it was amazing Shelves: Best book on theoretic and Computer Science-applied mathematics. Dec 01, Dragify rated it it was amazing Shelves: The instances in which I was not very satisfied with this book have been very, very little so far.

Or perhaps there was never one of them. Quite frankly, differently from what I heard from several other reviews, I never had particular issues at either reading or following this book, except for the not uncommon overtime sometimes needed for finishing some of the exercises. Aug 12, Samuel Gompers rated it it was amazing.

Discrete Mathematics is a field of study integral to the Computer Sciences. It lays the foundations for mathematical thinking in its coverage of proofs, it dives into relevant aspects of application ranging from recursive algorithm structure to modelling networks and efficient systems architecture for modern computing. It was a fascinating read half of the chapters and performed many of the exercises. I found myself more confident in my reasoning skills and also a more rounded citizen of the worl Discrete Mathematics is a field of study integral to the Computer Sciences.

I found myself more confident in my reasoning skills and also a more rounded citizen of the world. Great book! Dec 14, Mahmoud Attia rated it really liked it Shelves: Dec 24, Nathan added it.

Go to textbook for Discrete Math. Jul 23, Parsa rated it liked it. The book has 13 chapters and, There are better books describing each chapter algorithms being a clear example. However, if you are short on time or money, or just want a good introduction to each topic before you explore them at full length later, this book will do.

There is nothing really "bad" about this book. The applications, although certainty not bad, are far too simplified or s read through most of this the past month. The applications, although certainty not bad, are far too simplified or short on explanations to be very useful. All in all, i think this serves its purpose as an introduction.

But be prepared do dive deeper into some of these topics. May 02, Brandon Stumpf rated it it was amazing Shelves: While this book is dense, I absolutely love the examples and applications Rosen covers. We covered chapters 3, 5, 6, 8, 9, 11, and 13 for my Discrete Mathematics class, and wow there are some interesting topics covered in this book. The algorithm and modeling computation chapters won't make much sense unless you have a computer science backgro While this book is dense, I absolutely love the examples and applications Rosen covers.

The algorithm and modeling computation chapters won't make much sense unless you have a computer science background. Strongly recommend for interesting math applications! Feb 19, Shamim Hafiz rated it it was amazing. The best text for transitioning to Computer Science with foundational knowledge on mathematics and basic coding skills.

I recommend students to read this book prior to studying Algorithms and other advanced topic in Computer Science. Mathematical concepts are illustrated through formal reasoning while computational problems are presented with relevant pseudocode.

May 20, Yehudis rated it it was amazing. I don't usually rate mathbooks, but as I begin my second descrete mathematics course with a crappy textbook, I am overwhelmed by love and appreciation for this facinating and easy to read textbook complete with detailed solutions!

Aug 29, Sarah rated it really liked it. This book required me to supplement with other resources, but went far deeper than any others I found. A great book with examples from contemporary times.

The build up through the book is sustained with motivational examples and relevant references. As there are examples from various disciplines of computer science, it really helps you to draw parallels. A must read on this topic. Oct 14, Katherine rated it really liked it. I studied Discrete Mathematics this year. This book helped me in my exams.

Must read if you will study Computer Science. Rosen does a fine job of going over the mathematical underpinnings of computation theory. He begins with an exploration of logic and proceeds into basic algorithms, set theory, probability, counting problems, basic graph theory, and eventually the foundations of computing. The best use of this book is as a primer to get one ready for the new vocabularies, symbol sets, and basic proofs that form the foundation of computer science, graph theory, number theory, and set theory.

It feels as if this cl Rosen does a fine job of going over the mathematical underpinnings of computation theory. It feels as if this class should be taught before calculus in high school. And, on that note, they both out be taught before trig. The problem sets are a good study tool and answers to odd number problems are conveniently included; a nice, but incomplete, glossary comes at end of each chapter; and the programming challenges provide a useful chore for anyone beginning their first forays into a language.

I bought this book as a university textbook and I have to say the author has done a great job. This was by far the best book for a beginner with a basic understanding of mathematics and programming. I think however that the first chapters were wordy and a bit boring ,however since the book is trying to give an introductory account of Discrete mathematics maybe there is a need to wordy sometimes.

Overall I think even beginners would really like this book and I strongly recommend this book to them I bought this book as a university textbook and I have to say the author has done a great job. Overall I think even beginners would really like this book and I strongly recommend this book to them. The first chapter is amazing and easy..

I was doing very bad in the exams but at the end I was always getting the highest mark in the class, I don't know HOW????!!! I hate the proof methods It was really difficult and hard.. All my colleagues believe that it's an awful course where I don't.. Rosen is suitable for an introductory course in an undergraduate computer science and mathematics curriculum.

Topics include mathematical reasoning, combinatorial analysis, discrete structures, algorithmic thinking, and enhanced problem-solving skills through modelling. This was by far the best book for a beginner with a basic understanding of mathematics and trying to get into a career which involves computer programming. Lots of examples to illustrate the mathematical concepts and its applications in computing and a lot of exercises as well. Nov 12, Yu Song rated it really liked it.

Well written, an excellent beginner's course to discrete mathematics, however, not comparable to the classics such as 'concrete mathematics', 'an introduction to combinatorics'.

But definately a good choice for beginners. I consider 'concrete mathematics' somewhat difficult for beginners and 'an introduction to combinatorics' more suitable for beginners who are interested in math puzzles and enjoy the fun of solving interesting combinatorial games.

Jul 06, Joey Andres rated it it was amazing. This book should be the computer science standard for introductory to formal logic. It is easy to follow and ALOT of exercises to really create thicken those neural path in your head. Although, Calculus is not a prerequisite for this book, I still recommend having a concrete understanding of limits, differentiation, and integrals to really extract the most out of the exercises.

After all, some important topics are also discussed in exercises. Excellent book for it's field. Has been a bible of mine. And it is fun well, if you like this kind of math. The historical facts are also an interesting factor. There might be sections that seem too long and over-explained. The answers to this exercise are not unique; there are many ways of expressing the same propositions sym- bolically.

Note that C x, y and C y, x say the same thing. Our domain of discourse for persons here consists of people in this class. We need to make up a predicate in each case. We let P s, c, m be the statement that student s has class standing c and is majoring in m. The variable s ranges over students in the class, the variable c ranges over the four class standings, and the variable m ranges over all possible majors.

It is true from the given information. This is false, since there are some mathematics majors. This is true, since there is a sophomore majoring in computer science.

This is false, since there is a freshman mathematics major. This is false. It cannot be that m is mathematics, since there is no senior mathematics major, and it cannot be that m is computer science, since there is no freshman computer science major.

Nor, of course, can m be any other major.

The best explanation is to assert that a certain universal conditional statement is not true. We need to use the transformations shown in Table 2 of Section 1. The logical expression is asserting that the domain consists of at most two members. It is saying that whenever you have two unequal objects, any object has to be one of those two. Note that this is vacuously true for domains with one element.

Therefore any domain having one or two members will make it true such as the female members of the United States Supreme Court in , and any domain with more than two members will make it false such as all members of the United States Supreme Court in In each case we need to specify some predicates and identify the domain of discourse. In English, everybody in this class has either chatted with no one else or has chatted with two or more others.

In English, some student in this class has sent e-mail to exactly two other students in this class. In English, for every student in this class, there is some exercise that he or she has not solved.

Word order in English sometimes makes for a little ambiguity. In English, some student has solved at least one exercise in every section of this book. This x provides a counterexample.

The domain here is all real numbers. This statement says that there is a number that is less than or equal to all squares. We need to show that each of these propositions implies the other. By our hypothesis, one of two things must be true. Either P is universally true, or Q is universally true. Next we need to prove the converse. Otherwise, P x0 must be false for some x0 in the domain of discourse.

Since P x0 is false, it must be the case that Q y is true for each y. Logic and Proofs c First we rewrite this using Table 7 in Section 1. This is modus tollens. Modus tollens is valid. This is, according to Table 1, disjunctive syllogism. See Table 1 for the other parts of this exercise as well. We want to conclude r. We set up the proof in two columns, with reasons, as in Example 6.

Note that it is valid to replace subexpressions by other expressions logically equivalent to them. Step Reason 1. Alternatively, we could apply modus tollens. Another application of modus tollens then tells us that I did not play hockey. We could say using existential generalization that, for example, there exists a non-six-legged creature that eats a six-legged creature, and that there exists a non-insect that eats an insect.

Now modus tollens tells us that Homer is not a student. There are no conclusions to be drawn about Maggie.

Universal instantiation and modus ponens therefore tell us that tofu does not taste good. The third sentence says that if you eat x, then x tastes good. No conclusions can be drawn about cheeseburgers from these statements. Therefore by modus ponens we know that I see elephants running down the road. In each case we set up the proof in two columns, with reasons, as in Example 6. In what follows y represents an arbitrary person. After applying universal instantiation, it contains the fallacy of denying the hypothesis.

We know that some s exists that makes S s, Max true, but we cannot conclude that Max is one such s. We will give an argument establishing the conclusion. We want to show that all hummingbirds are small. Let Tweety be an arbitrary hummingbird. We must show that Tweety is small.

Therefore by universal modus ponens we can conclude that Tweety is richly colored. The third premise implies that if Tweety does not live on honey, then Tweety is not richly colored. Therefore by universal modus tollens we can now conclude that Tweety does live on honey.

Finally, the second premise implies that if Tweety is a large bird, then Tweety does not live on honey. Therefore again by universal modus tollens we can now conclude that Tweety is not a large bird, i. Notice that we invoke universal generalization as the last step. Thus we want to show that if P a is true for a particu- lar a, then R a is also true. The right-hand side is equivalent to F. Let us use the following letters to stand for the relevant propositions: As we noted above, the answer is yes, this conclusion is valid.

This conditional statement fails in the case in which s is true and e is false. If we take d to be true as well, then both of our assumptions are true. Therefore this conclusion is not valid. This does not follow from our assumptions. If we take d to be false, e to be true, and s to be false, then this proposition is false but our assumptions are true. We noted above that this validly follows from our assumptions. The only case in which this is false is when s is false and both e and d are true.

Therefore, in all cases in which the assumptions hold, this statement holds as well, so it is a valid conclusion. We must show that whenever we have two even integers, their sum is even.

Suppose that a and b are two even integers. We must show that whenever we have an even integer, its negative is even. Suppose that a is an even integer. This is true. We give a proof by contradiction. By Exercise 26, the product is rational. We give a proof by contraposition. If it is not true than m is even or n is even, then m and n are both odd.

By Exercise 6, this tells us that mn is odd, and our proof is complete. Assume that n is odd. But this is obviously not true. Therefore our supposition was wrong, and the proof by contradiction is complete.

Therefore the conditional statement is true. This is an example of a trivial proof, since we merely showed that the conclusion was true. Then we drew at most one of each color. This accounts for only two socks. But we are drawing three socks. Therefore our supposition that we did not get a pair of blue socks or a pair of black socks is incorrect, and our proof is complete.

Since we have chosen 25 days, at least three of them must fall in the same month. Since n is even, it can be written as 2k for some integer k. This is 2 times an integer, so it is even, as desired. So suppose that n is not even, i. This is 1 more than 2 times an integer, so it is odd. That completes the proof by contraposition. There are two things to prove. Now the only way that a product of two numbers can be zero is if one of them is zero. We write these in symbols: It is now clear that all three statements are equivalent.

We give direct proofs that i implies ii , that ii implies iii , and that iii implies i. These are therefore the only possible solutions, but we have no guarantee that they are solutions, since not all of our steps were reversible in particular, squaring both sides.

Therefore we must substitute these values back into the original equation to determine whether they do indeed satisfy it. But these each follow with one or more intermediate steps: We claim that 7 is such a number in fact, it is the smallest such number. The only squares that can be used to contribute to the sum are 0 , 1 , and 4. Thus 7 cannot be written as the sum of three squares. By Exercise 39, at least one of the sums must be greater than or equal to Example 1 showed that v implies i , and Example 8 showed that i implies v.

The cubes that might go into the sum are 1 , 8 , 27 , 64 , , , , , and We must show that no two of these sum to a number on this list. Having exhausted the possibilities, we conclude that no cube less than is the sum of two cubes.

There are three main cases, depending on which of the three numbers is smallest. In the second case, b is smallest or tied for smallest. Since one of the three has to be smallest we have taken care of all the cases. The number 1 has this property, since the only positive integer not exceeding 1 is 1 itself, and therefore the sum is 1. This is a constructive proof. Therefore these two consecutive integers cannot both be perfect squares. This is a nonconstructive proofâ€”we do not know which of them meets the requirement.

In fact, a computer algebra system will tell us that neither of them is a perfect square. Of these three numbers, at least two must have the same sign both positive or both negative , since there are only two signs. It is conceivable that some of them are zero, but we view zero as positive for the purposes of this problem. The product of two with the same sign is nonnegative. In fact, a computer algebra system will tell us that all three are positive, so all three products are positive.

We know from algebra that the following equations are equivalent: This shows, constructively, what the unique solution of the given equation is. Given r , let a be the closest integer to r less than r , and let b be the closest integer to r greater than r. In the notation to be introduced in Section 2. We follow the hint. This is clearly always true, and our proof is complete. This is impossible with an odd number of bits.

Clearly only the last two digits of n contribute to the last two digits of n2. So we can compute 02 , 12 , 22 , 32 ,. From that point on, the list repeats in reverse order as we take the squares from to , and then it all repeats again as we take the squares from to Thus our list which contains 22 numbers is complete.

Clearly there are no integer solutions to these equations, so there are no solutions to the original equation. One proof that 3 2 is irrational is similar to the proof that 2 is irrational, given in Example 10 in Section 1. Thus p3 is even. Now we play the same game with q. Since q 3 is even, q must be even. We have now concluded that p and q are both even, that is, that 2 is a common divisor of p and q. The solution is not unique, but here is one way to measure out four gallons.

Fill the 5-gallon jug from the 8-gallon jug, leaving the contents 3, 5, 0 , where we are using the ordered triple to record the amount of water in the 8-gallon jug, the 5-gallon jug, and the 3-gallon jug, respectively. Pour the contents of the 3-gallon jug back into the 8-gallon jug, leaving 6, 2, 0. Without loss of generality, we number the squares from 1 to 25, starting in the top row and proceeding left to right in each row; and we assume that squares 5 upper right corner , 21 lower left corner , and 25 lower right corner are the missing ones.

We argue that there is no way to cover the remaining squares with dominoes. By symmetry we can assume that there is a domino placed in using the obvious notation. If square 3 is covered by , then the following dominoes are forced in turn: Therefore we must use along with If we use all of , , and , then we are again quickly forced into a sequence of placements that lead to a contradiction. Therefore without loss of generality, we can assume that we use , which then forces , , , , , , and , and we are stuck once again.

This completes the proof by contradiction that no placement is possible. The barriers shown in the diagram split the board into one continuous closed path of 64 squares, each adjacent to the next for example, start at the upper left corner, go all the way to the right, then all the way down, then all the way to the left, and then weave your way back up to the starting point.

Because each square in the path is adjacent to its neighbors, the colors alternate. Therefore, if we remove one black square and one white square, this closed path decomposes into two paths, each of which starts in one color and ends in the other color and therefore has even length. Clearly each such path can be covered by dominoes by starting at one end. This completes the proof. Supplementary Exercises 31 Therefore the same argument as was used in Example 22 shows that we cannot tile the board using straight triominoes if any one of those other 60 squares is removed.

The following drawing rotated as necessary shows that we can tile the board using straight triominoes if one of those four squares is removed. Assume that 25 straight tetrominoes can cover the board. Some will be placed horizontally and some vertically.

Because there is an odd number of tiles, the number placed horizontally and the number placed vertically cannot both be odd, so assume without loss of generality that an even number of tiles are placed horizontally.

Color the squares in order using the colors red, blue, green, yellow in that order repeatedly, starting in the upper left corner and proceeding row by row, from left to right in each row. Then it is clear that every horizontally placed tile covers one square of each color and each vertically placed tile covers either zero or two squares of each color. It follows that in this tiling an even number of squares of each color are covered. But this contradicts the fact that there are 25 squares of each color.

Therefore no such coloring exists. The truth table is as follows. Logic and Proofs 6. Since both knights and knaves claim that they are knights the former truthfully and the latter deceivingly , we know that A is a knave. Thus all three are knaves. If S is a proposition, then it is either true or false.

Hence it has a true conclusion modus ponens , and so unicorns live. But we know that unicorns do not live. It follows that S cannot be a proposition. The given statement tells us that there are exactly two elements in the domain. Therefore the statement will be true as long as we choose the domain to be anything with size 2 , such as the United States presidents named Bush. Let us assume the hypothesis.

This means that there is some x0 such that P x0 , y holds for all y. Here is an example. Supplementary Exercises 33 Let W r means that room r is painted white. Let I r, b mean that room r is in building b.

Let L b, u mean that building b is on the campus of United States university u.