Discrete Mathematics and Its Applications, 8th Edition by Kenneth Rosen Integrated course eBook; Supporting how-to videos, interactives and extra practice. Discrete Mathematics and Its Applications Seventh Edition 7th Edition. by Kenneth H Rosen (Author). out of 5 stars customer reviews. ISBN -. Editorial Reviews. About the Author. Ken Rosen (Middletown, NJ) is a distinguished member of Kindle Store · Kindle eBooks · Science & Math.
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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, SIXTH EDITION .. Kenneth H. Rosen has had a long career as a Distinguished Member of the Technical. Discrete Mathematics And Its Applications [ 7th Edition] Kenneth H. Rosen. Identifier-arkark://t58d42v5r. OcrABBYY FineReader
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Pricing subject to change at any time. After completing your transaction, you can access your course using the section url supplied by your instructor. Skip to main content x Sign In. Sign in to shop, sample, or access your account information. Alternatively, all students in the school have visited North Dakota. No student in the school has visited North Dakota.
Alternatively, there does not exist a student in the school who has visited North Dakota. Alternatively, there exists a student in the school who has not visited North Dakota. It is not true that every student in the school has visited North Dakota. Alternatively, not all students in the school have visited North Dakota.
This is technically the correct answer, although common English usage takes this sentence to mean—incorrectly—the answer to part e. To be perfectly clear, one could say that every student in this school has failed to visit North Dakota, or simply that no student has visited North Dakota. Note that part b and part c are not the sorts of things one would normally say.
Alternatively, every rabbit hops. Alternatively, some rabbits hop. Alternatively, some hopping animals are rabbits.
See Examples 11 and The other parts of this exercise are similar. Many answer are possible in each case. If the domain were all residents of the United States, then this is certainly false. If the domain consists of all United States Presidents, then the statement is false. In all of these, we will let Y x be the propositional function that x is in your school or class, as appropriate. In each case we need to specify some propositional functions predicates and identify the domain of discourse.
There are many ways to write these, depending on what we use for predicates.
For example, we can take P x to mean that x is an even number a multiple of 2 and Q x to mean that x is a multiple of 3. Thus both sides of the logical equivalence are true hence equivalent. Now suppose that A is false.
If P x is true for all x , then the left-hand side is true. On the other hand, if P x is false for some x, then both sides are false. Therefore again the two sides are logically equivalent.
If P x is true for at least one x, then the left-hand side is true. On the other hand, if P x is false for all x , then both sides are false. If A is false, then both sides of the equivalence are true, because a conditional statement with a false hypothesis is true.
If A is false, then both sides of the equivalence are true, because a conditional statement with a false hypothesis is true and we are assuming that the domain is nonempty. It is saying that one of the two predicates, P or Q , is universally true; whereas the second proposition is simply saying that for every x either P x or Q x holds, but which it is may well depend on x. As a simple counterexample, let P x be the statement that x is odd, and let Q x be the statement that x is even.
Let the domain of discourse be the positive integers. The second proposition is true, since every positive integer is either odd or even. P x is true, so we form the disjunction of these three cases. So the response is no. So the response is yes. Following the idea and syntax of Example 28, we have the following rule: The unsatisfactory excuse guaranteed by part b cannot be a clear explanation by part a. If x is one of my poultry, then he is a duck by part c , hence not willing to waltz part a.
Or, more simply, a nonnegative number minus a negative number is positive which is true. 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.