# DISCRETE MATHEMATICS WITH GRAPH THEORY 3RD EDITION PDF

Instructor's Solutions Manual to accompany. This manual contains complete solutions to all exercises in Discrete Mathematics with Graph Theory, Third Edition, by Edgar G. Goodaire and Michael M. Parmenter. Edgar G. Goodaire and Michael M. Parmenter Department of Mathematics and. Discrete Mathematics With Graph Theory 3rd Edition - Ebook download as PDF File .pdf), Text File .txt) or read book online. Discrete Mathematics textbook. Discrete mathematics with graph theory / Edgar G. Goodaire, Michael M. Parmenter.- . We believe that writing skills are terribly important, so, in this edition, we.

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The first and third premises give. Together with the second premise double negation and a second application of modus tollens , we get r. So assume the premises are true but the conclusion is false.

Assume p V qis false.

This means both p and q are false. Since p V r is true and p is false, r must be true. Since q V. There are five rows when the premises are all true and in each case the conclusion is also true. The argument is valid.

In every case, the conclusion is true. Now -,p follows by modus tollens. If q and r are false, p and 8 are true, and t takes on any truth value, then all premises are true, yet the conclusion is false.

The first of these is logically equivalent to q -- p while the third is r -- p. Using Exercise 3 a , we get q V r -- p, so certainly q V r -- p V r. To see this, note that the second premise is r -- t V 8 , so the chain rule gives r -- p. Also, the first premise is q -- p. Exercise 3 a tells us that q V r -- p, so certainly qVr -- pVr.

If p and r are false while q and 8 are true, all premises are true, yet the conclusion is false. The first premise is qV [ -,p V 8] while the second is -,q Yr. Now -,p V 8Vr follows by resolution, and this is the conclusion.

Since q V r, resolution gives p V r. Using r -- p and the result of part f , we get p V p, which is p. Using g twice, we have p -- [8 V -,p ] which is -,p V 8 V -,p , which is -,p V 8, which is the conclusion. I stay up late at night I am tired in the morning. The given argument is p-tq q p This is not valid, as the truth table shows. In row three, the two premises are true but the conclusion is false. In row one, the two premises are true but the conclusion is false.

Solutions to Exercises I stay up late at night I am tired in the morning. I wear a red tie I wear blue socks. In row five, the two premises are true but the conclusion is false. I work hard h Let p, q, and r be the statements q: I earn lots of money r: I pay high taxes. I work hard i Let p, q, and r be the statements q: I like mathematics j Let p, q, and r be the statements q: I study r: I like football.

The given argument is p: I like mathematics k Let p, q, and r be the statements q: I like mathematics q: I study I [BB] Let p, q, and r be the statements r: I pass mathematics s: I graduate. I study m Let p, q, and r be the statements r: We will prove by contradiction that no such conclusion is possible. Say to the contrary that there is such a conclusion e. Since e is not a tautology, some set of truth values for p and q must make e false. This contradicts e being a valid conclusion for this argument.

By Exercise 8 a , the final premise is equivalent to the list of premises ql, qz, Thus the given premises imply which, again using 8 a , are logically equivalent to the single premise This is a reflection of the fact that modus tollens has a negative -,p as its conclusion, while modus ponens affirms the truth of a statement q.

Chapter 1 Review 1.

## Discrete Mathematics with Graph Theory, 3rd Edition

Hence the entire statement has truth value T. Assume that some set of truth values on the variables makes A true. So 'B must be true also. Similarly, if 'B is true, then A must also be true. We conclude that A is true if and only if'B is true. This means A and 'B are logically equivalent. But Property 12 also says If p is false and q is true, the premises are true but the conclusion is not.

If p, r and t are true while q is false and s takes on either truth value , the hypotheses are true while the conclusion is false. This argument is valid. The hypotheses can never both be true at the same time, so there can be no case when the hypotheses are true while the conclusion is false. We can write it as shown. When s and r are true and p is false, the hypotheses are true, while the conclusion is false.

But this is not true. The empty set is a subset of every set. The empty set does not contain any elements. Section 2.

See Exercise 15 in Section 5. Therefore, x is an element of C which is not in A, proving A f C. A E B means that A belongs to the set B. Since B is a subset of C, any element of B also belongs to C. Hence, A E C. Then A is not a subset of B and B is not a proper subset of A. Thus A is not a subset of B.

## Discrete Mathematics with Graph Theory (3rd Edition)

For this, let X E P A. Therefore, X is a subset of A; that is, every element of X is an element of B. The two sets are equal. See Exercise 11 of Section 2. There are four. T; Converse: Since 1, 1 E A, 2,1 and 2,2 are inA.

Certainly x is also in A orin AC. This suggests cases. First let b E B. A Section 2.

Here we have bE B n A and, hence, b In either case, we obtain bE C. It follows that B c;;: A similar argument shows C c;;: Since a E A and A c;;: Thus, A c;;: Similarly, we have B c;;: Therefore, A n B x C c;;: Therefore, A x C n B x C c;;: Therefore, A, B x C c;;: This means that x, y E A x C, but x, y Therefore, AxC , BxC c;;: But 1,4 AxC and 1,4 BxD, so 1,4 AxC U BxD.

Then, since 3 However, because 2 E B, A, B, so 2,3 George Boole was one of the greatest mathematicians of the nineteenth century. He was the first Professor of Mathematics at University College Cork then called Queen's College and is best known today as the inventor of a subject called mathematical logic.

Indeed he introduced much of the symbolic language and notation we use today. Like Charles Babbage and Alan Turing, Boole also had a great impact in computer science, long before the computer was even a dream.

He invented an algebra of logic known as Boolean Algebra, which is used widely today and forms the basis of much of the internal logic of computers. Exercises 2. A x B is the set of all ordered pairs a, b where a is a street and b is a person. The answer is yes and the only such binary relations are subsets of the equality binary relation.

To see why, let R: This need not be the case: See Exercise 5 g. Every word has at least one letter in common with itself. If a and b have at least one letter in common, then so do b and a.

Not antisymmetric: Not transitive: Let a be a person. If a is not enrolled at Miskatonic University, then a, a E R: On the other hand, if a is enrolled at MU, then a is taking at least one course with himself, so again a,a En. If a, b E 'R, then either it is the case that neither a nor b is enrolled at MU so neither is b or a, hence, b, a E n or it is the case that a and b are both enrolled and are taking at least one course together in which caseb and a are enrolled and taking a common course, so b, a En.

In any case, if a, b En, then b, a En. If a and b are two different students in the same class at Miskatonic University, then a, b En and b, a En, but a f- b.

At most universities, this is not a transitive relation. Let a, b and e be three students enrolled at MU such that a and b are enrolled in some course together and b and e are enrolled in some other course together, but a and e are taking no courses together.

Then a, b and b,e are in n but a,e tI- R: Not symmetric: It is never the case that for two different elements a and b in A we have both a, b and b, a in R: Transitive vacuously; that is, there exists no counterexample to disprove transitivity: The situation a, b En and b, e En never occurs. For any a E Z, it is true that a2 2: Thus, a, a E R: If a, b En, then ab 2: If a, b and b, e are both in R: For example, 0,7 E R: If n E N, then n of- n is not true.

If nl of- n2, then n2 of- nl. G Reflexive: Since a and b are positive, so are n and m. The argument given in Example 24 for Z works the same way for N. As before. As shown in Example It is not transitive because, for example, 2,0 E R: It is not antisymmetric since, for example, 0,1 E nand 1,0 E R; but 0 1.

Let a, bE S. Antisymmetric "vacuously": Recall that an implication is false only when the hypothesis is true and the conclusion is false. Y, Z E n because the price of Y is greater than the price of Z and the length of Y is greater than the length of Z, but for these same reasons, Z, Y If a, b and b, a are both in R: If a, b and b, c are in R; then the price of a is; Also the length of a is ; Hence, a, c E R: For any book a, the price ofa is; One could also use a similar argument concerning length.

Z, U En because the length of Z is; So Mike is now clearly identified as the one who shot , and Pippy Park is where that occurred. Hence, Mike's round of 74 was at Clovelly. Since Edgar has only one entry in binary relation two, he must have shot 72 at both courses. Finally, Bruce's 74 must have been at Clovelly and hence his 72 was at Pippy Park.

All information has been retrieved in this case. For any citizen a of New York City, either a does not own a cell phone in which case a , Suppose a , If a does not have a cell phone, then neither does b and, since b , On the other hand, if a does have a cell phone then so does b and a's and b's exchanges are the same. Since b , It follows that a and c have the same exchange and so, in this case as well, a rv c. There is one equivalence class consisting of all residents of New York who do not own a cell phone and one equivalence class for each New York City exchange consisting of all residents who have cell phones in that exchange.

It would also be acceptable to not that n is not transitive. It would also be acceptable to note that this relation is not transitive: If a E S, then a and a have the same number of elements, so a rv a. If a rv b, then a and b have the same number of elements, so b and a have the same number of elements. Thus b rv a. If a rv b and b rv c, then a and b have the same number of elements, and band c have the same number of elements, so a and c have the same number of elements.

Thus a , Therefore, b rv a. It follows that either a and b are both even or both are odd.

## 45825864 Discrete Mathematics With Graph Theory 3rd Edition

The quotient set is the set of equivalence classes. Now 2. Thus, a ""' c. Thus b '" a. Thus a '" c. Hence, a '" c. If a '" b, then a2 - b2 is divisible by 3, so b2 - a2 is divisible by 3, so b '" a. Note that if a is any triangle, a f'V a because a is congruent to itself. Assume a f'V b. Then a and b are congruent. Therefore, b and a are congruent, so b f'V a.

If a f'V b and b f'V c, then a and b are congruent and band c are congruent, so a and c are congruent.

## Related titles

Thus, a f'V c. If a is a circle, then a f'V a because a has the same center as itself. Then a and b have the same center.

Thus, b and a have the same center, so b f'V a. Assume a f'V b and b f'V c. Then a and b have the same center and band c have the same center, so a and c have the same center.

If a is a line, then a is parallel to itself, so a f'V a. If a f'V b, then a is parallel to b. Thus, b is parallel to a.

Hence, b f'V a. If a f'V b and b f'V c, then a is parallel to b and b is parallel to c, so a is parallel to c. The reflexive property does not hold because no line is perpendicular to itself.

Here they are: Therefore, rv is reflexive. Much better is to state symmetry like this: Here is the correct argument: Logical arguments consist of a sequence of implications but here it is not clear where these implications start.

Certainly the first sentence is not an implication. The quotient set is the set of lines with slope -!. Hence, a, b is the union of the z-axis and the y-axis. The relation is not symmetric. Since the relation is reflexive, symmetric and transitive, it is an equivalence relation.

The quotient set is the set of vertical lines. For example, it is not reflexive: Remembering that x is just the set of elements equivalent to x, we are given that a rv b, c rv d and d rv b.

By Proposition 2. This is the given relation. If a E A, then a2 is a perfect square, so a rv a. If a rv b, then ab is a perfect square. If a rv band b rv e, then ab and be are each perfect squares. Because ae is an integer, so also x: Therefore, a rv e.

We have to prove that the given sets are disjoint and have union S. For the latter, we note that since n is reflexive, for any a E S, a, a E n and so a and a are elements of the same set Si; that is, a E Si for some i. To prove that the sets are disjoint, suppose there is some x E Sk n St. Since Sk rz. Sj for any j i- k. By transitivity, y, z E R: But the only set to which y belongs is Sk.

Since z does not belong to Sk, we have a contradiction: No x E Sk n Se exists. This partial order is a total order because for any a, b E R, either a This is not a total order; for example, 1,4 and 2,5 are incomparable. Helmut Hasse was one of the more important mathematicians of the twentieth century.

He grew up in Berlin and was a member of Germany's navy during the first World War. He received his PhD from the University of Gottingen in for a thesis in number theory, which was to be the subject of his life's work. He is known for his research with Richard Brauer and Emmy Noether on simple algebras, his proof of the Riemann Hypothesis one of today's most famous open problems for zeta functions on elliptic curves, and his work on the arithmetical properties of abelian number fields.

Hasse's career started at Kiel and continued at Halle and Marburg. When the Nazis came to power in , all Jewish mathematicians, including eighteen at the University of Gottingen, were summarily dismissed from their jobs. It is hard to know the degree of ambivalence Hasse may have had when he received an offer of employment at Gottingen around this time, but he accepted the position.

While some of Hasse's closest research collaborators were Jewish, he nonetheless made no secret of his support for Hitler's policies.

In , he was dismissed by the British, lost his right to teach and eventually moved to Berlin. In May , he was appointed professor at the Humboldt University in East Berlin but he moved to Hamburg the next year and worked there until his retirement in If a is not maximal, there is an element al such that al - a.

If al is not maximal, there is an element a2 such that a2 - al. Since A is finite, eventually this process must stop, and it stops at a maximal element.

A similar argument shows that A, For example, R, Since al Similarly, b This partial order is not a total order: Then a We must prove that A n B has these properties. We must prove that Au B has these properties.

Thus b is an upper bound for a and b. Thus a is a lower bound for a and b. In a poset which is not totally ordered, they don't necessarily, however.

But 0 is a minimum because 0 is a subset of any set and the set S itself is a maximum because any of its subsets is contained in it. Thus, a is a maximum. By definition of n, x is in both A and B, in particular, x E A.

Conversely, let x E A. Let bE B and let a be any element of A. Thus b E C. A n C "B Region 3: An B n C Region 4: B" C "A consists of region 6.

On the other hand, A g; C. Chapter 2 51 Symmetric by definition. Please be aware of copyright issues and fair use. To share files, ge. In an effort to encourage the free distribution of knowledge, please also consider using these open access links for your research:. List of open-access journals.

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I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns. Case ii: a, b odd. Case iii: a even, b odd. Case iv: a odd, b even. This is similar to Case iii, and the result follows. Case 1: k is even. Case 2: k is odd. Since each case leads to one of the desired conclusions, the result follows. In all cases, the desired conclusion is true. So the statement must be true. We give a proof by contradiction.

Note that the negation of an "or" statement is an "and" statement.

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Now suppose that A is invertible. This is false.Szekely Bowling Green State University and those who prefer to remain anonymous.

More than worked examples and problems, as well as over exercises are included. Solutions Manuals are available for thousands of the most popular college and high school textbooks in subjects such as Math, Science Physics , Chemistry , Biology , Engineering Mechanical , Electrical , Civil , Business and more.

Finally, Bruce's 74 must have been at Clovelly and hence his 72 was at Pippy Park. This textbook is written to engage your mind and to offer a fun way to learn some mathematics.

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