The notes form the base text for the course ”MAT Graph Theory”. computational methods given by the mathematical combinatoric and linear- algebraic. 4 Traversal: Eulerian and Hamiltonian Graphs. 5 Graph Optimization. 6 Planarity and Colorings. MAT (Discrete Math). Graph Theory. Fall 2 / PDF | On Apr 1, , Bhavanari Satyanarayana and others published Discrete mathematics and graph theory. 2nd ed.
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Definition. A planar graph is one which can be drawn in the plane without any edges crossing. Such a drawing is called an embedding of the graph in the plane . CSH: Discrete Mathematics Introduction to Graph Theory. 2/ Directed Graphs. ▷ All graphs we considered so far are undirected. ▷ In undirected graphs. Discrete mathematics - Graph theory and algorithms. credits. h + h. 1q. Teacher(s): Delvenne Jean-Charles ; Blondel Vincent ;. Language.
Here are some statements which employ both types of quantifiers. It's usually helpful to begin 1. Surely, "There exists a real number which does not have a real square root.
Specifically, The negation of "For all something, p" is the statement "There exists something such that -p. We won't use this notation in this book, but it is so common that you should know about it. What may I assume? In our experience, when asked to prove something, students often wonder just what they are allowed to assume. For the rest of this book, the answer is any fact, including the result of any exercise, stated earlier in the book.
This chapter is somewhat special because we are talking "about" mathematics and endeavoring to use only "familiar" ideas to motivate our discussion. In addition to basic college algebra, here is a list of mathematical definitions and facts which we assume and which the student is free also to assume in this chapter's exercises.
The product of nonzero real numbers is nonzero. The square of a nonzero real number is a positive real number.
The product of two even integers is even; the product of two odd integers is odd; the product of an odd integer and an even integer is even. A real number is rational if it is a common fraction, that is, the quotient 'n of integers mnand n with n 4 0. A real number is irrational if it is not rational. For example, 7r and V5 are irrational numbers.
An irrational number has a decimal expansion which neither repeats nor terminates. This is false. The contrapositive of the implication in Pause 2 is "If 1, then 42 7 The contrapositive of the implication in Pause 3 is "If 42 5 16, then 4 1. These answers are the same as in Pauses 2 and 3. This is always the case, as we shall see in Section 1. You are urged to read the final paragraphof this section-What may I assume? Classify each of the following statements as true or false and explain your answers.
Write down the negation of each of the following statements in clear and concise English. Do not use the expression "It is not the case that" in your answers. Every integer is divisible by a prime. Write down the converse and the contrapositive of each of the following implications. Rewrite each of the following statements using the quantifiers "for all" and "there exists" as appropriate.
Is it possible for both an implication and its converse to be false? Explain your answer. The number x is called a counterexample to Statement 4; that is, a specific example which proves that an implication is false.
M. O. Albertson and J. P. Hutchinson
There is a very important point to note here. To show that a theorem, or a step in a proof, is false, it is enough to find a single case where the implication does not hold.
However, as we saw with Statement 3, to show that a theorem is true, we must give a proof which covers all possible cases. Theorems in mathematics don't have to be about numbers. For example, a very famous theorem proved in asserts that if G is a planar graph, then g can be colored with at most four colors. The definitions and details are in Chapter This is an implication of the form "A -X 3," where the hypothesis A is the statement that g is a planar graph and the conclusion 'B is the statement that g can be colored with at most four colors.
The Four-Color Theorem states that this implication is true. M For students who have studied linear algebra. The statement given in Pause 8 is a theorem in linear algebra; that is, the implication is true.
Case i: Case ii: Case iii: Case iv: This is similar to Case iii, and the result follows. Note that x: Case 1: 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.
Now suppose that A is invertible. This is false. Since a and b are integers, it follows that a2 2: There are two possibilities: We offer a proof by contradiction. This contradicts the fact that b is not rational. Thus, the original assumption has led us to a contradiction. If n is a positive integer, then n The result is false. A square and a rectangle which is not a square have equal angles but not pairwise proportional sides.
The right hand side is divisible by the prime p, hence f pn is divisible by p. Suppose all the digits occur just a finite number of times. Then there is a number n1 which has the property that after n1 digits in the decimal expansion of tt, the digit 1 no longer occurs. Similarly, there is a number n2 such that after n2 digits, the digit 2 no longer occurs, and so on. Let N be the largest of the numbers n1, n2, Then after N digits in the decimal expansion of 7r, the only digit which can appear is O.
This contradicts the fact that the decimal expansion of 7r does not terminate. Chapter 0 9 We have proven in the text that J2 is irrational. Chapter 0 Review 1. If ab is an integer, then a and b are integers. If ab is not an integer, then either a or b is not an integer.
There exist integers a and b such that ab is not an integer. If x2 is an even integer, then x is an even integer. If x2 is an odd integer, then x is an odd integer. There exists an even integer x such that x2 is odd.
Every graph which can be colored with at most four colors is planar. Every graph which cannot be colored with at most four colors is not planar. There exists a planar graph which cannot be colored with at most four colors. A matrix which equals its transpose is symmetric.
If a matrix does not equal its transpose, then it is not symmetric. There exists a symmetric matrix which is not equal to its transpose. A set of at least n vectors is a spanning set. A set of less than n vectors is not a spanning set. There exists a spanning set containing less than n vectors.
There exists an x The contrapositive of A is false since A is false. It is true. The conclusion would have us believe that every two real numbers are equal. This means that n must be odd. Thus b2 - 5a is the difference of even integers and hence even as well.
The rectangle that remains has dimensions 1 by 7 - 1. Suppose, to the contrary, that I b mtegers. We use a proof by contradiction which mimics that proof of the irrationality of v'2 given in Problem 8.
This says that b is also a multiple of 3, contradicting our assumption that a and b have no factors in common.
Section 1. On a standard checker board, there are 32 squares of one color and 32 of another. Since squares in opposite comers have the same color, the hint shows that our defective board has 32 squares of one color and 30 of the other. Since each domino covers one square of each color, the result follows. Exercises 1. We construct a partial truth table. We must show that the given "or" statement can be both true and false. We construct truth tables for each part of the "or" and show that certain identical values for the variables make both parts T so that the "or" is true and other certain identical values for the variables make both parts F so that the "or" is false.
We are given that A is false for any values of its variables. Since A is always false, A t 13 is always true. So it is a tautology. Since A is false and the tautology 13 is true for any values of the variables they contain, 13 t A is always false. So it is a contradiction. Thus these statements have different truth tables and hence are not logically equivalent. The truth table shows that p y: So the truth values of p y: Thus these statements are logically equivalent.
Since p V [-, p A q ] is true for all values of p and q, this statement is a tautology. By associativity, this is logically equivalent to [ So the negation is a contradiction. By associativity, no further parentheses are required here. So the given statement is equivalent to By commutativity, this is The given statement is a tautology!
So these are both logically equivalent to -,p V q. This requires four rows of a truth table. Commutativity still holds. Just one distributive law holds. The truth table shows that p y.. Finally, neither law of absorption holds: The definition permits just a single midterm. August De Morgan was born in India in , died in England in J and lived his life without the sight of his right eye, which was damaged at birth.
He was apparently quite a man of principle. He refused to study for the MA degree because of a required theological exam and twice resigned his chair at University College, on matters of principle. He was Section 1. De Morgan's mathematical contributions include the definition and introduction of "mathematical induction", the most important method of proof in mathematics today.
See Section 5. His definition of a limit was the first attempt to define the idea in precise mathematical terms. Nonetheless, it is the area of mathematical logic with which De Morgan's name is most closely associated. De Morgan also developed a system of notation for symbolic logic that could denote converses and contradictions.
Graph and Graph Models - Discrete Mathematics
In row one, the premises are true but the conclusion is not. The argument is not valid. In row three, the premises are true but the conclusion is not. Alternatively, we can proceed as follows. Assume that the argument is not valid.
This means that we can find truth values for p, q, r and s such that the premises are true but the conclusion is false. But this means both r and q are false. But then q V Hence we have a contradiction, so the argument is valid. The second and third premises give q by Modus Ponens. Together with the first premise double negation and dijunctive syllogism , we get. 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. 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.
GraphTheory.pdf - GRAPH THEORY PROJECT Discrete Mathematics...
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. 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. 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. 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. Not antisymmetric because! Not transitive.
We have! However, a rf c because Since the relation is not transitive, it is not an equivalence relation and it is not a partial order. This is not an equivalence relation because it's not reflexive or symmetric. This is not a partial order because it's not reflexive or antisymmetric. Since the hypothesis is always false, this implication is true.
Discrete Mathematics With Graph Theory (3rd Edition)
The relation is antisymmetric. Thus bRa. Thus aRc. The relation is not anti symmetric, so it is not a partial order. Then t rv x so t rv a by transitivity. Thus tEa. Since x E x, by symmetry, x E a. Thus x rv a. Now d tj. S is reflexive, antisymmetric and transitive on A. The relation is reflexive: For any a, b E A, a, b rv a, b because a It is antisymmetric: It is transitive: Thus q rv p.
Suppose p rv q and q rv r. If the points p, q, r are different, then the line through p and q passes through the origin, as does the line through q and r.
Since the line through the origin and q is unique, p and r lie on this line, p rv r. The equivalence classes are lines through the origin.In every case, the conclusion is true.
This is Fermat's Last Theorem 4. Now -,p follows by modus tollens. Then n has no prime factor smaller than v'9. For example, 0,7 E R: Adhitya Kamakshidasan. Not transitive.