Discrete mathematics with graph theory pdf

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PDF | On Apr 1, , Bhavanari Satyanarayana and others published Discrete mathematics and graph theory. 2nd ed. 4 Traversal: Eulerian and Hamiltonian Graphs. 5 Graph Optimization. 6 Planarity and Colorings. MAT (Discrete Math). Graph Theory. Fall 2 / The notes form the base text for the course ”MAT Graph Theory”. computational methods given by the mathematical combinatoric and linear- algebraic.

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Library of Congress Cataloging-in-Publication Data. Goodaire, Edgar G. Discrete mathematics with graph theory / Edgar G. Goodaire, Michael M. Parmenter Download as PDF, TXT or read online from Scribd . As a textbook used for discrete mathematics and graph theory, there are many topics which must be. 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.

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. Hence, a '" c. Planar Graphs and Colorings The contrapositive of A is false since A is false. There Are Proofs!

P2 by C and so on until we replace Pn by C. Suppose A and 3 are logically equivalent statements involving variables PI. Show that t -[ -p S. C are statements. In Problems 14 and 15 we used. The fact that we didn't really think about this at the time tells us that the theorem is easily understandable and quite painless to apply in practice. There are times.. I A primary application of the work in this section is reducing statements to logically equivalent simpler forms U The next problem illustrates clearly why employing the basic logical equivalences discussed in this section is often more efficient than working simply with truth tables.

Associativity and idempotence.. Express Solution. A which is in disjunctive normal form on the variables p.. In this way. A am2 A. This statement is logically equivalent to p A q V r A -q. As shown in Example A compound statement based on variables xI. We construct a truth table A a2n V. This gives the minterm -p A q A r.

V aml X A amn where. In row 4. One reason is that the minterms. We leave it to you to decide for yourself which method you prefer. This could be done with truth tables. Since A 3. A -q A r omitting the second occurrence of -p A q A r at the last step. A and 3 have the same truth tables. In applying the distributive property. J3 V C for any other statement C. The claim is that if A 9 then A v C 3. A similar argument shows that if A 4z. This is just a restatement of Property Disjunctive normal form is useful in applications of logic to computer science.

Different methods of proof were discussed informally in Section 1. An called premises or hypotheses followed by a statement 13 called the conclusion. Express each of the following statements in disjunctive normal form. What simpler statement is logically equivalent to both of them? Such an argument is valid if. An are all true. Taken in its entirety.

Using the properties in the text together with the absorption properties given in Exercise 3. Verify each of the 13 properties of logical equivalence which appear in this section [BB.

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I S An argument is a finite collection of statements Al. Simplify each of the following statements. Using truth tables. Which of the following are in disjunctive normal form on the appropriate set of variables? Find out what you can about Augustus De Morgan and write a paragraph or two about him. Now we relate these ideas to some of the more formal concepts introduced in Sections 1.

In Exercise 10 of Section 1. The steps required in the proof generally consist of showing that if certain statements are true. For each of the properties discussed in this section including those of absorption given in Exercise 3 determine whether or not the property holds with V replacing v wherever it occurs [BB. Either I study biology or I fail the course. If I like biology. In this row. If I fail the course. Determine whether or not the following argument is valid.

We construct a truth table. In particular. Thus -p is true as desired. Show that the argument p r -q q r -p is valid. Arguments can be shown to be valid without the construction of a truth table. Thus -q is false and.

Assume that all premises are true. Thus the argument is valid. I In Problem In general. B is a tautology. Let p be "I like biology..

Pm are replaced by statements C I. In the same spirit as Theorem The argument is not valid. This can be analyzed by a truth table. Rules of Inference Because of Theorem 1. Here is a list of some of the most common rules of inference. AA is true. In row I The theorem which follows relates the idea of a valid argument to the notions introduced in Sections We illustrate how the rules of inference can be applied.

Modus ponens: Property 13 of logical equivalence as given in Section 1. Modus tollens: Show that the following argument is valid. One of the laws of De Morgan and the principle of double negationsee Section 1. Chain rule: Disjunctive syllogism: The chain rule now tells us that our argument is valid. If a truth table were used to answer Problem If I do not go to a movie. Determine the validity of the following argument. Verify that each of the five rules of inference given in this section is a valid argument.

Determine whether or not each of the following arguments is valid. While this can be shown with a truth table. There are five variables. You are encouragedto use the result of any exercise in this set to assist with the solution of any other. Verify that each of the following arguments is valid.

If I study. Since -q is true. I went to a movie. Hence -p is true and we are done. Let p. I failed. If the argument is one of those listed in the text. If I stay up late at night. Either I wear a red tie or I wear blue socks. If I work hard. I did not stay up late last night. If I don't graduate. I am not wearing a red tie. If I like mathematics. Either I don't study or I pass mathematics.

I am not tired this morning. If I pay high taxes. I am wearing pink socks. If I earn lots of money. I am wearing blue socks. I am wearing a red tie. I stayed up late last night. Test the validity of each of the following arguments. If I do not work hard. I will not study. I like football. If I don't study. If I graduate. I am tired this morning. Either I like mathematics or I like football. Either I study or I like football. Determine the validity of each of the following arguments.

If I like football. A r V p V q A -p in disjunctive. Explain 4. Prove that n3 is odd if and only if n is odd. Determine the truth value of [p V q. Construct a truth table for the compound statement [p A q - -r ] -[ -q V r]- 9. Write down the converse and contrapositive of each of the following implications. Exercise 8 a. Let n be an integer greater than 1. With a proof by contradiction.

Establish the logical equivalence [ p [ p V r A - q A -r ]. Express p V q normal form. Translate these expressions into English and explain. What can you conclude about A and l3? Discuss the validity of the argument p A q -'p A r Purple toads live on Mars.

Either I wear a green hat or I do not wear blue socks. If I don't pass mathematics. Either I wear a red tie or I wear a green hat. Review Exercises Determine whether or not each of the following argu ments is valid. Most of mathematics is based upon the single undefined concept of set.

Primitive humans discovered the set of natural numbers with which they learned to count. What would it be like to delve into a dictionary if you didn't already know the meanings of some of the words in it? Most people. Mathematics is the same way. There are a few basic terms which we accept without definitions. Some reflection indicates that a dictionary can be of no use unless there are some words which are so basic that we can understand them without definitions.

Without constant review of the meanings of words.

It is. We advise readers of this book to maintain and constantly review a mathematical vocabulary list. Students of French or Spanish know that memory work is a fundamental part of their studies. The set of natural numbers. The expression "k E N" is read "k belongs to N.

Our convention. The set of all real numbers is denoted R or IR. J is the set of natural numbers. The set of common fractions-numbers like 3. There are various ways to describe sets. We might describe this set by.

The full set of integers. For our purposes. Sometimes it is possible to list the elements of a set within braces. To define the real numbers properly requires considerable mathematical maturity Sometimes people are surprised to discover that a set can be an element of another set.

An irrational number is a number which cannot be written in the form nn with m and n both integers. True or false? The unique set which contains no elements is called the empty set. Such numbers are called irrational. In addition to the rational numbers. Equality of Sets a 0 Sets A and B are equal.

Set theorists originally used 0 zero to denote this set. These sets are all equal since none of them contains any elements. Upon encountering it. When A C B. We occasionally see "B D A. If 0 C A is false. For this reason. Proof If a E A. This is an absurdity since there is no x E 0. The proof that 0 C A is a classic model of proof by contradiction.

S For any set A. I True or false? Note the distinction between membership. This statement is false: The empty set is a subset of any set. By the former statement. Both statements are true. List the distinct elements in each of the following sets: I 0 The power set of a set A. There is just one element in the set namely. Is x C A also possible? Find the power sets of each of the following sets: List five elements in each of the following sets: Determine which of the following are true and which are false.

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Determine whether each of the following statements is true or false. Suppose A. Prove or give a counterexample which disproves each of the following assertions. Justify your answers. How many elements are in P P A? For each of the following statements either prove it is true or give a counterexample to show that it is false. Are the following statements true or false? Suppose A and B are sets. While parentheses are not required in expressions like 1 or 2.

The intersection of A and B. The two different interpretations corresponding to different insertions of parentheses agree. The intersection of AI. As with addition and multiplication of real numbers.

The last term-An-indicates that the last set in the union is An. This is probably most easily seen by the use of the Venn diagram shown in Fig 2. The union of n sets Al. U An that n is actually greater than 3 since the first part of this expression-AI U A2 U A 3-is present only to make the general pattern clear. The diagram indicates that A consists of the points in the regions labeled 1. While pictures can be helpful in making certain statements seem plausible.

In Fig 2. Here this just amounts to expressing the meaning of U and n in words. Venn diagram. A sets construct a specific counterexample. Let A. Verify equation 3 without the aid of a Solution. As observed in Proposition 2. These are just the points of A n B U C as observed previously. Notice that A n B U C consists of the points in regions 2. B consists of those points in regions 3.

B and C be sets. The set B U C consists of points in the regions labeled 3. The region A n B consists of the points in regions 3 and 4.

This suggests cases. The diagram enables us to see that. In either case. I For sets A and B. By definition of union. Here x is in A as well as in C. For sets A and B.

To prove the equality of A n B and A. This completes the proof. You may have previously encountered standard notation to describe various types of intervals of real numbers. By definition of intersection. This proves A C B. For this. Since A C B. To describe infinite intervals. We illustrate by showing the equivalence of the first law of De Morgan and the rule for negating "A or 3.

The Four-Color Theorem. The first interval here is half open. What is AC? The following two laws. As indicated. See Section Readers should be struck by the obvious connection between these laws and the rules for negating and and or compound sentences described in Section 1. A square bracket indicates that the adjacent end point is in the interval. Let A be the statement "x c A" and 93 be the statement "x C B. A eD B A C also consists of the points in regions 1.

More generally. Notice that the symmetric difference of sets can be expressed in terms of previously defined operations.

The Cartesian Product of Sets There is yet another way in which two sets can be combined to obtain another. Use a Venn diagram to illustrate the plausibility of the fact that G is an associative operation. Regions 1. See Exercise 21 of Section 5. A D B consists of the points in the regions labeled With reference to Fig 2. I As a consequence of 4. More generally.. The elements of A' are called n-tuples.

This example illustrates that. The elements a and b are the coordinates of the ordered pair a. The elements of A x B are called orderedpairs because their order is important: Since the elements in A x B U C are ordered pairs.. It means that x.

I 4 Rene Descartes The adjective "Cartesian" is derived from Descartes. Elements of A x B are equal if and only if they have the same first coordinates and the same second coordinates: List the elements in each of the following sets.

Let A and B be nonempty sets. If the latter. The reverse subset relation was established in Problem Then x is in A U B by definition of U. What do you observe? E A for. So we begin by letting x. Sox e B. To prove the sets A U B and B are equal. So suppose x c A. What do you notice? We conclude that the two sets in question are equal. Then x is certainly in A U B. Since B: Then x is either in A or in B. In the first case.

Every prime is both a natural number and an integer. M denote the set of math majors CS denote the set of computer science majors T denote the set of students who had a test on Friday P denote those students who ate pizza last Thursday. Justify your answer. Find each of the following sets. What can you conclude? Using only the set theoretical notation introduced in this chapter.

Every prime except 2 is odd. Using only the set theoretical notation we have introduced in this chapter. Find equivalent formulations of each of the following statements using the notation of set theory that has been introduced in this section. What set is represented by the region labeled 2? By that labeled 3? By that labeled 4? Z and N denote the integers and natural numbers.

Find a. Let P denote the set of primes and E the set of even integers. The universal set for this problem is the set of students attending Miskatonic University. Use the first law of De Morgan to prove the second: As always. A picture of A in the xy-plane might help. List the elements in the sets A a. Explain your answers. Find out what you can about George Boole and write a paragraph or two about him in good English.

There are occasions when we are interested in a certain subset of A x B. If A is the set of students who were registered at the University of Toronto during the Fall semester and B is the set [History. A binary relation on A is a subset of A x A. OD Hank Aaron. I S Let A and B denote sets. Explain your answers with Venn diagrams if you wish.

Let A and B be sets. A binary relationfrom A to B is a subset of A x B. For each of the following. If xy 7. This example reminds us that a reflexive relation must all pairs of the form a. Most is not enough. EE A binary relation 7? The statement " a. Suppose 7? A Suppose R is a binary relation on A. Criticize and then Solution. Our primary intent in this section is to identify special properties of binary relations on a set. To prove that 7?

A binary relation on the real numbers or on any subset of R is symmetric if. I If a set A has n elements and n is reasonably small. This relation is reflexive all points on the main diagonal-top left corner to lower right-are present. A Figure 2. For any x. A Note that "antisymmetric" is not the same as "not symmetric.

Let A denote the set of names of streets in St. A binary relation 7? Determine which of the properties reflexive. Is the relation? Interpret the Cartesian product A x B. Newfoundland and B denote the names of all the residents of St. The answer is yes. Give a sensible example of a binary relation from S to B.

Interpret the Cartesian product S x B. Give a sensible example of a binary relation from A to B. If it is yes. Since now we have both a. Let A be the power set of S. List the ordered pairs in a relation on A which is d. Then for any b with a. Is K reflexive? Is it possible for a binary relation to be both symmetric and antisymmetric? If the answer is no. If a relation has a certain property.

Give a proof or counterexample as appropriate. Is KZ reflexive? Answer Exercise 8 for each of the following relations: Let A be the set of books for sale in a certain university bookstore and assume that among these are books with the following properties.

Let S be a set which contains at least two elements a and b. Determine whether or not each of the binary relations K defined on the given sets A are reflexive. With a table like that in Fig 2. Determine which of the properties-reflexivity. Suppose K is a symmetric and transitive relation on a set A and let a E A. The student should mentally confirm that.

For a. These three particular relationships are reflexive. Such relationships occur everywhere. A Congruence mod 3 5 Define. Certainly a. Two people may be of the same sex. Section 4.

Since 7? It may be because of examples like this that it is common to say "a is related to b. If R is a binary relation on a set A and a. S S An equivalence relation on a set A is a binary relation 7? Call a and b equivalent if a and b are residents of the same state or district. In this section.

All drop-off points in a given neighborhood of town may be the "same" to the driver of a newspaper truck. An equivalence relation changes our view of the universe the underlying set A. A Let A be the set of all people. The groups into which an equivalence relation divides the underlying set are called equivalence classes. The set of all equivalence classes is called the quotient set of A mod.

The set of things related to a is the same as the set of things to which a is related. The equivalence relation "same parents" groups people into families.

For the equivalence relation in Example Since an equivalence relation is symmetric. Surely the three most fundamental properties of equality are reflexivity: For this equivalence relation. Little children may think of their brothers and sisters as the same and other children as "different. We have in mind a certain characteristic or property of elements and wish only to consider as different. The equivalence class of an element is the collection of all things related to it.

While it is reflexive and transitive. The set of all students has been grouped into. What is 1? The residents of Colorado. The even integers. How many elements does the quotient set contain? For general a. The relation is reflexive. The equivalence class of a. The relation is symmetric. The quotient set is the set of forty-nine American states on the North American continent.

In Example For x. So this equivalence class is the circle with center. We shall denote this set 3Z. What are the equivalence classes for the equivalence relation which is congruence mod 3? What is 0. Prove that. If a person is a resident of one state. Since x C a-.

Then there is an element x E a-n b. Since we also have a. First suppose y c Y. The cells are said to partition A. We know x E x because x. We say that the equivalence classes "partition" A or "form a partition of" A. Canada is partitioned into ten provinces and three territories. Suppose a-n b: It is the implication which is the substance of this proposition.

We must prove that the two sets x and a. In each of the examples of equivalence relations which we have discussed in this section.

In conjunction with Proposition 2. These examples are suggestive of a result which is true in general. Then y. A student number cannot begin with 79 and also with Then for any x E A. Proof Suppose a 0 b. The word partition is used as both verb and noun. These disjoint sets are called cells or blocks.

Not only does an equivalence relation determine a partition. Students are partitioned into groups according to the first two digits of their student numbers. The human race is partitioned into groups by eye color. The equivalence classes associated with an equivalence relation on a set A form a partition of A. The suits "heart. Describe the corresponding equivalence relation on a deck of cards.

The partition of the integers into "evens" and "odds" corresponds to the equivalence relation which says two integers are equivalent if and only if they are both even or both odd. It sure does. The correspondence between equivalence relations and partitions provides a simple way to exhibit equivalence relations on small sets. A We have seen that the equivalence classes of an equivalence relation on a set A are disjoint sets whose union is A.

A deck of playing cards is partitioned into four suits. The quotient set has 51 elements. Since every integer r is either a multiple of 3. Then either a and b are residents of the same U.

The equivalence class of a consists of those people equivalent to a in the sense of -. If a does not live in any state of the United States. Two cards are equivalent if and only if they have the same suit.

Let A be the set of all citizens of New York City. What is the usual name for this equivalence relation? If a does live in a U. The equivalence class of 0. It follows that either all three of a. Show that. What's the equivalence class of -5? Find the equivalence class of 0. How many are there altogether? Find the equivalence class of 2.

For integers a and b. Make a guess about the quotient set. Describe geometrically the equivalence class of 1. There are 52 partitions of a set of five elements. For natural numbers a and b.

The number of partitions of a set of n elements grows rather rapidly. What is the equivalence class of 5 2 9 c What is the quotient set determined by this equivalence relation? Describe geometrically the equivalence class of 0. We mentioned that we view equivalence as a weak form of equality and employed a symbol. In Exercise Prove that [SI.

Suppose that for each i. Determine which of the following define equivalence relations in R2.

Discrete Mathematics With Graph Theory (3rd Edition)

For those which do. Assume a. Si i Uj5. A partial order on a set A is a reflexive. I be a collection of subsets of S with the property that a. A partially ordered set. For any set S. By "word. A The adjective "partial" This is the situation where word a. Here then are a few examples of partial orders. This ordering of words is called lexicographic because when the basic alphabet is the English alphabet.

If X and Y are subsets of a set 5. Segal of the ambiguous position in which Hasse found himself during the Nazi period. There is a fascinating account by S. Suppose that in some Hasse diagram. Two Hasse diagrams are shown in Fig 2. The article. The effect of the last property here is to remove redundant lines. Partial orders are often pictured by means of a diagram named after Helmut Hasse 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.

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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. 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. S band b:: There is no maximum element. Section 3. Two elements of a poset can have at most one least upper bound and here's why. Interchanging i1. Exercises 3. There's no pair of the form 3, -. There is more than one pair of the form 1, -: In fact, there are four! If y E Y, then y is a person living in one of the countries in the British Commonwealth. Thus, the domicile of y is a uniquely determined element x E X.

On the other hand, Prime Minister is not onto since there are people in a country who are not the Prime Minister. If y is any person living in that country, then Domicile: On the other hand, if Y1 and Y2 are different people living in country z, then Domicile: Y1 ff X and Domicile: Y2 ff x, so Domicile is not one-to-one. We know that r x 1 is an integer greater than or equal to x and that r y 1 is an integer greater than or equal to y.

This gives the result. Neither is 9 onto: For any x E Z, Ixl2: Thus, 1 is one-to-one, but it is not onto: But it is onto since for any y 2: The function is, therefore, onto. It is also one-to-one: Both Xl and X2 are nonnegative. Both Xl and X2 are negative. One of Xl, X2 is nonnegative and the other is negative. Since the left side is at least 0 and the right side is less than 0, the equation cannot be true. R, 1 is not onto. Since f: Solution 2. This solution mimics that given in Problem 8 in our discussion of discrete functions in this section.

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On the other hand, f is not onto. But the only pairs of integers whose product is 1 are the pairs 1,1 and -1, We must prove that f is one-to-one.

B were onto, then there would exist elements aI, a2, Assume that f: Therefore, f is onto. Conversely, suppose f is onto.

Again there is no contradiction since N is not finite. Since y E domf, y ; Since y E domf, y:: Then 5 - Thus f is one-to-one. If Xl and X2 are both ; Next we show that f is onto. Suppose that n E N. Since f is one-to-one and onto, it has an inverse. It contains both 1,2 and 2, 2 ; 9 is not onto: We call fog an even function since it is symmetric with respect to the y-axis: In particular, the graph of go f lies entirely on or below the x-axis.

Let then a be some element of A. Consider g b E C. Since 9 0 f: We prove that 9 0 f: Let then c E C. Then 9 0 f: Thus 1 IS one-to-one. Now it is straightforward to see that t is one-to- one: The left side is an integer; hence, ""- so is the right. Thus m.

To show that g is onto, let a, b E N x Z. By part a , the function g: This is not a partial order because it is not antisymmetric. This time, in each case, 8, S is a partial order. Certainly A S Band B S C, then A If A and B have the same cardinality, then there is a one-to-one onto function f: Such a function has an inverse r: B -t A which is one-to-one and onto because it has an inverse , so B and A have the same cardinality. Then there is a one-to-one onto function f: A -t B and a one-to-one onto function g: Since the composition of one-to-one functions is one-to-one and the composition of onto functions is onto Exercises 23 and 24 of Section 3.

A -t C is one-to-one and onto. Thus, A and C have the same cardinality. By transitivity of "same cardinality", 1,00 and 3,00 have the same cardinality too. Thus go f-1 provides a one-to-one correspondence a, b -t c, d.

Thus, the composition 9 0 x f: In Problem 27, p. Furthermore, the function f: See the remarks preceding Problem Thus 9 0 f: Thus, f is one-to-one.

To show that f is onto, let r E R. Start at 0,0 and move as illustrated. We include an extra row across the bottom as follows: The first few terms would be 0,0 , 1,0 , 0,1 , 0,2 , 1,1 , 2,0 , 3,0 , 2,1 , 1,2 , 0,3. Just follow the sequence given in the text for the set of all positive rationals, but omit any rational number not in 1,2. In fact, it contains at most elements since there are 99 possible numerators and, for each numerator, 99 possible denominators.

In Exercise 3 c we showed it is in one-to-one correspondence with N x N. The interval [-1,1] contains 0,1 which we showed in the text to be uncountable. There is some minimum volume V which a grain of sand must occupy. On the other hand, there is a finite value M for the volume of all the sand. So the number of grains is Here is a listing: So we have to decide if the set of sentences is countably infinite or uncountable.

In fact, it is countably infinite. If ak,l, ak,2, ak,3, To the contrary, suppose that the union were a finite set 8. Since 8 has only finitely many subsets the precise number is , there could not have been infinitely many sets at the outset. The maximum number of elements in the union of sets AI, If even one infinite set is contained in a union, then the union must be infinite. Rays emanating from this common point establish a one-to-one correspondence between the points on 81 and the points on Define f: Then f is a one-to-one correspondence.

We employ a concept known as stereographic projection. Imagine the sphere sitting on the Cartesian plane with south pole at the origin. Any line from the north pole to the plane punctures the sphere at a unique point and the collection of such lines establishes a one-to-one correspondence between the points of the plane and the sphere except for the north pole. A small modification of this correspondence finishes the job.

Suppose PO,Pl,P2, Map the north pole to 0,0 , the origin to 1,0 , PI to 2,0 , and so forth and let all other points of the sphere go to the same points as before. Then A U B is countably infinite because it is infinite and its elements can be listed as follows: The function f: Since A is countable, we can list its elements: Since B is countable, we can list its elements: Thus, f is a one-to-one correspondence. Thus the given set of polynomials is countable too. Thus, for each x E X, f x is a subset of X.

Define a subset Y of X as follows: On the other hand, if y rJ. We reach the absurd situation that y is neither in or not in Y. It follows that there can exist no onto function f: Define the sequence bi, bz, b3, This contradiction gives the result.

Assume, to the contrary, that S is countable and that its elements can be listed al,a2,a3,.. Note that the choice of al guarantees that b -I- 0. The Fields Medal is the highest honour which can be bestowed upon a mathematician, there being no Nobel Prize in mathematics.

One of the three Canadian mathematics institutes is also named after Fields. Michael Monastyrsky wrote an informative article on the history of the Fields Medal for the March Notes of the Canadian Mathematical Society 33, no. More information can be also obtained from http.: Chapter 3 Review 1. This function is not onto because 1 is not in the range of I.

This is not true. Yes, I is one-to-one. I is not bijective since it is not onto. I is onto because if b is any rational numb. I' bii. IS ijective since It IS one-to-one an onto: To find the range, suppose y is in the range.

This leads to x 2y Certainly, the identity function is reflexive. Conversely, if lis reflexive, then for any a E A, a, a E f.

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Lots offunctions R - R are symmetric, for example, the function defined by! The greatest integer function is one example. There is no x. Since go! Hence e, d '" a, b. The function is one-to-one because if! Clearly f is onto, so it is a one-to-one correspondence. For each i ; Since each bi is 2 or 7, b is in 8. But for any i, we have b i- ai since b and ai differ in the ith decimal place.

This contradicts 8 being countable. We conclude from the first remark at the end of Section 3. Since 8' contains the infinite sequence 0. First assume 8 is finite with n elements Xl, X2, In the case that 8 is infinite, we offer a proof by contradiction.

Suppose then that f is onto. Thus every subset of 8 is of the form f x , for some X E 8.

Reflecting on the question "is x E T? Exercises 4. Section 4. If a and b are real numbers, certainly a - b is a real number. Case 3: Multiplying by the negative number b, we obtain a The parenthetical comment follows from the fact that N has n digits in base b. Since the gcd of any two elements of this poset is, in each case, still an element of the poset, every pair of elements has a glb.

The lcm of 4 and 6 is Since 12 is not in the poset, 4 and 6 have no lub. There are no minimum nor maximum elements. There is no maximum. We derive a contradiction. Since y2 - x is an integer, this says 4 I 2, which is impossible. In either case we reach a contradiction, so 4 I n2 - 2 can never be true. Thus, n 1 a - b as required. Since -xy is an integer, ale.

Since xc is an integer, a 1 be. Since xy is an integer, ae 1 bd. Now, since b 1 e , ab 1 rae and since ale , ab 1 sbe.

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The pairs 93, , , , , and , are each relatively prime. Then, by Proposition 4. Again, by Proposition 4. The result follows. Suppose this is the case. As in the proof of Lemma 4. We prove that 91 S 92 and 92 S Thus, 91 divides both a and b, hence, 91 S 92, the greatest common divisor of a and b. The left side of this equation is even while the right side is odd, a contradiction. Since the left side is divisible by 4, the right side would have to be divisible by 4, which is not the case.

By Theorem 4. First note that e91 I ae since 91 I a, and, similarly, e91 I be. Since x I band b I e, we have x I e by transitivity of I.

Thus the smallest k such that n I ks is the smallest k such that n1 I ks1. First we prove that 9 1 a. Similarly, 9 1 b. Similarly, b 1 r so that r is a common multiple of a and b.

See Proposition 4. Similarly, it is a multiple of b. Similarly, 9 I b. Since also 9 I c, we see that 9 is a common divisor of a, b, c. Next suppose that d is any common divisor of a, b, c. Since in particular, d is a common divisor of a and b, d I 91 by Corollary 4. Since also d I c, we see that d is a common divisor of 91 and c, so d:: This proves that 9 is the largest of the common divisors of a, b and c.

Thus, Xl,X2 j Xl,X2. Suppose Xl. X2 and YI. Y2 and yI. Y2 j Xl, X2. Then Xl jl YI. X2 , YI. X2 j YI. Y2 and Yl, Y2 j ZI. Thus, Xl. X2 j Zl, Z2. Then Z x Z is not totally ordered since, for example, 1,2 and 0,3 are not comparable. Now we show that any two elements a, b E A have a glb and a lub. Certainly lcm a, b has the properties required of a least upper bound and gcd a, b of a greatest lower bound.

The key point is to show that these elements lie in A, and this follows from the definitions of gcd and lcm. Finally we must show that a' and b' are unique. Now Section 4. The uniqueness of a' and b' follows from the uniqueness of a and b. Uniqueness can also be proved directly as in i.

Then n has no prime factor smaller than v'9. Since a prime factor of nip is a prime factor of n, q would then be a prime factor of n not exceeding VnlP, contradicting the fact that the smallest prime factor of n is bigger than this. Since , the smallest prime dividing 38,, is larger than , 38, is prime, by part a. We must prove that rv is reflexive, symmetric and transitive.