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# Mathematical proofs chartrand pdf

Gary Chartrand More on Direct Proof and Proof by Contrapositive Proofs Prove or Disprove Conjectures in Mathematics Revisiting Quantified. Names: Chartrand, Gary, author. | Polimeni Title: Mathematical proofs: a transition to advanced mathematics / Gary Chartrand,. Western .. Presentation slides in PDF and LaTeX formats have been created to accompany. Gary Chartrand is Professor Emeritus of Mathematics at Western Michigan University. He received his Ph.D. in mathematics from Michigan State University. Author: JENI MATHYS Language: English, Spanish, Arabic Country: Slovenia Genre: Environment Pages: 601 Published (Last): 08.12.2015 ISBN: 420-5-51349-770-7 ePub File Size: 24.51 MB PDF File Size: 11.82 MB Distribution: Free* [*Regsitration Required] Downloads: 43406 Uploaded by: FRITZ

Mathematical Proofs. A Transition to. Advanced Mathematics. Gary Chartrand. Western Michigan University. Albert D. Polimeni. State University of New York at . Mathematical Proofs - 3rd Edition - Chartrand - Ebook download as PDF File .pdf ), Text File .txt) or read book online. textbook. MATHEMATICS. Third Edition. SEYMOUR LIPSCHUTZ, Ph.D. Temple University. MARC LARS LIPSON, Ph.D Sc Third Edition Mathematical Proofs A.

Then 3x3 is even if and only if x is even. Congruence Modulo n 8. If 2 3a, then 2 a. Similarly, d R c. A real number r is irrational if and only if r has a nonrepeating decimal expansion. That is.

Answer for Exercise 2. True 2. If 17 is even, then 19 is prime. If 2 is rational and 23 is rational, then 3 is rational. If 2 is rational and 32 is rational, then 3 is not rational. If 2 is not rational and 23 is rational, then 3 is rational. If 2 is rational or 32 is rational, then 3 is not rational. True Exercises for Section 2. If 13 is prime, then 15 is prime. If 33 is prime, then 43 is prime. True for all x. That x is odd is a necessary and sufficient condition for x2 to be odd 2. See the truth table below. If P and P implies Q, then Q. Note that the last two columns in the truth table are not the same. See the truth table. Some Fundamental Properties of Logical Equiv- alence 2.

F Figure There exists an odd integer x such that x2 is even. Characterizations of Statements 2. A real number is rational if and only if it has a repeating decimal expansion. Pythagorean theorem e Not a characterization.

Every positive number is the area of some rectangle. Additional Exercises for Chapter 2 2. At least two of the three elements a, b, and c are the same. This is impossible for a statement. Hence the statement is true trivially. Thus the statement is true trivially. Thus the statement is true vacuously. Exercises for Section 3. Let x be an odd integer. Let x be an even integer. Assume that a and c are odd integers. Assume that x is odd. If 15n is even, then n is even. Use a proof by contrapositive to verify this lemma.

Then use this lemma to prove the result. Proof of Result. Assume that 15n is even. Since 9a is an integer, 9n is even. Assume first that x is odd. For the converse, assume that x is even. The converse of the implication must also be verified. The lemma used to prove the converse depends on whether a direct proof or a proof by contrapositive of the converse is used.

One possibility is to prove the following lemma: We consider two cases. Case 1. Case 2. Assume that x or y is even, say x is even. Since ay is an integer, xy is even. By Exercise 3. For the converse of this implication, use a proof by contrapositive and consider two cases, where say Case 1.

Assume that a or b is odd, say a is odd. Since ay is an integer, ab is even. Then x2 is even if and only if x even. Case 3. Proof Evaluations 3. This can also be restated as: No proof has been given of the result itself. A proof of this result requires a proof of an implication and its converse. Nowhere in the proposed proof is it indicated which implication is being considered and what is being assumed.

If 3x3 is even, then x is even. If 5x2 is even, then x is even. Both lemmas can be proved using a proof by contrapositive. Use Lemma 1 to show that if 3x3 is even, then 5x2 is even; and use Lemma 2 to show that if 5x2 is even, then 3x3 is even.

One possible choice with a single lemma is: Then 3x3 is even if and only if x is even. Assume that x and y are of the same parity. Thus x and y are both even or both odd. Consider these two cases.

Assume that x and y are of opposite parity. Assume first that x is odd or y is even. We consider these two cases. For the converse, assume that x is even and y is odd. We then consider two cases, according to whether y is even or y is odd.

Assume that some pair, say a, b, of integers of S are of opposite parity. Hence we may assume that a is even and b is odd. There are now four possibilities for c and d. Case 4. This second case was never considered and it was never stated that we could consider the first case only without loss of generality. Assume that a and b are even integers. Assume that n is an odd integer. By Result A, n3 is an odd integer.

Solving for abc 2 gives us the desired result. Assume that a b. Since c2 is an integer, a2 b2. Assume that a b and b a. Assume that 3 m.

Since 3q 2 is an integer, 3 m2. Assume that 3 6 m. Then 3 m if and only if 3 m2. We then consider the following four cases. Use similar arguments for the remaining cases. Assume that a b or a c, say the latter. Since bk is an integer, a bc.

Use an argument similar to that in Case 1. Then consider these two cases.

## Third Edition Mathematical Proofs A Transition to Advanced Mathematics Gary Chartrand - PDF Drive

By Theorem 3. We consider these three cases. Exercises for Section 4. There are two subcases. Subcase 1. The proof is similar to that of Subcase 1. Subcase 2. The proof of each subcase is similar to that of Subcase 1. Then n is congruent to one of 0, 1, 2, 3, 4, 5, or 6 modulo 7. If n is congruent to one of 0, 1, 2, or 3 modulo 7, then n2 is congruent to one of 0, 1, 2, or 4 modulo 7 by a - d.

Three cases remain. Proofs Involving Real Numbers 4. There are three cases. It remains to verify the converse. By Theorem 4. Proofs Involving Cartesian Products of Sets 4. First, we assume that A and B are not disjoint. Since 5k 2 is an integer, 5 n2. We consider four cases.

The remaining three cases are proved in a manner similar to Case 1. Since cb is an integer, 3 ab. There are four cases. The remaining cases are proved in a manner similar to Case 1. Let n be an odd integer. We consider three cases. If a2 is even, then a is even. A change in the order of the steps in the first paragraph could make for a clearer proof.

See below. The proof is complete after the first paragraph. Exercises for Section 5. Assume, to the contrary, that there exists a largest negative rational number r. Assume, to the contrary, that can be written as the sum of an odd integer a and two even integers b and c. There are two cases. None of a, b, and c is even. Exactly two of a, b, and c are even, say a and b are even and c is odd.

The argument is similar to that in Case 1. Assume, to the contrary, that there exist an irrational number a and a nonzero rational number b such that ab is rational. Let a be an integer. Then 3 a2 if and only if 3 a. Assume to the contrary, that 3 is rational.

Since 3 p2 , it follows by the lemma that 3 p. Since x2 is an integer, 3 q 2. Assume, to the contrary, that 6 is rational.

Because 3b2 is an integer, a2 is even. Because c2 is an integer, 3b2 is even. Since 3 is not even, b2 is even and so b is even by Theorem 3. However, we saw in Exercise 5. This is impossible since there is no integer between 2 and 3. We consider two cases, according to whether x and y are of the same parity or of opposite parity.

Produce a contradiction in each case. Consider the irrational numbers 3 and 2. Hence c is a solution. Disproving Existence Statements 5. Let a and b be odd integers. Then n is even or n is odd. Consequently, there are three possibilities: Now, if the second suitor had seen a silver crown on the third suitor, then the second suitor would have known that his crown was gold; for had it been silver, then, as we saw, the first suitor would have known his crown was gold.

This meant that 1 did not occur and that the third suitor had a gold crown. Since neither the first suitor nor the second suitor could determine what kind of crown he had, only 2 or 3 was possible and, in either case, the third suitor knew that his crown must be gold. By Exercise 4.

Case 1 is not described well. It would be better if Case 1 were written as: Exactly two of x, y, and z are odd. Assume, without loss of generality, that x and y are odd and z is even. This is not the desired result. Assume, to the contrary, that the sum of the irrational numbers 2, 3, and 5 is rational. This is a contradiction. The Principle of Mathematical Induction 6.

Let S be a nonempty subset of B. We show that S has a least element. Since A is well-ordered, S has a least element. Therefore, B is well-ordered. Let S be a nonempty set of negative integers. Hence T is a nonempty set of positive integers. By the Well-Ordering Principle, T has a least element m. We proceed by induction. We use induction. The result then follows by the Principle of Mathematical Induction.

We verify this formula by mathematical induction. Exercises for Section 6. We need only show that every nonempty subset of S has a least element. So let T be a nonempty subset of S. Hence we may assume that T is not a subset of N. Since 4! Suppose that k! By the Principle of Mathematical Induction, n!

We employ mathematical induction. Assume for some positive integer k that if 3 2k a, then 3 a. By the induction hypothesis, 3 a. By the Principle of Mathematical Induction, it follows that for every positive integer n, if 3 2n a, then 3 a. Assume, for any k sets A1 , A2 ,. Assume that for some positive integer k and any 2k integers a1 , a2 ,. Now let c1 , c2 ,. By Result 4. Assume that: By the Principle of Mathematical Induction, the result is true. Certainly, the only element of a set with one element is the largest element of this set.

By the induction hypothesis, T has a largest element, say a. In either case, S has a largest element. By the Principle of Mathematical Induction, every finite nonempty set of real numbers has a largest element.

Let S be a finite nonempty set of real numbers. Proof by Minimum Counterexample 6. Let m be this integer. Let m be the smallest such integer. Thus 20 , 21 ,. We proceed by mathematical induction. The result follows by the Principle of Mathematical Induction. We proceed by the Strong Principle of Mathematical Induction. The result then follows by the Strong Principle of Mathematical Induction.

By the Strong Principle of Mathematical Induction, 2 Fn if and only if 3 n for every positive integer n. By Result 6. We use the Strong Principle of Mathematical Induction. By a and Result 9. We verify this formula in b by induction.

The proof for the formula in c is similar. By the Strong Principle of Mathematical Induction, every element of S either belongs to P or can be expressed as a product of elements of P. The result follows by the Strong Principle of Mathematical Induction.

We proceed by the Principle of Finite Induction. The result follows by the Principle of Finite Induction. Proof by minimum counterexample. The first equation is what we actually need to prove. By writing this equation, it appears that we already knew that the equation is true.

An acceptable proof can be constructed by proceeding down the left side of the equations. Now let Pk be the k-gon such that whose vertices are v1 , v2 ,. Revisiting Quantified Statements 7. This can be proved by a direct proof with two cases, namely n even and n odd. Thus we may assume that x is not an integer. Exercises for Section 7. Testing Statements 7. If 7 is prime, then 5 is prime.

If 2 is prime, then 7 is prime. If 28 is prime, then 9 is prime. If 8 is prime, then 11 is prime. Consider the integer Now apply the Intermediate Value Theorem of Calculus.

Let a be an odd integer. Let A be a nonempty set. Thus A, B, and C form a counterexample. Observe that at least two of a, b, and c are of the same parity, say a and b are of the same parity. Let p be an odd prime. Additional Exercises for Chapter 7 7. Further- more, a positive odd integer is not the sum of any two distinct positive odd integers. Assume that 3 a. Since 2x is an integer, 3 2a. Then 3 2a if and only if 3 a.

If 2 3a, then 2 a. This statement is true. We show that if 3 4a, then 3 a. Assume that 3 4a. This statement is false. Relations 8. Exercises for Section 8. Properties of Relations 8.

The possible choices for y, z in R are a, a , a, b , and a, c. Thus R is not transitive. Since R is reflexive, R contains a, a , b, b , c, c , and d, d. So the answer is 1. Thus b R a and R is symmetric. Thus R is reflexive. Therefore, b R a and R is symmetric. Therefore, a R c and R is transitive. Since a R b and R is symmetric, b R a. Similarly, d R c.

Because b R a, a R d, and R is transitive, b R d. Finally, since b R d and d R c, it follows that b R c, as desired. First assume that R is an equivalence relation on A. It remains only to show that R is circular. Assume that x R y and y R z. Since R is transitive, x R z.

Since R is symmetric, z R x. Thus R is circular. For the converse, assume that R is a reflexive, circular relation on A. Since R is reflexive, it remains only to show that R is symmetric and transitive. Since R is reflexive, y R y. Since R is circular, z R x. Now because R is symmetric, we have x R z. Thus R is transitive. Therefore, R is an equivalence relation on A. Properties of Equivalence Classes 8. There are two distinct equivalence classes: Thus R is not transitive, and so R is not an equivalence relation.

First, we show that R is reflexive. Next, we show that R is symmetric. Thus y R x and R is symmetric. Finally, we show that R is transitive. Thus x R x and R is reflexive. So y R x and R is symmetric. Therefore, x R z and R is transitive. Suppose that R1 and R2 are two equivalence relations defined on a set S.

Hence R is symmetric. Therefore, R is transitive. Congruence Modulo n 8. Hence a R a and R is reflexive. Hence b R a and R is symmetric. Hence R is reflexive.

## Mathematical Proofs - 3rd Edition - Chartrand

The distinct equivalence classes are , , , , and . In fact, the set of distinct equivalence classes is Z5. The Integers Modulo n 8. Similarly, c R d. Additional Exercises for Chapter 8 8. Next suppose that a R b.

Finally, suppose that a R b and b R c. Hence R is transitive. Thus b R a and so R is symmetric. Thus R is symmetric. Hence both R1 and R2 are reflexive. Then b R a. Since R is transitive, c R a. Thus a R a for every integer a and so R is reflexive. Thus a R c and R is transitive. Then b R a and so R is symmetric. Thus a R c and so R is transitive. Let R be a symmetric, sequential relation on some set A.

Consider the sequence a, a, a. Since R is sequential, a R a and so R is reflexive. We now show that R is transitive. We show that a R c. Consider the sequence a, c, a. Since R is sequential, either a R c or c R a. We show that a, b R e, f. A symbol should not be used in the statement of a theorem or in its proof exactly once. Although what you intended may seem clear.

The theorem does not depend on what the function is called. If it is useful to have a name for an arbitrary bijective function in the proof as it probably will be.

It should be: Every integer exceeding 1 is a prime or is composite. We wrote two equal signs on one line since that line would have contained very little material otherwise. In other words. If the expressions are rather lengthy. If there are several mathematical expressions that are linked by equal signs and inequality symbols.

A possible display is given next: This is illustrated below. Notice how the equal signs are lined up. Writing Mathematical Expressions 9 these expressions are relatively short. The reason for doing this is that ending the line with one of these symbols alerts the reader that more will follow. This word also brings the reader into the discussion with the author and gives the impression of a team effort.

The integer n is now shown to be even. This gives us even more reasons to avoid these words. It can give the impression that the author is putting the reader down. We will now show that n is even. One now shows that n is even. Common Words and Phrases in Mathematics There are some words and phrases that appear so often in mathematical writing that it is useful to discuss them. When the word any is encountered. If by any.

These words should be used sparingly and with caution. There is also the possibility that the writer a student? Any Each Every This statement is true for any integer n.

There is a danger of being too casual. Does this mean that the statement is true for some integer n or all integers n? Since the word any can be vague. Which of these sounds the best to you?

If they are used. These are four ways that we might write a sentence in a proof. Often within a proof. Common Words and Phrases in Mathematics 11 4. The word that introduces a restrictive clause and. That is. The simple answer is: Both are correct—or. Sometimes they are interchangeable. That Which These words are often confused with each other. In sentence 3. Sentence 3 could be the response to the question: Which of the numbers 4. In this context. Which of the numbers 2. A nonrestrictive clause only provides additional information that is not essential to the meaning of the sentence.

There are many many! Mathematicians cannot survive without these words. The solution to the equation is the number less than 5 that is positive. It introduces a nonrestrictive or parenthetical clause. For another illustration. Use good English. Hence the advice here is: While we are discussing the word that.

The nonrestrictive clause in 5 should be set off by commas: I always keep the math text. No sentence begins with a symbol! Some Closing Comments about Writing 1.

Many mathematicians prefer to include it rather than omit it. Omitting that leaves us with an implied that. Write in complete sentences. The solution to the equation is the number less than 5. Among these are: This clause may have been added because the solution to an earlier equation is negative. Assume N is a normal subgroup. Capitalize theorem and lemma as in Theorem 1 and Lemma 4.

A possible guideline to follow as you seek to determine whether that or which is the proper word to use is to ask yourself: I always keep the math text that I like with me.

Many words that occur often in mathematical writing are commonly misspelled. We present solutions. We pose problems. We prove theorems. We solve problems. We ask questions. Some Closing Comments about Writing feasible suitable. When you did this. If a set consists of a small number of elements.

Although there is a formal subject called set theory in which the properties of sets follow from a number of axioms. It is almost a certainty that portions of this chapter will be familiar to you. The objects that make up a set are called its elements or members. The elements of a softball team are the players. They describe the same As a small child.

Y to designate sets and lower case letters for example. It is customary to use capital upper case letters such as A. If a is an element of the set A. The most important requirement when describing a set is that the description makes it clear precisely which elements belong to the set.

A set is a collection of objects. In addition to the example given above. The elements of a set may in fact be sets themselves.

Perhaps even the set B just given contains too many elements to describe in this manner. Some sets contain too many elements to be listed this way. Example 1. There is only one set that contains no elements. A set need not contain any elements. In the sets S and T that we have just described.

Although it may seem peculiar to consider sets without elements. In such cases. In this case. The set of real numbers is denoted by R. A real number that can be expressed in the form mn. For a set S. The set of all integers positive. We use N to denote the set of all positive integers or natural numbers.

The number S is also referred to as the cardinal number or cardinality of S. Although the notation is identical for the cardinality of a set and the absolute value of a real number. In Chapter If B denotes the set consisting of those elements of A that are less than 8. The real numbers 2. For S4. Thus every real number is a complex number.

Let C denote the set of complex numbers. Describe the set H by listing its elements. Describe the set J in another manner. H and J. Determine the cardinality of each set D. Give an example of three elements that belong to E but do not belong to D. If 3 and 4 belong to the same set. This is a prelude to logic. If this were not the case. A contains none of 2. Since 1 is the only element of A. If A is a subset of B. This property of subsets might remind you of the property of real numbers where if a.

What are the possibilities for the set A? Solution By a and b. We can therefore conclude that 3 and 4 belong to different sets. To see why this is so. U may not even be a set of numbers. For a. This fact will be very useful to us. But this would mean there is some element. Two sets A and B are equal. While 2. Figure 1. The diagram on the left represents two sets A and B that have no elements in common. We will follow the notation introduced above. It is often convenient to represent sets by diagrams called Venn diagrams.

The element x belongs to A but not to B. In each case. Later we will explain why this is true. Since every element under consideration belongs to the universal set. A rectangle in a Venn diagram represents the universal set in this case. If A is a proper subset of B. The set consisting of all subsets of a given set A is called the power set of A and is denoted by P A.

A Figure 1. For the set C above. The union of two sets A and B. In symbols. An empty box and a box containing an empty box are not the same. To see why this is true. If two sets A and B have no elements in common. Then x belongs to both A and B. Q and I are disjoint.

A1 and A3 are not disjoint since 2 and 8 belong to both sets. A2 and A3 described in Example 1. A Venn diagram for A is shown in Figure 1. This fact will be established later. Solution Example 1.

Determine each of the following: B and C using interval notation. For a set A. B and C. In the case of three sets A. B A C Figure 1.

Because it is often useful to consider the union of several sets. Determine the following: We can also describe this set by means of an index set. Using N as an index set. Such a collection is called an indexed collection of sets. For this reason. We will often have the occasion especially in Chapter 8 to encounter.

Determine which of these sets are partitions of A. A collection S of subsets of a set A is called pairwise disjoint if every two distinct subsets that belong to S are disjoint. This union can be described in a number of ways. Such a collection is called a partition of A. These elements are: When we speak of the ordered pair x. As the word partition probably suggests. The partition S1 of the set A in Example 1. The set S3 is not a partition of A either since the element 5 belongs to two distinct subsets in S3.

R can be partitioned into the set Q of rational numbers and the set I of irrational numbers. Describing a Set 1. Which of the following are sets? Exercises for Chapter 1 29 however.

The ordered pair x. Cartesian products will be explored in more detail in Chapter 7. Describe the following sets in a similar manner. List the elements of the following sets in a similar manner. Write each of the following sets by listing its elements within braces. Describe the set C by listing its elements. C and D. Determine the cardinality of each of the sets A. Give an example of three elements that belong to B but do not belong to A.

Determine the cardinality of each of the following sets: Describe the set D in another manner. Which of the following sets are equal? If a set B has one more element than a set A. Let a. Three subsets A. If four sets A. Subsets 1. Determine whether the following statements are true or false. Give examples of three sets A. Determine C. Section 1. Exercises for Chapter 1 31 1. Draw the accompanying Venn diagram. B and C differ by 1. Give an example of four different sets A.

Set Operations 1. Give an example of a universal set U. Draw a Venn diagram for each of the following set operations. What can you say about parts a and b? Let A. What is B? Draw a Venn diagram for each of the following sets. Let U be a universal set and let A and B be two subsets of U. B and C be nonempty subsets of a universal set U. For each of the following collections of sets.

For each of the following. Give examples of a universal set U and sets A. Give an example of four different subsets A. For a real number r. Explain why your set S has the required properties.

Exercises for Chapter 1 33 1. B and C such that each of the following sets contains exactly one element: Cartesian Products of Sets 1. Give an example of a partition S of A satisfying the following requirements: Partitions of Sets 1. One of the sets in P1 is then partitioned into two subsets. Give an example of a partition of N into three subsets. This is continued until a partition P6 of S is produced.

A set S is partitioned into two subsets S1 and S2. Give an example of a partition of Q into three subsets. For each collection of subsets that is not a partition of A. A total of P1 sets in P2 are partitioned into two subsets each. Give an example of a partition of Z into four subsets. What is P6? S1 and S2 such that S1 is a partition of A. Give an example of three sets A. Determine the cardinality of each of the following sets.

Plot the corresponding points in the Euclidean x y. Determine the following. Describe each of the following sets by listing its elements within braces. Additional Exercises for Chapter 1 35 1. A and B. For the sets S and T described below. Determine U. S are the sets in a. Let T be the set of 2-element subsets of C.

Determine the sum of the elements of C. Give an example of four sets A1. Let I denote the interval [0. Let A and B be subsets of some unknown universal set U. The integer 57 is prime. Pn if there Under what conditions does an object possess a particular property? Finding answers to questions such as these is important. Q and R to denote statements. Every statement has a truth value.

By a statement we mean a declarative sentence or assertion that is true or false but not both. Although it is possible to spend a great deal of time studying logic. Statements therefore declare or assert the truth of something. We often use P. Are there connections between two given mathematical concepts? Example 2. It may not be immediately clear whether a statement is true or false. An open sentence that contains a variable x is typically represented by P x.

We have seen that P1: The integer 3 is odd. This is an example of what is often referred to as an open sentence. Without this additional information. Sentences that are imperative commands such as Substitute the number 2 for x. If P x is an open sentence. Q and R. Pn contains 2n possible combinations of truth values for these statements and a truth table showing these combinations would have n columns and 2n rows.

This is displayed in Figure 2. Much of the time. If a third statement R is involved. Since there are two possible truth values for each of P and Q. The truth table showing all these combinations is also given in Figure 2.

The truth tables for two statements P and Q are given in Figure 2. It is not the case that the integer 3 is odd. The integer 3 is even. The integer 57 is not prime. The integer 3 is not odd. The negation of a statement P is the statement: Figure 2. This is summarized in Figure 2. P1 is true. For two statements P and Q. Either 3 is odd or 57 is prime.

The integer 57 is prime.. If 3 is an odd integer.. For P1: For statements P and Q. Did his instructor tell the truth? I think we would all agree that she did.

We illustrate this with an example. How could this happen? Perhaps his instructor was lenient. The instructor did not lie here either. In any case. So if P and Q are both true. If 57 is prime. What she said was false. This agrees with the third row of the table in Figure2.

Perhaps the instructor made a mistake. Once again. In mathematics. Thus it says that if P is true. P only if Q. We can then form the following open sentences: These sentences are. Just as new statements can be formed from statements P and Q by negation. P implies Q. The values assigned to the variables come from their respective domains. Q is necessary for P. We will see that open sentences which result in true statements for all values of the domain will be especially interesting to us.

You may recall that a triangle is called equilateral if the lengths of its three sides are the same. For an equilateral triangle T1. A triangle T is isosceles if T is equilateral. If T is an equilateral triangle. P T and Q T are open sentences over the domain S of all triangles. A triangle T being equilateral implies that T is isosceles.

T2 is isosceles as well. A triangle T is equilateral only if T is isosceles. If T2 is not an equilateral triangle. For a triangle T to be equilateral. T is equilateral. T is isosceles. For a triangle T to be isosceles. We now investigate the truth or falseness of implications involving open sentences for values of their variables.

For x. Determine the truth or falseness of the implication P x. Solution For x. Solution In this case. Then T is isosceles. Suppose that T is an equilateral triangle. This allows us to identify the hypothesis and conclusion more easily. Whenever a triangle is equilateral. Every equilateral triangle is isosceles. The converse of an implication will often be of interest to us. If 3 is an odd integer.

This is given in Figure 2. From this table. For all other real numbers x. T is equilateral if and only if T is isosceles. Determine three distinct elements a. Q 1 and Q 4 are false. If T is isosceles. If T is equilateral. Consider the following open sentences over the domain S: Solution Observe that P 0: From given statements. Thus P 0 and P 4 are false. Hence the value 4 in S is the only choice for c.

More generally. A compound statement S is called a tautology if it is true for all possible combinations of truth values of the component statements that comprise S.

Hence the statement 3 is odd and 3 is not odd. Letting P1: Let R be a mathematical statement that we would like to show is true and suppose that R and some statement S are logically equivalent.

If R and S are logically equivalent. If we can show that S is true. Then we know that R and S have the same truth values for all possible combinations of truth values of their component statements. Suppose that R and S are logically equivalent compound statements. Let R and S be two compound statements involving the same component statements. Then R and S are called logically equivalent if R and S have the same truth values for all combinations of truth values of their component statements.

The corresponding columns of these compound statements are identical. Theorem 2. In Figure 2. There are other fundamental logical equivalences that we often encounter as well. Each part of Theorem 2.

The laws given in Theorem 2. We mentioned in Figure 2. As an additional example. If this instructor was not truthful. Let P x be an open sentence over a domain S. There are other ways that an open sentence can be converted into a statement. We illustrate this again. Each of the phrases there exists. Consider the open sentence P x: If it were not the case that for every real number x.

## Third Edition Mathematical Proofs A Transition to Advanced Mathematics Gary Chartrand

Suppose now that we were to consider the open sentence Q x: Q 1 is false. The statement 2. The negation of the statement 2. Q x that is. We know that for an open sentence P x over a domain S. For every real number x. Let P x. Suppose that Q x is an open sentence over a domain S. Hence the negation of 2. Q x is not true. Q x is false. As we saw in 2. For an open sentence containing two variables.

While we present examples of these now. The negation of 2. This statement. In words. This means that for some element x in S. Recall that whenever you see P if and only if Q. Suppose that some concept or object is expressed in an open sentence P x over a domain S and Q x is another open sentence over the domain S concerning this concept. This provides a characterization of irrational numbers: A real number r is irrational if and only if r has a nonrepeating decimal expansion.

They are characterized however as triangles whose angles are equal. Therefore, we have the characterization: A triangle T is equilateral if and only if T has three equal angles. You might think that equilateral triangles are also characterized as those triangles having three equal sides but the associated biconditional: A triangle T is equilateral if and only if T has three equal sides.

A characterization of a concept then gives an alternative, but equivalent, way of looking at this concept. Characterizations are often valuable in studying concepts or in proving other results. We will see examples of this in future chapters.

We mentioned that the following biconditional, though true, is not a characterization: Consequently, a triangle with three equal sides is equilateral but a triangle that does not have three equal sides is not equilateral. Statements 2. Which of the following sentences are statements? For those that are, indicate the truth value.

The integer is prime. The integer 0 is even. What an impossible question! Consider the sets A, B, C and D below. Which of the following statements are true? Give an explanation for each false statement. For the open sentence P x: For the open sentence P A: Let P n: Find six positive integers n for which P n is true.

Let P n:.

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Section 2. The Negation of a Statement 2. State the negation of each of the following statements. Complete the truth table in Figure 2.

At least two of my library books are overdue. One of my two friends misplaced his homework assignment. No one expected that to happen. The Disjunction and Conjunction of Statements 2. Determine which of the following statements are true.

Let P: State each of the following in words, and determine whether they are true or false. The Implication 2. Consider the statements P: Write each of the following statements in words and indicate whether it is true or false. Consider the statements: Write each of the following statements in words and indicate whether the statement is true or false.

Which of the following are true? Two sets A and B are nonempty disjoint subsets of a set S. If x is not an element of A, then x must be an element of B. A college student makes the following statement: For each of the following, determine whether this statement is true or false. The student sees his advisor both days.

The student sees his advisor on one of the two days. The instructor of a computer science class announces to her class that there will be a well-known speaker on campus later that day. Four students in the class are Alice, Ben, Cindy and Don. Don says that he will go to the lecture if Cindy does. That afternoon exactly two of the four students attend the talk.

Which two students went to the lecture? Consider the statement implication: If Bill takes Sam to the concert, then Sam will take Bill to dinner. Which of the following implies that this statement is true? Sam takes Bill to dinner only if Bill takes Sam to the concert.

Bill takes Sam to the concert. Bill takes Sam to the concert and Sam takes Bill to dinner. The concert is canceled. Let P and Q be statements. More on Implications 2. Consider the open sentences P n: Is this statement true or false? In each of the following, two open sentences P x and Q x over a domain S are given. In each of the following, two open sentences P x, y and Q x, y are given, where the domain of both x and y is Z.

Each of the following describes an implication. The Biconditional 2. Let P x: For the open sentences P x: Consider the open sentences: Tautologies and Contradictions 2.

This is an important logical argument form, called modus ponens. Then state this compound statement in words. This is another important logical argument form, called syllogism. Let R and S be compound statements involving the same component statements. If R is a tautology and S is a contradiction, then what can be said of the following?

Logical Equivalence 2. For statements P, Q and R, use a truth table to show that each of the following pairs of statements is logically equivalent. If S and T are not logically equivalent, then what can we conclude from this?

Some Fundamental Properties of Logical Equivalence 2. Verify the following laws stated in Theorem 2.

Write negations of the following open sentences: Consider the implication: If x and y are even, then x y is even. State the implication as a disjunction see Theorem 2. State the negation of the implication as a conjunction see Theorem 2.

For a real number x, let P x: For which implication is its negation the following? For which biconditional is its negation the following? Quantified Statements 2. Let S denote the set of odd integers and let P x: Determine, with explanations, whether the following statements are true: