A transition to advanced mathematics solutions pdf

Toggle navigation Menu. Home » Bellevue » A transition to advanced mathematics solutions manual pdf smith. Previous: A course in ordinary differential equations second edition solution manual. 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.

Exercises for Section 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. 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. Hence a R a and R is reflexive. Hence b R a and R is symmetric. Hence R is reflexive. The distinct equivalence classes are [0], [1], [2], [3], and [4]. In fact, the set of distinct equivalence classes is Z5. 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.

Therefore, R is an equivalence relation. These are two equivalence classes consist of the set of all points in the plane that lie on the diamond-shaped figure shown in Figure Figure The equivalence classes in Exercise 8. Exercises for Section 9. By interchanging x and y in f1 , f2 , f3 , f4 , we obtain f5 , f6 , f7 , f8. One way to show that f is onto is to use the Intermediate Value Theorem.

Method 1. Method 2. First, we show that f is one-to-one. Next, we show that f is onto. We first show that f is one-to-one. Therefore, f is one-to-one. Therefore f is bijective. Thus f is one-to-one. Thus f is onto. We show that f is one-to-one. Assume that f is not one-to-one. This is Corollary 9. Consider the functions f and g in b.

Then f is a function from A to A. Next we show that f is onto. The proposed proof that f is onto is not written properly, beginning with the second sentence.

Sentences 2—5 result in solving for x in terms of r, which is not a part of the proof; however, these sentences supply the necessary information to provide a proof. The information provided in the display is critical in a proper proof. Then it is not possible to list elements of A as in a. Assume that f is onto. Assume that g is one-to-one. Then f is onto if and only if g is one-to-one.

Thus gf is a function from A to C. The answer is no. First, we prove two lemmas. Lemma 1. We proceed by induction on m. Lemma 2. Let m be an odd positive integer. Let m be a nonnegative integer. Let n be a positive even integer. Let m be a positive even integer. Next, let m be a positive odd integer. Therefore, the set A is countable. The result follows by Theorem Construct a table where aij is in row i, column j.

Denote the set of irrational numbers by I. Assume, to the contrary, that I is denumerable. Hence f is one-to-one. Therefore, f is onto. Thus f is a bijective function from 0, 1 to 0, 2. Then g is bijective and so 0, 1 and a, b have the same cardinality. Exercises for Section Since B is an infinite subset of the denumerable set C, it follows that B is denumerable. The function f is bijective. See d. Consider R. We use induction on n. The function f1 is the restriction of f to C.

Additional Exercises for Chapter 10 First we show that f is one-to-one. Observe that 1 is the only positive integer whose image under f is 0. First, a b we show that f is one-to-one. The proof that f is bijective is similar to that in a. Assume first that S is countable. Then S is either finite or denumerable. If S is denumerable, then there exists a bijective function from N to S. Let A be a finite nonempty set. Since A is not denumerable, there is no bijective function from A to N.

We now show that f is injective. By Result By the induction hypothesis, A is denumerable. Assume that a b and c d. Assume that ac bc. We employ induction. Thus these n numbers are composite. By Theorem Assume for some positive integer k, that if a1 , a2 ,. By the induction hypothesis, d b. Hence in any case, S is nonempty. By Theorem 6. The remainder of the proof is identical to the proof of Theorem Let p be a prime different from 2 and 5.

Let n be an integer that is not a multiple of 3. Thus n2 is not a multiple of 3. Let n be an integer that is not a multiple of 5. We consider these four cases. The other three cases are handled similarly. We consider two cases, depending on whether a is even or a is odd. Thus this case cannot occur. Assume that if a1 , a2 ,.

Now let b1 , b2 ,. Assume that an even number of a, b, and c are congruent to 1 modulo 3. None of a, b, and c is congruent to 1 modulo 3. We consider two subcases. None of a, b, and c is congruent to 0 modulo 3. Then all of a, b, and c are congruent to 2 modulo 3.

Applying Result 4. Exactly two of a, b, and c are congruent to 1 modulo 3, say a and b are congruent to 1 modulo 3 and c is not congruent to 1 modulo 3.

The proof is similar to that of Case 1. Show, for every n integers a1 , a2 ,. Assume first that n is a linear combination of a and b. For the converse, assume that d n. Since xc and yc are integers, n is a linear combination of a and b. It then follows by Theorem Assume, to the contrary, that 3 is rational.

Since b2 is an integer, 3 a2. It then follows by Corollary Since x2 is an integer, 3 b2 and so 3 b by Corollary Since 3b2 is an integer, a2 is even. Since c2 is an integer, 2 3b2. Since bn is an integer, p an. Since p is a prime, it follows by Corollary By Corollary We give a proof by contrapositive. Assume that p is not a prime. Thus p ab. Bettina Richmond and Tom Richmond. Upload a Thing! Customize a Thing. Download All Files. Select a Collection.

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