A transition to advanced mathematics solution manual

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Click here to download the table of contents in pdf. Not feeling ready for this? This was overwhelmed when. 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. 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. 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. 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. There are three cases. It remains to verify the converse.

By Theorem 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.

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. Let a and b be odd integers. Then n is even or n is odd. Consequently, there are three possibilities: 1 the second suitor had a gold crown and the third suitor had a silver crown; 2 the second and the third suitors had gold crowns; 3 the second suitor had a silver crown and the third suitor had a gold crown. 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. 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: If a1 , a2 ,. 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. 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 ,. 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. 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. 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. Existence and Proof by Contradiction 6. Mathematical Induction 7. Reviewing Proof Techniques 8. Prove or Disprove 9. Equivalence Relations Functions Cardinalities of Sets Proofs in Number Theory Proofs in Combinatorics Proofs in Calculus Proofs in Group Theory. This is NOT the textbook.

All orders are in downloadable format ONLY. Instant Download After Purchase. The full files will be in the same format as the free sample. Table of Contents 0.