To prove: That 7 is an accumulation point for [ 3 , 7 ) .

Question

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Answer 1

Question

Chapter 7.3, Problem 1E

(a)

To determine

To prove: That 7 is an accumulation point for [3,7) .

(b)

To determine

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

(c)

To determine

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

(d)

To determine

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

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- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?

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Answer 2

Question

Chapter 7.3, Problem 1E

(a)

To determine

To prove: That 7 is an accumulation point for [3,7) .

(b)

To determine

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

(c)

To determine

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

(d)

To determine

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

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Answer 3

Question

Chapter 7.3, Problem 1E

(a)

To determine

To prove: That 7 is an accumulation point for [3,7) .

(b)

To determine

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

(c)

To determine

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

(d)

To determine

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

Expert Solution & Answer

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Answer 4

Question

Chapter 7.3, Problem 1E

(a)

To determine

To prove: That 7 is an accumulation point for [3,7) .

(b)

To determine

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

(c)

To determine

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

(d)

To determine

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

Expert Solution & Answer

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Answer 5

Chapter 7.3, Problem 1E

(a)

To determine

To prove: That 7 is an accumulation point for [3,7) .

(b)

To determine

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

(c)

To determine

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

(d)

To determine

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

Answer 6

Chapter 7.3, Problem 1E

(a)

To determine

To prove: That 7 is an accumulation point for [3,7) .

Answer 7

Chapter 7.3, Problem 1E

(a)

To determine

To prove: That 7 is an accumulation point for [3,7) .

Answer 8

(b)

To determine

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

Answer 9

(b)

To determine

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

Answer 10

(c)

To determine

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

Answer 11

(c)

To determine

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

Answer 12

(d)

To determine

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

Answer 13

(d)

To determine

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Answer 17

Ch. 7.1 - Prob. 1E Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

To prove: That 7 is an accumulation point for [ 3 , 7 ) .

Solution Summary: The author proves that 7 is an accumulation point for left[3,7).

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To prove: That 7 is an accumulation point for [3,7) .

To Prove: That 5 is an accumulation point for (4,5)∪(5,6) .

To prove: That 0 is an accumulation point for {1+(−1)nn:n∈N}

To prove: That e is an accumulation point for {(1+1n)n:n∈N} .

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Chapter 7 Solutions