To prove: That there exists an integer k such that x = x 0 + 5 k and y = y 0 − 2 k . It is given that 2 x 0 + 5 y 0 = 19 = 2 x + 5 y for certain integers x 0 , y 0 , x , y .

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Answer 1

Question

Chapter 4.3, Problem 15E

(a)

To determine

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

(b)

To determine

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

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Problem 11 (a) A tank is discharging water through an orifice at a depth of T meter below the surface of the water whose area is A m². The following are the values of a for the corresponding values of A: A 1.257 1.390 x 1.50 1.65 1.520 1.650 1.809 1.962 2.123 2.295 2.462|2.650 1.80 1.95 2.10 2.25 2.40 2.55 2.70 2.85 Using the formula -3.0 (0.018)T = dx. calculate T, the time in seconds for the level of the water to drop from 3.0 m to 1.5 m above the orifice. (b) The velocity of a train which starts from rest is given by the fol- lowing table, the time being reckoned in minutes from the start and the speed in km/hour: | † (minutes) |2|4 6 8 10 12 14 16 18 20 v (km/hr) 16 28.8 40 46.4 51.2 32.0 17.6 8 3.2 0 Estimate approximately the total distance ran in 20 minutes.

- 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?

Answer 2

Question

Chapter 4.3, Problem 15E

(a)

To determine

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

(b)

To determine

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 3

Question

Chapter 4.3, Problem 15E

(a)

To determine

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

(b)

To determine

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

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Answer 4

Question

Chapter 4.3, Problem 15E

(a)

To determine

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

(b)

To determine

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Expert Solution & Answer

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Answer 5

Chapter 4.3, Problem 15E

(a)

To determine

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

(b)

To determine

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Answer 6

Chapter 4.3, Problem 15E

(a)

To determine

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Answer 7

Chapter 4.3, Problem 15E

(a)

To determine

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Answer 8

(b)

To determine

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Answer 9

(b)

To determine

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Answer 10

Expert Solution & Answer

Answer 11

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Answer 12

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Answer 13

Ch. 4.1 - 227 Ch. 4.1 - The Well-Ordering Principle states that any...Ch. 4.1 - Prob. 3TFQ Ch. 4.1 - Prob. 4TFQ Ch. 4.1 - Prob. 5TFQ Ch. 4.1 - Prob. 6TFQ Ch. 4.1 - Prob. 7TFQ Ch. 4.1 - Prob. 8TFQ Ch. 4.1 - Prob. 9TFQ Ch. 4.1 - Prob. 10TFQ

To prove: That there exists an integer k such that x = x 0 + 5 k and y = y 0 − 2 k . It is given that 2 x 0 + 5 y 0 = 19 = 2 x + 5 y for certain integers x 0 , y 0 , x , y .

Solution Summary: The author explains how to prove that there exists an integer k such that x=x_0+5k.

Concept explainers

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To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

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Chapter 4 Solutions

To prove: That there exists an integer k such that x = x 0 + 5 k and y = y 0 − 2 k . It is given that 2 x 0 + 5 y 0 = 19 = 2 x + 5 y for certain integers x 0 , y 0 , x , y .

Solution Summary: The author explains how to prove that there exists an integer k such that x=x_0+5k.

Concept explainers

Videos

To prove: That there exists an integer k such that x=x0+5k and y=y0−2k. It is given that 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k. Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.

Want to see the full answer?

Chapter 4 Solutions

To prove: If 2x+5y=19, show that there exists an integer k such that x=2+5k and y=3−2k.

Also, 2x0+5y0=19=2x+5y for certain integers x0,y0,x,y.