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By associativity of addition and (b2 + a) + bị = bz. Since b2 + a = 0, this gives 0 + b1 = b2, %3D or b1 = b2, %3D as desired. + b= 0. Hence, there is a unique b e R such that a The point of this example was to illustrate a standard uniqueness proof, and It is necessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | b and a c, then a | (bm + cn). 2. Prove that for all real numbers a and b, if 0 < a < b, then 0 < a² < b². 3. Prove that for all integers m, if m is odd, then there exists k e Z such that m² = 8k + 1. 4. Using definitions, prove by cases that for every integer n, n2 +n + 5 is odd. 5. Determine whether each statement is true or false. If true, then prove it. If false, then provide a counterexample. (a) For all positive integers n, n is divisible by 3 is necessary for n to be divisible by 6. (b) For all positive integers n, n is divisible by 3 is sufficient for n to be divisible by 6. (c) For all real numbers x, x2 - 2x-3 = 0 only if a = 3. (d) For all real numbers x, x2 -2x-3 = 0 if x = 3. (e) For all integers a, b, c, if a | bc, then a | b or a | c. (f) For all integers a, b, c, if a | (b+c), then a | b or a | c. (g) For all even integers m and n, 4 | mn. (h) For all integers n, if n² is a multiple of 4, then n is a multiple of 4. (i) There exist integers m and n such that 15m + 12n = -6. %3D

By associativity of addition and (b2 + a) + bị = bz. Since b2 + a = 0, this gives 0 + b1 = b2, %3D or b1 = b2, %3D as desired. + b= 0. Hence, there is a unique b e R such that a The point of this example was to illustrate a standard uniqueness proof, and It is necessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | b and a c, then a | (bm + cn). 2. Prove that for all real numbers a and b, if 0 < a < b, then 0 < a² < b². 3. Prove that for all integers m, if m is odd, then there exists k e Z such that m² = 8k + 1. 4. Using definitions, prove by cases that for every integer n, n2 +n + 5 is odd. 5. Determine whether each statement is true or false. If true, then prove it. If false, then provide a counterexample. (a) For all positive integers n, n is divisible by 3 is necessary for n to be divisible by 6. (b) For all positive integers n, n is divisible by 3 is sufficient for n to be divisible by 6. (c) For all real numbers x, x2 - 2x-3 = 0 only if a = 3. (d) For all real numbers x, x2 -2x-3 = 0 if x = 3. (e) For all integers a, b, c, if a | bc, then a | b or a | c. (f) For all integers a, b, c, if a | (b+c), then a | b or a | c. (g) For all even integers m and n, 4 | mn. (h) For all integers n, if n² is a multiple of 4, then n is a multiple of 4. (i) There exist integers m and n such that 15m + 12n = -6. %3D

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter8: Polynomials
Section8.4: Zeros Of A Polynomial
Problem 22E
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By associativity of addition and (b2 + a) + bị = bz. Since b2 + a = 0, this gives 0 + b1 = b2, %3D or b1 = b2, %3D as desired. + b= 0. Hence, there is a unique b e R such that a The point of this example was to illustrate a standard uniqueness proof, and It is necessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | b and a c, then a | (bm + cn). 2. Prove that for all real numbers a and b, if 0 < a < b, then 0 < a² < b². 3. Prove that for all integers m, if m is odd, then there exists k e Z such that m² = 8k + 1. 4. Using definitions, prove by cases that for every integer n, n2 +n + 5 is odd. 5. Determine whether each statement is true or false. If true, then prove it. If false, then provide a counterexample. (a) For all positive integers n, n is divisible by 3 is necessary for n to be divisible by 6. (b) For all positive integers n, n is divisible by 3 is sufficient for n to be divisible by 6. (c) For all real numbers x, x2 - 2x-3 = 0 only if a = 3. (d) For all real numbers x, x2 -2x-3 = 0 if x = 3. (e) For all integers a, b, c, if a | bc, then a | b or a | c. (f) For all integers a, b, c, if a | (b+c), then a | b or a | c. (g) For all even integers m and n, 4 | mn. (h) For all integers n, if n² is a multiple of 4, then n is a multiple of 4. (i) There exist integers m and n such that 15m + 12n = -6. %3D
Transcribed Image Text:By associativity of addition and (b2 + a) + bị = bz. Since b2 + a = 0, this gives 0 + b1 = b2, %3D or b1 = b2, %3D as desired. + b= 0. Hence, there is a unique b e R such that a The point of this example was to illustrate a standard uniqueness proof, and It is necessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | b and a c, then a | (bm + cn). 2. Prove that for all real numbers a and b, if 0 < a < b, then 0 < a² < b². 3. Prove that for all integers m, if m is odd, then there exists k e Z such that m² = 8k + 1. 4. Using definitions, prove by cases that for every integer n, n2 +n + 5 is odd. 5. Determine whether each statement is true or false. If true, then prove it. If false, then provide a counterexample. (a) For all positive integers n, n is divisible by 3 is necessary for n to be divisible by 6. (b) For all positive integers n, n is divisible by 3 is sufficient for n to be divisible by 6. (c) For all real numbers x, x2 - 2x-3 = 0 only if a = 3. (d) For all real numbers x, x2 -2x-3 = 0 if x = 3. (e) For all integers a, b, c, if a | bc, then a | b or a | c. (f) For all integers a, b, c, if a | (b+c), then a | b or a | c. (g) For all even integers m and n, 4 | mn. (h) For all integers n, if n² is a multiple of 4, then n is a multiple of 4. (i) There exist integers m and n such that 15m + 12n = -6. %3D
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