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(b2 + a) + b1 = 02. Since b2 + a = 0, this gives %3D 0 + b1 = b2, or bị = b2, %3D as desired. Hence, there is a unique b ER such that a + b= 0. The point of this example was to illustrate a standard uniqueness proof, and 2o Is hecessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. 0 Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | 6 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 x = 3. (d) For all real numbers x, x2 – (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. 2x-3 0 if x = 3.

(b2 + a) + b1 = 02. Since b2 + a = 0, this gives %3D 0 + b1 = b2, or bị = b2, %3D as desired. Hence, there is a unique b ER such that a + b= 0. The point of this example was to illustrate a standard uniqueness proof, and 2o Is hecessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. 0 Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | 6 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 x = 3. (d) For all real numbers x, x2 – (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. 2x-3 0 if x = 3.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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(b2 + a) + b1 = 02. Since b2 + a = 0, this gives %3D 0 + b1 = b2, or bị = b2, %3D as desired. Hence, there is a unique b ER such that a + b= 0. The point of this example was to illustrate a standard uniqueness proof, and 2o Is hecessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. 0 Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | 6 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 x = 3. (d) For all real numbers x, x2 – (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. 2x-3 0 if x = 3.
Transcribed Image Text:(b2 + a) + b1 = 02. Since b2 + a = 0, this gives %3D 0 + b1 = b2, or bị = b2, %3D as desired. Hence, there is a unique b ER such that a + b= 0. The point of this example was to illustrate a standard uniqueness proof, and 2o Is hecessary to prove statements of this sort in an abstract algebra course. We'll see more uniqueness statements later in this textbook. 0 Exercises 2.1 1. Let a, b, and c be integers. Prove that for all integers m and n, if a | 6 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 x = 3. (d) For all real numbers x, x2 – (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. 2x-3 0 if x = 3.
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