Consider the following statement. If a and b are any odd integers, then a + b is even. Construct a proof for the statement by selecting sentences from the following scrambled list and putting them in the correct order. Suppose a and b are any odd integers. By definition of odd integer, a = 2r + 1 and b = 2r + 1 for any integers r and s. Thus, a2 + b2 = 2k, where k is an integer. Suppose a and b are any integers. By definition of odd integer, a = 2r + 1 and b = 2s + 1 for some integers r and s. By substitution and algebra, a2 + b2 = (2r + 1)2 + (2s + 1)2 = 2[2(r2 + s2) + 2(r + s) + 1]. Let k = 2r2 + 2s². Then k is an integer because sums and products of integers are integers. By substitution and algebra, a2 + b2 = (2r)2 + (2s)2 = 2(2r2 + 2s2). Hence a2 + b² is even by definition of even. Let k = 2(r2 + s²) + 2(r + s) + 1. Then k is an integer because sums and products of integers are integers. Proof: 1. |---Select-- 2. ---Select--- 3. ---Select--- 4. ---Select--- 5. ---Select--- 6. ---Select--- << < < <

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.2: Mathematical Induction
Problem 49E: Show that if the statement is assumed to be true for , then it can be proved to be true for . Is...
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Consider the following statement.
If a and b are any odd integers, then a + b' is even.
Construct a proof for the statement by selecting sentences from the following scrambled list and putting them in the correct order.
Suppose a and b are any odd integers.
By definition of odd integer, a = 2r + 1 and b = 2r + 1 for any integers r and s.
Thus, a2 + b2 = 2k, wherek is an integer.
Suppose a and b are any integers.
By definition of odd integer, a = 2r + 1 and b = 2s + 1 for some integers r and s.
By substitution and algebra, a² + b2 = (2r + 1)2 + (2s + 1)2 = 2[2(r2 + s²) + 2(r + s) + 1].
Let k = 2r2 + 2s². Then k is an integer because sums and products of integers are integers.
By substitution and algebra, a² + b² = (2r)² + (2s)² = 2(2r2 + 2s²).
Hence a2 + b2 is even by definition of even.
Let k = 2(r2 + s²) + 2(r + s) + 1. Then k is an integer because sums and products of integers are integers.
Proof:
1.
|---Select---
2.
|---Select---
3.
|---Select---
4.
|---Select---
5.
---Select---
6.
---Select---
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Transcribed Image Text:Consider the following statement. If a and b are any odd integers, then a + b' is even. Construct a proof for the statement by selecting sentences from the following scrambled list and putting them in the correct order. Suppose a and b are any odd integers. By definition of odd integer, a = 2r + 1 and b = 2r + 1 for any integers r and s. Thus, a2 + b2 = 2k, wherek is an integer. Suppose a and b are any integers. By definition of odd integer, a = 2r + 1 and b = 2s + 1 for some integers r and s. By substitution and algebra, a² + b2 = (2r + 1)2 + (2s + 1)2 = 2[2(r2 + s²) + 2(r + s) + 1]. Let k = 2r2 + 2s². Then k is an integer because sums and products of integers are integers. By substitution and algebra, a² + b² = (2r)² + (2s)² = 2(2r2 + 2s²). Hence a2 + b2 is even by definition of even. Let k = 2(r2 + s²) + 2(r + s) + 1. Then k is an integer because sums and products of integers are integers. Proof: 1. |---Select--- 2. |---Select--- 3. |---Select--- 4. |---Select--- 5. ---Select--- 6. ---Select--- Need Help? Read It Watch It
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9781285463230
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Cengage Learning,