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Consider the following statement. If a and b are any odd integers, then a2 + b2 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. 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. Thus, a2 + b2 = 2k, where k is an integer. Suppose a and b are any integers. Let k = 2(r2 + s²) + 2(r + s) + 1. Then k is an integer because sums and products of integers are integers. Hence a? + b2 is even by definition of even. By substitution and algebra, a2 + b2 = (2r)2 + (2s)² = 2(2r2 + 2s²). Proof: 1. ---Select--- 2. ---Select-- 3. ---Select-- 4. ---Select-- 5. ---Select-- 6. ---Select--

Consider the following statement. If a and b are any odd integers, then a2 + b2 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. 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. Thus, a2 + b2 = 2k, where k is an integer. Suppose a and b are any integers. Let k = 2(r2 + s²) + 2(r + s) + 1. Then k is an integer because sums and products of integers are integers. Hence a? + b2 is even by definition of even. By substitution and algebra, a2 + b2 = (2r)2 + (2s)² = 2(2r2 + 2s²). Proof: 1. ---Select--- 2. ---Select-- 3. ---Select-- 4. ---Select-- 5. ---Select-- 6. ---Select--

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