Consider the following statement. If a and b are any odd integers, then a 2 + b 2 is even. Construct a proof for the statement by selecting sentences from the following scrambled list and putting them...

• Letk = 2r² + 2s². Thenk is an integer because sums and products of integers are integers.


• By definition of odd integer,a = 2r + 1 andb = 2r + 1 for any integersr ands.


• Hencea² +b² is even by definition of even.


• By definition of odd integer,a = 2r + 1 andb = 2s + 1 for some integersr ands.


• Letk = 2(r² +s²) + 2(r +s) + 1. Thenk is an integer because sums and products of integers are integers.


• Supposea andb are any odd integers.


• Thus,a² +b² = 2k, wherek is an integer.


• By substitution and algebra,a² +b² = (2r)² + (2s)² = 2(2r² + 2s²).


• Supposea andb are any integers.


• By substitution and algebra,a² +b² = (2r + 1)² + (2s + 1)² = 2[2(r² +s²) + 2(r +s) + 1].

Extracted 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--