Question

UUIDOR Quiz 2 - Ma Consider the following theorem. Theorem: The sum of any even integer and any odd integer is odd. Six of the sentences in the following scrambled list can be used to prove the theorem. By definition of even and odd, there are integers rands such that m = 2r and n = 2s + 1. By substitution and algebra, m + n = 2r + 25 + 1) = 2(r + s) + 1. Suppose m is any even integer and n is any odd integer. Hence, the sum is twice an integer plus one. So by definition of odd, m + nis odd. So by definition of event is even Let t = r + s. Then, t is an integer because it is a sum of integers. By definition of even and odd, there is an integer r such that m = 2r and n = 2r + 1. By substitution, m + n = 2t + 1. Let m + n be any odd integer. Proof: We construct a proof by selecting appropriate sentences from the list and putting them in the correct order. 1. ---Select- ---Select- ---Select-- 4. --Select- ---Select- 6. ---Select-

Answer 1

We have constructed the proof by selecting appropriate sentences.

1. Suppose m is any even integer and n is any odd integer. 2. By definition of even and odd, there are integers r and s such that m = 2r and n = 2s + 1 3. By substitution and algebra, m+n=2r + (2s + 1) = 2(r + s) +1 4. Let t=r + s. then t is an integer because it is a sum of integers. 5. By substitution, m+n= 2t +1 6. Hence the sum is twice an integer plus one. So by definition of odd, m +n is odd