Prove that if n is an integer and 3n + 2 is even, then n is even using
a) a proof by contraposition.
b) a proof by contradiction
a) Assume the negation of conclusion:
Let "n" is an integer. Assume that "n" is odd. By definition of odd numbers, exists integer "k" such that "n=2k+1."
Hence
So by definition of odd numbers, "3n+2" is odd number. Thus we prove the negation of hypothesis.
Therefore we prove that if "n" is an integer and "3n+2" is even, then "n" is even using a proof by contraposition.
b)
Let "n" is an integer. Assume that "3n+2" is even and "n" is odd. Then by definition of odd numbers, there exists an integer "k" such that "n=2k+1." Hence
So by definition of odd numbers, "3n+2" is odd number, that contradicts our assumption that "3n+2" is even.
Hence, it is not the case that "3n+2" is even and "n" is odd.
Therefore we prove that if "n" is an integer and "3n+2" is even, then "n" is even using a proof by contradiction.
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