Show that if n is an integer and n3 + 5 is odd, then n is even using a proof by

Question

Show that if n is an integer and n3 + 5 is odd, then n is even using a proof by contraposition. State the contrapositive statement Proof proper

Explanation / Answer

a) Contraposition: Convert the statement from if A, then B to if not B then not A--> If n is not even, then n^3 + 5 is not odd or in other word, if n is odd, then n^3 + 5 is even Proof: that if n is odd, n^3 + 5 is even for integer n Since n is odd as given, let k be an integer such that n = 2k + 1 n^3 + 5 = (2k + 1)^3 + 5 = 8k^3 + 6k^2 + 6k + 1 + 5 = 8k^3 + 6k^2 + 6k + 6 = 2(4k^3 +3k^2 + 3k + 3) = an even number So n^3 + 5 is even. Hence by contraposition, if n^3 + 5 is odd, then n is even b) by contradiction Assume there is an integer n such that n^3 + 5 is odd and n is odd. Since n is odd, let k be an integer so that n = 2k + 1 Then n^3 + 5 = (2k + 1)^3 + 5 = 8k^3 + 6k^2 + 6k + 1 + 5 = 8k^3 + 6k^2 + 6k + 6 = 2(4k^3 +3k^2 + 3k + 3) = an even number This contradicts n^3 + 5 is odd. Hence by contradiction n is even if n^3 + 5 is odd

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