3 years ago
Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contradiction.
1 Answers
Best Answer
kkimmel78 Staff answered 3 years ago
kkimmel78 Staff answered 3 years ago
Proof by contradiction
Suppose that 3n + 2 is even and 7 is not even. Because 7 is not even then 7 is odd. By the definition of odd numbers, there is an integer k such that
n=2k+1.
Substituting n=2k+1 into 3n+2, we get
3n+2=3(2k+1)+2=6k+3+2=6k+4+1=2(3k+2)+1.
Thus, we can find an integer l=3k+2 such that
3n+2=2l+1.
That means, 3n+2 is odd.
So far, our assumption, ""n is odd"", leads us to the contradiction that 3n + 2 is both even and odd. This is a contradiction. Hence, ""n is odd"" must be a false statement. This completes the proof that n must be an even number.
Result:
Assume that 3n+1 is even and n is not even, and then derive a contradiction.
Suppose that 3n + 2 is even and 7 is not even. Because 7 is not even then 7 is odd. By the definition of odd numbers, there is an integer k such that
n=2k+1.
Substituting n=2k+1 into 3n+2, we get
3n+2=3(2k+1)+2=6k+3+2=6k+4+1=2(3k+2)+1.
Thus, we can find an integer l=3k+2 such that
3n+2=2l+1.
That means, 3n+2 is odd.
So far, our assumption, ""n is odd"", leads us to the contradiction that 3n + 2 is both even and odd. This is a contradiction. Hence, ""n is odd"" must be a false statement. This completes the proof that n must be an even number.
Result:
Assume that 3n+1 is even and n is not even, and then derive a contradiction.