Question

Suppose n is an integer. Using the definitions of even and odd, prove that n is odd if and only if 3n+1 is even.

Discrete math
ANSWERED
asked 2021-02-09

Suppose n is an integer. Using the definitions of even and odd, prove that n is odd if and only if \(3n+1\) is even.

Expert Answers (1)

2021-02-10

Suppose that n is odd. Then \(\displaystyle{n}={2}{k}+{1}\), where k is some integer. Furthermore, \(\displaystyle{3}{n}+{1}={3}{\left({2}{k}+{1}\right)}+{1}={6}{k}+{4}={2}{\left({3}{k}+{2}\right)}\)
Since \(3k+2\) is an integer, we conclude that \(3n+1\) is even.
Suppose that \(3n+1\) is even. Then \(3n+1=2l,\)
for some integer l. This also means that
\(\displaystyle{3}{n}={2}{l}-{1}={2}{\left({l}-{1}\right)}+{1}\)
Since \(l-1\) is an integer, we conclude that 3n is odd. Now, if n was even then clearly 3n would also be even, so we get a contradiction. Therefore, n must be odd.

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