Even Integer Plus 5 is Odd/Proof by Contradiction

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Let $x \in \Z$ be an even integer.

Then $x + 5$ is odd.


Let $x$ be an even integer.

Then by definition:

$x = 2 n$

for some integer $n$.

Aiming for a contradiction, suppose $y = x + 5 = 2 m$ for some integer $m$.


\(\ds x\) \(=\) \(\ds 2 m - 5\)
\(\ds \) \(=\) \(\ds \paren {2 m - 6} + 1\)
\(\ds \) \(=\) \(\ds 2 \paren {m - 3} + 1\)
\(\ds \) \(=\) \(\ds 2 r + 1\) where $r = m - 3 \in \Z$

Hence $x$ is odd.

But this contradicts our premise that $x$ is even.

Hence the result by Proof by Contradiction.


Historical Note

There is nothing profound about this result.

Gary Chartrand used it as a simple demonstration of the construction of various kinds of proof in his Introductory Graph Theory of $1977$.

It is questionable whether the indirect proof and the Proof by Contradiction actually constitute different proofs of this result, but both are included on $\mathsf{Pr} \infty \mathsf{fWiki}$ anyway, in case they are found to be instructional.

He sets a similar theorem as an exercise:

Prove the implication "If $x$ is an odd integer, then $y = x - 3$ is an even integer" using the three proof techniques: ...

but it has been considered not sufficiently different from this one to be actually included on $\mathsf{Pr} \infty \mathsf{fWiki}$ as a separate result to be proved.

For similar reasons, several other of the trivial exercises in applied logic that he sets have also been omitted from this site.