For every $n\in \Bbb{N}$, either $n$ is even or $n-1$ is even (Must use Induction)

Question

I have gotten a lot of the proof done so far but I am now stuck: Pf: Let $n\in\Bbb{N}$. $P(n)$ is the statement, "$n$ is even or $n-1$ is even" Base Case ($n=1$): $n=1$ and $1-1=1$ so $n-1$ is even. $P(2)$ holds. Induction: Let $k\in\Bbb{N}$. We will assume $P(k)$ is true, $k$ is even or $k-1$ is even, in order to prove $P(k+1)$. Then $k+1$ is even or $(k+1)-1$ is even. My next step was going to be to either multiply or add $(k+1)$ and $(k+1)-1$ together since with natural numbers you can do that but I am not sure what to do after that...

Answer 1

Let's go through this line by line.

Base Case ($n=1$): $n=1$ and $1-1=\color{red}{1}$ so $n-1$ is even.

I think this is just a typo. $1-1=0$, and that is even.

$P(2)$ holds.

You probably mean $P(1)$ holds. That is what you proved.

Induction: Let $k∈N$. We will assume $P(k)$ is true, [that is,] $k$ is even or $k-1$ is even, in order to prove $P(k+1)$.

Good. I added in that bracketed part because it emphasizes that “$P(k)$ is true” and “$k$ is even or $k-1$ is even” are equivalent statements.

Then $k+1$ is even or $(k+1)-1$ is even.

Hold on. This statement is equivalent to $P(k+1)$, which has not been proved yet. When you say “Then ... ” it sounds like you are saying it follows from previous work.

Well, what do you know at this point? You know (or rather, you're assuming) that either $k$ or $k-1$ is even. You want to show that either $k+1$ or $k$ is even. Try to prove this first:

Lemma. If $n$ is an even integer, then $n+2$ is even.

Then see if that lemma helps you with the inductive step.