How to get rid of the inductive step while proving $n!+2$ is divisible by $2$ for all integers $n \ge 2$ by induction?

Question

In the inductive step we assume that $k!+2$ is divisible by $2$. So $k!+2=2r$ for some $r \in \Bbb Z$.

To show that $(k+1)!+2$ is divisible by $2$.

Consider $(k+1)!+2=(k+1)k!+2=k\cdot k!+ k!+2=k\cdot k!+2r.$

From here I can only think that since $k \ge 2$, the number $k!$ is even. Thus $k!=2s$ for some $r \in \Bbb Z$.

But I feel that by this logic, we don't need Principle of Mathematical Induction at all to solve this problem. Just apply this logic to the given statement.

So should I just write $k\cdot k!+2r=k\cdot k\cdot (k-1)\cdot \cdot \cdot 2\cdot 1+2r=2(k\cdot (k-1) \cdot \cdot \cdot 1 + r)$ and finish it off? are there any alternate ways?

Answer 1

You're right in thinking that you don't really need induction here since the statement can be proven directly.

But if you want to use induction, try this:

$$\begin{align} (k+1)!+2&=(k+1)k!+2\\ &=(k+1)(k!+2-2)+2\\ &=(k+1)(2r-2)+2\\ &=\ldots \end{align}$$

Answer 2

Maybe you can follow $$(k+1)!+2=(k+1)k!+2=k\cdot k!+k!+2=k(2r-2)+(2r-2)+2$$ and then the conclusion follows.

Answer 3

You can do this problem directly as for $n\ge 2, n! $ is always even, then so is $n! +2$, and you are done.

If you want to use Principle of Mathematical Induction, then consider $(k+1)!$ which can be written as $(k+1) k!$ and substitute value of $k!$ from your induction hypothesis and you are done.

$(k+1)!= (k+1)k! = (k+1) (2r-2)$

Thus $(k+1)!+2= (k+1) (2r-2)+ 2$, which you can observe is divisible by $2$.