# False statement proven by induction? $n \geq a \Rightarrow n! \geq a^n$

I will prove the false statement, that $$n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}$$ with induction

For $$n=1$$ $$1\geq a\Rightarrow 1!\geq a^1\Rightarrow 1 \geq a$$ which is true.

Suppose that $$n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}$$

Then,
$$n\geq a\Rightarrow (n+1)!\geq nn!\geq aa^n=a^{n+1}$$ which yields that $$n+1\geq a\Rightarrow(n+1)!\geq a^{n+1}$$

Therefore, $$n\geq a\Rightarrow n!\geq a^n$$
But for $$n=3,a=2$$ using the inequality we just proved $$3\geq 2\Rightarrow3!\geq 2^3\Leftrightarrow 6\geq 8$$ Impossible!!. Where is my mistake?

[EDIT] Don't bother answering. I have highlighted the mistake I made that rendered the inequality invalid for a greater than 1

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Stephen Tashi

Suppose that $$n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \}$$

That's not the hypothesis used in induction.

Ordinary induction would use the hypothesis:
$n!\geq a^n$ where we think of $n$ as a particular integer, not as "all integers".

and you would have to prove $(n+1)! \geq a^{n+1}$

So-called "strong induction" would use the hypothesis:
For each integer $i: 0 < i \leq n , i! \geq a^i$

and you would still have to prove $(n+1)! \geq a^{n+1}$

That's not the hypothesis used in induction.

Ordinary induction would use the hypothesis:
$n!\geq a^n$ where we think of $n$ as a particular integer, not as "all integers".

and you would have to prove $(n+1)! \geq a^{n+1}$

So-called "strong induction" would use the hypothesis:
For each integer $i: 0 < i \leq n , i! \geq a^i$

and you would still have to prove $(n+1)! \geq a^{n+1}$

I am sorry, I messed up with latex. That was not my hypothesis...

Stephen Tashi

I was having trouble too. The world will probably end because of clerical errors.

disregardthat

The problem is that $n+1 \geq a$ does not imply that $n \geq a$ which you seem to have used.

mathman