# Prove False: n\geq a\Rightarrow n!\geq a^n

• 3.1415926535
In summary, the conversation discussed the use of induction to prove the statement n\geq a\Rightarrow n!\geq a^n, n\in \mathbb{N}-\left \{ 0 \right \} for a specific value of n. However, the incorrect use of the hypothesis led to an invalid inequality. It was also mentioned that the allowed values of a increase with n, which is not compatible with the use of induction.

#### 3.1415926535

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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3.1415926535 said:
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 $(n+1)! \geq a^{n+1}$

Stephen Tashi said:
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...

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

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

The basic problem seems to be that allowed values of a increase with n, while induction would work only if a is constant.