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

[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

 

[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}

 

[3.1415926535]

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]

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