Prove that $10 \times n^2 > n!$ is false for all n in the natural numbers I know this statement can be easily disproven by using a counter example (like n = 6) but I was wondering if there was a more rigorous proof for this problem.
 A: I don't know what you mean by rigorously.

If the statement is:
$$\forall x\in X: P(x)$$
(where $P(x)$ is some statement about the element $x$), then the negation of this statement is
$$\exists x\in X: \neg P(x).$$
A proof of any statement that begins with $\exists$ can be done rigorously by finding one such $x$ that satisfies the statement following the quantifier.
This means that if $\neg P(x_0)$ is true for some $x_0\in X$, then this very fact proves that $$\exists x\in X: \neg P(x)$$ is a true statement, and since this is a strue statement, its negation is a false statement.

In your case, the original statement is
$$\forall n\in\mathbb N: 10n^2>n!$$
and its negation is
$$\exists n\in\mathbb N: 10n^2\leq n!.$$
Setting $n_0=6$ means that the statement $$10n_0^2 \leq n!$$ is true, meaning that the statement
$$\exists n\in\mathbb N: 10n^2\leq n!$$
is also true, and this means that the original statement is false.

Every logical step I explained is correct, so I would call this proof rigorous.
A: For $ n \geq 6 $ it's possible to prove that the statement is false
$10 n^² \leq n!$
$10 n^² - n! \leq 0$
$ n(10n- (n-1)!) \leq 0$
$ (10n- (n-1)!) \leq 0$  since n is natural
it's easy to prove that $(n-2)(n-1) \geq n $ for $n\geq 4$
And that 1.2.3.4 $\geq 10=1.2.5$
.....you can show then that n must be $\geq 6$
So for $ n \geq 6 $ , $n! \geq 10 n^2 $
A: It's important to distinguish between the following:


*

*$10n^2>n!$ is false for all $n\in\mathbb N$ (what you wrote) and  

*The statement "$10n^2>n!$ for all $n\in\mathbb N$" is false


The first version would be equivalent to "$10n^2\leq n!$ for all $n\in\mathbb N$", and you would need to prove this for every $n$. But you can't, because it isn't true for all $n$, as Toby Mak points out. (It's true for all $n\geq 6$, and you could prove this e.g. by induction.)
For the second version, a counterexample is enough, as 5xum says.
A: Claim: 
For every $6 \leq n \in \mathbb{N}$ , 
we have $10 \times n^2 < n!$ .  
Proof by induction: 
The claime is true for $n=6$ . 
$ \ \ \ \ \ \checkmark \checkmark \checkmark$  
Suppose that the assertion holds for $k=n$; 
i.e. $\color{Blue}{10 \times n^2 < n!}$ ,  
on the other hand we know that : $ 22 < n! $;
also we know that $ 10 < 2n$ , which implies $20n+10 < 22n$;
so we can conclude that: 
$$ 
22 < n! 
\Longrightarrow 
22  n < n! \cdot n 
\Longrightarrow 
20n+10 < n! \cdot n 
\Longrightarrow 
\color{Red}{10 \times (2n+1) < n! \cdot n} 
\Longrightarrow 
\\ 
10 \times (n+1)^2 = 
\color{blue}{ 10 \times n^2} + 
\color{Red}{10 \times (2n+1)} 
< 
\color{Blue}{n!} + 
\color{Red}{n! \cdot n} = 
n!(1+n) = 
(n+1)! 
\ \ 
; 
$$ 
which implise the claime for $k=n+1$ .
