Q:

disprove : $\forall n \in \mathbb{N} \exists m \in \mathbb{N}$ such that $n<m<n^2$

I figured this would be easier to solve if I thought of it as proving and existential statement and ended up with  $\exists n \in \mathbb{N}\ \forall m \in \mathbb{N}$ such that $n\ge m \ge$. Now I'm stuck.  Is the only way that statement can be true is if $n =m$, but that doesn't really disprove the original statement. At least I don't think it does. Any suggestion about how to go about this problem?

A:

It's not true for $n = 1$. (because $n = n^2$)  
But if $n>1$ then:
$n \ge 2$.  Let $m = n + 1$ so
$ n < m =n + 1  < n + n = 2*n \le n*n = n^2 $.
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'nother proof.
Let $n \ge 2$.  Then $n^2 - n = n(n - 1) \ge 2*1 = 2$.  So there is at least one natural number between $n$ and $n^2$.
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third proof:
If there is no such m.  Then either:
1:  $n^2 = n + 1$.  But that is impossible ans $n^2 - n - 1 = 0$ has no rational roots.
2: $n^2 = n$.  Then $n(n - 1) = 0$ so $n = 1$ or $n = 0$.
3: $n^2 < n$.  As $0^2 = 0$, $n \ne 0$. So as N is natural number $n \gt 0$.  So $n^2/n < n/n$ so $n < 1$ which contradicts that $n$ is a natural number.  So this is impossible.
Conclusion:  $n = 1$ or $n = 0$ (If you are using the definition that 0 is a natural number. Most mathematicians use the definition that 0 isn't a natural number.)  

