Show that $n^2 -10\notin O(n)$
My friend keeps on insisting that my proof is incorrect. Could you please tell me if there is a mistake? (made suggested edit)
Suppose to the contrary that $n^2 -10\in O(n)$. By definition this means there exist $c >0$ and $n_0 > 0$ such that $n^2 - 10 \leq cn, \forall n>n_0$. Let $n = n_0 + c + 10$,
then $(n_0 + c + 10)^2 - 10 \leq c(n_0 + c + 10)$
$\Rightarrow$ $(n_0 + c)^2 + 2(10)(n_0+c) + 90 - cn_0 -c^2 - 10c \leq 0$
$\Rightarrow$ $(n_0^2 + 2n_0c + c^2) + 20n_0+20c + 90 - cn_0 -c^2 - 10c \leq 0$
$\Rightarrow$ $n_0^2 + n_0c + 20n_0+10c + 90 \leq 0$. Since $n_0$ and $c >0$ The left side is always greater than $0$. A contradction. Thus $n^2 -10\notin O(n)$.