Is it true that $ f(n) = O(g(n))$ implies $g(n) = O(f(n))$ So I have this is an assignment for algorithms. I've googled a lot, read the chapter in the book about big Oh notation, and I understand the concept. I do not however understand how to prove it. I know I need to work with a constant and put that out of the equation. And I know that the answer I've found for $f(n) = O(g(n))$ implies $g(n) = O(f(n))$ is

NO. $f(x)=x$,$g(x)=x^2$ is a counterexample.

But I do not understand this answer. If $g(x)$ could be $x^2$ then $f(n)$ would not be equal the $O(g(n))$.
Can someone explain in a simple way why this is the case though? :(
 A: The definition of the big-oh notation is as follows :

$f(x)=O(g(x))$ if $|f(x)| \leq c|g(x)|$ for every big enough $x$ and some constant $c$ 

This is why $f(x)=x$ and $g(x)=x^2$ is a counter-example :


*

*$x=O(x^2)$ because for example taking $c=1$ we have $x \leq x^2$ for every $x \geq 1$

*$x^2$ can't be $O(x)$ because that would mean that $x^2 \leq cx$ for every $x \geq x_0$ or $x^2-cx \leq 0$ but this is false because : $$\lim_{x \to \infty} x^2-cx=\infty$$
The usual intuition is that : 

$f(x)=O(g(x))$ when $g(x)$ grows faster than $f(x)$

This explains why $x=O(x^2)$ but the reverse isn't true .
A: Recall the definition of $g = O(f)$: 

We say that $g = O(f)$ iff there are $c \ge 0$ and $N \in \mathbf N$ such that 
  $$\def\abs#1{\left|#1\right|}\abs{g(n)} \le c\abs{f(n)}, \quad n \ge N. $$

Suppose now $f(n) = n$ and $g(n) = n^2$, we would have to show that 
$$ n^2 \le cn $$
is true for "$n$ large" and some $c$. But regardless of which $c$ we choose, if $n$ is larger this does not hold (note that $c$ has to be fixed). Hence $g \ne O(f)$.
