For any $q\in [0,1]$ there exist $p>0$ such that $q(1+p)=1$ I was reading what is mathematics and to prove that if $q$ is a number between $0$ and $1$ then $q^n$ tends to $0$, they use that $q$ can be written as $q=\frac{1}{1+p}$ with $p>0$. 

It's equivalent to: 
  for any $q\in (0,1)$ there exist $p>0$ such that $q(1+p)=1$. 

If I have a number, such as $8/10$ I can find such a $p$, but not sure how can I proceed to prove it in 'general'.

How about just solving the equation?
$$q(1+p)=1$$
$$1+p = \frac{1}{q}$$
$$ p=\frac{1}{q}-1$$
Of course this doesn't work for $q=0$ (division by zero) or $q=1$ (then $p =0 \not \gt 0$) , but in those cases, the statement is clearly false.
Thus, for $q \in (0,1)$, we know that $\frac{1}{q}>1$, so $p=\frac{1}{q}-1>0$.
