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'.

For $q\in[0,1)$, we can write it as $\frac1k$ for $k\in(1,\infty)$, and when we exponentiate, it is easy to see that $k^{n+1}$ is greater than $k^{n}$ and thus $k\rightarrow \infty \implies q\rightarrow0$.
