prove by induction; integer division by $5$ I'm having trouble with my math project.
 I have to prove that for very natural, $n$, there is an integer $q$ , and an integer $r$ such that:
$n=5q+r$   and $0\le r<5$   using induction
I have tried using the axiom of Archimedes but I can't really get around the problem. Sorry if this seems basic , I still have a really hard time with demonstrations 

Start with the base case: $n=0$. Then, clearly $n=5\cdot 0+0$, so we are okay. 
Next, suppose that the result holds for some fixed integer $n\geq 0$. We want to prove that the result holds for $n+1$. I.e., we want to prove that there exist integers $q$ and $0\leq r< 5$ satisfying $n+1=5q+r$.  
By the inductive hypothesis, there are integers $q'$ and $0\leq r'<5$ such that $n=5q'+r'$. But this means that $n+1=5q'+r'+1$. Are we finished? If $r'+1<5$, then yes (take $1=1'$ and $r=r'+1$). But what if $r'+1=5$. This happens precisely when $r'=4$. Then you have $n+1=5q'+5$. Do you see what you can do here?
