> 
> Let b denote a fixed positive integer, for every integer $n≥0$ there
> exist nonnegative integers q and r such that $n=qb+r$$,0≤r<b$
> 
> 
> 

So I'm trying to solve it with induction, the same way I have with other problems of my book :

* $A(0):0=0b+0$
* $A(1):1=0b+1$

...

* $A(k):k=qb+r$

now I evaluate for k+1

* $A(k+1):k+1=qb+r$

$A(k)+1:k+1=qb+r+1$

So, at this point, I try to do, as I do with other proofs by induction, and try to prove:

$qb+r=qb+r+1$

but as you can see, this makes no sense, because $qb+r<qb+r+1$, so, I think I might be doing it the wrong way. Could you help me with this one?

Thanks in advance.