Can we prove the converse of the division algorithm? The Division Alogrithm states that $\forall a, b \in \mathbb{N}$ where $b \neq 0,  \exists q,r\in \mathbb{N}$ such that $a=qb+r$ with $0 \leq r \lt b$. 
I want to check whether it is true that
 $ \forall q,r\in \mathbb{N}$, $\exists a, b \in \mathbb{N}$ such that $a=qb+r$ with $0 \leq r \lt b$. 
Next,
I am interested in the case where $b$ is a prime number and $a$ is any natural number, whether the statement is true, i.e.
$ \forall q,r\in \mathbb{N},  \forall b \in \mathbb{P},\exists a \in \mathbb{N}$, such that $a=qb+r$ with $0 \leq r \lt b$. 
Lastly,
I am interested in whether $ \forall r\in \mathbb{N},  \forall b \in \mathbb{P},\exists a,q \in \mathbb{N}$, such that $a=qb+r$ with $0 \leq r \lt b$

The first claim is true, you can just pick any $b>r$ and choose $a=bq+r$.
If you suppose that $r<b$, then the second claim holds too since you can, again just write $a=bq+r$ because $bq+r \in \mathbb N$ clearly.
For the third, one, consider $a-r$, then you can always find a suitable $a$ such that $b$ divides $a-r$ thanks to the prime decomposition, hence, there exists a $q$, such that:
$a=bq+r$
