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 $0 \leq r \lt b$
 A: 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:
$a=bq+r$
A: The first assertion is true. Given $r,q \in \mathbb{N}$, just choose $b > r$ and then choose $a$ to be equal to $bq + r$.
The next two are both false. Given $r,b \in \mathbb{N}$, if $r > b$, then these cannot hold true.  
In case you assume $r < b$, then in all the cases, whenever you are given $b,q,r \in \mathbb{N}$, you are free to choose $a$ to equal $bq+r$, and thus the claims are true.
A: One important distinction is that for $a,b$ then $a = qb + r; 0 \le r < b$ than $q$ and $r$ are unique.
For $q,r$ let $b$ be any $b > r$ then let $a = qb +r$ then $a$ is dependent upon $b$ which can be any value satisfying a very weak condition; $a,b$ are far from unique.
The next statement is bizarrely stated:  You say for any prime $b$ but then you claim $r < b$.  Well obviously it isn't true for any $r$ and for  ALL prime $b$ that $b > r$.  Just let $r = 10$ then $b = 2,3,5,7 \le r$.
But for any $q,r$ and any prime $b$ so that $b > r$ we can simply let $a = qb + r$ to find such a number.
So $\forall q,r$ and $\forall$ prime $b > r$ then $\exists a$ so that $a = qb + r$.
The last one is exactly the same as the third one except with the labels "b" and "q" reversed.  Labels are just labels.  Is it is true so long as you require that $b > r$.
