How do I make a primitive recursive function that does division? I am trying to define a primitive recursive function that does division. I looked at this answer but it seems wrong to me, because according to Wikipedia:

The primitive recursive functions are among the number-theoretic functions, which are functions from the natural numbers (nonnegative integers) {0, 1, 2, ...} to the natural numbers

So the inequality x−t⋅y≥0 will always be true and the function will always keep adding +1. The function given in the answers seems right but only assuming that I have negative numbers. Now how could I build a PRF with just natural numbers?
EDIT: I found a way to either make a division that always rounds up or always rounds down. But I haven't found one yet that always does the correct thing. So far:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,S(m)))))

where S(m) is successor, A is addition, V is 0 if 0 and 1 otherwise, D is subtraction and M is multiplication.
Now the above always rounds down and the next one always rounds up:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,m))))

 A: Definition by cases is a valid, derived principle of definition for primitive recursive functions. So is subtraction, and so is equality. I will therefore use them freely.
Moreover, it is a good idea to define not just integer division $d(m,n)$ but also the remainder function $r(m,n)$. One can then write
\begin{align}
r(0,n) & = 0 \\
r(m+1,n) & = \begin{cases} 0 & \text{if}\; n-1 = r(m,n) \\ r(m,n) + 1 &\text{otherwise} \end{cases}
\end{align}
and define integer division by
\begin{align*}
d(0,n) & = 0 \\
d(m+1,n) & = \begin{cases} d(m,n) + 1 & \text{if}\; r(m,n) = n-1 \\
d(m,n) & \text{otherwise} \end{cases}
\end{align*}
A: I know I'm late to the party, but here goes formulation given by N. Cutland in Computability: An Introduction to Recursive Function Theory (p. 36):
First, let's define
$$
sg(x) = \begin{cases}
0 &\mbox{if } x = 0,\\
1 &\mbox{if } x \neq 0.
\end{cases}
$$
By recursion we have $sg(0) = 0$ and $sg(x + 1) = 1$, hence $sg$ is computable. Analogically, we define $\overline{sg}$ which returns $1$ if $x = 0$ and 0 otherwise. Next, let's define reminder when $y$ is divided by $x$ (to obtain a total function we adopt the convention $rm(0, y) = y)$}. We have:
$$
rm(x, y + 1) = \begin{cases} rm(x, y) + 1 &\mbox{if } rm(x, y) + 1 \neq x \\
0 & \mbox{if } rm(x, y) + 1 = x \end{cases}
$$
This gives the following definition by recursion:
$$
rm(x, 0) = 0,\\
rm(x, y + 1) = (rm(x, y) + 1) \times sg(|x - (rm(x, y) + 1)|).
$$
The second equation can be written as $rm(x, y + 1) = g(x, rm(x, y))$
where $g(x, z) = (z + 1) \times sg(|x - (z + 1)|)$; and $g$ is computable by several applications of substitution. Hence, $rm(x, y)$ is computable. Then, we can define quotient when $y$ is divided by $x$ (to obtain a total function we define $qt(0, y) = 0$):
$$
qt(x, y + 1) = \begin{cases}
qt(x, y) + 1 &\mbox{if } rm(x, y) + 1 = x,\\
qt(x, y) &\mbox{if } rm(x, y) +1 \neq x.
\end{cases}
$$
By recursion, we have:
$$
qt(x, 0) = 0,\\
qt(x, y + 1) = qt(x, y) + \overline{sg}(|x - (rm(x, y) + 1)|).
$$
Hence, $qt$ is computable.
A: If y has to divide x such that x>y,
then, initially assuming i=0;
x/y=f(x,y,0);
f(x,y,i)=f(x-y,y,i+1)
quotient =i and remainder =x-y (finally, such that when 
y>(x-y)) ,
 the recursive step has to be stopped.
Example 1:
25/4=f(25,4,0)
=f(21,4,1)
=f(17,4,2)
=f(13,4,3)
=f(9,4,4)
=f(5,4,5)
=f(1,4,6)
Hence quotient=i=6 and reaminder=x-y=1.
i.e, 25=6*4+1!
Example 2:
30/8=f(30,8,0)
=f(22,8,1)
=f(14,8,2)
=f(6,8,3)
Since 6<8, we stop here.
Therefore 30=3*8+6!!
A: I found a solution to my problem. The main issue was to figure out if $a - b \geq 0.$ One has to use a little trick to figure this out. Below is the function I used for it and the final primitive recursive function that does division.
$a-b \geq 0$
$BOE(a,0) = 1$
$BOE(a,S(m)) = A(V(D(a,S(m))),E(a,S(m)))$
with $A$ being Addition, $V(0) = 0$ else $1$, $D$ being subtraction, $E$ being equals, $S$ being successor and $M$ being multiplication.
$$ Div(x,y,0) = 0$$
$$Div(x,y,S(m)) = A(A(Div(x,y,m),V(D(x,M(y,S(m))))),E(D(x,M(m,y)),D(M(m,y),x)))$$
