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))))


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.
