Proof that $f(a)$ = $a$ div $d$ is onto

Solution 1:

Your reasoning (insofar as I can understand it) is circular. First you say

$$𝑎 \text{ div } 𝑑=𝑦 \text{ implies that } 𝑎=𝑑𝑦+𝑟$$

without specifying any particular $a$ whose existence you want to prove.

Then you end with $$ \text{ for an arbitrary } 𝑦, \text{ there is an element }𝑑𝑦+𝑟 \ldots $$

but you never did actually point out such an element.

Here's a hint. What is the remainder when you divide $a=3$ by $d=7$?