Prove that for all $d,n \in \mathbb{N}$ and all $x \in \mathbb{Z}$, if $d | n$ then $(x \bmod n) \bmod d = x \bmod d$

Question

Prove that for all $d,n \in \mathbb{N}$ and all $x \in \mathbb{Z}$, if $d | n$ then $(x \bmod n) \bmod d = x \bmod d$. I'm confused on where to even start. My initial thought is to perform a proof by contradiction by letting $d \nmid n$, resulting in $0 < n \bmod d < d$. I'm not sure where to go from here or if it's the right approach. Any help is appreciated!

Answer 1

By definition, we have that $x \text{ (mod } n)$ is the least non-negative residue of $x$ modulo $n,$ i.e., it is the remainder (as obtained by the Division Algorithm) of $x$ when divided by $n.$ Consequently, if we write $x = qn + r$ for some integers $q$ and $0 \leq r \leq n - 1,$ then $x \text{ (mod } n)$ is precisely $r.$ Likewise, the integer $[x \text{ (mod } n)] \text{ (mod } d)$ is the remainder (as obtained by the Division Algorithm) of $r = x \text{ (mod } n)$ when divided by $d.$ Put another way, if we write $r = q' d + r'$ for some integers $q'$ and $0 \leq r' \leq d - 1,$ then $[x \text{ (mod } n)] \text{ (mod } d) = r \text{ (mod } d)$ is precisely $r'.$

Let us assume now that $d \,|\, n,$ i.e., assume that there exists an integer $a$ such that $n = ad.$ Can you put this together with the content of the first paragraph to finish the proof?