In the ring $\mathbb Z_n$,we have $kx=k.x$.

Question

Let $\mathbb Z_n$ be the ring of integers modulo $n$.Now,suppose for $x\in \mathbb Z_n$,define $kx=x+x+...+x(k$ times$)$.And $k.x$ denote the product of $k\in \mathbb Z_n$ and $x\in \mathbb Z_n$.Is it true that $kx=k.x$?It is although very trivial still relevant thing to note.

Answer 1

Write $k = k_0 + nb$ and $x = x_0 + nc$ where all numbers are integers. Then

\begin{align} kx &= (k_0 + nb)(x_0+nc)\\ &= k_0 x_0 + n\left( bx_0 + ck_0 + nbc\right) \end{align} where all calculations are in $\mathbb{Z}$. Thus, modulo $n$, you have $[kx] =[k][x] = \underbrace{[x]+[x]+\ldots + [x]}_{k \text{times}}$. So the answer is yes.