Why does $\frac{X}{b}=d_n b^{n-1}+\ldots+ d_1+ \frac{d_0}{b}$ imply $d_0$ is the remainder?

Question

Suppose you have an integer $X$ written in a base $b>1$ as:

$$X=d_n b^{n}+\ldots+d_1 b^1+d_0.$$

Dividing both sides by $b$ we get:

$$\frac{X}{b}=d_n b^{n-1}+\ldots+ d_1+ \frac{d_0}{b}.$$ Why does this imply that $d_0$ is the remainder of the division of $X$ by $b$?

Thanks.

Answer 1

If $D,d\in\mathbb N$ then, by definition. the quotient and the remainder of the division of $D$ by $d$ are the only numbers $q,r\in\mathbb{Z}^+$ such that $D=q\times d+r$ and that $0\leqslant r<d$.

So, since$$X=b\times(d_nb^{n-1}+b_{n-1}b^{n-2}+\cdots+d_1)+d_0$$and since $0\leqslant d_0<b$, $d_0$ is the remainder of the division of $X$ by $b$, by the definition of remainder.