Except the Dividend Itself — Any Divisor is Less than Half of the Dividend

Question

Postulate $d \neq n$ is a divisor, $n$ is a dividend. Why $d \le n/2$? I know the dividend itself is a divisor.

$d|n$ is defined as $\exists \; c\in \mathbb{Z}$ such that $dc = n$.

$\color{blue}{|c| \ge 1}$ therefore $d\color{blue}{|c| \ge 1}d.$ Take absolute value of this
$|n|=|dc| \quad \ge |d| \iff |n| \ge |d| $.

(1) How does this result in $d \le n/2$?
(2) Any intuiton?

Answer 1

If $d>\frac{n}{2}$ and $k\ge 2$ then $kd>n$.

Answer 2

If $d$ is a divisor of $n$, it means that $n$ is a multiple of $d$, so $n = kd$ for some integer $k$. If further $d < n$, then $k > 1$.

So, as $k \ge 2$, $$d = \frac{n}{k} \le \frac{n}{2}$$

Further you can things like: if $d < n/2$, then $d \le n/3$, and so on.

Answer 3

If $n=cd$ and $c\neq 1$ then $c\geq 2$ hence $n\geq 2d$. This all under assumption that $d$ and $n$ are positive.