A Universe of Sorts

It's not that all divisors are smaller than $\sqrt n$ . For example, consider $14 = 7 \times 2$ . $\sqrt{14} \sim 4$ , but one of its diviors ( $7$ ) is greater than 4.
Rather, it is that if there is a divisor $l$ (for large) which is larger than $\sqrt n$ , there will be another divisor $s$ which is smaller than $\sqrt n$ .
Proof: Suppose $l \div n$ , $l \geq \sqrt n$ . So there exists an $s$ such that $ls = n$ , or $s = n / l$ .
Since $l \geq \sqrt n$ , $n / l \leq n / \sqrt n = \sqrt n$ . Thus $s \leq \sqrt n$ .
So if we wish to find some factor of $n$ , we can simply search within the range $\sqrt n$ .
If $n$ has no factors in the range $\sqrt n$ , then $n$ must be prime, for if $n$ did have a larger factor, $n$ would also have a smaller factor we would have found.

§ Why Searching for Divisors Upto sqrt(n) Works