I'll argue that, during trial division (checking whether $n$ is a prime, for example) it is not necessary to test if $n$ is divisible by $n-1$, or $n -2$, down to a given number.

Suppose 17, for example (a prime). Using $x \nmid y$ to mean x does not divide y, and testing whether 17 is divisible by numbers between 1 and 17, suppose we get to 8; and we realise $8 \nmid 17$; naturally we would test $9 \mid 17$.

But if 9 divided 17, that would imply $17 = 9 * k, k \in \mathbb{N}$. This follows from the definition of division. To be divisible (loosely) means that a given number can be expressed as a product of two other numbers. But the least possible value of $k$ is 2, and $9 * 2 = 18 > 17$, and so we conclude no number bigger than 8 can be a divisor to 17 — this, together with knowning no number up to 8 divides 17, means 17 is a prime number.

A possible upper bound for checking division of primes could be $n / 2$, but, in fact, the best bound is $sqrt(n)$; the reason why is left as an exercise.