Why do we check up to the square root of a prime number to determine if it is prime?
If a number n is not a prime, it can be factored into two factors a and b:
n = a * b
Now a and b can't be both greater than the square root of n, since then the product a * b would be greater than sqrt(n) * sqrt(n) = n. So in any factorization of n, at least one of the factors must be smaller than the square root of n, and if we can't find any factors less than or equal to the square root, n must be a prime.
Let's say m = sqrt(n) then m × m = n. Now if n is not a prime then n can be written as n = a × b, so m × m = a × b. Notice that m is a real number whereas n, a and b are natural numbers.
Now there can be 3 cases:
- a > m ⇒ b < m
- a = m ⇒ b = m
- a < m ⇒ b > m
In all 3 cases, min(a, b) ≤ m. Hence if we search till m, we are bound to find at least one factor of n, which is enough to show that n is not prime.
Because if a factor is greater than the square root of n, the other factor that would multiply with it to equal n is necessarily less than the square root of n.