Because square roots are usually irrational, we generally don’t want to convert them into decimal form until the very last step, if at all. However, it is typical to simplify square roots by taking out any perfect square factors. To do this, we need to identify these.

One method is to create factor trees. The method in this post, however, involves creating a list of prime factors. We test the target number against each prime number in turn. The first few prime numbers are 2, 3, 5, 7, 11, and 13.

For instance, what are the prime factors of 924?

Can 924 be divided by 2? Yes, 924/2 = 462. Can 462 be divided by 2? Yes, 462/2 = 231.

231 can’t be divided by 2, so now we try 3. 231/3 = 77.

77 can’t be divided by 3, so we try 5. That doesn’t work either, so we try 7: 77/7 = 11, which we also know is a prime.

This gives us our list of factors: 924 = 2 * 2 * 3 * 7 * 11. Of these, only 2 * 2 represents a perfect square, so \(\sqrt{924} = 2\sqrt{231}\).

When we’re testing prime numbers, it’s important to know when we can stop. Let’s say we want to know if 113 is prime. Do we need to test all numbers smaller than 101? That’s a lot of work.

It turns out we only need to test all the prime numbers up to \(\sqrt{113}\). Why is this?

To see why, look at 115. The prime factors of 115 are 5 and 23. While there is a prime factor that is greater than \(\sqrt{115}\), there is also a prime factor less than it.

In general, if a number \(n\) is composite, it has at least two factors, \(a\) and \(b\). Since \(n = ab\), then \(a = n/b\). Let \(m = \sqrt{n}\), so \(n = m\cdot m\) and \(m = n/m\). If \(a > m\), then \(n/b > n/m\).

Solve this for \(b\): \(n > nb/m \Rightarrow nm > nb \Rightarrow m > n\).

In other words, if there is a factor that is greater than \(\sqrt{n}\), there is another factor that is less than \(\sqrt{n}\). So we only need to try the prime numbers less than the square root of a number to see if it’s prime.

For 113 specifically: Since 113 is less than 121, we only need to test prime numbers less than 11, that is, 2, 3, 5, and 7. None of these are factors of 113, so we can conclude that 113 is prime (which it is).

Since 17 * 17 = 289, that means that the list of primes provided above (2, 3, 5, 7, 11, 13) are enough to test any number less than 289 for factors. Adding 17 and 19 to the list lets us test less that 529.