Ruby - determine if a number is a prime

Ruby comes with predefined classes such as Prime. All you have to do is to require that class into your project.

require 'prime'

Than, you can use some of the Prime methods such as first to get first x prime elements:

Prime.first(5) # Ret => [2, 3, 5, 6, 11]

Or you could do something like this:

Prime.each(100) do |prime|
  p prime # Ret => [2, 3, 5, 7, 11, ..., 97]
end

I hope you find this useful.

It's because = is of higher precedence than or. See Ruby's operator precedence table below (highest to lowest precedence):

[ ] [ ]=
**
! ~ + -
* / %
+ -
>> <<
&
^ |
<= < > >=
<=> == === != =~ !~
&&
||
.. ...
? :
= %= { /= -= += |= &= >>= <<= *= &&= ||= **=
defined?
not
or and
if unless while until
begin/end

The problematic line is being parsed as...

(foundDivider = ((n % d) == 0)) or foundDivider

...which is certainly not what you mean. There are two possible solutions:

Force the precedence to be what you really mean...

foundDivider = (((n % d) == 0) or foundDivider)

...or use the || operator instead, which has higher precedence than =:

foundDivider = ((n % d) == 0) || foundDivider

def prime? n
  (2..Math.sqrt(n)).none? {|f| n % f == 0}
end

The range of factors should start at 2 and end at the square root of n because every number is divisible by one and no number is divisible by two numbers greater than its square root.

Explanation: A non-prime number is the product of two numbers.

n = f1 * f2

n is always divisible by its square root so both f1 and f2 cannot be greater than the square root of n, otherwise f1 * f2 would be greater than n. Therefore, at least one factor is less than or at most equal to Math.sqrt(n). In the case of finding prime numbers its only necessary to find one factor so we should loop from 2 to the square root of n.

def prime(n)
  return false if n < 2

  (2..n/2).none?{|i| n % i == 0}
end

A prime number is any number that has no positive divisors other than itself and 1.