Calculating the square root of 2 longhand

There are many numerical methods to calculate square roots. This is the long hand method I learned in junior high. It produces one (accurate) digit at a time, but the working numbers get larger each iteration. Eventually, it bogs down because it is doing math with integers hundreds, then thousands of digits long.

Still, it is fun for tinkering. If you run this Ruby script as is, it will calculate the square root of 2 to 1,000 digits. To adjust the precision, change the number of iterations on this line:

while iterations < 1000

Here is the entire script...

#!/usr/bin/ruby
# calculate a square root of 2 using longhand

def newroot(divisor, doubleroot)
  # calculate new root
  # formula is doubleroot _ * _ = closest to divisor
  # try 9, then 8, then 7 ...

  i = 9
  while i >= 0
    multiple1 = (doubleroot.to_s + i.to_s).to_i
    multiple2 = i
    product = multiple1 * multiple2

    if product <= divisor
      # found new root
      root = i
      # get modulus
      modulus = divisor - product
      break
    end
    i = i - 1
  end


  return root, modulus
end

roots = Array.new

# the nearest root to 2 is 1

root = 1
roots.push(root)
remainder = 2 - root

doubleroot = root * 2
divisor = remainder * 100

iterations = 0

while iterations < 1000
  root,remainder = newroot(divisor, doubleroot)
  roots.push(root)

  # compute new doubleroot 

  return root, modulus
end

roots = Array.new

# the nearest root to 2 is 1

root = 1
roots.push(root)
remainder = 2 - root

doubleroot = root * 2
divisor = remainder * 100

iterations = 0

while iterations < 1000
  root,remainder = newroot(divisor, doubleroot)
  roots.push(root)

  # compute new doubleroot
  doubleroot = roots.to_s.to_i * 2

  # compute divisor
  divisor = remainder * 100

  iterations = iterations + 1

  print "iteration " + iterations.to_s + "\n"
end

print "final roots are " + roots.to_s + "\n"