Square root loop in python

Question

I need to take an input of a number greater than 2, and take the square root until the square root is less than two. I need a print statement that includes the count of the times the square root of the number was taken as well as the output. What I have so far is:

import math

input_num = float(input("Enter a number greater than two: "))

while input_num < 2:
    input_num = float(input("Enter a number greater than two: "))
else:
    sqrt_num = math.sqrt(input_num)
    count = 1
    while sqrt_num > 2:
        sqrt_num = math.sqrt(sqrt_num)
        count += 1
        print(count, ": ", sqrt_num, sep = '')

With the output of:

Enter a number greater than two: 20
2: 2.114742526881128
3: 1.4542154334489537

I want to include that first iteration of count 1. How do I write a proper loop so it looks like:

Enter a number greater than two: 20
1: 4.47213595499958
2: 2.114742526881128
3: 1.4542154334489537

Dyspro Dyspro · Accepted Answer · 2019-03-19T01:25:22

Its kind of a hacky way to do it, or at least doesn't make as much sense since it makes the variable sqrt_num not the square root, but I would initialize count to 0 and initialize sqrt_num to input_num, like so:

import math

input_num = float(input("Enter a number greater than two: "))

while input_num < 2:
    input_num = float(input("Enter a number greater than two: "))
else:
    sqrt_num = input_num
    count = 0
    while sqrt_num > 2:
        sqrt_num = math.sqrt(sqrt_num)
        count += 1
        print(count, ": ", sqrt_num, sep = '')

Square root loop in python

1 Answers