Java: Find the square root of a number using Babylonian method

Last update on May 15 2025 13:08:34 (UTC/GMT +8 hours)

Babylonian Square Root

Write a Java program to find the square root of a number using the Babylonian method.

Sample Solution:

Java Code:

import java.util.*;
public class solution {	
  public static float square_Root(float num) 
    { 
        float a = num; 
        float b = 1; 
        double e = 0.000001; 
        while (a - b > e) { 
            a = (a + b) / 2; 
            b = num / a; 
        } 
        return a; 
    } 
 
   public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        System.out.print("Input an integer: ");
        int num = scan.nextInt();
        scan.close(); 
		System.out.println("Square root of a number using Babylonian method: "+square_Root(num));		
		}
 }

Sample Output:

 Input an integer:  25
Square root of a number using Babylonian method: 5.0

Flowchart:

For more Practice: Solve these Related Problems:

• Write a Java program to calculate the square root of a number using the Babylonian method with a given tolerance.
• Write a Java program to implement the Babylonian method recursively and count the iterations needed for convergence.
• Write a Java program to compare the Babylonian square root result with Math.sqrt over various test cases.
• Write a Java program to simulate the Babylonian square root algorithm using tail recursion.

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