# Square root of a number without using sqrt() function

Given a number N, the task is to find the square root of N without using sqrt() function.

Examples:

Input: N = 25
Output: 5

Input: N = 3
Output: 1.73205

Input: N = 2.5
Output: 1.58114

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Start iterating from i = 1. If i * i = n, then print i as n is a perfect square whose square root is i.
• Else find the smallest i for which i * i is strictly greater than n.
• Now we know square root of n lies in the interval i – 1 and i and we can use Binary Search algorithm to find the square root.
• Find mid of i – 1 and i and compare mid * mid with n, with precision upto 5 decimal places.
1. If mid * mid = n then return mid.
2. If mid * mid < n then recur for the second half.
3. If mid * mid > n then recur for the first half.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Recursive function that returns square root ` `// of a number with precision upto 5 decimal places ` `double` `Square(``double` `n, ``double` `i, ``double` `j) ` `{ ` `    ``double` `mid = (i + j) / 2; ` `    ``double` `mul = mid * mid; ` ` `  `    ``// If mid itself is the square root, ` `    ``// return mid ` `    ``if` `((mul == n) || (``abs``(mul - n) < 0.00001)) ` `        ``return` `mid; ` ` `  `    ``// If mul is less than n, recur second half ` `    ``else` `if` `(mul < n) ` `        ``return` `Square(n, mid, j); ` ` `  `    ``// Else recur first half ` `    ``else` `        ``return` `Square(n, i, mid); ` `} ` ` `  `// Function to find the square root of n ` `void` `findSqrt(``double` `n) ` `{ ` `    ``double` `i = 1; ` ` `  `    ``// While the square root is not found ` `    ``bool` `found = ``false``; ` `    ``while` `(!found) { ` ` `  `        ``// If n is a perfect square ` `        ``if` `(i * i == n) { ` `            ``cout << fixed << setprecision(0) << i; ` `            ``found = ``true``; ` `        ``} ` `        ``else` `if` `(i * i > n) { ` ` `  `            ``// Square root will lie in the ` `            ``// interval i-1 and i ` `            ``double` `res = Square(n, i - 1, i); ` `            ``cout << fixed << setprecision(5) << res; ` `            ``found = ``true``; ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``double` `n = 3; ` ` `  `    ``findSqrt(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Recursive function that returns  ` `// square root of a number with  ` `// precision upto 5 decimal places ` `static` `double` `Square(``double` `n,  ` `                     ``double` `i, ``double` `j) ` `{ ` `    ``double` `mid = (i + j) / ``2``; ` `    ``double` `mul = mid * mid; ` ` `  `    ``// If mid itself is the square root, ` `    ``// return mid ` `    ``if` `((mul == n) ||    ` `        ``(Math.abs(mul - n) < ``0.00001``)) ` `        ``return` `mid; ` ` `  `    ``// If mul is less than n,  ` `    ``// recur second half ` `    ``else` `if` `(mul < n) ` `        ``return` `Square(n, mid, j); ` ` `  `    ``// Else recur first half ` `    ``else` `        ``return` `Square(n, i, mid); ` `} ` ` `  `// Function to find the square root of n ` `static` `void` `findSqrt(``double` `n) ` `{ ` `    ``double` `i = ``1``; ` ` `  `    ``// While the square root is not found ` `    ``boolean` `found = ``false``; ` `    ``while` `(!found)  ` `    ``{ ` ` `  `        ``// If n is a perfect square ` `        ``if` `(i * i == n)  ` `        ``{ ` `            ``System.out.println(i); ` `            ``found = ``true``; ` `        ``} ` `         `  `        ``else` `if` `(i * i > n)  ` `        ``{ ` ` `  `            ``// Square root will lie in the ` `            ``// interval i-1 and i ` `            ``double` `res = Square(n, i - ``1``, i); ` `            ``System.out.printf(``"%.5f"``, res); ` `            ``found = ``true``; ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``double` `n = ``3``; ` ` `  `    ``findSqrt(n); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` `import` `math ` ` `  `# Recursive function that returns square root ` `# of a number with precision upto 5 decimal places ` `def` `Square(n, i, j): ` ` `  `    ``mid ``=` `(i ``+` `j) ``/` `2``; ` `    ``mul ``=` `mid ``*` `mid; ` ` `  `    ``# If mid itself is the square root, ` `    ``# return mid ` `    ``if` `((mul ``=``=` `n) ``or` `(``abs``(mul ``-` `n) < ``0.00001``)): ` `        ``return` `mid; ` ` `  `    ``# If mul is less than n, recur second half ` `    ``elif` `(mul < n): ` `        ``return` `Square(n, mid, j); ` ` `  `    ``# Else recur first half ` `    ``else``: ` `        ``return` `Square(n, i, mid); ` ` `  `# Function to find the square root of n ` `def` `findSqrt(n): ` `    ``i ``=` `1``; ` ` `  `    ``# While the square root is not found ` `    ``found ``=` `False``; ` `    ``while` `(found ``=``=` `False``): ` ` `  `        ``# If n is a perfect square ` `        ``if` `(i ``*` `i ``=``=` `n): ` `            ``print``(i); ` `            ``found ``=` `True``; ` `         `  `        ``elif` `(i ``*` `i > n): ` ` `  `            ``# Square root will lie in the ` `            ``# interval i-1 and i ` `            ``res ``=` `Square(n, i ``-` `1``, i); ` `            ``print` `(``"{0:.5f}"``.``format``(res))  ` `            ``found ``=` `True` `        ``i ``+``=` `1``; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `3``; ` ` `  `    ``findSqrt(n); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `// Recursive function that returns  ` `// square root of a number with  ` `// precision upto 5 decimal places ` `static` `double` `Square(``double` `n,  ` `                     ``double` `i, ``double` `j) ` `{ ` `    ``double` `mid = (i + j) / 2; ` `    ``double` `mul = mid * mid; ` ` `  `    ``// If mid itself is the square root, ` `    ``// return mid ` `    ``if` `((mul == n) ||  ` `        ``(Math.Abs(mul - n) < 0.00001)) ` `        ``return` `mid; ` ` `  `    ``// If mul is less than n,  ` `    ``// recur second half ` `    ``else` `if` `(mul < n) ` `        ``return` `Square(n, mid, j); ` ` `  `    ``// Else recur first half ` `    ``else` `        ``return` `Square(n, i, mid); ` `} ` ` `  `// Function to find the square root of n ` `static` `void` `findSqrt(``double` `n) ` `{ ` `    ``double` `i = 1; ` ` `  `    ``// While the square root is not found ` `    ``Boolean found = ``false``; ` `    ``while` `(!found)  ` `    ``{ ` ` `  `        ``// If n is a perfect square ` `        ``if` `(i * i == n)  ` `        ``{ ` `            ``Console.WriteLine(i); ` `            ``found = ``true``; ` `        ``} ` `         `  `        ``else` `if` `(i * i > n)  ` `        ``{ ` ` `  `            ``// Square root will lie in the ` `            ``// interval i-1 and i ` `            ``double` `res = Square(n, i - 1, i); ` `            ``Console.Write(``"{0:F5}"``, res); ` `            ``found = ``true``; ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``double` `n = 3; ` ` `  `    ``findSqrt(n); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```1.73205
```

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