Convert given Decimal number into an irreducible Fraction
Given a decimal number as N, the task is to convert N into an equivalent irreducible fraction.
An irreducible fraction is a fraction in which numerator and denominator are co-primes i.e., they have no other common divisor other than 1.
Examples:
Input: N = 4.50
Output: 9/2
Explanation:
9/2 in decimal is written as 4.5Input: N = 0.75
Output: 3/4
Explanation:
3/4 in decimal is written as 0.75
Approach: Follow the steps given below to solve the problem.
- Fetch integral value and fractional part of the decimal number ‘n’.
- Consider the precision value to be 109 to convert the fractional part of the decimal to an integral equivalent.
- Calculate GCD of the integral equivalent of fractional part and precision value.
- Calculate numerator by dividing the integral equivalent of fractional part by GCD value. Calculate the denominator by dividing the precision value by GCD value.
- From the obtained mixed fraction, convert it into an improper fraction.
For example N = 4.50, integral value = 4 and fractional part = 0.50
Consider precision value to be (109) that is precision value = 1000000000
Calculate GCD(0.50 * 109, 109) = 500000000
Calculate numerator = (0.50 * 10^9) / 500000000 = 1 and denominator = 10^9/ 500000000 = 2
Convert mixed fraction into improper fraction that is fraction = ((4 * 2) + 1) / 2 = 9/2
Below is the implementation of the above approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Recursive function to// return GCD of a and blong long gcd(long long a, long long b){ if (a == 0) return b; else if (b == 0) return a; if (a < b) return gcd(a, b % a); else return gcd(b, a % b);}// Function to convert decimal to fractionvoid decimalToFraction(double number){ // Fetch integral value of the decimal double intVal = floor(number); // Fetch fractional part of the decimal double fVal = number - intVal; // Consider precision value to // convert fractional part to // integral equivalent const long pVal = 1000000000; // Calculate GCD of integral // equivalent of fractional // part and precision value long long gcdVal = gcd(round(fVal * pVal), pVal); // Calculate num and deno long long num = round(fVal * pVal) / gcdVal; long long deno = pVal / gcdVal; // Print the fraction cout << (intVal * deno) + num << "/" << deno << endl;}// Driver Codeint main(){ double N = 4.5; decimalToFraction(N); return 0;} |
C
// C implementation of the above approach#include <math.h>#include <stdio.h>int gcfFinder(int a, int b){ // gcf finder int gcf = 1; for (int i = 1; i <= a && i <= b; i++) { if (a % i == 0 && b % i == 0) { gcf = i; } } return gcf;}int shortform(int* a, int* b){ for (int i = 2; i <= *a && i <= *b; i++) { if (*a % i == 0 && *b % i == 0) { *a = *a / i; *b = *b / i; } } return 0;}// Driver Codeint main(void){ // converting decimal into fraction. double a = 4.50; int c = 10000; double b = (a - floor(a)) * c; int d = (int)floor(a) * c + (int)(b + .5f); while (1) { if (d % 10 == 0) { d = d / 10; c = c / 10; } else break; } int* i = &d; int* j = &c; int t = 0; while (t != 1) { int gcf = gcfFinder(d, c); if (gcf == 1) { printf("%d/%d\n", d, c); t = 1; } else { shortform(i, j); } } return 0;}// this code is contributed by harsh sinha username-// harshsinha03 |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Recursive function to// return GCD of a and bstatic long gcd(long a, long b){ if (a == 0) return b; else if (b == 0) return a; if (a < b) return gcd(a, b % a); else return gcd(b, a % b);} // Function to convert decimal to fractionstatic void decimalToFraction(double number){ // Fetch integral value of the decimal double intVal = Math.floor(number); // Fetch fractional part of the decimal double fVal = number - intVal; // Consider precision value to // convert fractional part to // integral equivalent final long pVal = 1000000000; // Calculate GCD of integral // equivalent of fractional // part and precision value long gcdVal = gcd(Math.round( fVal * pVal), pVal); // Calculate num and deno long num = Math.round(fVal * pVal) / gcdVal; long deno = pVal / gcdVal; // Print the fraction System.out.println((long)(intVal * deno) + num + "/" + deno);} // Driver Codepublic static void main(String s[]){ double N = 4.5; decimalToFraction(N);} }// This code is contributed by rutvik_56 |
Python3
# Python3 program for the above approachfrom math import floor# Recursive function to# return GCD of a and bdef gcd(a, b): if (a == 0): return b elif (b == 0): return a if (a < b): return gcd(a, b % a) else: return gcd(b, a % b)# Function to convert decimal to fractiondef decimalToFraction(number): # Fetch integral value of the decimal intVal = floor(number) # Fetch fractional part of the decimal fVal = number - intVal # Consider precision value to # convert fractional part to # integral equivalent pVal = 1000000000 # Calculate GCD of integral # equivalent of fractional # part and precision value gcdVal = gcd(round(fVal * pVal), pVal) # Calculate num and deno num= round(fVal * pVal) // gcdVal deno = pVal // gcdVal # Print the fraction print((intVal * deno) + num, "/", deno)# Driver Codeif __name__ == '__main__': N = 4.5 decimalToFraction(N)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{ // Recursive function to// return GCD of a and bstatic long gcd(long a, long b){ if (a == 0) return b; else if (b == 0) return a; if (a < b) return gcd(a, b % a); else return gcd(b, a % b);} // Function to convert decimal to fractionstatic void decimalToFraction(double number){ // Fetch integral value of the decimal double intVal = Math.Floor(number); // Fetch fractional part of the decimal double fVal = number - intVal; // Consider precision value to // convert fractional part to // integral equivalent long pVal = 1000000000; // Calculate GCD of integral // equivalent of fractional // part and precision value long gcdVal = gcd((long)Math.Round( fVal * pVal), pVal); // Calculate num and deno long num = (long)Math.Round(fVal * pVal) / gcdVal; long deno = pVal / gcdVal; // Print the fraction Console.WriteLine((long)(intVal * deno) + num + "/" + deno);}// Driver Codepublic static void Main(String []s){ double N = 4.5; decimalToFraction(N);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to implement// the above approach// Recursive function to// return GCD of a and bfunction gcd(a, b){ if (a == 0) return b; else if (b == 0) return a; if (a < b) return gcd(a, b % a); else return gcd(b, a % b);} // Function to convert decimal to fractionfunction decimalToFraction(number){ // Fetch letegral value of the decimal let letVal = Math.floor(number); // Fetch fractional part of the decimal let fVal = number - letVal; // Consider precision value to // convert fractional part to // letegral equivalent let pVal = 1000000000; // Calculate GCD of letegral // equivalent of fractional // part and precision value let gcdVal = gcd(Math.round( fVal * pVal), pVal); // Calculate num and deno let num = Math.round(fVal * pVal) / gcdVal; let deno = pVal / gcdVal; // Print the fraction document.write((letVal * deno) + num + "/" + deno);} // Driver Code let N = 4.5; decimalToFraction(N);</script> |
Output:
9/2
Time complexity: O(log min(a, b))
Auxiliary space: O(1)
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