Program to add two fractions

Add two fraction a/b and c/d and print answer in simplest form.

Examples :

Input:  1/2 + 3/2
Output: 2/1

Input:  1/3 + 3/9
Output: 2/3

Input:  1/5 + 3/15
Output: 2/5



Algorithm to add two fractions

  • Find a common denominator by finding the LCM (Least Common Multiple) of the two denominators.
  • Change the fractions to have the same denominator and add both terms.
  • Reduce the final fraction obtained into its simpler form by dividing both numerator and denominator by there largest common factor.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to add 2 fractions
#include<bits/stdc++.h>
using namespace std;
  
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}
  
// Function to convert the obtained fraction
// into it's simplest form
void lowest(int &den3, int &num3)
{
    // Finding gcd of both terms
    int common_factor = gcd(num3,den3);
  
    // Converting both terms into simpler 
    // terms by dividing them by common factor 
    den3 = den3/common_factor;
    num3 = num3/common_factor;
}
  
//Function to add two fractions
void addFraction(int num1, int den1, int num2, 
                 int den2, int &num3, int &den3)
{
    // Finding gcd of den1 and den2
    den3 = gcd(den1,den2);
  
    // Denominator of final fraction obtained
    // finding LCM of den1 and den2
    // LCM * GCD = a * b 
    den3 = (den1*den2) / den3;
  
    // Changing the fractions to have same denominator
    // Numerator of the final fraction obtained
    num3 = (num1)*(den3/den1) + (num2)*(den3/den2);
  
    // Calling function to convert final fraction
    // into it's simplest form
    lowest(den3,num3);
}
  
// Driver program
int main()
{
    int num1=1, den1=500, num2=2, den2=1500, den3, num3;
    addFraction(num1, den1, num2, den2, num3, den3);
    printf("%d/%d + %d/%d is equal to %d/%d\n", num1, den1,
                                   num2, den2, num3, den3);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to add 2 fractions 
  
class GFG{
// Function to return gcd of a and b 
static int gcd(int a, int b) 
    if (a == 0
        return b; 
    return gcd(b%a, a); 
  
// Function to convert the obtained fraction 
// into it's simplest form 
static void lowest(int den3, int num3) 
    // Finding gcd of both terms 
    int common_factor = gcd(num3,den3); 
  
    // Converting both terms into simpler 
    // terms by dividing them by common factor 
    den3 = den3/common_factor; 
    num3 = num3/common_factor;
    System.out.println(num3+"/"+den3);
  
//Function to add two fractions 
static void addFraction(int num1, int den1, 
                        int num2, int den2) 
    // Finding gcd of den1 and den2 
    int den3 = gcd(den1,den2); 
  
    // Denominator of final fraction obtained 
    // finding LCM of den1 and den2 
    // LCM * GCD = a * b 
    den3 = (den1*den2) / den3; 
  
    // Changing the fractions to have same denominator 
    // Numerator of the final fraction obtained 
    int num3 = (num1)*(den3/den1) + (num2)*(den3/den2); 
  
    // Calling function to convert final fraction 
    // into it's simplest form 
    lowest(den3,num3); 
  
// Driver program 
public static void main(String[] args) 
    int num1=1, den1=500, num2=2, den2=1500
    System.out.print(num1+"/"+den1+" + "+num2+"/"+den2+" is equal to ");
    addFraction(num1, den1, num2, den2);
}
// This code is contributed by mits

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to add 2 fractions 
  
# Function to return gcd of a and b 
def gcd(a, b):
    if (a == 0): 
        return b; 
    return gcd(b % a, a); 
  
# Function to convert the obtained 
# fraction into it's simplest form 
def lowest(den3, num3): 
  
    # Finding gcd of both terms 
    common_factor = gcd(num3, den3); 
  
    # Converting both terms 
    # into simpler terms by 
    # dividing them by common factor 
    den3 = int(den3 / common_factor); 
    num3 = int(num3 / common_factor);
    print(num3, "/", den3);
  
# Function to add two fractions 
def addFraction(num1, den1, num2, den2): 
  
    # Finding gcd of den1 and den2 
    den3 = gcd(den1, den2); 
  
    # Denominator of final 
    # fraction obtained finding 
    # LCM of den1 and den2 
    # LCM * GCD = a * b 
    den3 = (den1 * den2) / den3; 
  
    # Changing the fractions to 
    # have same denominator Numerator 
    # of the final fraction obtained 
    num3 = ((num1) * (den3 / den1) + 
            (num2) * (den3 / den2)); 
  
    # Calling function to convert 
    # final fraction into it's 
    # simplest form 
    lowest(den3, num3); 
  
# Driver Code 
num1 = 1; den1 = 500
num2 = 2; den2 = 1500
  
print(num1, "/", den1, " + ", num2, "/"
      den2, " is equal to ", end = ""); 
addFraction(num1, den1, num2, den2);
  
# This code is contributed by mits

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to add 2 fractions 
  
class GFG{
// Function to return gcd of a and b 
static int gcd(int a, int b) 
    if (a == 0) 
        return b; 
    return gcd(b%a, a); 
  
// Function to convert the obtained fraction 
// into it's simplest form 
static void lowest(int den3, int num3) 
    // Finding gcd of both terms 
    int common_factor = gcd(num3,den3); 
  
    // Converting both terms into simpler 
    // terms by dividing them by common factor 
    den3 = den3/common_factor; 
    num3 = num3/common_factor;
    System.Console.WriteLine(num3+"/"+den3);
  
//Function to add two fractions 
static void addFraction(int num1, int den1, int num2, int den2) 
    // Finding gcd of den1 and den2 
    int den3 = gcd(den1,den2); 
  
    // Denominator of final fraction obtained 
    // finding LCM of den1 and den2 
    // LCM * GCD = a * b 
    den3 = (den1*den2) / den3; 
  
    // Changing the fractions to have same denominator 
    // Numerator of the final fraction obtained 
    int num3 = (num1)*(den3/den1) + (num2)*(den3/den2); 
  
    // Calling function to convert final fraction 
    // into it's simplest form 
    lowest(den3,num3); 
  
// Driver program 
public static void Main() 
    int num1=1, den1=500, num2=2, den2=1500; 
    System.Console.Write(num1+"/"+den1+" + "+num2+"/"+den2+" is equal to ");
    addFraction(num1, den1, num2, den2);
}
// This code is contributed by mits

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to add
// 2 fractions
  
// Function to return 
// gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
  
// Function to convert the 
// obtained fraction into 
// it's simplest form
function lowest(&$den3, &$num3)
{
    // Finding gcd of both terms
    $common_factor = gcd($num3, $den3);
  
    // Converting both terms  
    // into simpler terms by 
    // dividing them by common factor 
      
    $den3 = (int)$den3 / $common_factor;
    $num3 = (int) $num3 / $common_factor;
}
  
// Function to add
// two fractions
function addFraction($num1, $den1, $num2
                     $den2, &$num3, &$den3)
{
    // Finding gcd of den1 and den2
    $den3 = gcd($den1, $den2);
  
    // Denominator of final 
    // fraction obtained finding 
    // LCM of den1 and den2
    // LCM * GCD = a * b 
    $den3 = ($den1 * $den2) / $den3;
  
    // Changing the fractions to 
    // have same denominator Numerator
    // of the final fraction obtained
    $num3 = ($num1) * ($den3 / $den1) + 
            ($num2) * ($den3 / $den2);
  
    // Calling function to convert 
    // final fraction into it's 
    // simplest form
    lowest($den3, $num3);
}
  
// Driver Code
$num1 = 1; $den1 = 500; 
$num2 = 2; $den2 = 1500; 
$den3; $num3;
addFraction($num1, $den1, $num2
            $den2, $num3, $den3);
echo $num1, "/", $den1, " + "
     $num2,"/", $den2, " is equal to "
               $num3, "/", $den3, "\n";
              
// This code is contributed by aj_36
?>

chevron_right



Output :

1/500 + 2/1500 is equal to 1/300

See below for doing the same using library functions.
Ratio Manipulations in C++ | Set 1 (Arithmetic)

This article is contributed by Rahul Agrawal .If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : jit_t, Mithun Kumar



Article Tags :
Practice Tags :


4


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.