# 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++

`// 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;` `}` |

## Java

`// 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` |

## Python3

`# 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` |

## C#

`// 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` |

## PHP

`<?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` `?>` |

## Javascript

`<script>` `// Javascript program to add 2 fractions` ` ` `// Function to return gcd of a and b` `const gcd = (a, b) => {` ` ` `if` `(a == 0)` ` ` `return` `b;` ` ` `return` `gcd(b % a, a);` `}` `// Function to convert the ` `// obtained fraction into ` `// it's simplest form` `const lowest = (den3, num3) => {` ` ` `// Finding gcd of both terms` ` ` `let common_factor = gcd(num3, den3);` ` ` ` ` `// Converting both terms ` ` ` `// into simpler terms by ` ` ` `// dividing them by common factor ` ` ` ` ` `den3 = parseInt(den3 / common_factor);` ` ` `num3 = parseInt(num3 / common_factor);` ` ` ` ` `document.write(`${num3}/${den3}`)` `}` `// Function to add two fractions` `const addFraction = (num1, den1, num2, den2) => {` ` ` `// Finding gcd of den1 and den2` ` ` `let 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` ` ` `let num3 = ((num1) * (den3 / den1) +` ` ` `(num2) * (den3 / den2));` ` ` ` ` `// Calling function to convert ` ` ` `// final fraction into it's ` ` ` `// simplest form` ` ` `lowest(den3, num3);` `}` `// Driver Code` `let num1 = 1;` `let den1 = 500; ` `let num2 = 2;` `let den2 = 1500; ` `document.write(`${num1}/${den1} + ${num2}/${den2} is equal to `);` `addFraction(num1, den1, num2, den2);` ` ` `// This code is contributed by _saurabh_jaiswal` `</script>` |

**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)

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