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# Fraction

• Difficulty Level : Easy
• Last Updated : 06 Jul, 2022

A fraction is a ratio of two values. Fractions have the form a/b where a is called the numerator, b is called the denominator and b cannot equal 0 (since division by 0 is undefined). The denominator gives how many equal parts are there. The numerator represents how many of these are taken. For example, one-half, eight-fifths, three-quarters (1/2, 8/5, 3/4). 1. Fractions can be reduced if the numerator and denominator have the greatest common divisor(gcd) greater than 1.
2. Addition and Subtraction of Fractions: When adding or subtracting fractions, they must have the same denominator. If they do not have the same denominator, we must find a common one for both. To do this, we first need to find the lowest common multiple(lcm) of the two denominators or multiply each fraction by the proper integers so that there will be the same denominator.

3. Multiplication and Division of Fractions: When multiplying two fractions, simply multiply the two numerators and multiply the two denominators. When dividing two fractions, the first fraction must be multiplied by the reciprocal of the second fraction.
4. There are three types of fractions :
• Proper Fractions: The numerator is less than the denominator. For Example, 1/3, 3/4, 2/7
• Improper Fractions: The numerator is greater than (or equal to) the denominator. For Example, 4/3, 11/4, 7/7.
• Mixed Fractions: A whole number and proper fraction together. For Example, 1 1/3, 2 1/4, 16 2/5.

Add two fractions a/b and c/d and print the answer in the 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```

• Find a common denominator by finding the LCM (The 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 their largest common factor.

## C++

 `// C++ program to add 2 fractions``#include ``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``import` `java.util.*;` `class` `GFG``{``static` `int` `den3, num3;` `// 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()``{``    ``// 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``static` `void` `addFraction(``int` `num1, ``int` `den1,``                        ``int` `num2, ``int` `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();``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `num1 = ``1``, den1 = ``500``,``        ``num2 = ``2``, den2 = ``1500``;` `    ``addFraction(num1, den1, num2, den2);` `    ``System.out.printf(``"%d/%d + %d/%d is equal to %d/%d\n"``,``                      ``num1, den1, num2, den2, num3, den3);``}``}` `// This code is contributed by Rajput-Ji`

## 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)`

## C#

 `// C# program to add 2 fractions``using` `System;``    ` `class` `GFG``{``static` `int` `den3, num3;` `// 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()``{``    ``// 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``static` `void` `addFraction(``int` `num1, ``int` `den1,``                        ``int` `num2, ``int` `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();``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `num1 = 1, den1 = 500,``        ``num2 = 2, den2 = 1500;` `    ``addFraction(num1, den1, num2, den2);` `    ``Console.Write(``"{0}/{1} + {2}/{3} is equal to {4}/{5}\n"``,``                        ``num1, den1, num2, den2, num3, den3);``}``}` `// This code is contributed by PrinciRaj1992`

## PHP

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## Javascript

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Output :

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

Time Complexity: O(log(min(a, b)))

Auxiliary Space: O(log(min(a, b)))

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