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).
Fact about Fraction :
- Fractions can be reduced if the numerator and denominator have the greatest common divisor(gcd) greater than 1.
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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.
- 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.
- 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.
How to add two fractions?
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
Algorithm to add two fractions
- 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++ 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 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 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# 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 // 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" ;
?> |
<script> // JavaScript 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
var 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);
return [den3, num3];
}
// 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
return lowest(den3, num3);
}
// Driver Code
var num1 = 1,
den1 = 500,
num2 = 2,
den2 = 1500,
den3,
num3;
var [den3, num3] = addFraction(num1, den1, num2, den2, num3, den3);
document.write(
num1 +
"/" +
den1 +
" + " +
num2 +
"/" +
den2 +
" is equal to " +
num3 +
"/" +
den3 +
"<br>"
);
</script>
|
Output :
1/500 + 2/1500 is equal to 1/300
Time Complexity: O(log(min(a, b)))
Auxiliary Space: O(log(min(a, b)))
More problems related to Fraction:
- LCM and HCF of fractions
- Represent the fraction of two numbers in the string format
- Program to compare two fractions
- Convert Binary fraction to Decimal
- Convert decimal fraction to binary number
- Fractional Knapsack Problem
- Find Recurring Sequence in a Fraction