Thread: Rational Numbers Assignment Adding

We have everything done except we do not know how to do the adding of rational numbers. can you help?

Code:

#include "Rational.h"
#include "math.h"
#include <iostream>

using namespace std;

Rational::Rational()
{
     numerator = denominator = 1;
}

Rational::Rational(int N, int D)
{
    numerator = N;
    denominator = D;
} 

int Rational::getN()
{
              return numerator;
}

int Rational::getD()
{
              return denominator;
}

int Rational::multiply(Rational rational2)
{
    int finalN =getN()*rational2.getN();
    int finalD =getD()*rational2.getD();
    Rational rational(finalN,finalD);
    simplify(rational);
} 

int Rational::divide(Rational rational2)
{
    int finalN =getN()*rational2.getD();
    int finalD =getD()*rational2.getN();
    Rational rational(finalN,finalD);
    simplify(rational);
}

int Rational::add(Rational rational2)
{
int tempN = getN()*rational2.getD();
int tempD = getD()*rational2.getN();
int finalN = tempN*tempD 
int finalD = 
Rational rational(finalN,finalD);
simplify(rational);
}

void Rational::simplify(Rational rational)
{
     int x=rational.getN();
     int y=rational.getD();
     int z;
     while (y!=0)
     {
           z=x%y;
           x=y;
           y=z;
     }
     int finalN=rational.getN()/x;
     int finalD=rational.getD()/x;
     Rational rationalnew (finalN,finalD);
     cout <<finalN<<"/"<<finalD<<endl;
}
     
        
bool Rational::compare(Rational rational2)
{
     float initial=(float)getN()/getD();
     float final= (float)rational2.getN()/rational2.getD();
     if (initial==final) return true;
     else return false;
}

Simply put, multiply the nominators by the cross denominator and get a common denominator by multiplying the denominators together.

Adding and Subtracting Rational Functions - Free Math Help

An easy way to find a common denominator is by multiplying the denominators together:

Code:

1/4 + 3/9
9/36 + 12/36 // 1 times 9 = 9. 3 times 4 = 12. 4 times 9 equal 36

Knowing this, you should be able to find or make a function that does this for you.
If you need more help, just reply and I'll write some code.

Code:

int Rational::multiply(Rational rational2)
{
    int finalN =getN()*rational2.getN();
    int finalD =getD()*rational2.getD();
    Rational rational(finalN,finalD);
    simplify(rational);
}

Why not make simplify() a function that operates on "this" rational? That makes more sense to me. Unless the interface is part of the assignment that you can't change, of course.

Code:

if (initial==final) return true;

This is dangerous because
1. the difference can be smaller than representable in a float
2. floats (and doubles) should generally not be compared with == because not all values can be represented with infinite precision

For this function, I would just simplify both functions, and see if the denominators and numerators are equal.

Also, for simplify(), I would "float" the negative sign to the top, if there is one. So "1/(-4) == (-1)/4" will be true.

If you need more help, just reply and I'll write some code.

Just in case... please don't write people's homework for them.

Yes (to both) of course .