how do i add fractions?

[airfrog19]

I'm trying to create a method to add two fractions together, but i''m not sure on how it would look like. 

This is how the method looks like:

public class Fraction implements FractionInterface, Comparable<Fraction>{private int num; // Numerator private int den; // Denominator  public Fraction(){// set fraction to default = 0/1num = 0;                den = 1;} // end default constructor public FractionInterface add(FractionInterface secondFraction)//blahblah

i'm suppose to use secondFraction, but not sure how though. 

[AlexStriz]

Make sure they have the same denominator and then add the numerators together

[ilikemacandpc]

create a common denominator by multiplying their denominators,

 

the top half will be the the product of the numerators with the opposite denominators added. 

 

@airfrog19

ex. 2/3 + 4/7 = ?

 

denominator = 3*7 = 21

 

numerator = 2*7 + 4*3 = 26

 

final fraction = 26/21

[ItsAFeature]

create a common denominator by multiplying their denominators,

 

the top half will be the the product of the numerators with the opposite denominators added. 

 

@airfrog19

ex. 2/3 + 4/7 = ?

 

denominator = 3*7 = 21

 

numerator = 2*7 + 4*3 = 26

 

final fraction = 26/21

 

Exactly this. If your program requires you to simplify then you have to write some more code for that, but this is the way I'd approach this problem.

[olbaze]

a/b + c/d = ad + cb / bd

 

Have your fraction method take two integers as input and return a pair. Then make an addition method that takes two fractions as parameters, manipulates them as per the above formula (which is the definition of addition of fractions, btw), as returns the result.

 

Naturally, if you want to simplify the fractions, that's going to add a bit extra.

 

Here's a quick and ugly implementation on Python without simplifying. Note that if I wanted simplifying, I could just implement it into the fraction method, since fraction_addition returns a fraction. However, implementing proper simplifying means adding an equivalence relation.

def fraction(numerator, denominator):    return [numerator, denominator]def fraction_addition(first_fraction, second_fraction):    first_fraction = fraction(first_fraction[0], first_fraction[1])    second_fraction = fraction(second_fraction[0], second_fraction[1])    numerator = first_fraction[1] * second_fraction[0] + second_fraction[1] * first_fraction[0]    denominator = first_fraction[1] * second_fraction[1]    return fraction(numerator, denominator)

And here's an example of how it can be used:

first = fraction(1, 2)second = fraction(2, 3)print fraction_addition(first, second)

[airfrog19]

 

a/b + c/d = ad + cb / bd

 

Have your fraction method take two integers as input and return a pair. Then make an addition method that takes two fractions as parameters, manipulates them as per the above formula (which is the definition of addition of fractions, btw), as returns the result.

 

Naturally, if you want to simplify the fractions, that's going to add a bit extra.

 

Here's a quick and ugly implementation on Python without simplifying. Note that if I wanted simplifying, I could just implement it into the fraction method, since fraction_addition returns a fraction. However, implementing proper simplifying means adding an equivalence relation.

def fraction(numerator, denominator):    return [numerator, denominator]def fraction_addition(first_fraction, second_fraction):    first_fraction = fraction(first_fraction[0], first_fraction[1])    second_fraction = fraction(second_fraction[0], second_fraction[1])    numerator = first_fraction[1] * second_fraction[0] + second_fraction[1] * first_fraction[0]    denominator = first_fraction[1] * second_fraction[1]    return fraction(numerator, denominator)

And here's an example of how it can be used:

first = fraction(1, 2)second = fraction(2, 3)print fraction_addition(first, second)

okay i'm not sure if this works so this is what i got:

Fraction add = new Fraction();        int newden = den * secondFraction.den;        add.den = newden;        add.num = num * secondFraction.den + secondFraction.num * den;

[airfrog19]

 

@olbaze okay i'm not sure if this works so this is what i got:

Fraction add = new Fraction();        int newden = den * secondFraction.den;        add.den = newden;        add.num = num * secondFraction.den + secondFraction.num * den;

 

[madknight3]

 

okay i'm not sure if this works

 

Did you try running it? If it gives you a wrong answer then it doesn't work, if it gives you the right answer in multiple cases it (probably) does.

[airfrog19]

Did you try running it? If it gives you a wrong answer then it doesn't work, if it gives you the right answer in multiple cases it (probably) does.

yea. doesnt work. need to fix it 

[KuroSetsuna29]

 

okay i'm not sure if this works so this is what i got:

Fraction add = new Fraction();        int newden = den * secondFraction.den;        add.den = newden;        add.num = num * secondFraction.den + secondFraction.num * den;

 

 

yea. doesnt work. need to fix it 

 

Just curious, what results did you get vs what was expected? It looks right to me, although you may want to modify it to not return and instead add to current instance.

public void add(Fraction secondFraction){    this.den *= secondFraction.den;    this.num = this.num * secondFraction.den + secondFraction.num * this.den;    // Reduce fraction if needed}