Merced College;  Don Power

 

Adding and Subtracting Fractions

 

by Don Power, Mar 2002

 

Here is a way to add or subtract two fractions that you can succeed at!  It is based on reducing fractions, not on tedious calculations of the lowest common denominator (LCD).  In fact, you don't even have to find the LCD to set up your fractions. 

 

I'll illustrate each step with this example:  Add

 

Any step-by-step mathematical procedure is called an "algorithm," so here we go:

 

Algorithm:

  1. Write both numbers as proper or improper fractions (not mixed numbers or whole numbers).

    Example: 

  2. In a side calculation, form a new fraction with the denominators of both original fractions.

    Example:  Side calculation: 

  3. Reduce the new fraction.  Write the result as a proportion, with the unreduced fraction set equal to the reduced fraction.  (If the fraction cannot be reduced, both sides of the proportion will be the same).

    Example:  Side calculation: 

  4. Write the cross products, without actually multiplying them out -- you have to be able to see the factors.

    Example:  The cross products are 32(5) and 20(8), both of which equal 160.

                Optional step:  If you multiply out the cross products, you get the LCD.

                Example:  32(5) and 20(8) both equal the LCD, which is160.


  5. For each original fraction, identify the "cross factor" for its denominator.  Multiply both the numerator and denominator by that "cross factor."

    Example:  The cross factor of 32 is 5, and the cross factor of 20 is 8, so you should rewrite the original problem as
                   

 


  1. Carry out the multiplications to get

  2. The fractions resulting from step 6 now have a common denominator, and can be added or subtracted.

    Example: 

 

Notice that multiplying in step 4 is optional, and may even be a distraction:  it is not necessary to calculate the LCD before setting up the multiplications in step 5.




Here is the complete example, the way I would write it up:

 

 Add       Side calculation:     Cross products 32(5), 20(8)

 

 

 

See additional examples following the proof.

 

 

Comment for Teachers:

 

This algorithm for adding and subtracting fractions is designed to capitalize on a fraction skill that most students can handle well.  Most students find that they can reduce a fraction easily, whether they use the greatest common factor (GCF) to reduce the fraction in one step, or whether they take several steps.  It is much harder for most students to use traditional techniqes for finding and using the lowest common denominator (LCD).  This algorithm has the added benefit that a single technique is usable for all fraction addition/subtraction problems involving two fractions.

 

 

Proof:  Consider the fraction addition problem  for integers a, b, c, d, with b>0, d>0.

 

Let gcd(b,d) = e, so that b = ef  and d = eg for some positive integers f and g.

 

The side calculation is     The cross products are bg and df, both of which equal efg.

 

Since the cross products are equal multiples of b and d, they form a common denominator.

 

To show that they are the LCD, that is, the lowest common multiple of the denominators:

 

Calculate:   (from elementary number theory)

 

Alternately, let h>0  be any common multiple of the denominators b and d.

 

b divides h and d divides h.

 

ef divides h and eg divides h, and no common factors of f and g exist other than 1 (since e = gcd(f,g)).

 

e divides h, so  is an integer.

 

f divides  and g divides

Since f and g have no common factors (other than 1), fg divides 

 

efg divides h

 

Therefore efg £ h

 

Therefore efg is the least of the common multiples of the denominators.

 

 

Additional Examples:

 

(Teachers:  Once students have gotten used to the side calculation, you can use it to justify the simplified procedures that exist for special cases.)

 

 

One denominator divides the other:

 

Subtract        Side calculation:     Cross products 7(13), 91(1)

   which reduces to

 

 

No common factor:

 

Simplify       Side calculation:       Cross products 7(15), 15(7)

 

       

 

 

Integer with fraction:

 

Simplify       

 

Write    Side calculation:       Cross products 1(11), 11(1)

 

           

 

 

Sidelight:  This type of problem can also be handled like a mixed number conversion (another process students are comfortable with):


The numerator of the answer is the denominator times the integer plus or minus (as indicated) the numerator.
In this example, the numerator is 11(7)-32 = 45.

The denominator is, of course, the original denominator, 11.

 

 

Common denominator, with variables:

 

Simplify       Side calculation:      Cross products 3(1), 3(1)

 

The cross factors are both 1, so you do not change either fraction before adding.

 

In this case, the numerators are not like terms, so the result is

 

 

More variables:

 

Simplify       Side calculation:       Cross products 4a2(3b), 6ab(2a)

 

           

 

 

 

Return to:  Merced College;  Don Power                                  Updated 3/11/02