Adding Fractions

A Case Approach

Everyone, or nearly everyone it seems, hates fractions, especially their addition and the other operations. Of particular note is the fact that the rules for addition and multiplication are often confused and interchanged, and the rule for division is just plain forgotten.

This article attempts to present WTM’s own approach to one aspect of this situation, the addition of just two fractions. The thesis is: that if this is thoroughly understood, then the other operations will cause less misunderstanding and frustration; and later the student will operate with “algebraic” fractions in an Algebra I or II course with
greater comprehension of the basic structures involved.


This method, here called a “case approach”, takes the a priori assumption that the student already knows (or can acquire soon) the simple fundamentals of elementary number theory, e.g. such terms as least common multiple (LCM), greatest comon factor (GCF), and relatively prime numbers. Also it is assumed that the basic concepts and terminology of fractions (numerator, denominator, reduce, equivalent fractions) are likewise understood.

The key idea in this approach is to focus our attention first on the denominators, according to the following outline:

Case 1. Same denominators
Case 2. Different denominators
a) Relatively prime denominators
b) Not relatively prime denominators
c) Special case


Case 1 is practically self-explanatory. Both denominators are the same number, hence the addition comes naturally. Few individuals have much trouble here. Examples:

                                    2       5       7
                without reduction: ---  +  ---  =  ---
                                    9       9       9

                                  1        7        8       2
                with reduction: ----  +  ----  =  ----  =  ---
                                 12       12       12       3

However, Case 2 is the situation where the most difficulties arise. Let’s examine it more carefully now.


It is a fact in number theory that when two numbers are relatively prime (i.e. their GCF is 1), that their LCM is the product of the given numbers. Example:

The LCM of 5 and 8 is 5 × 8, or 40.

Therefore, when the two numbers are then used as denominators of fractions, the lowest common denominator (LCD) is, in effect, the aforementioned LCM. So, for Case 2a, the LCD is fairly automatic. Note:

                         2       3
                        ---  +  ---  =  ----  +  ----
                         5       8       40       40

In this problem the numerators are 2×8, or 16, and 3×5, or 15, respectively, yielding

                         2       3       16       15
                        ---  +  ---  =  ----  +  ----
                         5       8       40       40

The problem has now been converted to Case-1 style, hence the answer, 31/40 is now obvious. It is interesting to point out that if in a Case-2a problem the original fractions are given in “reduced” form to start with, then the sum itself will also be in “reduced” form, or as is often said, “in lowest terms”.

The foregoing example could be written in a condensed form as follows:

                         2       3       16 + 15      31
                        ---  +  ---  =  --------- =  ----
                         5       8          40        40

This mode of presentation suggests the formula for the addition of two fractions in Case-2a:

                         a       c       ad + bc
                        ---  +  ---  =  --------- 
                         b       d         bd

[As will soon be pointed out, this formula can technically be used for ANY two fractions, though this may not necessarily be the most desirable or practical thing to do in specific instances.]


Now let’s examine the most intriguing case: Case 2b. Here the denominators are NOT relatively prime, that is, their GCF is greater than 1. This means that their LCD (alias LCM) is some value less than their product. An example will clarify this:

			 3	 5
			---  +  ---  =   ?
			 4	 6

Here the LCD is not 24, but rather 12. So, we have (step by step):

		   3	   5
	Step 1:   ---  +  ---  =  ----  +  ----
		   4	   6	   12	    12

		   3	   5	    9	    10
	Step 2:   ---  +  ---  =  ----  +  ----
		   4	   6	   12	    12

		   3	   5	    9	    10	     19
	Step 3:   ---  +  ---  =  ----  +  ----  =  ----
		   4	   6	   12	    12	     12

Although the example above usually causes no great difficulty for most individuals (due to the smallness of the denominators), in other problems the search for the appropriate LCD can produce a certain degree of anxiety. [For example, try this problem and see how much time you spend finding the LCD, then using it:

			  7        8
			----  +  ----.]
			 26	  39

This then raises the question: What about the formula? Can it still be used? As was suggested earlier, the answer is “yes.” Note:

		   3	   5	   18 + 20	 38	  19
	          ---  +  ---  =  ---------  =  ----  =  ----
		   4	   6	    4 × 6	 24	  12

The only difference now is that a reduction step will always be necessary in order to give the answer in lowest terms, as is normally required.

Now it can be stated: Case 2b can be done in either of two ways: 1) using the LCD, or 2) using the formula. There are advantages and disadvantages for both methods.

A few words are in order for the formula’s disadvantages. In this modern era, the easy access to electronic calculators make the large number aspect no problem at all. The cross products are quickly done and (if you are using the simple 4-function models with memory keys) stored in the memory, where they are even added for you as well.
Note our big-number example earlier:

		  7        8      273 + 208	   481
		----  +  ----  = -----------  =  ------
		 26	  39       26 × 39        1014

The reduction step is also not as difficult as it often used to be before calculators either. This is because the GCF of the two original denominators (13 in the example above) is usually easier to find, or at least easier (and smaller) than the LCD. Dividing both parts of the fraction 481/1014 by 13 gives its reduced form: 37/78.

So when should one use a certain method? I suggest that it’s largely a matter of personal preference. We need to show students that there are times when a problem can be done in more than one way. This should tend to relieve a lot of the math anxiety prevalent in people’s minds.


Lastly, we turn our attention to Case 2c, the Special Case. This arises when one of the given denominators is already a multiple of the other, or equivalently the former is a factor of the latter.
Example:

		 3	 7	 6	 7	 13
		---  +  ---  =  ---  +  ---  =  ----
		 4	 8	 8	 8	  8

Note that here only one fraction had to be changed to an equivalent one, thereby keeping the numbers as small as possible. This case
should be stressed more than it normally is because it’s the one that probably occurs more frequently in “real life” situations.


In this brief article, I have not pretended to be exhaustive in my presentation of all the ramifications normally encountered while doing the variety of textbook problems that students are expected to master. Nor has mention been made of a) improper-vs-mixed-number answers, or b) that with the calculator* all the fractions could be converted to decimal form and added in the memory even more easily still. Rather it has been to offer a basic structure than can be readily used to organize one’s thoughts along the lines of number theory and that the types of denominators govern what strategy must or might be taken in a particular situation.


*Footnote: The original text of this article I wrote at a time when most young students might have access only to simple 4-function models with memory keys. Now scientific models are very common. And lately there have even appeared models that “do fractions” in pure fraction form! My, how the world is changing!


Appendix

An interesting fact from number theory not well known by the average student regarding LCM’s and GCF’s greatly aids the search for the LCD (LCM) in fraction work. It is that the product of the LCM and GCF of two numbers is equal to the product of the two numbers themselves. Symbolically this means:

LCM(a, b) × GCF(a, b) = a × b

This permits us to state a formula for LCD’s (LCM’s):

			                  a × b
			LCM (a, b)  =  -----------
			                GCF(a, b)

For our “big number” problem above, we can write:

                                   26  ×  39       26 × 39
                LCM(26, 39)  =  --------------  =  -------
                                 GCF (26, 39)         13

                             =  2 × 39  (or  26 × 3)

                             =  78
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