Fractions are Hard!

2.8 Adding and subtracting fractions

We have four beliefs about how fractions should work (accompanied with all the usual rules of arithmetic). We are choosing to believe that these statements hold for all types of numbers.

BELIEF 1: \(\dfrac{a}{a} = 1\)

BELIEF 2: \(\dfrac{a}{1}=a\)

BELIEF 3: \(x \times \dfrac{a}{b} = \dfrac{xa}{b}\)           (Basic Multiplication Belief)

BELIEF 4: \(\dfrac{a}{b}=\dfrac{xa}{xb}\)                           (Key Fraction Belief)

 

And we have two consequence of these beliefs

CONSEQUENCE 5: \(\dfrac{a}{b} \times b = a\).

CONSEQUENCE 6: \(\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd}\).

 

EXERCISE:   Establish that  \(\dfrac{1}{n}\times a = \dfrac{a}{n}\) , that is, that “multiplying by \(\dfrac{1}{n}\)  has the same effect as dividing by \(n\).”

(HINT: By Belief 2 we have \(\dfrac{1}{n} \times a = \dfrac{1}{n}\times \dfrac{a}{1}\) .)

 

The addition of fractions follows as a logical consequence of these beliefs as well.

 

EXAMPLE: Give a meaningful value to \(\dfrac{2}{3}+\dfrac{7}{5}\) as a logical consequence of these beliefs.

 

Answer: By Belief 2 we have

 

\(\dfrac{2}{3}+\dfrac{7}{5}=\dfrac{\dfrac{2}{3}+\dfrac{7}{5}}{1}\).

 

Now use the Key Faction property twice:

 

\(\dfrac{\dfrac{2}{3}+\dfrac{7}{5}}{1} = \dfrac{\left(\dfrac{2}{3}+ \dfrac{7}{5}\right) \times 3 \times 5}{1\times 3 \times 5}\).

 

Using the Basic Multiplication Property, this is

 

\( \dfrac{ \left(\dfrac{2}{3} + \dfrac{7}{5}\right) \times  3 \times  5}{1 \times  3 \times 5} =\dfrac{\dfrac{2}{3} \times 3 \times 5 + \dfrac{7}{5} \times 3 \times 5}{3 \times 5} = \dfrac{2 \times 5 + 7 \times 3}{3 \times 5}\).

 

Arithmetic gives this as \(\dfrac{31}{15}\).

 

EXERCISE:   Follow either of these arguments again to, more abstractly, show that \(\dfrac{a}{b}+\dfrac{c}{d}\) equals \(\dfrac{ad+bc}{bd}\) .

Also show that \(\dfrac{a}{b}-\dfrac{c}{d}\) equals \(\dfrac{ad-bc}{bd}\) .

 

 

A LESS ABSTRACT APPROACH FOR THOSE WHO DESIRE IT

We can go back to grade-school thinking and see how our pies-per-boy work connects directly  the addition of fractions there if this helps.

 

Consider first a simpler addition problem, say, \(\dfrac{4}{7} + \dfrac{2}{7}\).

 

But note, it is not immediately clear how to interpret this in the pies per boy model. I can think of two interpretations:

 

Four pies are shared among seven boys and two pies are shared among seven girls. That makes for a total of six pies shared among 14 people. The combined effect is \(\dfrac{6}{14}\).

 

Four pies are shared among seven boys and then two pies are shared among the seven boys. That makes for a total of six pies shared among those boys. The combined effect is \(\dfrac{6}{7}\).

 

Can we use other considerations from grade-school days to help guide us with which interpretation to follow?

 

PARTS OF A WHOLE: Adding portions of pie suggests that the answer to \(\dfrac{4}{7}+\dfrac{2}{7}\)  is \(\dfrac{6}{7}\) .

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NUMBER LINE MODEL: This model too suggests the answer \(\dfrac{6}{7}\) .

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EVERYDAY LANGUAGE: “4 sevenths plus 2 sevenths sounds like it should give 6 sevenths! (Just as 4 apples and 2 apples give 6 apples.)

 

It seems compelling to declare then:

 

ADDITION OF BASIC FRACTIONS:  \(\dfrac{a}{N}+\dfrac{b}{N}\) as to equal \(\dfrac{a+b}{N}\), at least for positive counting numbers \(a\), \(b\), and \(N\).

 

Comment: Some people might say that “Mrs. G brought in 4 pies for the 7 girls in a team and Mr. H brought in 2 pies for those 7 girls. That makes for 6 pies for those 7 girls.” Thus \(\dfrac{2}{7} + \dfrac{4}{7}=\dfrac{6}{7}\).

 

Exercise: Write  \(\dfrac{4}{7} + \dfrac{2}{7}\) as \(\dfrac{\dfrac{4}{7}+\dfrac{2}{7}}{1}\) and multiply numerator and denominator each by \(7\). Does this too suggest \(\dfrac{6}{7}\) is the answer to go with?

 

We can certainly add more than two fractions sharing a common denominator in the same way

 

\(\dfrac{1}{10}+\dfrac{8}{10}+\dfrac{4}{10} = \dfrac{13}{10}\)

 

and subtract fractions with a common denominator:

 

\(\dfrac{6}{11} – \dfrac{2}{11}= \dfrac{4}{11}\).

 

A rub comes in trying to add or subtract fractions with different denominators.

 

Example: What should be the answer to \(\dfrac{2}{5}+\dfrac{1}{3}\)?

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EPIPHANY: We can rewrite each of the expressions  \(\dfrac{2}{5}\) and \(\dfrac{1}{3}\) in an infinitude of equivalent forms using the Key Fraction Property. By luck, we might find two expressions with a common denominator and thus able to invoke our basic addition definition.

Here goes

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We do luck out and see that \(\dfrac{2}{5}\) and \(\dfrac{1}{3}\) is actually equivalent to the problem \(\dfrac{6}{15}\) and \(\dfrac{5}{15}\), which has answer \(\dfrac{11}{15}\). This work shows we could slice each of the two amounts of pie we have in our picture into fifteenths if we like and all should then be clear. (And it is!)

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As another example:

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Comment: Of course, there is an efficient way to find a “common denominator” of two fractions. Let students discover this for themselves.

 

Question:  a) What is \(\dfrac{1}{2}+\dfrac{1}{3}+ \dfrac{1}{6}\) ?       b) What is \(\dfrac{1}{2}-\dfrac{1}{4} – \dfrac{1}{8} – \dfrac{1}{16}\)?

 

Exercise:  Put \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\) over a common denominator and use the it to show that this approach also shows \(\dfrac{a}{b}+\dfrac{c}{d}\) should equal \(\dfrac{ad+bc}{bd}\).

 

 

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