Fractions are Hard!

2.6 Dividing fractions

As the quantities \(\dfrac{a}{b}\)  are defined through the notion of “division as sharing” it is not at all surprising that the operation of division is straightforward and natural for these quantities, especially with the key fraction property in hand.

 

Here’s the challenge we offered earlier in lesson 2.4.

 

Two-and-a-half pies are to be shared equally among four-and-a-half boys! How much pie does an individual (whole) boy receive? 

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We’re being asked to make sense of  \(\dfrac{2\frac{1}{2}}{4\frac{1}{2}}\).

 

Epiphany: If we double the number of pies and double the number of boys, then we see that this sharing task is equivalent to the sharing task \(\dfrac{5}{9}\). That is, each boy ends up with five-ninths of a pie.

f

We have  \(\dfrac{2\frac{1}{2}}{4\frac{1}{2}}=\dfrac{\left(2+\frac{1}{2}\right)\times 2}{\left(4+\frac{1}{2}\right)\times 2}=\dfrac{5}{9}\) .

 

Comment: We need to be flexible about how two halves of boys combine to make the equivalent of a whole boy!

 

 

Another example:

 

\(7\dfrac{2}{3}\) pies are to be shared among \(5\dfrac{3}{4}\) girls. How many pies does each individual girl receive?

 

 

We want to make

\(\dfrac{7+\dfrac{2}{3}}{5+\dfrac{3}{4}}\)

more manageable. Using the Key Fraction Property, let’s multiply the numerator and denominator of this fraction each through by \(3\). (Why three?)

 

\(\dfrac{\left(7+\frac{2}{3}\right)\times 3}{\left(5 + \frac{3}{4}\right) \times 3} = \dfrac{21+2}{15+\frac{9}{4}}\).

 

Let’s now multiply numerator and denominator each by \(4\). (Why four?)

 

\(\dfrac{\left(21+2\right) \times 4}{\left(15+\frac{9}{4}\right) \times 4} = \dfrac{84+8}{60+9}=\dfrac{92}{69}\).

 

 

This shows that sharing \(7\dfrac{2}{3}\) pies among \(5\dfrac{3}{4}\) girls is equivalent to sharing \(92\) pies among \(69\) girls. (Is that friendlier?) Each girl receives \(1\dfrac{23}{69}\) pies. Oh, that’s \(1\dfrac{1}{3}\) pies.

 

 

As another example, consider \(\dfrac{3\dfrac{1}{2}}{1\dfrac{1}{2}}\).

 

Multiplying the numerator and denominator each by \(2\) should be enough to make the expression look friendlier:

 

\(\dfrac{3\dfrac{1}{2}}{1\dfrac{1}{2}}=\dfrac{3+\frac{1}{2}}{1+\frac{1}{2}}=\dfrac{\left(3+\frac{1}{2}\right) \cdot 2}{\left(1+\frac{1}{2}\right) \cdot 2} = \dfrac{6+1}{2+1}= \dfrac{7}{3}\) .

 

 

Now look at each of these examples. What would you do to make each expression look more manageable?

 

a) Make \(\dfrac{4\dfrac{2}{3}}{5\dfrac{1}{3}}\) look friendlier.                        b) Make \(\dfrac{2\dfrac{1}{5}}{2\dfrac{1}{4}}\) look friendlier.

 

c) Make \(\dfrac{1\dfrac{4}{7}}{2\dfrac{3}{10}}\) look friendlier.                        d) Make \(\dfrac{\dfrac{3}{5}}{\dfrac{4}{7}}\) look friendlier.

 

 

Let’s work through the fourth one here.

 

Multiply numerator and denominator each by \(5\).

 

\(\dfrac{\frac{3}{5}\times 5}{\frac{4}{7}\times 5}=\dfrac{3}{\frac{20}{7}}\).

 

Now multiply the numerator and denominator each by \(7\):

 

\(\dfrac{3\times 7}{\frac{20}{7}\times 7}=\dfrac{21}{20}\).

 

That looks more manageable.

 

Comment: Most people would say that we have just computed \(\dfrac{3}{5}\div \dfrac{4}{7}\)  as \(\dfrac{21}{20}\) .

 

 

As another example, let’s compute \(\dfrac{5}{9} \div \dfrac{8}{11}\), that is, let’s make

 

\(\dfrac{\dfrac{5}{9}}{\dfrac{8}{11}}\)

 

more manageable.

 

 

Multiply numerator and denominator each by \(9\)  and by \(11\)  at the same time. (Why not?)

 

\(\dfrac{\dfrac{5}{9}\times 9 \times 11}{\dfrac{8}{11}\times 9 \times 11}= \dfrac{5 \times 11}{8 \times 9}\)

 

(do you see what happened here?)

 

and so

 

\(\dfrac{\dfrac{5}{9}}{\dfrac{8}{11}} = \dfrac{5 \times 11}{8 \times 9}= \dfrac{55}{72}\).

 

 

Comment: NEVER ONCE DO WE NEED A RULE ABOUT MULTIPLYING BY THE RECIPROCAL. To divide fractions – follow your nose and JUST DO IT!

 

 

Question: Compute each of the following:

 

a)   \(\dfrac{1}{2} \div \dfrac{1}{3}\)        b)  \(\dfrac{4}{5} \div \dfrac{3}{7}\)         c) \(\dfrac{2}{3} \div \dfrac{1}{5}\)

 

d)  Abstractly compute  \(\dfrac{a}{b}\div \dfrac{c}{d}\).

 

Notice that dividing mixed numbers first naturally slides us into the division of fractions with no fuss.

 

 

Question: Compute \(\dfrac{45}{45}\div \dfrac{902}{902}\). Do you see what the answer simply must be?

 

Question: Compute \(\dfrac{10}{13}\div \dfrac{2}{13}\).  Any general comments about this one?

 

Question: Compute \(\dfrac{3/4}{2/\left(\frac{3}{5}\right)}\).

 

Question: Compute \(\dfrac{1}{a/b}\).

 

There are all sorts of tricks and rules one can read off the internet. One can use them as great meta-questions to ponder upon.

 

Question:  

Some teachers have students solve division problems by having students rewrite terms to have a common denominator. For example, to compute

\(\dfrac{3}{4}\div \dfrac{2}{3}\)

rewrite the problem as

\(\dfrac{9}{12}\div \dfrac{8}{12}\).

The claim is then made that the answer to the original problem is \(9 \div 8 = \dfrac{9}{8}\).

 

a)       Does \(\dfrac{3}{4}\div \dfrac{2}{3}\) indeed equal \(\dfrac{9}{8}\)?

 

b)      Work out \(\dfrac{5}{4}\div \dfrac{7}{9}\) via the method of this lesson, and then again by the method described above. Are the answers indeed the same?

 

Why does this “common denominator method” work?

 

Question: Work out \(\dfrac{12}{15}\div \dfrac{3}{5}\) and show that it equals \(\dfrac{4}{3}\).

 

Now notice that

\(12\div 3 = 4\)

\(15 \div 5 = 3\)

and

\(\dfrac{12}{15} \div \dfrac{3}{5}  = \dfrac{4}{3}\).

 

Is this a coincidence or does \(\dfrac{a}{b}\div \dfrac{c}{d}\) always equal \(\dfrac{a \div c}{b \div d}\)?

 

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