## Fractions are Hard!

### 2.7 Multiplying 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)

One consequence of these beliefs is

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

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

* *

**Answer: **We have the task of sharing 7 pies among 5 girls, but want to reduce the amount of pie each girl receives by a factor of two-thirds. Belief 3 says to adjust the number of pies by two-thirds.

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

The fraction of “thirds” appearing in the numerator is irritating. Let’s use the Key Fraction Property (Belief 4) and change the numerator and denominator each by a factor of 3. This does not change the quantity.

* *

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

Within the numerator (using the usual commutativity of arithmetic) we see \(\dfrac{2}{3} \times 3\), which equals \(2\) by Consequence 5.

\(\dfrac{\dfrac{2}{3}\times 7 \times 3}{5 \times 3}=\dfrac{2\times 7}{5 \times 3}\)

and this is \(\dfrac{14}{15}\).

Alternatively, we could have argued

\(\dfrac{c}{d}\times \dfrac{a}{b} = \dfrac{\dfrac{c}{d}\times a}{b}\) (by Belief 3)

\( = \dfrac{\dfrac{ac}{d}}{b}\) (by Belief 3)

\(=\dfrac{\dfrac{ac}{d}\times d}{b \times d} = \dfrac{ac}{bd}\) (by Belief 4 and then Belief 3 again)

We have established the usual multiplication rule from grade school.

\(\dfrac{c}{d}\times \dfrac{a}{b}\)is the product of the “numerators over the product of the denominators.”

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

One can work to be efficient with the arithmetic.

Question: Ibrahim was asked to compute \(\dfrac{39}{7}\times \dfrac{14}{13}\) and within three seconds he said that the answer was \(6\). How did he see this so quickly? |

Question: What is the value of \(\dfrac{39}{35}\times \dfrac{14}{13}\)? |

**SO****ME HAPPY COINCIDENCES**

The word “of” is used in the first introduction of fractions to very young students we way “parts __of__ a whole.”

What if we look at parts of a parts of a whole?

**EXAMPLE: ***Using the parts of a whole thinking, draw a picture of \(\dfrac{2}{3}\) of \(\dfrac{4}{5}\).*

First, here’s a picture of \(\dfrac{4}{5}\) of a pie.

Now, in the early grades, \(\dfrac{2}{3}\) is two copies of one-third. Here’s a one-third of \(\dfrac{4}{5}\) of a pie.

Thus two thirds of \(\dfrac{4}{5}\) is two copies of this.

If we draw in the extra lines in the picture we can see the shaded region as matching 8 parts out of 15.

Thus \(\dfrac{2}{3}\) of \(\dfrac{4}{5}\) matches a picture of \(\dfrac{8}{15}\), sharing 8 pies among 15 boys (or 8 parts out of a whole of 15).

It is a happy coincidence that this matches the result of our multiplication formula: \(\dfrac{2}{3}\times \dfrac{4}{5}=\dfrac{8}{15}\).

**Comment: **It is a further happy coincidence that the picture of parts of a parts of a whole looks something akin to a picture of the area model of multiplication, giving a sense that we really have the right definition of the product of two fractions.

SUMMARY: We found with the pies per boy model that \(\dfrac{a}{b}\times \dfrac{c}{d}\) is \(\dfrac{ac}{bd}\). This formula happens to match the pictures we draw for “\(\dfrac{a}{b}\) of \(\dfrac{c}{d}\),” namely, a picture of \(ac\) parts to a total of \(bd\) parts, that is, the quanitity \(\dfrac{ac}{bd}\). And, moreover, if we happen to draw square pies, this seems to psychologically match the pictures we draw for multiplication via the area model.
Notice this this sequence of ideas is the reverse of the typical textbook approach: Textbooks draw pictures akin to the area model first, attempt to make some statement that “of means multiply” and then derive the formula \(\dfrac{a}{b}\times \dfrac{c}{d} = \dfrac{ac}{bd}\) . We got the formula first and |

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