## Fractions are Hard!

### 3.1 Mixed numbers

A mixed number is a number of the form \(a+\dfrac{b}{c}\), usually just written \(a\dfrac{b}{c}\), with \(a\), \(b\), and \(c\) integers.

**Question: **Is this right? Do textbooks allow \(2\dfrac{7}{3}\) and \(3\dfrac{-5}{6}\) and \(5\dfrac{0}{9}\) as examples of mixed numbers? Or must these particular numbers be written as \(4\dfrac{1}{3}\) and \(2\dfrac{1}{6}\) and \(5\), respectively?

Often fractions of the form \(\dfrac{a}{b}\) with \(a>b\) are dubbed “improper” and people prefer to express such fractions as mixed numbers.

**Example: ***Write \(\dfrac{39}{8}\) as a mixed number.*

**Answer: **\(\dfrac{39}{8} = \dfrac{32+7}{8} = 5 + \dfrac{7}{8} = 5\dfrac{7}{8}\).

Of course, we can convert mixed numbers into (improper) fractions too.

**Example: ***Write \(20\dfrac{1}{20}\) as fraction of the form \(\dfrac{a}{b}\) with \(a\) and \(b\) whole numbers. *

**Answer:**

\(20\dfrac{1}{20}= \dfrac{20\dfrac{1}{20}}{1}\).** **

** **

Now multiply top and bottom by \(20\) to get:

\(\dfrac{20\dfrac{1}{20}}{1} = \dfrac{20\dfrac{1}{20}\times 20}{1\times 20}= \dfrac{400+1}{20}=\dfrac{401}{20}\).** **

** **

** **

**Comment: **Alternatively, \(20\dfrac{1}{20} = 20+\dfrac{1}{20}=\dfrac{400}{20}+\dfrac{1}{20}=\dfrac{401}{20}\).

Question: Write each of the following as a mixed number.
a) \(\dfrac{8}{5}\) b) \(\dfrac{100}{13} \) c) \(\dfrac{200}{199}\) d) \(\dfrac{199\frac{1}{2}}{199}\)
(Can the answer to part d) simply be \(1\dfrac{1/2}{199}\)? Should we indeed have a strict definition of what a “mixed number” should be?)
Write each of the following in the form \(\dfrac{a}{b}\) with \(a\) and \(b\) integers.
e) \(7\dfrac{2}{9}\) f) \(2\dfrac{3}{4} + 5\dfrac{2}{7}\) g) \(300\dfrac{299}{300}\) |

** **

**MULTIPLYING MIXED NUMBERS**

The area model for multiplication can be of visual help here.

**Example: ***Compute \(4\dfrac{2}{5}\times 7\dfrac{3}{8}\).*

**Answer: **This is the area of a rectangle, which naturally divides into four pieces:

We see that

\(4\dfrac{2}{5}\times 7\dfrac{3}{8} = 4\times 7 + 4 \times \dfrac{3}{8}+\dfrac{2}{5}\times 7 + \dfrac{2}{5}\times \dfrac{3}{8}\)

\(=48+\dfrac{12}{8}+\dfrac{14}{5}+\dfrac{6}{40}\)

\(=48+1+\dfrac{1}{2}+2+\dfrac{4}{5}+\dfrac{3}{20}\)

\(=51+\dfrac{10+16+3}{20}=51+\dfrac{29}{20}=52\dfrac{9}{20}\).

Alternatively, \(4\dfrac{2}{5}\times 7\dfrac{3}{8}=\dfrac{22}{5}\times \dfrac{59}{8}=\dfrac{1298}{40}\)*.*

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