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:

f

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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