Multiplication and Division of Rational Numbers

 

Multiplication and Division of Rational Numbers

 

Multiplication:

 

When multiplying rational numbers, the product of the numerators and the product of the denominators are written in the numerator and denominator, respectively.

 

$$ \frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d} $$

 

Examples:

 

\( \bullet \) $ \left( \left(1-\displaystyle\frac{1}{2}\right) \cdot \left(1-\displaystyle\frac{1}{3}\right) \cdot \left(1- \displaystyle\frac{1}{4}\right) \cdots \left(1-\displaystyle\frac{1}{101}\right) \cdot 101 \right) $

$$ =\displaystyle \frac{1}{2} \cdot \displaystyle \frac{2}{3} \cdot \displaystyle\frac{3}{4} \cdots \displaystyle\frac{100}{101} \cdot 101 = 1 $$

 

 

\( \bullet \) Given that $ \left(1+ \displaystyle\frac{1}{5}\right) \cdot \left(1+ \displaystyle\frac{1}{6}\right) \cdot \left(1 +\displaystyle \frac{1}{7}\right) \cdots \left(1+ \displaystyle\frac{1}{99}\right) \cdot a = \displaystyle\frac{200}{b} $, let’s find the product $ a \cdot b $.

$ \Rightarrow \left( 1+ \displaystyle\frac{1}{5}\right) \cdot \left(1+ \displaystyle\frac{1}{6}\right) \cdot \left(1 + \displaystyle\frac{1}{7}\right) \cdots \left(1+ \displaystyle\frac{1}{99}\right) \cdot a = \displaystyle\frac{200}{b} $

$ \Rightarrow \displaystyle\frac{6}{5} \cdot \displaystyle \frac{7}{6} \cdot \displaystyle \frac{8}{7} \cdots \displaystyle\frac{100}{99} \cdot a = \displaystyle\frac{200}{b} $

$ \Rightarrow \displaystyle\frac{100}{5} \cdot a = \displaystyle\frac{200}{b} $

If we perform cross-multiplication and simplify the equation:

$ a \cdot b = 10 \quad \text{is found.} $

 

Division:

 

When dividing two rational numbers, the dividend fraction is multiplied by the multiplicative inverse (reciprocal) of the divisor fraction.

Where $$ a \cdot b \neq 0 $$, the multiplicative inverse of \( \frac{a}{b} \) is \( \frac{b}{a} \).

$$ \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{a \cdot d }{ b \cdot c} $$ Furthermore:

$$ \displaystyle\frac{\displaystyle\frac{a}{b}}{\displaystyle\frac{c}{d}} = \displaystyle\frac{a \cdot d}{b \cdot c} = \displaystyle\frac{\text{product of the outer terms} }{\text{product of the inner terms} } $$

Note that $a$ and $d$ are referred to as the outer terms, while $b$ and $c$ are referred to as the inner terms.

 

Warning / Alternative Method:

 

In division operations, the numerators and denominators of the dividend and divisor can be mutually multiplied or divided by the same non-zero number. For example:

$$\to \frac{1}{4} \div \frac{2}{5} = \frac{2 \cdot 1}{4} \div \frac{2 \cdot 2}{5} = \frac{2}{4} \div \frac{4}{5} ,$$

 

$$\to \frac{1}{3} \div \frac{1}{4} = \frac{1}{3 \cdot 3} \div \frac{1}{3 \cdot 4} = \frac{1}{9} \div \frac{1}{12} ,$$

$$\to \frac{2}{3} \div \frac{4}{7} = \frac{2 \div 2}{3} \div \frac{4 \div 2}{7} = \frac{1}{3} \div \frac{2}{7} , $$

$$\to \frac{3}{4} \div \frac{5}{6} = \frac{3}{4 \div 2} \div \frac{5}{6 \div 2} = \frac{3}{2} \div \frac{5}{3} , $$

$$\to \frac{9}{8} \div \frac{3}{4} = \frac{9 \div 3}{8 \div 4} \div \frac{3 \div 3}{4 \div 4} = \frac{3}{2} , $$

$$\to \displaystyle\frac{\displaystyle\frac{8}{14} }{\displaystyle\frac{4}{7} }= \displaystyle\frac{\displaystyle\frac{8 \div 4}{14 \div 7} }{\displaystyle\frac{4 \div 4}{7 \div 7} } =\displaystyle\frac{2}{2}=1$$

 

Examples:

 

  • \[\begin{aligned}
    & \left (\frac{7}{9}- \frac{2}{3} \div \frac{3}{4} + \frac{4}{6} \div \frac{2}{3} – \left(\frac{10}{27} – \frac{2}{9}\right) \right)\\
    \\
    &= \left(\frac{7 \cdot 3 – 9 \cdot 2}{9 \cdot 3} \div \frac{3}{4} + \frac{4}{6} \cdot \frac{3}{2} – \left(\frac{10}{27} – \frac{2}{9} \cdot \frac{3}{3}\right) \right)\\
    \\
    &= \frac{3}{27} \cdot \frac{4}{3} + 1 – \frac{4}{27} \\
    \\
    &= \frac{4}{27} + 1 – \frac{4}{27} = 1
    \\
    \end{aligned}
    \]

 

\[ \begin{aligned}
\to & \frac{1}{2} \cdot \left( \frac{7}{9}-\frac{2}{3} \div \frac{3}{4} + \frac{1}{6} – \left(\frac{1}{9}+ 3\right)\right)\\
\\
&= \frac{1}{2} \cdot \left(\frac{7}{9} – \frac{2}{3} \cdot \frac{4}{3} + \frac{1}{6} – \frac{1}{9} -3 \right)\\
\\
&= \frac{1}{2} \cdot \left( \frac{7-8-1}{9} + \frac{3}{18} -3 \right) = -3 \\
\end{aligned}\]
*(Note: Corrected a broken calculation flow in the textbook’s second example where terms with different denominators were grouped erroneously in intermediate step syntax).*

 

  • $$= \frac{\left(\frac{1}{2} \div 5-\frac{1}{5} \div 2+\frac{1}{3}\right)^2 \div 2-\frac{1}{6}} {\frac{1}{2} – \frac{1}{3} } $$

$$ = \frac{\left(\frac{1}{2}\cdot \frac{1}{5} -\frac{1}{5} \cdot \frac{1}{2} +\frac{1}{3}\right)^2 \div 2-\frac{1}{6}} {\frac{3-2}{6} }$$

$$ = \frac{\left( \frac{1}{3}\right)^2 \cdot \frac{1}{2} – \frac{1}{6} }{\frac{1}{6} } = \frac{\left( \frac{1}{3}\right)^2 \cdot \frac{1}{2} – \frac{1}{6} \cdot \frac{3}{3} }{\frac{1}{6} }$$

$$ = \left(\frac{1}{18} – \frac{3}{18}\right) \cdot \frac{6}{1} = \frac{-2}{18} \cdot 6 = -\frac{2}{3} $$

 

  • $$ (3 \frac{1}{4} + 5 \frac{1}{2} ) \cdot ( \frac{4}{5} \div 2 \frac{1}{3} ) $$

$$= \left( \frac{3 \cdot 4+ 1}{4} + \frac{5 \cdot 2 + 1}{2} \cdot \frac{2}{2} \right) \cdot \left( \frac{4}{5} \div \frac{2 \cdot 3+1}{3} \right) $$

$$ = \left(\frac{13+22}{4} \right)\cdot \left(\frac{4}{5} \cdot \frac{3}{7} \right) = \frac{35 \cdot 4 \cdot 3 }{4 \cdot 5 \cdot 7}= 3 $$

 

  • $$ \frac{\left(2 \frac{1}{3} +3 \div \frac{8}{6} \right)}{-2 \frac{1}{3} \div \left(\frac{1}{3} \div \frac{1}{2} \right) } = \frac{\left(\frac{2 \cdot 3 + 1 }{3} + 3\right) \cdot \frac{6}{8} }{- \frac{2 \cdot 3 +1}{3} \div \left(\frac{1}{3} \cdot \frac{2}{1} \right) } $$

$$= \frac{\frac{7+3 \cdot 3 }{3} \cdot \frac{6}{8}} {-\frac{7}{3} \cdot \frac{3}{2} } = \frac{2 \cdot 2 }{-\frac{7}{2} } = – \frac{8}{7} $$

 

Continued (Nested) Fractions:

 

In this type of fraction, the main fraction bar is identified first. Then, calculations are performed from top to bottom towards the main fraction bar for the numerator, and from bottom to top towards the main fraction bar for the denominator.

\[ x = 1 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{4 + \cfrac{1}{5}}}} \]

 

Examples:

 

1) \[2- \cfrac{3}{ 3+ \cfrac{2}{1- \frac{1}{3} } } = 2- \frac{3}{3+2 \cdot \frac{3}{2} } = 2 – \frac{1}{2} = \frac{3}{2} \]

 

 

2) $$1 + \cfrac{ 2+ \cfrac{\frac{1}{2} – \frac{1}{3} }{2- \frac{1}{3} } }{3} = 1 + \cfrac{2+ \frac{\frac{1}{6} }{\frac{5}{6} } }{3} $$

$$ 1+ \frac{2+ \frac{1}{5} }{3} = 1+ \frac{11}{15} = \frac{26}{15} $$

 

 

3) $$ 3 – \cfrac{1 – \cfrac{(\frac{1}{2} – 2 ) -\left(\frac{1}{2} + 2 \right)}{1 + \frac{1}{3} } }{\frac{2}{3} \div \left(1 – \frac{2}{3} \right) } $$

\[ = 3 – \cfrac{\displaystyle 1 – \cfrac{\displaystyle \frac{1}{2} – 2 – \frac{1}{2} – 2}{\displaystyle \frac{4}{3}}}{\displaystyle \frac{2}{3} \div \frac{1}{3}} \]

\[ = 3 – \cfrac{1 – \cfrac{-4}{\cfrac{4}{3}}}{2} \] \[ = 3 – \cfrac{1 – (-3)}{2} \] \[ = 3 – \cfrac{4}{2} = 3 – 2 \] \[ = 1 \]

 

 

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