1.5: Equivalent Fractions
- Page ID
- 49341
Overview
- Equivalent Fractions
- Reducing Fractions To Lowest Terms
- Raising Fractions To Higher Terms
Fractions that have the same value are called equivalent fractions
For example, \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\) represent the same part of a whole quantity and are therefore equivalent. Several more collections of equivalent fractions are listed below:
\(\dfrac{15}{25}, \dfrac{12}{20}, \dfrac{3}{5}\)
\(\dfrac{1}{3}, \dfrac{2}{6}, \dfrac{3}{9}, \dfrac{4}{12}\)
\(\dfrac{7}{6}, \dfrac{14}{12}, \dfrac{21}{18}, \dfrac{28}{24}, \dfrac{35}{30}\)
Reducing Fractions to Lowest Terms
Reduced to Lowest Terms
It is often useful to convert one fraction to an equivalent fraction that has reduced values in the numerator and denominator. When a fraction is converted to an equivalent fraction that has the smallest numerator and denominator in the collection of equivalent fractions, it is said to be reduced to lowest terms. The conversion process is called reducing a fraction.
We can reduce a fraction to lowest terms by
- Expressing the numerator and denominator as a product of prime numbers. (Find the prime factorization of the numerator and denominator. See Section 1.3 for this technique.)
- Divide the numerator and denominator by all common factors. (This technique is commonly called “cancelling.”)
Sample Set A:
\(\begin{aligned}
&\dfrac{6}{18}=\dfrac{2 \cdot 3}{2 \cdot 3 \cdot 3}\\
&=\dfrac{\not{2} \cdot \not{3}}{\not{2} \cdot \not{3} \cdot 3} \quad 2 \text { and } 3 \text { are common factors. }\\
&=\dfrac{1}{3}
\end{aligned}
\)
\(
\begin{aligned}
\dfrac{16}{20} &=\dfrac{2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 5} \\
&=\dfrac{\not{2} \cdot \not{2} \cdot 2 \cdot 2}{\not{2} \cdot \not{2} \cdot 5} \quad 2 \text { is the only common factor. } \\
&=\dfrac{4}{5}
\end{aligned}
\)
\(
\begin{aligned}
&\dfrac{56}{70}=\dfrac{2 \cdot 4 \cdot 7}{2 \cdot 5 \cdot 7}\\
&=\dfrac{\not{2} \cdot 4 \cdot \not{7}}{\not{2} \cdot 5 \cdot \not{7}} \quad 2 \text { and } 7 \text { are common factors. }\\
&=\dfrac{4}{5}
\end{aligned}
\)
\(
\dfrac{8}{15}=\dfrac{2 \cdot 2 \cdot 2}{3 \cdot 5}
\) There are no common factors.
Thus, \(\dfrac{8}{15}\) is reduced to lowest terms.
Raising a Fraction to Higher Terms
Equally important as reducing fractions is raising fractions to higher terms. Raising a fraction to higher terms is the process of constructing an equivalent fraction that has higher values in the numerator and denominator. The higher, equivalent fraction is constructed by multiplying the original fraction by 1.
Notice that \(\dfrac{3}{5}\) and \(\dfrac{9}{15}\) are equivalent, that is \(\dfrac{3}{5}\) = \(\dfrac{9}{15}\). Also,
\(
\begin{array}{l}
\dfrac{3}{5} \cdot 1=\dfrac{3}{5} \cdot \dfrac{3}{3}=\dfrac{3 \cdot 3}{5 \cdot 3}=\dfrac{9}{15} \\
1=\dfrac{3}{3}
\end{array}
\)
This observation helps us suggest the following method for raising a fraction to higher terms.
A fraction can be raised to higher terms by multiplying both the numerator and denominator by the same nonzero number.
For example, \(\dfrac{3}{4}\) can be raised to \(\dfrac{24}{32}\) by multiplying both the numerator and denominator by 8, that is, multiplying by 1 in the form \(\dfrac{8}{8}\).
\(
\dfrac{3}{4}=\dfrac{3 \cdot 8}{4 \cdot 8}=\dfrac{24}{32}
\)
How did we know to choose 8 as the proper factor? Since we wish to convert 4 to 32 by multiplying it by some number, we know that 4 must be a factor of 32. This means that 4 divides into 32. In fact, \(32 \div 4=8\). We divided the original denominator into the new, specified denominator to obtain the proper factor for the multiplication.
Sample Set B
Determine the missing numerator or denominator.
\(\dfrac{3}{7}=\dfrac{?}{35} . \quad \text{Divide the original denominator, } 7, \text{ into the new denominator }35\)
\(35 \div 7=5\)
\(\text{Multiply the original numerator by } 5.\)
\(\dfrac{3}{7}=\dfrac{3 \cdot 5}{7 \cdot 5}=\dfrac{15}{35}\)
\(\dfrac{5}{6}=\dfrac{45}{?} . \quad \text{Divide the original denominator, } 5, \text{ into the new denominator }45\)
\(45 \div 5=9\)
\(\text{Multiply the original numerator by } 9.\)
\(\dfrac{5}{6}=\dfrac{5 \cdot 9}{6 \cdot 9}=\dfrac{45}{54}\)
\(\dfrac{6}{8}\)
- Answer
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\(\dfrac{3}{4}\)
\(\dfrac{5}{10}\)
\(\dfrac{6}{14}\)
- Answer
-
\(\dfrac{3}{7}\)
\(\dfrac{4}{14}\)
\(\dfrac{18}{12}\)
- Answer
-
\(\dfrac{3}{2}\)
\(\dfrac{3}{2}\)
\(\dfrac{20}{8}\)
\(\dfrac{10}{6}\)
- Answer
-
\(\dfrac{5}{3}\)
\(\dfrac{14}{4}\)
\(\dfrac{10}{12}\)
- Answer
-
\(\dfrac{5}{6}\)
\(\dfrac{32}{28}\)
\(\dfrac{36}{10}\)
- Answer
-
\(\dfrac{18}{5}\)
\(\dfrac{26}{60}\)
\(\dfrac{12}{18}\)
- Answer
-
\(\dfrac{2}{3}\)
\(\dfrac{18}{27}\)
\(\dfrac{18}{24}\)
- Answer
-
\(\dfrac{3}{4}\)
\(\dfrac{32}{40}\)
\(\dfrac{11}{22}\)
- Answer
-
\(\dfrac{1}{2}\)
\(\dfrac{17}{51}\)
\(\dfrac{27}{81}\)
- Answer
-
\(\dfrac{1}{3}\)
\(\dfrac{16}{42}\)
\(\dfrac{6}{8}\)
- Answer
-
\(\dfrac{3}{4}\)
\(\dfrac{39}{13}\)
- Answer
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3
\(\dfrac{44}{11}\)
\(\dfrac{121}{132}\)
- Answer
-
\(\dfrac{11}{12}\)
\(\dfrac{30}{105}\)
\(\dfrac{108}{76}\)
- Answer
-
\(\dfrac{29}{19}\)
For the following problems, determine the missing numerator or denominator.
\(
\dfrac{1}{3}=\dfrac{?}{12}
\)
\(
\dfrac{1}{5}=\dfrac{?}{30}
\)
- Answer
-
6
\(
\dfrac{3}{3}=\dfrac{?}{9}
\)
\(
\dfrac{3}{4}=\dfrac{?}{16}
\)
- Answer
-
12
\(
\dfrac{5}{6}=\dfrac{?}{18}
\)
\(
\dfrac{4}{5}=\dfrac{?}{25}
\)
- Answer
-
20
\(
\dfrac{1}{2}=\dfrac{4}{?}
\)
\(
\dfrac{9}{25}=\dfrac{27}{?}
\)
- Answer
-
75
\(
\dfrac{3}{2}=\dfrac{18}{?}
\)
\(
\dfrac{5}{3}=\dfrac{80}{?}
\)
- Answer
-
48