# 4.3 Equivalent fractions, reducing fractions to lowest terms, and

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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses equivalent fractions, reducing fractions to lowest terms, and raising fractions to higher terms. By the end of the module students should be able to recognize equivalent fractions, reduce a fraction to lowest terms and be able to raise a fraction to higher terms.

## Section overview

• Equivalent Fractions
• Reducing Fractions to Lowest Terms
• Raising Fractions to Higher Terms

## Equivalent fractions

Let's examine the following two diagrams. Notice that both $\frac{2}{3}$ and $\frac{4}{6}$ represent the same part of the whole, that is, they represent the same number.

## Equivalent fractions

Fractions that have the same value are called equivalent fractions . Equiva­lent fractions may look different, but they are still the same point on the number line.

There is an interesting property that equivalent fractions satisfy. ## A test for equivalent fractions using the cross product

These pairs of products are called cross products . If the cross products are equal, the fractions are equivalent. If the cross products are not equal, the fractions are not equivalent.

Thus, $\frac{2}{3}$ and $\frac{4}{6}$ are equivalent, that is, $\frac{2}{3}=\frac{4}{6}$ .

## Sample set a

Determine if the following pairs of fractions are equivalent.

$\frac{3}{4}\text{and}\frac{6}{8}$ . Test for equality of the cross products.  The cross products are equals.

The fractions $\frac{3}{4}$ and $\frac{6}{8}$ are equivalent, so $\frac{3}{4}=\frac{6}{8}$ .

$\frac{3}{8}\text{and}\frac{9}{\text{16}}$ . Test for equality of the cross products.  The cross products are not equal.

The fractions $\frac{3}{8}$ and $\frac{9}{\text{16}}$ are not equivalent.

## Practice set a

Determine if the pairs of fractions are equivalent.

$\frac{1}{2}$ , $\frac{3}{6}$ , yes

$\frac{4}{5}$ , $\frac{\text{12}}{\text{15}}$ , yes

$\frac{2}{3}$ , $\frac{8}{\text{15}}$

$\mathrm{30}\ne \mathrm{24}$ , no

$\frac{1}{8}$ , $\frac{5}{\text{40}}$ , yes

$\frac{3}{\text{12}}$ , $\frac{1}{4}$ , yes

## Reducing fractions to lowest terms

It is often very useful to conver t one fraction to an equivalent fraction that has reduced values in the numerator and denominator. We can suggest a method for doing so by considering the equivalent fractions $\frac{9}{\text{15}}$ and $\frac{3}{5}$ . First, divide both the numerator and denominator of $\frac{9}{\text{15}}$ by 3. The fractions $\frac{9}{\text{15}}$ and $\frac{3}{5}$ are equivalent.

(Can you prove this?) So, $\frac{9}{\text{15}}=\frac{3}{5}$ . We wish to convert $\frac{9}{\text{15}}$ to $\frac{3}{5}$ . Now divide the numerator and denominator of $\frac{9}{\text{15}}$ by 3, and see what happens.

$\frac{9÷3}{\text{15}÷3}=\frac{3}{5}$

The fraction $\frac{9}{\text{15}}$ is converted to $\frac{3}{5}$ .

A natural question is "Why did we choose to divide by 3?" Notice that

$\frac{9}{\text{15}}=\frac{3\cdot 3}{5\cdot 3}$

We can see that the factor 3 is common to both the numerator and denominator.

## Reducing a fraction

From these observations we can suggest the following method for converting one fraction to an equivalent fraction that has reduced values in the numerator and denominator. The method is called reducing a fraction .

A fraction can be reduced by dividing both the numerator and denominator by the same nonzero whole number. Consider the collection of equivalent fractions

$\frac{5}{\text{20}}$ , $\frac{4}{\text{16}}$ , $\frac{3}{\text{12}}$ , $\frac{2}{8}$ , $\frac{1}{4}$

## Reduced to lowest terms

Notice that each of the first four fractions can be reduced to the last fraction, $\frac{1}{4}$ , by dividing both the numerator and denominator by, respectively, 5, 4, 3, and 2. When a fraction is converted to the fraction that has the smallest numerator and denomi­nator in its collection of equivalent fractions, it is said to be reduced to lowest terms . The fractions $\frac{1}{4}$ , $\frac{3}{8}$ , $\frac{2}{5}$ , and $\frac{7}{\text{10}}$ are all reduced to lowest terms.

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