Skip to main content

A University-Level Understanding of Number with an eye towards elementary education

Shannon Ezzat

Section 5.2 Arithmetic With Fractions

Subsection 5.2.1 Adding and Subtracting Fractions

In this subsection, we will discuss adding and subtracting fractions, starting with fractions with the same denominator and then moving on to fractions with unlike denominators. Adding and subtracting fractions is similar to adding and subtracting units, like apples or other objects, but instead, we are working with parts of a whole.
🔗
First, let’s discuss adding and subtracting fractions with the same denominator. If we understand that the denominator tells us how large each equal part is, we already know how to do this.
🔗

Checkpoint 5.2.1.

Consider adding 3 apples and 5 apples. How many apples do you have? What about 3 pens and 5 pens. What about 3 eighths and 5 eighths? How many eighths do you have?
🔗
🔗

Example 5.2.2. Adding and Subtracting Fractions with the Same Denominator.

Suppose we have two fractions \(\dfrac{3}{8}\) and \(\dfrac{5}{8}\) that we want to add. Since both fractions have the same denominator of 8, we can think of the numerators as the number of "eighths" we have. To add these two fractions, we simply add the numerators and keep the same denominator:
🔗
\begin{equation*} \dfrac{3}{8} + \dfrac{5}{8} = \dfrac{3 + 5}{8} = \dfrac{8}{8} \end{equation*}
Similarly, if we want to subtract the fractions, we subtract the numerators and keep the same denominator:
🔗
\begin{equation*} \dfrac{3}{8} - \dfrac{5}{8} = \dfrac{3 - 5}{8} = \dfrac{-2}{8} \end{equation*}
Again, this is like adding or subtracting units: if we have 3 apples and add 5 more apples, we end up with 8 apples. The same goes for subtraction.
🔗
🔗
Now let’s discuss adding and subtracting fractions with unlike denominators. To do this, we remind ourselves that we can find equivalent fractions with the same denominator. Note that below we do use the least common denominator, but we can indeed do this with any common denominator of the two fractions.
🔗

Example 5.2.3. Adding and Subtracting Fractions with Unlike Denominators.

Let’s say we want to add the fractions \(\dfrac{3}{4}\) and \(\dfrac{5}{6}\text{.}\) First, we need to find equivalent fractions with a common denominator, which in this case is the least common denominator (LCD) of 4 and 6. The LCD is 12, so we will find equivalent fractions with a denominator of 12:
🔗
\begin{equation*} \dfrac{3 \times 3}{4 \times 3} = \dfrac{9}{12}, \quad \dfrac{5 \times 2}{6 \times 2} = \dfrac{10}{12} \end{equation*}
Now that we have equivalent fractions with the same denominator, we can add them just as we did in the previous example:
🔗
\begin{equation*} \dfrac{9}{12} + \dfrac{10}{12} = \dfrac{9 + 10}{12} = \dfrac{19}{12} \end{equation*}
Similarly, if we want to subtract the fractions, we can do so by subtracting the numerators while keeping the same denominator:
🔗
\begin{equation*} \dfrac{9}{12} - \dfrac{10}{12} = \dfrac{9 - 10}{12} = \dfrac{-1}{12} \end{equation*}
🔗
🔗

Subsection 5.2.2 Mixed Fractions

Many times, we want to make as many wholes from fractions as possible. This is especially useful when we want to easily estimate the place of a fraction on a number line, or equivalently the size of a fraction compared to integers. We call these numbers with both whole and fractional parts mixed fractions. In this subsection, we will discuss mixed fractions and how to convert between mixed and fractions where the numerator is larger in magnitude than the denominator, called improper fractions. We give definitions of these concepts below. In all of thee definitions we let \(a,b,c \in \mathbb{N}.\)
🔗

Definition 5.2.4.

A proper fraction is a fraction where the numerator is less than the denominator, that is, \(\dfrac{a}{b}\) is a proper fraction if \(a < b\text{.}\)
🔗
🔗

Definition 5.2.5.

An improper fraction is a fraction where the numerator is equal to or greater than the denominator, that is, \(\dfrac{a}{b}\) is an improper fraction if \(a \geq b\text{.}\)
🔗
🔗

Definition 5.2.6.

A mixed number is a number that is represented as the sum of a whole number and a proper fraction. It is usually written in the form \(a\dfrac{b}{c}\) which really means \(a + \dfrac{b}{c}\) but written without the addition sign between the whole number and fraction, where \(a\) is a whole number, and \(\dfrac{b}{c}\) is a proper fraction, meaning \(b < c\text{.}\)
🔗
🔗
Mixed fractions are usually written in the form \(a\dfrac{b}{c}\text{,}\) . To convert a mixed fraction to an improper fraction, we can convert the whole number part \(a\) to the fraction \(\dfrac{a}{1}\) and then use equivalent fractions and addition of fractions.
🔗
To convert a mixed fraction to an improper fraction, we can use the idea of turning a whole number into a fraction with a denominator of 1, and the use the idea of equivalent fractions with the same denominator that we can easily add:
🔗
  1. Convert the whole number part \(a\) to the fraction \(\dfrac{a}{1}\text{.}\)
    🔗
  2. Find equivalent fractions for \(\dfrac{a}{1}\) and \(\dfrac{b}{c}\) with a common denominator.
    🔗
  3. Add the numerators of the equivalent fractions, and keep the common denominator.
    🔗

Example 5.2.7. Converting a Mixed Fraction to an Improper Fraction.

Let’s convert the mixed fraction \(3\dfrac{2}{5}\) to an improper fraction. First, we convert the whole number part 3 to the fraction \(\dfrac{3}{1}\text{.}\)
🔗
Now, we need to find equivalent fractions for \(\dfrac{3}{1}\) and \(\dfrac{2}{5}\) with a common denominator. In this case, the least common denominator is 5. We multiply the numerator and denominator of \(\dfrac{3}{1}\) by 5 to get the equivalent fraction \(\dfrac{15}{5}\text{.}\)
🔗
Now, we can add the numerators of the equivalent fractions: \(15 + 2 = 17\text{.}\) Keeping the common denominator, we get the improper fraction \(\dfrac{17}{5}\text{.}\) So, \(3\dfrac{2}{5} = \dfrac{17}{5}\text{.}\)
🔗
🔗

Example 5.2.8. Converting a Mixed Fraction to an Improper Fraction.

Let’s convert the mixed fraction \(4\dfrac{3}{8}\) to an improper fraction. First, we convert the whole number part 4 to the fraction \(\dfrac{4}{1}\text{.}\)
🔗
Now, we need to find equivalent fractions for \(\dfrac{4}{1}\) and \(\dfrac{3}{8}\) with a common denominator. In this case, the least common denominator is 8. We multiply the numerator and denominator of \(\dfrac{4}{1}\) by 8 to get the equivalent fraction \(\dfrac{32}{8}\text{.}\) Since both the numerator and denominator are divisible by 8 we can reduce this fraction to \(\dfrac{4}{1} = 4.\)
🔗
🔗
🔗

Subsection 5.2.3 Converting Improper Fractions to Mixed Numbers

To convert an improper fraction to a mixed number, we need to make as many wholes (or units) from the number of equal sized pieces of the whole that we have. For example, if we have 4 thirds, three thirds makes one whole and we have one third left ungrouped. So we have that \(\dfrac{4}{3}=1\dfrac{1}{3}.\) We can do this via the following algorithm:
🔗
  1. Divide the numerator by the denominator. This gives you the whole number part of the mixed number.
    🔗
  2. The remainder of the division becomes the numerator of the fractional part of the mixed number.
    🔗
  3. The denominator remains the same in the fractional part of the mixed number.
    🔗

Example 5.2.9. Converting an Improper Fraction to a Mixed Number.

Let’s convert the improper fraction \(\dfrac{23}{5}\) to a mixed number. First, divide the numerator 23 by the denominator 5. This gives us 4 with a remainder of 3. So, the whole number part of the mixed number is 4, and the numerator of the fractional part is 3.
🔗
The denominator remains the same in the fractional part of the mixed number. So, the fractional part is \(\dfrac{3}{5}\text{.}\)
🔗
Thus, the improper fraction \(\dfrac{23}{5}\) is equal to the mixed number \(4\left(\dfrac{3}{5}\right)\text{.}\)
🔗
🔗
🔗

Subsection 5.2.4 Multiplication of Fractions

Let’s delve into the concept of multiplication of fractions. This operation may appear more complex than addition or subtraction of fractions, but it is fundamentally straightforward. To illustrate, let’s consider a few examples.
🔗

Example 5.2.10.

What is \(\dfrac{1}{2} \times 3\text{,}\) or in other words, half of 3?
🔗
Solution.
If we take three units and split everything in half, we essentially have three halves, or \(\dfrac{3}{2}\text{.}\)
🔗
🔗
🔗

Example 5.2.11.

What is half of \(\dfrac{2}{3}\text{,}\) or in other words, \(\dfrac{1}{2} \times \dfrac{2}{3}\text{?}\)
🔗
Solution.
If we take two thirds and divide them in half, we get \(\dfrac{1}{3}\text{.}\) Therefore, half of \(\dfrac{2}{3}\) is \(\dfrac{1}{3}\text{.}\)
🔗
🔗
🔗

Example 5.2.12.

What is \(\dfrac{2}{5} \times \dfrac{2}{3} \text{,}\) or in other words, two-fifths of two-thirds?
🔗
Solution.
If we take two thirds and divide them into fifths, and then take two of those fifths, we essentially have \(\dfrac{4}{15}\text{.}\)
🔗
🔗
🔗
The examples above illustrate how multiplication of fractions works. Essentially, we multiply the numerators to get the numerator of the result, and multiply the denominators to get the denominator of the result. As you can see by the diagrams, the area model of multiplication is a good lens in which to view this idea. If we are multiplying \(\dfrac{a}{b} \times \dfrac{c}{d}\text{,}\) the denominator of the product is the product of the denominators of each fraction (the unit is broken up vertically into \(d\) equal pieces and horizontally into \(b\) equal pieces, and the numerator is the number of these equal pieces you have, so \(a \times c\) equal pieces. So, for any fractions \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\text{,}\) \(\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d} \text{.}\) This operation breaks up our units into \(b \times d\) equal pieces, and we have \(a \times c\) of these pieces.
🔗
To help us with division, we will need the concept of a fraction with the numerator and denominator "flipped". Like how the idea integers helped us turn all subtraction problems into addition problems (that is \(a - b = a + (-b))\text{,}\) this idea of the reciprocal helps us turn division problems into multiplication problems.
🔗

Definition 5.2.13.

The reciprocal of a fraction \(\dfrac{a}{b}\text{,}\) where \(a\) and \(b\) are both nonzero, is \(\dfrac{b}{a}\) - essentially, we swap the numerator and the denominator.
🔗
🔗

Example 5.2.14.

What is the reciprocal of \(\dfrac{4}{5}\text{?}\)
🔗
Solution.
The reciprocal of \(\dfrac{4}{5}\) is \(\dfrac{5}{4}\text{.}\)
🔗
🔗
🔗

Checkpoint 5.2.15.

  1. Why does both the numerator and denominator have to be non-zero to meaningfully talk about a reciprocal?
    🔗
    🔗
🔗
Note that a non-zero fraction multiplied by its reciprocal always equals one. This is because \(\dfrac{a}{b} \times \dfrac{b}{a} = \dfrac{a \times b}{b \times a}\text{.}\) Since \(a \times b\) equals \(b \times a\text{,}\) this simplifies to \(\dfrac{1}{1}\text{,}\) which of course equals 1. In fact, you can use this as the definition of the reciprocal of a fraction instead of the definition above.
🔗
🔗

Subsection 5.2.5 Division of Fractions

Dividing fractions is a concept that, like addition and subtraction, can be visualized using the denominator of fractions as some unit or object. Let’s consider an example where we have 13 apples divided into groups of 4 apples. We take the 13 apples, group them into sets of 4, and get 3 groups with 1 apple left over. This can be represented as 3 and a quarter groups (using mixed number notation), where the 1 apple left over is a quarter of a group. ADD PIC HERE
🔗
Now, if we consider a fraction like thirteen-fifths divided by four-fifths, the principle is the same. We group them into sets of four fifths and find that we have 3 groups of four fifths with one fifth left over. This also gives us 3 and a quarter groups, or in other terms, thirteen quarters. ADD PIC HERE
🔗
The key idea here is that the two problems above are fundamentally the same problem, as we are dividing thirteen units into groups of four units; whether the units are apples or fifths, or something else does not make a difference to our solution. So, fundamentally if we understand the idea of division of whole numbers, we understand the idea of division of fractions.
🔗
This leads us to the general rule of dividing fractions: if we have \(\dfrac{a}{c} \div \dfrac{b}{c}\) (where \(a, b, c\) are any whole numbers where \(b,c \neq 0\)), our result will be \(\dfrac{a}{b}\text{.}\) The important thing to note is that our units (the things we’re grouping) are the same, which allows us to simply divide the first number by the second number. ADD PIC HERE
🔗
However, it’s not always the case that we have fractions with common denominators to divide. For this, we need to know how to divide fractions with any denominators. The rule here is that \(\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c} = \dfrac{ad}{bc}\text{.}\)
🔗
You have most likely learned in school that to divide to fractions, you multiply the first fraction by the reciprocal of the second fraction; for example \(\dfrac{2}{3} \div {4}{5} = \dfrac{2}{3} \times {5}{4}.\) Of course this is correct, but as we are in the business of understanding why these ideas are correct, let’s look more closely at why this idea is true.
🔗
Consider the division \(\dfrac{a}{b} \div \dfrac{c}{d}\text{.}\) We know how to divide fractions with the same denominator, so let’s find equivalent fractions to these two with the denominator of \(bd\text{.}\)
🔗
\begin{equation*} \dfrac{a}{b} = \dfrac{ad}{bd} \qquad \dfrac{c}{d} = \dfrac{bc}{bd}. \end{equation*}
Now, we are dividing \(ad\) \(bd\)ths into groups of size \(bc\) \(bd\)ths. From our discussion above we know that \(\dfrac{ad}{bd} \div \dfrac{bc}{bd} = \dfrac{ad}{bc}\text{.}\) But we can express this as the multiplication of two fractions: \(\dfrac{ad}{bc} = \dfrac{a}{b} \times \dfrac{d}{c}\) which is indeed the multiplication by the reciprocal result we wanted to show.
🔗
In addition, it’s important to note that all the properties that hold for integers under addition also hold for fractions. Because we can convert any division operation to a multiplication operation (by multiplying by the reciprocal), we can think of dealing with fractions as only involving addition and multiplication. This allows us to use the commutative, associative, and distributive properties as needed to simplify expressions involving fractions.
🔗

Example 5.2.16. Adding, Subtracting, Multiplying, and Dividing Fractions.

Let’s perform multiple operations on fractions in one problem. Consider the expression:
🔗
\begin{equation*} \dfrac{3}{4} - \dfrac{1}{2} + \dfrac{5}{6} \cdot \dfrac{2}{3} ÷ \dfrac{4}{5}. \end{equation*}
Solution.
To simplify the expression, we follow the order of operations (BEDMAS): Brackets, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
🔗
First, we handle the division by changing the operation to multiplication by the reciprocal of the divisor:
🔗
\begin{equation*} \dfrac{5}{6} \times \dfrac{2}{3} ÷ \dfrac{4}{5} = \dfrac{5}{6} \times \dfrac{2}{3} \cdot \dfrac{5}{4}. \end{equation*}
Multiplying these fractions and reducing to equivalent fractions where possible:
🔗
\begin{equation*} \dfrac{5}{6} \times \dfrac{2}{3} \cdot \dfrac{5}{4} = \dfrac{10}{18} \times \dfrac{5}{4} = \dfrac{5}{9} \times {5}{4} = \dfrac{25}{36}. \end{equation*}
So the original expression is equivalent to
🔗
\begin{equation*} \dfrac{3}{4}-\dfrac{1}{2}+\dfrac{25}{36} \end{equation*}
We have dealt with all of the multiplication and division. Now we can add and subtract from left to right. Noting that \(\dfrac{3}{4}-\dfrac{1}{2} = \dfrac{1}{4}\) and then having a common denominator of 36 we have
🔗
\begin{equation*} \dfrac{1}{4} + \dfrac{25}{36} = \dfrac{9}{36} + \dfrac{25}{36}= \dfrac{9 + 25}{24} = \dfrac{34}{36} = \dfrac{17}{18}. \end{equation*}
Thus, the simplified result of the given expression is \(\dfrac{17}{18}\text{.}\)
🔗
🔗
🔗
🔗

Subsection 5.2.6 Negative Exponents and Fractions

Let’s consider some important rules about negative exponents. Remember from the previous section on exponents, we had that, for a base \(a \in \mathbb{R}\) and exponents \(b,c \in \mathbb{N}\) we have that \(a^b \times a^c=a^{b+c}.\) We would like to extend this idea to having negative integers as our exponents. Now that we have developed the idea of fractions, we can indeed define what a negative exponent would mean.
🔗
Suppose we have a number \(a\) and we are indeed curious about what \(a\) raised to a negative exponent would mean. We want to keep the same rules that we’ve used before, where if you have the same base and different exponents, you add them. This would imply that \(a^1 \cdot a^{-1}\) should equal \(a^{1+(-1)}\text{,}\) which is \(a^0\text{.}\) And anything raised to the power of zero is one. So what could we multiply \(a\) by to get one? The answer is \(\dfrac{1}{a}\text{.}\) Hence, whenever we see a negative exponent like this, it implies the reciprocal of the base raised to the corresponding positive exponent. In general, \(a^{-n} = \dfrac{1}{a^n}\text{.}\)
🔗
Another rule deals with fractions raised to an exponent. The exponent means we multiply the base, in this case our base is the fraction \(\dfrac{a}{b}\text{,}\) by itself the exponent number of times. Since we multiply fractions by multiplying the numerators and denominators together, respectively, this means that \((\dfrac{a}{b})^n\) equals \(\dfrac{a^n}{b^n}\text{.}\)
🔗
Now, what happens when we have a negative exponent of a fraction? Let’s consider \((\dfrac{a}{b})^{-1}\text{.}\) If we multiply \(\dfrac{a}{b}^{-1} \cdot \dfrac{a}{b}^{1}\text{,}\) we’d like that to be \(\dfrac{a}{b}^{0}\text{,}\) and anything raised to the power of 0 is 1. So what can we multiply \(\dfrac{a}{b}\) by to get 1? The answer is \(\dfrac{b}{a}\text{.}\) So, this means that \((\dfrac{a}{b})^{-1} = \dfrac{b}{a}\text{.}\)
🔗
We will now explore some examples to further solidify our understanding of these concepts.
🔗

Subsubsection 5.2.6.1 Simplifying Compound Fractions

A compound fraction, also known as a complex fraction, is a fraction where the numerator, the denominator, or both are fractions themselves. For example, \(\dfrac{a}{b}/\dfrac{c}{d}\) is a compound fraction. To simplify such fractions, we follow the principle of dividing fractions which states that dividing by a fraction is the same as multiplying by its reciprocal. Hence, \(\dfrac{a}{b}/\dfrac{c}{d}\) simplifies to \(\dfrac{a}{b} \cdot \dfrac{d}{c}\) which further simplifies to \(\dfrac{ad}{bc}\text{.}\)
🔗
Let’s now consider an example that combines the ideas of positive and negative exponents, positive and negative exponents of fractions, the exponent 0, and compound fractions.
🔗
Example 5.2.17. Simplifying a Complex Fraction with Exponents.
Simplify the compound fraction \(\dfrac{2^{3} \cdot 3^{-1} \cdot 5^{2}}{(3^{2} \cdot 5^{-2})/ (2^{-4} \cdot 3^{4}/(3^{4} \cdot 5^{4}))}\) and express the result such that there are no negative exponents anywhere.
🔗
Solution.
First, we simplify the denominator using the compound fraction rule: \((3^{2} \cdot 5^{-2})/ (2^{-4} \cdot 3^{4}/(3^{4} \cdot 5^{4})) = (3^{2} \cdot 5^{-2}) \cdot (3^{4} \cdot 5^{4}/2^{-4} \cdot 3^{4})\text{.}\) This simplifies further to \(3^{4} \cdot 5^{2} \cdot 2^{4}\text{.}\) Now our expression becomes \(\dfrac{2^{3} \cdot 3^{-1} \cdot 5^{2}}{3^{4} \cdot 5^{2} \cdot 2^{4}}\text{.}\) We use the rules of exponents to simplify this to \(2^{3-4} \cdot 3^{-1-4} \cdot 5^{2-2} = 2^{-1} \cdot 3^{-5}\text{.}\) Finally, applying the negative exponent rule, we get \(\dfrac{1}{2^{1} \cdot 3^{5}} = \dfrac{1}{2 \cdot 243} = \dfrac{1}{486}\text{.}\)
🔗
🔗
🔗
🔗
🔗
🔗
PrevTopNext