Fraction Operations

Fractions show up whenever something is split into equal parts: a pizza, an hour, a measuring cup, an inch on a ruler. In this lesson you’ll see how every fraction is built from unit fractions, and you’ll practise adding, subtracting, multiplying and dividing fractions, including mixed numbers and negative fractions. You’ll need these skills for formulas and equations all through high school.

Key ideas

Unit fractions

A unit fraction has a numerator of $1$, like $\dfrac{1}{2}$, $\dfrac{1}{8}$ or $\dfrac{1}{16}$. The fraction $\dfrac{1}{n}$ is one of $n$ equal parts of a whole.

Every other fraction is a number of unit fractions put together:

$$\frac{3}{8} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = 3 \times \frac{1}{8}$$

The bigger the denominator, the smaller the unit fraction: $\dfrac{1}{16}$ is smaller than $\dfrac{1}{8}$, because the whole is cut into more pieces.

Fractions on measuring tools

In Canada we measure mostly in metric units, which use decimals. But some tools still use fractions:

• Rulers and tape measures in inches split each inch into $16$ equal parts. Each smallest space is $\dfrac{1}{16}$ of an inch. Longer marks show eighths, quarters and halves.
• Measuring cups and spoons for cooking come in sizes like $\dfrac{1}{4}$ cup, $\dfrac{1}{3}$ cup and $\dfrac{1}{2}$ cup. To measure $\dfrac{3}{4}$ cup, you can fill the $\dfrac{1}{4}$ cup three times.

Each small space is $\dfrac{1}{16}$ inch. The orange bar measures $1\dfrac{5}{16}$ inches.

To read a ruler, count the small spaces past the last whole inch. The bar above ends $5$ spaces past $1$, so it’s $1\dfrac{5}{16}$ inches long. If you land on a longer mark, simplify: $\dfrac{8}{16} = \dfrac{1}{2}$ and $\dfrac{12}{16} = \dfrac{3}{4}$.

Equivalent fractions

Multiplying or dividing the numerator and denominator by the same number (not zero) gives an equivalent fraction, the same amount written differently:

$$\frac{3}{4} = \frac{3 \times 4}{4 \times 4} = \frac{12}{16} \qquad\qquad \frac{18}{24} = \frac{18 \div 6}{24 \div 6} = \frac{3}{4}$$

A fraction is in lowest terms when the numerator and denominator have no common factor except $1$.

Mixed numbers

A mixed number like $2\dfrac{1}{3}$ means $2 + \dfrac{1}{3}$. To write it as an improper fraction, multiply the whole number by the denominator and add the numerator:

$$2\frac{1}{3} = \frac{2 \times 3 + 1}{3} = \frac{7}{3}$$

To go back, divide: $7 \div 3 = 2$ remainder $1$, so $\dfrac{7}{3} = 2\dfrac{1}{3}$.

Adding and subtracting

You can only add or subtract fractions with the same denominator, because then you’re counting the same-sized pieces. Rewrite them using a common denominator (the lowest common multiple of the denominators works best), then add or subtract the numerators:

$$\frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$$

The denominator stays the same: twelfths plus twelfths gives twelfths.

Multiplying

Multiply the numerators, and multiply the denominators. You can divide out common factors before you multiply to keep the numbers small:

$$\frac{3}{4} \times \frac{8}{9} = \frac{3 \times 8}{4 \times 9} = \frac{24}{36} = \frac{2}{3}$$

In word problems, “of” usually means multiply: $\dfrac{1}{2}$ of $\dfrac{3}{4}$ cup is $\dfrac{1}{2} \times \dfrac{3}{4} = \dfrac{3}{8}$ cup.

Dividing

To divide by a fraction, multiply by its reciprocal (flip the second fraction):

$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$$

Why? Division asks “how many of these fit?” How many $\dfrac{1}{4}$ cups fit in $3$ cups? Each cup holds four of them, so $3 \div \dfrac{1}{4} = 3 \times 4 = 12$.

Always turn mixed numbers into improper fractions before you multiply or divide.

Positive and negative fractions

Negative fractions follow the same sign rules as integers:

$$-\frac{2}{3} = \frac{-2}{3} = \frac{2}{-3}$$

When you add or subtract, put the sign on the numerator and work with the integers on top. When you multiply or divide, same signs give a positive answer and different signs give a negative answer.

Fractions in formulas and relations

Fractions appear in many formulas. The temperature in degrees Fahrenheit is $F = \dfrac{9}{5}C + 32$, where $C$ is the temperature in degrees Celsius. Linear relations can have fractions too, like $y = -\dfrac{2}{3}x + \dfrac{1}{2}$. To evaluate, substitute and follow the order of operations, just as with whole numbers.

Worked examples

Example 1: Reading a ruler

A bolt reaches $11$ small spaces past the $1$-inch mark on a ruler marked in sixteenths.

• (a) How long is the bolt?
• (b) How much longer is it than $1\dfrac{1}{2}$ inches?

Solution.

(a) Each space is $\dfrac{1}{16}$ inch, so the bolt is $1\dfrac{11}{16}$ inches long. ($\dfrac{11}{16}$ is already in lowest terms.)

(b) Write $1\dfrac{1}{2}$ in sixteenths: $\dfrac{1}{2} = \dfrac{8}{16}$, so $1\dfrac{1}{2} = 1\dfrac{8}{16}$.

$$1\frac{11}{16} - 1\frac{8}{16} = \frac{3}{16}$$

The bolt is $\dfrac{3}{16}$ inch longer. On the ruler, that’s $3$ small spaces.

Example 2: Adding and subtracting

Evaluate.

• (a) $\dfrac{5}{6} + \dfrac{3}{4}$
• (b) $3\dfrac{1}{6} - 1\dfrac{3}{4}$
• (c) $-\dfrac{2}{5} + \dfrac{1}{3}$

Solution.

(a) The lowest common denominator of $6$ and $4$ is $12$:

$$\frac{5}{6} + \frac{3}{4} = \frac{10}{12} + \frac{9}{12} = \frac{19}{12} = 1\frac{7}{12}$$

(b) Change to improper fractions, then use the denominator $12$:

$$\begin{aligned} 3\frac{1}{6} - 1\frac{3}{4} &= \frac{19}{6} - \frac{7}{4} \\ &= \frac{38}{12} - \frac{21}{12} \\ &= \frac{17}{12} = 1\frac{5}{12} \end{aligned}$$

Check with estimation: about $3.17 - 1.75 = 1.42$, and $1\dfrac{5}{12} \approx 1.42$. ✓

(c) Put the sign on the numerator and use the denominator $15$:

$$-\frac{2}{5} + \frac{1}{3} = \frac{-6}{15} + \frac{5}{15} = \frac{-1}{15} = -\frac{1}{15}$$

Example 3: Multiplying and dividing

• (a) $-\dfrac{3}{4} \times \dfrac{8}{9}$
• (b) $2\dfrac{1}{2} \div \left(-\dfrac{5}{6}\right)$
• (c) A muffin recipe needs $\dfrac{3}{4}$ cup of flour per batch. How many batches can you make with $3$ cups of flour?

Solution.

(a) Different signs, so the answer is negative. Divide out the common factors first: $3$ goes into $3$ and $9$, and $4$ goes into $4$ and $8$:

$$-\frac{3}{4} \times \frac{8}{9} = -\frac{3 \times 8}{4 \times 9} = -\frac{1 \times 2}{1 \times 3} = -\frac{2}{3}$$

(b) Change $2\dfrac{1}{2}$ to $\dfrac{5}{2}$, then multiply by the reciprocal of $-\dfrac{5}{6}$:

$$\frac{5}{2} \div \left(-\frac{5}{6}\right) = \frac{5}{2} \times \left(-\frac{6}{5}\right) = -\frac{30}{10} = -3$$

(c) How many $\dfrac{3}{4}$ cups fit in $3$ cups?

$$3 \div \frac{3}{4} = \frac{3}{1} \times \frac{4}{3} = \frac{12}{3} = 4$$

You can make $4$ batches. Check: $4 \times \dfrac{3}{4} = 3$ cups. ✓

Example 4: Fractions in formulas

• (a) Use $F = \dfrac{9}{5}C + 32$ to convert $-15\,^\circ\text{C}$ to degrees Fahrenheit.
• (b) For the linear relation $y = -\dfrac{2}{3}x + \dfrac{1}{2}$, find $y$ when $x = -\dfrac{3}{2}$.

Solution.

(a) Substitute $C = -15$:

$$F = \frac{9}{5}(-15) + 32 = -27 + 32 = 5$$

$-15\,^\circ\text{C}$ is $5\,^\circ\text{F}$.

(b) Substitute $x = -\dfrac{3}{2}$. Multiply first:

$$y = -\frac{2}{3}\left(-\frac{3}{2}\right) + \frac{1}{2} = \frac{6}{6} + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$$

Two negatives multiply to a positive, so $y = \dfrac{3}{2}$.

Common mistakes

Adding the denominators. $\dfrac{1}{2} + \dfrac{1}{3}$ is not $\dfrac{2}{5}$. Half a pizza plus a third of a pizza is more than half, but $\dfrac{2}{5}$ is less than half! Use a common denominator: $\dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}$.

Multiplying mixed numbers part by part. $2\dfrac{1}{2} \times 2\dfrac{1}{2}$ is not $4\dfrac{1}{4}$. Change to improper fractions first: $\dfrac{5}{2} \times \dfrac{5}{2} = \dfrac{25}{4} = 6\dfrac{1}{4}$.

Flipping the wrong fraction when dividing. Only the second fraction (the divisor) gets flipped. $\dfrac{2}{3} \div \dfrac{4}{5} = \dfrac{2}{3} \times \dfrac{5}{4} = \dfrac{5}{6}$.

Subtracting mixed numbers without enough to take away. In $3\dfrac{1}{6} - 1\dfrac{3}{4}$, you can’t take $\dfrac{3}{4}$ from $\dfrac{1}{6}$ directly. Changing both to improper fractions (as in Example 2) avoids the problem.

Losing track of negative signs. $-\dfrac{1}{2} - \dfrac{1}{3}$ is $-\dfrac{3}{6} - \dfrac{2}{6} = -\dfrac{5}{6}$, not $-\dfrac{1}{6}$. Moving left twice on a number line takes you further left.

Counting marks instead of spaces on a ruler. Start counting from the whole-inch mark, and count the spaces you move, not the lines you pass. Also, each space is $\dfrac{1}{16}$ inch, not $\dfrac{1}{10}$.

Practice

1. (Warm-up) A ruler has $16$ equal spaces in each inch.

• (a) What fraction of an inch is one space?
• (b) How many spaces make $\dfrac{3}{4}$ inch?
• (c) Write $\dfrac{5}{8}$ inch in sixteenths.

Solution

(a) $\dfrac{1}{16}$ inch.

(b) $\dfrac{3}{4} = \dfrac{12}{16}$, so $12$ spaces.

(c) $\dfrac{5}{8} = \dfrac{5 \times 2}{8 \times 2} = \dfrac{10}{16}$ inch.

2. (Warm-up)

• (a) Write $\dfrac{18}{24}$ in lowest terms.
• (b) Write $\dfrac{7}{3}$ as a mixed number.
• (c) Write $4\dfrac{2}{5}$ as an improper fraction.

Solution

(a) Divide both by $6$: $\dfrac{18}{24} = \dfrac{3}{4}$.

(b) $7 \div 3 = 2$ remainder $1$, so $\dfrac{7}{3} = 2\dfrac{1}{3}$.

(c) $4\dfrac{2}{5} = \dfrac{4 \times 5 + 2}{5} = \dfrac{22}{5}$.

3. (Warm-up) Put $\dfrac{3}{5}$, $\dfrac{2}{3}$ and $\dfrac{5}{8}$ in order from smallest to largest.

Solution

Use the common denominator $120$:

$$\frac{3}{5} = \frac{72}{120}, \qquad \frac{2}{3} = \frac{80}{120}, \qquad \frac{5}{8} = \frac{75}{120}$$

From smallest to largest: $\dfrac{3}{5}$, $\dfrac{5}{8}$, $\dfrac{2}{3}$. (As decimals: $0.6$, $0.625$, $0.\overline{6}$.)

4. (Core) Evaluate. Write answers in lowest terms.

• (a) $\dfrac{7}{8} - \dfrac{1}{6}$
• (b) $1\dfrac{2}{3} + 2\dfrac{5}{6}$
• (c) $-\dfrac{3}{10} - \dfrac{2}{5}$

Solution

(a) The common denominator is $24$: $\dfrac{21}{24} - \dfrac{4}{24} = \dfrac{17}{24}$.

(b) Add the whole numbers and the fractions separately, using sixths:

$$1\frac{4}{6} + 2\frac{5}{6} = 3\frac{9}{6} = 3 + 1\frac{3}{6} = 4\frac{1}{2}$$

(c) The common denominator is $10$:

$$-\frac{3}{10} - \frac{4}{10} = \frac{-3 - 4}{10} = -\frac{7}{10}$$

5. (Core) Evaluate. Write answers in lowest terms.

• (a) $-\dfrac{4}{9} \times \left(-\dfrac{3}{8}\right)$
• (b) $3\dfrac{1}{3} \times 1\dfrac{1}{5}$
• (c) $-\dfrac{7}{12} \div \dfrac{7}{4}$

Solution

(a) Same signs, so positive: $\dfrac{4 \times 3}{9 \times 8} = \dfrac{12}{72} = \dfrac{1}{6}$.

(b) Change to improper fractions: $\dfrac{10}{3} \times \dfrac{6}{5} = \dfrac{60}{15} = 4$.

(c) Multiply by the reciprocal: $-\dfrac{7}{12} \times \dfrac{4}{7} = -\dfrac{28}{84} = -\dfrac{1}{3}$.

6. (Core) A recipe needs $2\dfrac{1}{4}$ cups of milk, but the only measuring cup you can find holds $\dfrac{1}{3}$ cup. How many times must you fill it?

Solution

$$2\frac{1}{4} \div \frac{1}{3} = \frac{9}{4} \times \frac{3}{1} = \frac{27}{4} = 6\frac{3}{4}$$

You need $6$ full scoops plus another scoop filled $\dfrac{3}{4}$ of the way, so $7$ fills in all (the last one not quite full).

Check: $6\dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{27}{4} \times \dfrac{1}{3} = \dfrac{9}{4} = 2\dfrac{1}{4}$ cups. ✓

7. (Core) A board $2\dfrac{1}{2}$ m long is cut into pieces that are each $\dfrac{3}{8}$ m long. How many full pieces can be cut, and how much of the board is left over? (Ignore the width of the saw cuts.)

Solution

$$2\frac{1}{2} \div \frac{3}{8} = \frac{5}{2} \times \frac{8}{3} = \frac{40}{6} = 6\frac{2}{3}$$

So $6$ full pieces fit. They use $6 \times \dfrac{3}{8} = \dfrac{18}{8}$ m. The leftover is

$$\frac{5}{2} - \frac{18}{8} = \frac{20}{8} - \frac{18}{8} = \frac{2}{8} = \frac{1}{4} \text{ m}$$

8. (Challenge) For the linear relation $y = \dfrac{3}{4}x - \dfrac{5}{2}$:

• (a) Find $y$ when $x = -\dfrac{2}{3}$.
• (b) Find $x$ when $y = 1$.

Solution

(a)

$$y = \frac{3}{4}\left(-\frac{2}{3}\right) - \frac{5}{2} = -\frac{6}{12} - \frac{5}{2} = -\frac{1}{2} - \frac{5}{2} = -\frac{6}{2} = -3$$

(b) Substitute $y = 1$ and undo each step:

$$\begin{aligned} 1 &= \frac{3}{4}x - \frac{5}{2} \\ 1 + \frac{5}{2} &= \frac{3}{4}x && \text{add } \tfrac{5}{2} \text{ to both sides} \\ \frac{7}{2} &= \frac{3}{4}x \\ x &= \frac{7}{2} \times \frac{4}{3} = \frac{28}{6} = \frac{14}{3} && \text{multiply by the reciprocal of } \tfrac{3}{4} \end{aligned}$$

So $x = \dfrac{14}{3} = 4\dfrac{2}{3}$. Check: $\dfrac{3}{4} \times \dfrac{14}{3} = \dfrac{14}{4} = \dfrac{7}{2}$, and $\dfrac{7}{2} - \dfrac{5}{2} = 1$. ✓

9. (Challenge) About $3600$ years ago, Egyptian scribes wrote most fractions as sums of different unit fractions. (The Rhind Papyrus, an Egyptian math text, has tables of them.) For example, $\dfrac{3}{4} = \dfrac{1}{2} + \dfrac{1}{4}$. Write each fraction as a sum of two different unit fractions, and check your answer.

• (a) $\dfrac{5}{6}$
• (b) $\dfrac{7}{12}$
• (c) $\dfrac{9}{20}$

Solution

A good strategy: start with the largest unit fraction that fits, and see what’s left.

(a) $\dfrac{1}{2}$ fits, leaving $\dfrac{5}{6} - \dfrac{3}{6} = \dfrac{2}{6} = \dfrac{1}{3}$. So $\dfrac{5}{6} = \dfrac{1}{2} + \dfrac{1}{3}$. Check: $\dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}$. ✓

(b) $\dfrac{1}{2} = \dfrac{6}{12}$ fits, leaving $\dfrac{1}{12}$. So $\dfrac{7}{12} = \dfrac{1}{2} + \dfrac{1}{12}$. Check: $\dfrac{6}{12} + \dfrac{1}{12} = \dfrac{7}{12}$. ✓ (Another answer: $\dfrac{1}{3} + \dfrac{1}{4} = \dfrac{4}{12} + \dfrac{3}{12} = \dfrac{7}{12}$.)

(c) $\dfrac{1}{2} = \dfrac{10}{20}$ is too big. $\dfrac{1}{3}$ fits, but it leaves $\dfrac{27}{60} - \dfrac{20}{60} = \dfrac{7}{60}$, which isn’t a unit fraction. Try $\dfrac{1}{4} = \dfrac{5}{20}$: it leaves $\dfrac{4}{20} = \dfrac{1}{5}$. So $\dfrac{9}{20} = \dfrac{1}{4} + \dfrac{1}{5}$. Check: $\dfrac{5}{20} + \dfrac{4}{20} = \dfrac{9}{20}$. ✓