Primary & Middle School Mathematics · Grades 1–9

55Fractions: All Four Operations

This chapter completes the arithmetic of fractions: addition and subtraction with any denominators, multiplication, and — the newcomer — division, which turns out to be a multiplication in disguise. Signs from Chapter 54 come along for the ride.

55.1 Adding with any denominators

Method 55.1 (Common denominator)

To add or subtract $\frac ab$ and $\frac cd$ :

find a common multiple of $b$ and $d$ (the product $b \times d$ always works; a smaller one saves effort);
rewrite both fractions with that denominator;
add or subtract the numerators; simplify.

Example 55.2

Compute $\dfrac{5}{6} + \dfrac{3}{4}$ . A common multiple of $6$ and $4$ is $12$ :

\frac{5}{6} = \frac{10}{12}, \qquad \frac{3}{4} = \frac{9}{12}, \qquad \frac{5}{6} + \frac{3}{4} = \frac{10 + 9}{12} = \frac{19}{12}.

With signs: $\dfrac{2}{3} - \dfrac{7}{5} = \dfrac{10}{15} - \dfrac{21}{15} = -\dfrac{11}{15}$ .

55.2 Multiplying fractions

Theorem 55.3 (Product of fractions)

\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} .

Why, on a picture. Take $\frac34$ of $\frac25$ of a square: cut the square in $5$ vertical strips and keep $2$ ; cut horizontally in $4$ and keep $3$ rows of what remained. The kept part is a grid of $3 \times 2$ small cells out of $4 \times 5$ : fraction $\frac{6}{20}$ . ∎

3/4 × 2/5: the doubly-shaded region is 3 × 2 = 6 cells out of 4 × 5 = 20, i.e. 6/20 = 3/10 — numerators multiplied, denominators multiplied. — $\frac34 \times \frac25$ : the doubly-shaded region is $3 \times 2 = 6$ cells out of $4 \times 5 = 20$ , i.e. $\frac{6}{20} = \frac{3}{10}$ — numerators multiplied, denominators multiplied.

Example 55.4

Simplify before multiplying, crossing out common factors:

\frac{7}{12} \times \frac{9}{14} = \frac{7 \times 9}{12 \times 14} = \frac{7 \times 3 \times 3}{3 \times 4 \times 2 \times 7} = \frac{3}{8}

(the factor $7$ and one factor $3$ cancel between top and bottom). With signs, the rules of Theorem 54.1 apply first: $\left(-\frac23\right) \times \left(-\frac{5}{4}\right) = +\frac{10}{12} = \frac56$ .

55.3 Dividing fractions

Definition 55.5 (Inverse)

The inverse of a nonzero number $x$ is the number which multiplied by $x$ gives $1$ . The inverse of $\frac ab$ is $\frac ba$ , since $\frac ab \times \frac ba = \frac{ab}{ab} = 1$ ; the inverse of an integer $n$ is $\frac1n$ .

Theorem 55.6 (Dividing means multiplying by the inverse)

For $c \neq 0$ :

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

Proof. By definition, the quotient $q$ of $\frac ab$ by $\frac cd$ is the number with $q \times \frac cd = \frac ab$ . Check the candidate $q = \frac ab \times \frac dc$ :

\left(\frac ab \times \frac dc\right) \times \frac cd = \frac ab \times \left(\frac dc \times \frac cd\right) = \frac ab \times 1 = \frac ab . \qedhere

∎

Example 55.7

\frac{3}{5} \div \frac{7}{10} = \frac{3}{5} \times \frac{10}{7} = \frac{3 \times 10}{5 \times 7} = \frac{30}{35} = \frac{6}{7}.

A sanity check: dividing by $\frac{7}{10}$ , a number smaller than $1$ , must give a result larger than $\frac35$ — and $\frac67 > \frac35$ indeed ( $\frac{30}{35} > \frac{21}{35}$ ).

Example 55.8 (Fraction bars within fraction bars)

A “double-decker” is just a division written vertically:

\frac{\ \frac{2}{3}\ }{\ \frac{5}{6}\ } = \frac{2}{3} \div \frac{5}{6} = \frac{2}{3} \times \frac{6}{5} = \frac{12}{15} = \frac{4}{5}.

Locate the main bar first (the longest one), then apply Theorem 55.6.

Method 55.9 (Mixed computations)

In an expression mixing fractions and the four operations:

apply the usual priorities (Chapter 45);
convert every division into a multiplication by the inverse;
decide signs first (Method 54.6), then multiply or find common denominators;
simplify at the earliest opportunity.

Example 55.10

\begin{align*} \frac12 + \frac{2}{3} \times \frac{9}{4} &= \frac12 + \frac{2 \times 9}{3 \times 4} && \text{(product first!)} \\ &= \frac12 + \frac{18}{12} = \frac12 + \frac32 = \frac{4}{2} = 2 . \end{align*}

55.4 Exercises

Exercise 55.1 ★

Compute and simplify:

\frac{3}{4} + \frac{2}{5}, \qquad \frac{5}{6} - \frac{3}{8}, \qquad \frac{7}{10} + \frac{8}{15} .

Solution

Solution of Exercise 55.1.

$\dfrac34 + \dfrac25 = \dfrac{15}{20} + \dfrac{8}{20} = \dfrac{23}{20}$ .

$\dfrac56 - \dfrac38 = \dfrac{20}{24} - \dfrac{9}{24} = \dfrac{11}{24}$ .

$\dfrac{7}{10} + \dfrac{8}{15} = \dfrac{21}{30} + \dfrac{16}{30} = \dfrac{37}{30}$ .

Exercise 55.2 ★

Compute (signs first):

-\frac{2}{7} + \frac{3}{14}, \qquad \frac{1}{6} - \frac{5}{4}, \qquad -\frac{3}{5} - \frac{1}{2} .

Solution

Solution of Exercise 55.2.

$-\dfrac27 + \dfrac{3}{14} = -\dfrac{4}{14} + \dfrac{3}{14} = -\dfrac{1}{14}$ .

$\dfrac16 - \dfrac54 = \dfrac{2}{12} - \dfrac{15}{12} = -\dfrac{13}{12}$ .

$-\dfrac35 - \dfrac12 = -\dfrac{6}{10} - \dfrac{5}{10} = -\dfrac{11}{10}$ .

Exercise 55.3 ★

Compute, simplifying before multiplying:

\frac{5}{8} \times \frac{4}{15}, \qquad \frac{9}{14} \times \frac{7}{3}, \qquad \left(-\frac{6}{5}\right) \times \frac{10}{9} .

Solution

Solution of Exercise 55.3.

$\dfrac58 \times \dfrac{4}{15} = \dfrac{5 \times 4}{8 \times 15} = \dfrac{1}{2 \times 3} = \dfrac16$ (simplify by $5$ and by $4$ ).

$\dfrac{9}{14} \times \dfrac73 = \dfrac{9 \times 7}{14 \times 3} = \dfrac{3}{2}$ (simplify by $7$ and by $3$ ).

$\left(-\dfrac65\right) \times \dfrac{10}{9} = -\dfrac{60}{45} = -\dfrac43$ (opposite signs: negative).

Exercise 55.4 ★

Give the inverse of: $\dfrac{3}{7}$ ; $5$ ; $\dfrac{1}{4}$ ; $-\dfrac{2}{9}$ .

Solution

Solution of Exercise 55.4.

Inverses: $\dfrac73$ ; $\dfrac15$ ; $4$ ; $-\dfrac92$ (the inverse keeps the sign).

Exercise 55.5 ★

Compute:

\frac{4}{9} \div \frac{2}{3}, \qquad \frac{7}{5} \div 14, \qquad 6 \div \frac{3}{4} .

Solution

Solution of Exercise 55.5.

$\dfrac49 \div \dfrac23 = \dfrac49 \times \dfrac32 = \dfrac{12}{18} = \dfrac23$ .

$\dfrac75 \div 14 = \dfrac75 \times \dfrac{1}{14} = \dfrac{7}{70} = \dfrac{1}{10}$ .

$6 \div \dfrac34 = 6 \times \dfrac43 = 8$ .

Exercise 55.6 ★

Compute the double-deckers:

\frac{\ \frac{3}{4}\ }{\ \frac{9}{8}\ }, \qquad \frac{\ \frac{5}{6}\ }{\ 10\ }, \qquad \frac{\ 2\ }{\ \frac{4}{7}\ } .

Solution

Solution of Exercise 55.6.

$\dfrac{3/4}{9/8} = \dfrac34 \times \dfrac89 = \dfrac{24}{36} = \dfrac23$ .

$\dfrac{5/6}{10} = \dfrac56 \times \dfrac{1}{10} = \dfrac{5}{60} = \dfrac{1}{12}$ .

$\dfrac{2}{4/7} = 2 \times \dfrac74 = \dfrac72$ .

Exercise 55.7 ★

Compute, respecting priorities:

\frac{1}{3} + \frac{1}{2} \times \frac{4}{5}, \qquad \left(\frac{1}{3} + \frac{1}{2}\right) \times \frac{4}{5} .

Solution

Solution of Exercise 55.7.

Product first: $\dfrac13 + \dfrac12 \times \dfrac45 = \dfrac13 + \dfrac{4}{10} = \dfrac{10}{30} + \dfrac{12}{30} = \dfrac{22}{30} = \dfrac{11}{15}$ .

Brackets first: $\left(\dfrac13 + \dfrac12\right) \times \dfrac45 = \dfrac56 \times \dfrac45 = \dfrac{20}{30} = \dfrac23$ .

Exercise 55.8 ★★

A tank is $\frac{2}{5}$ full. One adds $\frac{1}{3}$ of the tank’s capacity. What fraction of the tank is now full? What fraction is still empty?

Solution

Solution of Exercise 55.8.

Full: $\dfrac25 + \dfrac13 = \dfrac{6}{15} + \dfrac{5}{15} = \dfrac{11}{15}$ of the tank. Empty: $1 - \dfrac{11}{15} = \dfrac{4}{15}$ .

Exercise 55.9 ★★

Three quarters of the students of a class passed a test; among those, two thirds got more than $14$ out of $20$ . What fraction of the class got more than $14$ ? If that represents $12$ students, how large is the class?

Solution

Solution of Exercise 55.9.

Fraction with more than $14$ : $\dfrac23 \times \dfrac34 = \dfrac{6}{12} = \dfrac12$ of the class. If that is $12$ students, the class has $24$ students.

Exercise 55.10 ★★

A rope of $\frac{15}{2}$ m must be cut into pieces of $\frac{3}{4}$ m each. How many pieces does one get? (A division of fractions — check that the answer is a whole number.)

Solution

Solution of Exercise 55.10.

$\dfrac{15}{2} \div \dfrac34 = \dfrac{15}{2} \times \dfrac43 = \dfrac{60}{6} = 10$ pieces — a whole number, no rope wasted.

Exercise 55.11 ★★

Compute

E = \frac{2}{3} - \frac{5}{6} \div \frac{10}{9}, \qquad F = \left(-\frac{1}{2}\right)^2 \times \frac{8}{3} .

Solution

Solution of Exercise 55.11.

$E$ : division first: $\dfrac56 \div \dfrac{10}{9} = \dfrac56 \times \dfrac{9}{10} = \dfrac{45}{60} = \dfrac34$ ; then $E = \dfrac23 - \dfrac34 = \dfrac{8}{12} - \dfrac{9}{12} = -\dfrac{1}{12}$ .

$F = \left(-\dfrac12\right)^2 \times \dfrac83 = \dfrac14 \times \dfrac83 = \dfrac{8}{12} = \dfrac23$ (the square is positive).

Exercise 55.12 ★★★

Simplify the expression

1 + \cfrac{1}{1 + \cfrac{1}{1 + 1}}

(start from the innermost fraction and work outward, one bar at a time).

Solution

Solution of Exercise 55.12.

From the inside out: $1 + 1 = 2$ ; then $1 + \dfrac12 = \dfrac32$ ; then

1 + \frac{1}{3/2} = 1 + \frac23 = \frac53 .

55.5 Problem: Egyptian fractions

Problem 55.1

Weekend problem — every fraction is a sum of distinct unit fractions

A unit fraction is a fraction with numerator $1$ . The scribes of ancient Egypt — the Rhind papyrus was copied around $1550$ BC — wrote every fraction as a sum of distinct unit fractions: never $\frac27 = \frac17 + \frac17$ , but $\frac27 = \frac14 + \frac{1}{28}$ . Such a sum is called an Egyptian writing of the fraction. This problem develops the addition and comparison techniques of this chapter (Method 55.1) into a method — published by Fibonacci in $1202$ — that produces an Egyptian writing of any fraction between $0$ and $1$ , and ends with a celebrated inheritance puzzle.

Part I — Unit fractions and the splitting identity.

Compute $\frac12 + \frac13$ and $\frac12 + \frac14$ , and deduce Egyptian writings of $\frac56$ and $\frac34$ .
Verify the scribes’ entry for $\frac27$ : show that $\frac14 + \frac{1}{28} = \frac27$ .
Check that $\frac13 + \frac16 = \frac12$ and that $\frac14 + \frac{1}{12} = \frac13$ . Then prove the splitting identity: for every whole number $n \geq 1$ ,

$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)} .$
Deduce an Egyptian writing of $1$ itself, as a sum of three distinct unit fractions.
Split the last term of $\frac56 = \frac12 + \frac13$ to obtain a second Egyptian writing of $\frac56$ , with three unit fractions. Explain why no fraction has a unique Egyptian writing.

Part II — Fibonacci’s greedy method. The method is: subtract the largest unit fraction that fits, and repeat with what remains.

Using common denominators, show that $\frac14 < \frac27 < \frac13$ . Why does this prove that $\frac14$ is the largest unit fraction smaller than $\frac27$ ?
Compute $\frac27 - \frac14$ . Which writing of $\frac27$ does the greedy method produce?
Now take $\frac57$ . Show that the largest unit fraction smaller than $\frac57$ is $\frac12$ , and compute $\frac57 - \frac12$ .
Show that the largest unit fraction smaller than $\frac{3}{14}$ is $\frac15$ (compare $\frac{3}{14}$ with $\frac14$ and with $\frac15$ ), compute $\frac{3}{14} - \frac15$ , and conclude:

$\frac57 = \frac12 + \frac15 + \frac{1}{70} .$

Verify this writing directly with the common denominator $70$ .
Run the greedy method on $\frac45$ , checking at each step that your unit fraction is the largest one that fits, and verify the final writing.

Part III — Why the method always stops, and the eighteenth camel.

In questions 7–10 the method was applied to $\frac27$ , $\frac57$ and $\frac45$ . For each, list the numerators of the successive remainders (before any simplification). What do you observe?
For a fraction $\frac ab$ and a whole number $n$ , show that

$\frac{a}{b} - \frac{1}{n} = \frac{a \times n - b}{b \times n} .$
Suppose the greedy method subtracts $\frac1n$ from $\frac ab$ : then $\frac1n$ fits but $\frac{1}{n-1}$ is already too big, $\frac{1}{n-1} > \frac ab$ . Putting both over the common denominator $b \times (n - 1)$ , deduce that $b > a \times n - a$ , hence that the new numerator satisfies

$a \times n - b < a .$

Explain why this proves that the method always stops.
An old tale: a father dies leaving $17$ camels, willing $\frac12$ of the herd to his eldest child, $\frac13$ to the second, $\frac19$ to the third — and none of the shares is a whole number of camels. A neighbour lends the family an eighteenth camel; the children take $9$ , $6$ and $2$ camels, and the borrowed camel is returned. Compute $\frac12 + \frac13 + \frac19$ and explain the trick: why could the herd be shared in whole camels, and why did no child receive less than the will promised?
Repair the will: propose three distinct unit fractions that add up to exactly $1$ , and give the smallest herd size for which all three shares are whole numbers of camels.

Solution

Solution of Problem 55.1.

1. $\frac12 + \frac13 = \frac36 + \frac26 = \frac56$ and $\frac12 + \frac14 = \frac24 + \frac14 = \frac34$ . So $\frac56 = \frac12 + \frac13$ and $\frac34 = \frac12 + \frac14$ are Egyptian writings (distinct unit fractions in each).

2. With the common denominator $28$ :

\frac14 + \frac{1}{28} = \frac{7}{28} + \frac{1}{28} = \frac{8}{28} = \frac{2}{7} .

3. $\frac13 + \frac16 = \frac26 + \frac16 = \frac36 = \frac12$ , and $\frac14 + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac13$ . In general, the common denominator of $\frac{1}{n+1}$ and $\frac{1}{n(n+1)}$ is $n(n+1)$ :

\frac{1}{n+1} + \frac{1}{n(n+1)} = \frac{n}{n(n+1)} + \frac{1}{n(n+1)} = \frac{n+1}{n(n+1)} = \frac{1}{n} .

4. Start from $1 = \frac12 + \frac12$ and split the second half by question 3:

1 = \frac12 + \frac13 + \frac16 ,

three distinct unit fractions.

5. Splitting the $\frac13$ in $\frac56 = \frac12 + \frac13$ gives

\frac56 = \frac12 + \frac14 + \frac{1}{12} .

No Egyptian writing is ever unique: the splitting identity can always replace the unit fraction with the largest denominator by two new, even smaller unit fractions — so from any writing one manufactures another.

6. With denominator $21$ : $\frac27 = \frac{6}{21} < \frac{7}{21} = \frac13$ . With denominator $28$ : $\frac14 = \frac{7}{28} < \frac{8}{28} = \frac27$ . Unit fractions shrink as their denominators grow, $1 > \frac12 > \frac13 > \frac14 > \dots$ ; since $\frac13$ (and everything above it) is too big and $\frac14$ fits, $\frac14$ is the largest unit fraction smaller than $\frac27$ .

7. $\frac27 - \frac14 = \frac{8}{28} - \frac{7}{28} = \frac{1}{28}$ : the greedy method produces $\frac27 = \frac14 + \frac{1}{28}$ , exactly the scribes’ entry of question 2.

8. With denominator $14$ : $\frac12 = \frac{7}{14} < \frac{10}{14} = \frac57$ , while the only larger unit fraction, $1$ , exceeds $\frac57$ . So $\frac12$ is the largest that fits, and

\frac57 - \frac12 = \frac{10}{14} - \frac{7}{14} = \frac{3}{14} .

9. With denominator $28$ : $\frac{3}{14} = \frac{6}{28} < \frac{7}{28} = \frac14$ , so $\frac14$ is too big. With denominator $70$ : $\frac15 = \frac{14}{70} < \frac{15}{70} = \frac{3}{14}$ , so $\frac15$ fits: it is the largest unit fraction smaller than $\frac{3}{14}$ . Then

\frac{3}{14} - \frac15 = \frac{15}{70} - \frac{14}{70} = \frac{1}{70}, \qquad\text{hence}\qquad \frac57 = \frac12 + \frac15 + \frac{1}{70} .

Check: $\frac{35}{70} + \frac{14}{70} + \frac{1}{70} = \frac{50}{70} = \frac57$ .

10. First step: $\frac12 = \frac{5}{10} < \frac{8}{10} = \frac45$ and $1 > \frac45$ , so subtract $\frac12$ : $\frac45 - \frac12 = \frac{8}{10} - \frac{5}{10} = \frac{3}{10}$ . Second step: $\frac13 = \frac{10}{30} > \frac{9}{30} = \frac{3}{10}$ is too big, and $\frac14 = \frac{5}{20} < \frac{6}{20} = \frac{3}{10}$ fits, so subtract $\frac14$ : $\frac{3}{10} - \frac14 = \frac{6}{20} - \frac{5}{20} = \frac{1}{20}$ . Conclusion:

\frac45 = \frac12 + \frac14 + \frac{1}{20}, \qquad \frac{10}{20} + \frac{5}{20} + \frac{1}{20} = \frac{16}{20} = \frac45 . \checkmark

11. The numerators of the successive remainders are: for $\frac27$ : $2$ , then $1$ ; for $\frac57$ : $5$ , then $3$ , then $1$ ; for $\frac45$ : $4$ , then $3$ , then $1$ . Each list is strictly decreasing.

12. The common denominator of $\frac ab$ and $\frac1n$ is $b \times n$ :

\frac{a}{b} - \frac{1}{n} = \frac{a \times n}{b \times n} - \frac{b}{b \times n} = \frac{a \times n - b}{b \times n} .

13. Over the common denominator $b \times (n-1)$ , the inequality $\frac{1}{n-1} > \frac ab$ compares the numerators $b$ and $a \times (n - 1)$ :

b > a \times (n - 1) = a \times n - a ,

and adding $a - b$ to both sides gives $a > a \times n - b$ . So after each greedy step the numerator of the remainder (question 12) is a whole number strictly smaller than the one before. Whole numbers cannot decrease strictly forever: after at most $a$ steps the remainder has numerator $1$ — a unit fraction — or is $0$ , and the method stops.

14. With the common denominator $18$ :

\frac12 + \frac13 + \frac19 = \frac{9}{18} + \frac{6}{18} + \frac{2}{18} = \frac{17}{18} ,

which is less than $1$ : the will forgot $\frac{1}{18}$ of the herd. That gap is the whole trick. Out of $18$ camels the shares $\frac12$ , $\frac13$ , $\frac19$ are the whole numbers $9$ , $6$ and $2$ , which add up to $9 + 6 + 2 = 17$ : exactly the herd, and the eighteenth camel goes home. And nobody can complain: the will promised, out of $17$ camels, $\frac{17}{2} = 8.5$ , then $\frac{17}{3} \approx 5.67$ , then $\frac{17}{9} \approx 1.89$ camels, and each child received strictly more ( $9$ , $6$ and $2$ ). The forgotten $\frac{1}{18}$ is what the neighbour’s camel temporarily filled.

15. Question 4 provides the repaired will:

\frac12 + \frac13 + \frac16 = 1 .

The shares are whole numbers when the herd size is divisible by $2$ , by $3$ and by $6$ — that is, divisible by $6$ . The smallest such herd is $6$ camels: shares $3$ , $2$ and $1$ , and nothing is left over.