---
title: "Fractions: All Four Operations"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 55
exercises: 12
source: https://one-course.com/books/math/1/en/chapter/55-fractions-all-four-operations
---

# Chapter 55 — Fractions: All Four Operations

This chapter completes the arithmetic of [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def): addition and subtraction with *any* [denominators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def), multiplication, and — the newcomer — division, which turns out to be a multiplication in disguise. Signs from [Chapter 54](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#ch-g8-negprod) come along for the ride.

## 55.1 Adding with any denominators

**Method 55.1 (Common denominator).**

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

1. find a common [multiple](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) of $b$ and $d$ (the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) $b \times d$ always works; a smaller one saves effort);
2. rewrite both [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) with that [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) ;
3. add or subtract the [numerators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) ; simplify.

**Example 55.2.**

Compute $\dfrac{5}{6} + \dfrac{3}{4}$. A common [multiple](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-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](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) $\frac{6}{20}$. ∎

![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.](https://one-course.com/images/onecourse/chapters/math-1/g8-fractions/fig-f7244915f4b2.svg)

*$\frac34 \times \frac25$: 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) multiplied, [denominators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#thm-g8-negprod-rules) 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $q$ 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](#thm-g8-fractions-div).

**Method 55.9 (Mixed computations).**

In an expression mixing [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) and the four operations:

1. apply the usual priorities ( [Chapter 45](https://one-course.com/books/math/1/en/chapter/45-priorities-of-operations#ch-g7-priorities) );
2. convert every division into a multiplication by the [inverse](#def-g8-fractions-inverse) ;
3. decide signs first ( [Method 54.6](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#met-g8-negprod-compute) ), then multiply or find common [denominators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) ;
4. 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 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 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 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](#def-g8-fractions-inverse) of: $\dfrac{3}{7}$; $5$; $\dfrac{1}{4}$; $-\dfrac{2}{9}$.

**Solution of Exercise 55.4.**

[Inverses](#def-g8-fractions-inverse): $\dfrac73$; $\dfrac15$; $4$; $-\dfrac92$ (the [inverse](#def-g8-fractions-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 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 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 of Exercise 55.7.**

[Product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) 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](https://one-course.com/books/math/1/en/chapter/12-money-and-measures#def-g2-measure-units)*. What [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) of the tank is now full? What [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) is still empty?

**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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of the students of a class passed a test; among those, two thirds got more than $14$ out of $20$. What [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) of the class got more than $14$? If that represents $12$ students, how large is the class?

**Solution of Exercise 55.9.**

[Fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) — check that the answer is a whole number.)

**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 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](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) and work outward, one bar at a time).

**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](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def)* is a [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) with [numerator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $1$. The scribes of ancient Egypt — the Rhind papyrus was copied around $1550$ BC — wrote every [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) as a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of *distinct* unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def): never $\frac27 = \frac17 + \frac17$, but $\frac27 = \frac14 + \frac{1}{28}$. Such a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is called an *Egyptian writing* of the [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def). This problem develops the addition and comparison techniques of this chapter ([Method 55.1](#met-g8-fractions-add)) into a method — published by Fibonacci in $1202$ — that produces an Egyptian writing of any [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) between $0$ and $1$, and ends with a celebrated inheritance puzzle.

**Part I — Unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) and the splitting identity.**

1. Compute $\frac12 + \frac13$ and $\frac12 + \frac14$ , and deduce Egyptian writings of $\frac56$ and $\frac34$ .
2. Verify the scribes’ entry for $\frac27$ : show that $\frac14 + \frac{1}{28} = \frac27$ .
3. Check that $\frac13 + \frac16 = \frac12$ and that $\frac14 + \frac{1}{12} = \frac13$. Then prove the *splitting identity*: for every whole number $n \geq 1$, $$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)} .$$
4. Deduce an Egyptian writing of $1$ itself, as a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of three distinct unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) .
5. Split the last term of $\frac56 = \frac12 + \frac13$ to obtain a *second* Egyptian writing of $\frac56$ , with three unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) . Explain why no [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) has a unique Egyptian writing.

**Part II — Fibonacci’s greedy method.** The method is: subtract the *largest* unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) that fits, and repeat with what remains.

6. Using common [denominators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) , show that $\frac14 < \frac27 < \frac13$ . Why does this prove that $\frac14$ is the largest unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) smaller than $\frac27$ ?
7. Compute $\frac27 - \frac14$ . Which writing of $\frac27$ does the greedy method produce?
8. Now take $\frac57$ . Show that the largest unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) smaller than $\frac57$ is $\frac12$ , and compute $\frac57 - \frac12$ .
9. Show that the largest unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $70$.
10. Run the greedy method on $\frac45$ , checking at each step that your unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) is the largest one that fits, and verify the final writing.

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

11. In questions 7–10 the method was applied to $\frac27$ , $\frac57$ and $\frac45$ . For each, list the [numerators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) of the successive [remainders](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) (before any simplification). What do you observe?
12. For a [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) $\frac ab$ and a whole number $n$, show that $$\frac{a}{b} - \frac{1}{n}  = \frac{a \times n - b}{b \times n} .$$
13. 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $b \times (n - 1)$, deduce that $b > a \times n - a$, hence that the new [numerator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) satisfies $$a \times n - b < a .$$ Explain why this proves that the method always stops.
14. 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?
15. Repair the will: propose three *distinct* unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) that add up to exactly $1$ , and give the smallest herd size for which all three shares are whole numbers of camels.

**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](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) in each).

**2.** With the common [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $28$:

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

**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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) by question 3:

$$
1 = \frac12 + \frac13 + \frac16 ,
$$

three distinct unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def).

**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](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) with the largest [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) by two new, [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) smaller unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) — so from any writing one manufactures another.

**6.** With [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $21$: $\frac27 = \frac{6}{21} <
\frac{7}{21} = \frac13$. With [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $28$: $\frac14 =
\frac{7}{28} < \frac{8}{28} = \frac27$. Unit [fractions](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) shrink as their [denominators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) grow, $1 > \frac12 > \frac13 > \frac14 >
\dots$; since $\frac13$ (and everything above it) is too big and $\frac14$ fits, $\frac14$ is the largest unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $14$: $\frac12 = \frac{7}{14} <
\frac{10}{14} = \frac57$, while the only larger unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def), $1$, exceeds $\frac57$. So $\frac12$ is the largest that fits, and

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

**9.** With [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $28$: $\frac{3}{14} = \frac{6}{28} <
\frac{7}{28} = \frac14$, so $\frac14$ is too big. With [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $70$: $\frac15 = \frac{14}{70} < \frac{15}{70} =
\frac{3}{14}$, so $\frac15$ fits: it is the largest unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) of the successive [remainders](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $b \times (n-1)$, the inequality $\frac{1}{n-1} > \frac ab$ compares the [numerators](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $b$ 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](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) of the [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-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](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) has [numerator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $1$ — a unit [fraction](https://one-course.com/books/math/1/en/chapter/39-fractions-first-steps#def-g6-fractions-def) — or is $0$, and the method stops.

**14.** With the common [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def) $18$:

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

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](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $2$, by $3$ and by $6$ — that is, [divisible](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $6$. The smallest such herd is $6$ camels: shares $3$, $2$ and $1$, and nothing is left over.