Primary & Middle School Mathematics · Grades 1–9

39Fractions: First Steps

Sharing three pizzas fairly among four people gives each person … not a whole number of pizzas. Fractions are the numbers invented for sharing. This chapter builds the picture: a fraction as a part of a whole, as a point on the number line, and as the exact result of a division.

39.1 What a fraction means

Definition 39.1 (Fraction)

Cut a unit into $b$ equal parts, and take $a$ of them: the quantity obtained is the fraction

\frac{a}{b} \qquad \begin{array}{l} a \text{: the \emph{numerator} --- how many parts we take;}\\ b \text{: the \emph{denominator} --- into how many parts the unit is cut.} \end{array}

Two fractions of the same bar. The denominator tells how fine the cutting is, the numerator how much we take.

Example 39.2

A fraction can be bigger than $1$ : $\frac74$ means seven quarters — one whole unit (four quarters) and three quarters more:

\frac74 = 1 + \frac34 .

Proposition 39.3 (Fractions on the number line)

To place $\frac ab$ on the number line, cut each unit interval into $b$ equal parts and count $a$ parts from $0$ .

The line graduated in quarters: 3/4 is before 1, 7/4 is between 1 and 2, and 22/4 = 5 + 2/4 is halfway between 5 and 6. — The line graduated in quarters: $\frac34$ is before $1$ , $\frac74$ is between $1$ and $2$ , and $\frac{22}{4} = 5 + \frac24$ is halfway between $5$ and $6$ .

Theorem 39.4 (A fraction is a division)

The fraction $\frac ab$ is the number which, multiplied by $b$ , gives $a$ :

\frac ab \times b = a .

In other words, $\frac ab$ is the exact quotient of $a$ by $b$ — even when the division “does not come out even”.

Proof. Admitted at this level. ∎

Example 39.5

Three pizzas for four people: each person gets $\frac34$ of a pizza. Check with the theorem: $4$ shares of $\frac34$ make $\frac34 \times 4 = 3$ pizzas. Some fractions are decimal numbers ( $\frac34 = 0.75$ ), others are not: $\frac13 = 0.333\dots$ never stops. The fraction is the exact value; $0.33$ is only an approximation.

39.2 Equal fractions

Proposition 39.6 (Equivalent fractions)

A fraction does not change when its numerator and denominator are both multiplied (or divided) by the same nonzero number:

\frac{a}{b} = \frac{a \times k}{b \times k} .

Idea of proof. Cutting each part of the sharing into $k$ smaller pieces multiplies both the number of pieces taken and the total number of pieces by $k$ , but the quantity taken is the same. A picture says it best: ∎

2/3 = 4/6: cutting each third in two doubles the counts, top and bottom, without changing the shaded amount. — $\frac23 = \frac46$ : cutting each third in two doubles the counts, top and bottom, without changing the shaded amount.

Method 39.7 (Simplifying a fraction)

Look for a number dividing the numerator and the denominator, and divide both. Repeat until no common divisor remains:

\frac{12}{18} = \frac{12 \div 2}{18 \div 2} = \frac{6}{9} = \frac{6 \div 3}{9 \div 3} = \frac{2}{3}.

39.3 Taking a fraction of a quantity

Method 39.8 (Fraction of a quantity)

To compute $\frac ab$ of a quantity, divide the quantity by $b$ (one part), then multiply by $a$ (the number of parts):

\frac ab \text{ of } Q = (Q \div b) \times a .

(Multiplying first by $a$ and then dividing by $b$ gives the same result — choose the order that makes the numbers nicer.)

Example 39.9

Compute $\frac35$ of $40$ euros, step by step:

one fifth of $40$ : $40 \div 5 = 8$ ;
three fifths: $8 \times 3 = 24$ .

So $\frac35$ of $40$ euros is $24$ euros. Check: the remaining $\frac25$ is $16$ euros, and $24 + 16 = 40$ .

39.4 Exercises

Exercise 39.1 ★

For each picture description, write the fraction: a cake cut in $8$ slices, $3$ eaten; a chocolate bar of $12$ squares, $7$ eaten; a pie cut in $6$ , all $6$ eaten.

Solution

Solution of Exercise 39.1.

Cake: $\frac38$ eaten. Chocolate: $\frac{7}{12}$ . Pie: $\frac66 = 1$ (the whole pie).

Exercise 39.2 ★

Draw a bar cut into $5$ equal parts and color $\frac45$ of it. Is $\frac45$ smaller or larger than $1$ ? And $\frac65$ ?

Solution

Solution of Exercise 39.2.

Four parts out of five are colored. $\frac45 < 1$ (four parts out of five is less than the whole bar); $\frac65 > 1$ (six fifths is one whole bar and one fifth more).

Exercise 39.3 ★

Place on a number line graduated in thirds: $\frac13$ ; $\frac53$ ; $2$ ; $\frac73$ . Which two of these numbers are equal distances from $2$ ?

Solution

Solution of Exercise 39.3.

On the line graduated in thirds: $\frac13$ is one tick after $0$ ; $\frac53$ is one tick before $2$ (since $2 = \frac63$ ); $\frac73$ is one tick after $2$ . So $\frac53$ and $\frac73$ are both at distance $\frac13$ from $2$ .

Exercise 39.4 ★

Write each fraction as a whole number plus a fraction smaller than $1$ (as in $\frac74 = 1 + \frac34$ ):

\frac{9}{4}, \qquad \frac{13}{5}, \qquad \frac{12}{6}, \qquad \frac{25}{8} .

Solution

Solution of Exercise 39.4.

$\frac94 = 2 + \frac14$ ; $\frac{13}{5} = 2 + \frac35$ ; $\frac{12}{6} = 2$ (exactly); $\frac{25}{8} = 3 + \frac18$ .

Exercise 39.5 ★

Complete so the fractions are equal:

\frac{2}{3} = \frac{?}{12}, \qquad \frac{15}{20} = \frac{3}{?}, \qquad \frac{7}{10} = \frac{21}{?}, \qquad \frac{?}{6} = \frac{10}{12}.

Solution

Solution of Exercise 39.5.

$\frac23 = \frac{8}{12}$ (multiply by $4$ ); $\frac{15}{20} = \frac34$ (divide by $5$ ); $\frac{7}{10} = \frac{21}{30}$ (multiply by $3$ ); $\frac56 = \frac{10}{12}$ (divide by $2$ ).

Exercise 39.6 ★

Simplify as much as possible: $\dfrac{6}{8}$ ; $\dfrac{15}{25}$ ; $\dfrac{18}{24}$ ; $\dfrac{35}{7}$ .

Solution

Solution of Exercise 39.6.

$\frac68 = \frac34$ ; $\frac{15}{25} = \frac35$ ; $\frac{18}{24} = \frac{9}{12} = \frac34$ ; $\frac{35}{7} = 5$ (a whole number).

Exercise 39.7 ★

Compute: $\frac12$ of $86$ ; $\frac34$ of $60$ ; $\frac25$ of $35$ ; $\frac{7}{10}$ of $250$ .

Solution

Solution of Exercise 39.7.

$\frac12$ of $86$ : $86 \div 2 = 43$ .

$\frac34$ of $60$ : $60 \div 4 = 15$ , then $15 \times 3 = 45$ .

$\frac25$ of $35$ : $35 \div 5 = 7$ , then $7 \times 2 = 14$ .

$\frac{7}{10}$ of $250$ : $250 \div 10 = 25$ , then $25 \times 7 = 175$ .

Exercise 39.8 ★

Which is larger: $\frac12$ or $\frac13$ (of the same cake)? $\frac25$ or $\frac35$ ? $\frac34$ or $\frac78$ (write both with denominator $8$ )?

Solution

Solution of Exercise 39.8.

$\frac12 > \frac13$ : halves are bigger parts than thirds. $\frac35 > \frac25$ : same parts, more of them. $\frac34 = \frac68 < \frac78$ : seven eighths beat six eighths.

Exercise 39.9 ★★

Give the decimal value of the fractions that have one, and say which fraction does not: $\frac12$ ; $\frac34$ ; $\frac13$ ; $\frac{7}{10}$ ; $\frac{9}{4}$ .

Solution

Solution of Exercise 39.9.

$\frac12 = 0.5$ ; $\frac34 = 0.75$ ; $\frac{7}{10} = 0.7$ ; $\frac94 = 2.25$ . The odd one out is $\frac13 = 0.333\dots$ , whose decimal writing never ends: it is not a decimal number.

Exercise 39.10 ★★

A class has $28$ students. Three quarters of them come to school by bus, and the rest walk. How many students walk? Solve step by step, and check that your two group sizes add up to $28$ .

Solution

Solution of Exercise 39.10.

Bus users: $\frac34$ of $28$ : $28 \div 4 = 7$ , then $7 \times 3 = 21$ students. Walkers: $28 - 21 = 7$ students (that is the remaining quarter: $28 \div 4 = 7$ ). Check: $21 + 7 = 28$ .

Exercise 39.11 ★★

Leo spent $\frac23$ of his savings on a game that cost $18$ euros. How much money did Leo have before? (One third of his savings is half of $18$ … think about why, or find the size of one part first.)

Solution

Solution of Exercise 39.11.

Two thirds of the savings are $18$ euros, so one third is $18 \div 2 = 9$ euros, and the whole savings (three thirds) were $9 \times 3 = 27$ euros. Check: $\frac23$ of $27$ is $18$ .

Exercise 39.12 ★★★

A tank is filled to $\frac58$ of its capacity. After using $10$ liters, it is half full. What is the capacity of the tank? (Which fraction of the tank do the $10$ liters represent?)

Solution

Solution of Exercise 39.12.

The used water went from $\frac58$ to $\frac48$ (a half is four eighths) of the tank: the $10$ liters represent $\frac58 - \frac48 = \frac18$ of the capacity. So the full tank holds $10 \times 8 = 80$ liters. Check: $\frac58$ of $80$ is $50$ , minus $10$ leaves $40$ , which is indeed half of $80$ .

39.5 Problem: The chocolate bar you can never finish

Problem 39.1

Weekend problem — halves of halves: the sum $\frac12 + \frac14 + \frac18 + \dots$ creeps up to $1$ without ever touching it

Take a chocolate bar. Eat half of it. Then eat half of what is left. Then again half of what is left, and so on. Two things seem to be true at once: you never finish the bar (something always remains), and yet you eat almost all of it. This problem turns both feelings into exact statements about fractions — and meets, on the way, a runner from ancient Greece and a sharing puzzle solved entirely with a knife that only cuts things in half.

Part I — Halves of halves. The bar is a rectangle of $16$ equal squares.

First bite: you eat half the bar. How many squares is that? Write the bite as a fraction of the bar in two ways, with denominator $2$ and with denominator $16$ (Proposition 39.6).
Second bite: half of what is left. How many squares? Show, with the picture or with equivalent fractions, that this bite is $\frac14$ of the whole bar: half of a half is a quarter.
Third and fourth bites, same rule: give each one in squares and as a fraction of the bar. What happens to the denominator from one bite to the next?
After the four bites, how many squares have been eaten in all? What fraction of the bar is that, and what fraction is left?
Question 4 says, written in sixteenths:

$\frac{8}{16} + \frac{4}{16} + \frac{2}{16} + \frac{1}{16} = \frac{15}{16}, \qquad\text{that is}\qquad \frac12 + \frac14 + \frac18 + \frac{1}{16} = \frac{15}{16} .$

Check each of the four rewritings ( $\frac12 = \frac{8}{16}$ , and so on).

Part II — The staircase towards $1$ . To keep biting, imagine a finer bar of $64$ squares.

List the sizes of the first six bites, in squares ( $32$ , then …), and check that after the sixth bite exactly one square is left. What fraction of the bar remains?
After each bite, what fraction of the bar remains? Write the list ( $\frac12$ , $\frac14$ , …) down to the sixth bite, and describe the pattern. If an imaginary knife could halve even the last square, what would remain after a seventh bite?
Explain why the eaten total can never reach $1$ , no matter how many bites are taken. (Compare with the twenty nines of Problem 38.1: same story?)
Yet the total passes any target below $1$ . A friend challenges you to eat more than $\frac{99}{100}$ of the bar. Which is smaller, $\frac{1}{128}$ or $\frac{1}{100}$ — and after which bite is the challenge won?
On a number line from $0$ to $1$ graduated in sixteenths, place the eaten totals after each of the first four bites: $\frac12$ , $\frac34$ , $\frac78$ , $\frac{15}{16}$ . Where does each new total land, compared with the previous one and with $1$ ?

Part III — Zeno’s runner, and a knife that only halves.

Twenty-four centuries ago, the Greek philosopher Zeno told this story: to reach a wall, a runner must first cover half the distance, then half of what remains, then half of what remains again … “so the runner never reaches the wall!” Using questions 8 and 9, say what is right in Zeno’s story, and where the trap is. (Does the runner really need forever to make all those steps?)
The sharing puzzle: share $15$ identical chocolate bars fairly among $16$ children, using only cuts into halves (of a bar, of a half-bar, of a quarter-bar …). Describe how to cut: how many bars are cut into halves, how many into quarters, into eighths, into sixteenths, so that every child receives one piece of each size.
Check your sharing: write each child’s share as a sum of fractions, compute it in sixteenths, and verify that the $16$ shares together use up exactly the $15$ bars.
In the class next door, $7$ bars are shared fairly among $8$ children. Which child receives more chocolate — one of the $16$ , or one of the $8$ ? (Write both shares with denominator $16$ , Exercise 39.8.)
The finale: a friend claims that with infinitely many bites one would eat exactly the whole bar. Draw a square and shade the bites inside it: half the square, then a quarter, then an eighth … What do you observe about the unshaded corner? State carefully the two true facts of this whole problem: what remains after every finite number of bites, and how small that remainder becomes. (What an “infinite sum” exactly means is a story for the High School volume.)

Solution

Solution of Problem 39.1.

1. Half of $16$ squares is $8$ squares. As a fraction: $\frac12$ of the bar, and since each half is $8$ squares out of $16$ , also $\frac{8}{16}$ — the same fraction written two ways (Proposition 39.6).

2. Half of the $8$ remaining squares is $4$ squares, which is $\frac{4}{16}$ of the bar. Simplifying by $4$ : $\frac{4}{16} = \frac14$ . On the picture: cutting the remaining half in two produces two quarters of the original bar — half of a half is a quarter.

3. Third bite: half of $4$ squares is $2$ squares, i.e. $\frac{2}{16} = \frac18$ of the bar. Fourth bite: $1$ square, i.e. $\frac{1}{16}$ . From one bite to the next, the denominator doubles: $2$ , $4$ , $8$ , $16$ .

4. Eaten: $8 + 4 + 2 + 1 = 15$ squares out of $16$ , that is $\frac{15}{16}$ of the bar. Left: $1$ square, $\frac{1}{16}$ .

5. $\frac12 = \frac{8}{16}$ (multiply top and bottom by $8$ ), $\frac14 = \frac{4}{16}$ (by $4$ ), $\frac18 = \frac{2}{16}$ (by $2$ ), and $\frac{1}{16}$ stays. Counting sixteenths: $8 + 4 + 2 + 1 = 15$ of them, so the total is $\frac{15}{16}$ .

6. Bites: $32$ , $16$ , $8$ , $4$ , $2$ , $1$ squares. Total eaten: $32 + 16 + 8 + 4 + 2 + 1 = 63$ squares, so $1$ square remains: $\frac{1}{64}$ of the bar.

7. After each bite there remains

\frac12,\quad \frac14,\quad \frac18,\quad \frac{1}{16},\quad \frac{1}{32},\quad \frac{1}{64} :

the remainder halves each time — its denominator doubles. A seventh, imaginary bite would leave $\frac{1}{128}$ .

8. Every bite eats only half of what remains, so the other half of the remainder is still there: after any number of bites, something is always left. The eaten total is therefore always below $1$ — exactly like the twenty nines of Problem 38.1, which come ever closer to $3$ without reaching it.

9. Both fractions have numerator $1$ , and cutting into $128$ parts makes smaller parts than cutting into $100$ : so $\frac{1}{128} < \frac{1}{100}$ . After the seventh bite the remainder is $\frac{1}{128}$ , so the eaten part exceeds $1 - \frac{1}{100} = \frac{99}{100}$ : the challenge is won at bite seven.

10. In sixteenths, the totals are $\frac{8}{16}$ , $\frac{12}{16}$ , $\frac{14}{16}$ , $\frac{15}{16}$ . Each new total lands exactly halfway between the previous total and $1$ : the staircase keeps halving its remaining distance to $1$ , without ever stepping on it.

11. What is right: at every stage of Zeno’s description, some distance indeed remains (question 8) — the list of stages never ends. The trap: the stages become extremely short, in distance and in running time; the runner does not spend equal time on each stage. Describing the run in infinitely many shrinking pieces does not make the run itself endless: the runner reaches the wall, and question 9 shows the description passing any point short of it.

12. Cut $8$ bars into halves: $16$ half-pieces. Cut $4$ bars into quarters: $16$ quarter-pieces. Cut $2$ bars into eighths: $16$ eighth-pieces. Cut the last bar into sixteenths: $16$ small pieces. That uses $8 + 4 + 2 + 1 = 15$ bars, and each child receives one piece of each size.

13. Each child’s share is

\frac12 + \frac14 + \frac18 + \frac{1}{16} = \frac{15}{16}

of a bar (question 5). Sixteen such shares make $16 \times \frac{15}{16} = 15$ bars (Theorem 39.4): exactly what was cut, with nothing left over — a fair sharing of $15$ bars among $16$ children, by halving alone.

14. The neighbours receive $\frac78$ of a bar each. With denominator $16$ : $\frac78 = \frac{14}{16}$ , while our children receive $\frac{15}{16}$ . So a child of the $16$ receives more: $\frac{15}{16} > \frac{14}{16}$ , by one sixteenth of a bar.

15. In the square picture, each shaded bite fills half of the still-unshaded corner, and the unshaded corner keeps shrinking: after each bite it is half as large as before, and it never disappears. The two true facts: (i) after every finite number of bites, a remainder is left — the total eaten is always strictly below $1$ ; (ii) that remainder becomes smaller than any fraction one cares to name ( $\frac{1}{100}$ , $\frac{1}{1000}$ , …), if one bites long enough. Giving the words “the infinite sum equals $1$ ” their exact meaning is the job of limits, in the High School volume.