What a Fraction Actually Means

Every division so far came out clean. Now split $3$ bars of clay among $4$ friends. Whole numbers cannot do it, since handing out one bar each leaves the fourth friend with nothing. But a fair share clearly exists somewhere between the whole-number marks, so we need a new kind of number.

Problem

Two campers share one strip of dried mango by making a single cut at the middle, giving each person an equal piece. The amount one camper holds is

1 \div 2

. Write that share as a fraction

a / b

Show a hint

A fraction $a / b$ records one simple idea, the bottom number $b$ is how many equal parts the whole was split into and the top number $a$ is how many of those parts the camper is holding. Look at the strip after the single cut and read off those two counts, how many equal pieces there are and how many one camper walked away with.

Show the full solution

One cut at the middle makes

2

equal pieces, and each camper takes

1

of them. That share is

1 / 2

of the strip. The bottom number says how many equal parts the whole was cut into, and the top says how many of those parts you hold. So

1 \div 2

has a finished answer, and the answer is

\frac{1}{2}

Problem

A workbench has

5

equal color bands, and blue tape covers

2

of them. The

5

counts equal pieces in the whole, and the

2

counts how many the tape covers. Write the fraction as

a / b

Show a hint

The two numbers do different jobs. One of them sets the size of a piece by saying how many equal pieces it takes to rebuild the whole bench. The other just counts how many of those equal pieces the tape actually sits on.
The bottom of the fraction is how many equal pieces make one whole, which is $5$ . The top is how many of those pieces you have, which is $2$ . Stack the count over the size.

Show the full solution

The bench is cut into

5

equal bands, so one band is one fifth of the bench. The tape sits on

2

of those bands, which is

2 / 5

. The two numbers do different jobs. The bottom one sets the size of a piece by saying how many equal pieces make the whole, and the top one just counts how many of those pieces you have.

Problem

Three clay bars are shared equally among

4

friends. Slice every bar into

4

equal strips, deal one strip from each bar to each friend. Write each friend's share as a fraction

a / b

Show a hint

Count what one friend actually walks away with. They took a strip from the first bar, a strip from the second, and a strip from the third. How many strips is that, and how big is each one compared to a whole bar?
Each friend ends up with $3$ strips, and each strip is $\frac{1}{4}$ of a bar. Three quarter-bars stacked together is $\frac{1}{4} + \frac{1}{4} + \frac{1}{4}$ . Write that as a single fraction $a / b$ .

Show the full solution

Cutting each of the

3

bars into

4

equal strips gives

3 \times 4 = 12

strips, and

12 \div 4 = 3

strips per friend. Each strip is

\frac{1}{4}

of a bar, so one friend's pile is

\frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} .

Each share is

3 / 4

of a bar. So

3 \div 4

never gets stuck with a leftover. The bar in

\frac{3}{4}

is the division sign, and the fraction is the finished answer.

Problem

A chocolatier pours

7

scoops across

4

trays. Each tray holds

7 \div 4

, and the same amount can be found by counting quarter-parts on one tray. Write that amount as a fraction

a / b

Show a hint

Try the two readings side by side and see if they really collide. Sharing means $7 \div 4$ . Counting quarter parts means asking how many pieces of size $\frac{1}{4}$ it takes to build up all 7 scoops on one tray. Both are describing one tray, so both should land on the same number.
Seven whole scoops, each cut into 4 quarter parts, gives $7 \times 4 = 28$ quarter parts in total, spread fairly over 4 trays. That leaves $28 \div 4 = 7$ quarter parts on a single tray. Seven pieces each of size $\frac{1}{4}$ is $7 \times \frac{1}{4}$ . Now write that as one fraction $a / b$ .

Show the full solution

Sharing gives

7 \div 4 = \frac{7}{4}

on each tray. Counting quarter parts gives the same amount, since

7

scoops cut into quarters make

7 \times 4 = 28

quarter parts, and

28 \div 4 = 7

of them land on one tray, so a tray holds

7 \times \frac{1}{4} = \frac{7}{4} .

One tray holds

7 / 4

. The two readings of a fraction always agree. And

\frac{7}{4}

is past

1

, so each tray really does hold more than one full scoop.

The same number, read two ways. On the left we cut one whole bar into

5

equal cells and shade

2

of them, which is the cut and take reading of

\frac{2}{5}

. On the right we take two whole bars, cut each into

5

equal cells, and deal one fifth from each bar into a single share, which is the share reading of

\frac{2}{5}

. The two summary bars at the bottom come out the same length, so both readings land on one number. In

\frac{2}{5}

the bottom number

5

is the denominator, naming how many equal pieces the bar is divided into, and the top number

2

is the numerator, counting how many of those pieces you keep.

Problem

A beetle starts at

0

and hops toward

1

in equal steps, landing exactly on

1

after

6

hops. Write the beetle's position after a single hop as a fraction $\frac{a}{b}$ .

Show a hint

The whole journey from $0$ to $1$ got split into $6$ equal pieces, one piece per hop. After one hop the beetle has covered just one of those $6$ equal pieces of the way to $1$ .
A fraction $\frac{a}{b}$ tells you to cut the trip to $1$ into $b$ equal parts and take $a$ of them. Here the trip is cut into $6$ equal parts and the beetle has taken $1$ of them.

Show the full solution

Six equal hops cover the whole distance from

0

1

, so a hop of length

h

satisfies

6 \times h = 1

. The number that stacks up

6

times to make

1

is one sixth, so after a single hop the beetle sits at

1 / 6 .

The

6

underneath records that the trip to

1

was cut into

6

equal steps, and the

1

on top records that one of those steps has been taken.

Problem

The beetle hops in steps of

\frac{1}{6}

and makes

5

hops. It ends up past the halfway mark but short of

1

. Write the beetle's position after

5

hops as

\frac{a}{b}

Show a hint

The denominator tells you the size of one hop, and the numerator counts how many of those hops the beetle has taken. So ask yourself, what is the step size here, and how many steps?
Each step is $\frac{1}{6}$ , and the beetle took $5$ of them. Five steps of size one-sixth is $\frac{5}{6}$ .

Show the full solution

Each step is

\frac{1}{6}

and the beetle takes

5

of them, so it lands at

5 \times \frac{1}{6} = \frac{5}{6} .

The position is

5 / 6

. Halfway is

3

steps, written

\frac{3}{6}

, and

5

steps is more than

3

but fewer than the full

6

, which matches a spot past the middle and short of

1

The bottom number, the denominator 6, sets the spacing. It chops each whole unit into 6 equal ticks, so one tick is

\frac{1}{6}

. The top number just counts ticks starting from 0. Five ticks lands you on

\frac{5}{6}

, still a little short of 1. Notice that nothing special happens at 1. The same equal steps walk right past it, and three more steps land on

\frac{8}{6}

. That is exactly why a fraction is allowed to be bigger than 1.

Problem

Walk

\frac{7}{4}

: four quarter-steps reach

1

, then

3

more land between

1

and

2

. Enter the larger whole number

\frac{7}{4}

falls between.

Show a hint

Walk in quarter steps from $0$ . Four of them stack to $\frac{4}{4} = 1$ , a whole number you pass through. Where you finally stop is wedged between two whole numbers in a row. The question asks for the larger one.
You take $7$ quarter steps in all. The first $4$ bring you to $1$ . The remaining $3$ carry you past $1$ but stop short of $2$ , because landing on $2$ would need $4$ more steps and you only had $3$ . So $\frac{7}{4}$ sits between $1$ and $2$ , and the larger of those is the answer.

Show the full solution

Quarter steps land on a whole number every

4

steps, since

\frac{4}{4} = 1

and

\frac{8}{4} = 2

. Seven steps clears the checkpoint at step

4

but falls one short of step

8

, so

\frac{7}{4}

sits between

1

and

2

. The larger of those is

2

. The denominator sets the step size and the numerator counts the steps, so you can place any fraction by asking which multiples of the bottom the top falls between.

Problem

A machine lights a lamp when the top divided by

8

is a whole number. It tries

\frac{8}{8}

\frac{24}{8}

\frac{40}{8}

\frac{52}{8}

. How many tops light the lamp?

Show a hint

A fraction $\frac{a}{8}$ is a whole number exactly when $8$ goes into the top $a$ evenly, with nothing left over. So instead of dividing, just ask of each top, is it a multiple of $8$ ?
Walk the multiples of $8$ and check each top against them. $8$ , $16$ , $24$ , $32$ , $40$ , $48$ , $56$ . Now look at $8$ , $24$ , $40$ , $52$ one at a time and count how many appear in that list.

Show the full solution

A fraction

\frac{a}{8}

is a whole number exactly when

8

divides the top. The multiples of

8

are

8, 16, 24, 32, 40, 48, 56

. Of the four tops,

8 = 8 \times 1

24 = 8 \times 3

, and

40 = 8 \times 5

are on that list, while

52 = 48 + 4

leaves a remainder of

4

. So

3

tops light the lamp. No real division is needed. A fraction is a whole number exactly when the bottom divides the top, which is last chapter's divisibility all over again.

Problem

A diver descends

7

kicks of

\frac{1}{3}

m, landing at

- \frac{7}{3}

m between two whole marks. Which whole number is deeper? Enter it with its minus sign.

Show a hint

Walking left from $0$ , every $3$ kicks of size $\frac{1}{3}$ add up to one whole meter down. How many full meters can you fit inside $7$ thirds before you run out, and how much is left over?
$7$ thirds is $6$ thirds plus $1$ more third. The $6$ thirds carry her to exactly $- 2$ , and the leftover third pushes her a little past $- 2$ toward $- 3$ . So she hangs between $- 2$ and $- 3$ . The deeper mark is the more negative one.

Show the full solution

Thirds clump into whole meters three at a time, since

\frac{3}{3} = 1

, so

- \frac{7}{3} = - \frac{3}{3} - \frac{3}{3} - \frac{1}{3} = - 1 - 1 - \frac{1}{3} .

The first six thirds carry her to exactly

- 2

, and the leftover third nudges her just past it, still short of

- 3

. She hangs between

- 2

and

- 3

, so the deeper mark is

- 3 .

Left of

0

the numbers get smaller as you go down, so the deeper of two marks is always the more negative one.

Problem

Ava:

- \frac{12}{5}

. Ben:

\frac{12}{- 5}

. Cleo:

- ⁣ (\frac{12}{5})

. All claim the same point left of

0

. How many distinct points do they name?

Show a hint

Don't trust the look of the writing, trust where the walking ends. Carry out each student's instruction as a real trip on the line and mark the spot it stops. Are those stopping spots in different places, or the same place written three ways?
The original mark is $12$ backward steps of size $\frac{1}{5}$ , which is $- \frac{12}{5}$ . Now check each student. Ava already wrote $- \frac{12}{5}$ . Ben's $\frac{12}{- 5}$ means $12$ copies of a backward fifth, also $- \frac{12}{5}$ . Cleo's $- (\frac{12}{5})$ reflects $\frac{12}{5}$ to the other side, again $- \frac{12}{5}$ . Count how many different landing spots that gives.

Show the full solution

All three write the same number. Ava has

- \frac{12}{5}

outright. Ben's

\frac{12}{- 5}

12

backward fifths, and flipping the sign of the bottom flips the sign of the quotient, so

\frac{12}{- 5} = - \frac{12}{5}

. Cleo's

- (\frac{12}{5})

reflects

\frac{12}{5}

across

0

, landing at

- \frac{12}{5}

too. They name

1

point. A single minus sign has three legal homes, out front, on top, or on the bottom, and none of them changes where you land.

Keep the top number fixed and let the bottom run through $1, 2, 3, \dots$ . Counting how many of those bottoms make the fraction a whole number is the same as counting the divisors of the top, so this is the divisor counting from last chapter.

Problem

A pogo stick hops

\frac{1}{8}

m, and a flag marks every whole-meter landing from hop

1

40

. How many of the $40$ hops plant a flag?

Show a hint

After $n$ hops the stick is at $\frac{n}{8}$ of a meter. A flag goes in only when that position is a whole number. So you are looking for the counts $n$ that make $\frac{n}{8}$ come out even, with nothing left over.
The fraction $\frac{n}{8}$ is a whole number exactly when $n$ is a multiple of $8$ . Just count the multiples of $8$ that sit in the range from $1$ to $40$ .

Show the full solution

After

n

hops the stick rests at

\frac{n}{8}

of a meter, which is a whole number exactly when

n

is a multiple of

8

. Between

1

and

40

those are

8, 16, 24, 32, 40

, so

5

hops plant a flag. You can skip the listing. The count of multiples of

8

up to

40

is just

40 \div 8 = 5

Problem

Machine places

\frac{60}{n}

for

n = 1

50

, with a green light when the result is whole. One divisor of

60

exceeds

50

. How many values of $n$ light the lamp?

Show a hint

The light comes on exactly when $n$ divides $60$ evenly, so your real job is to list the divisors of $60$ . Find them all first, then look for any that sit outside the range $1$ to $50$ .
Divisors come in pairs that multiply to $60$ , like $1$ with $60$ , $2$ with $30$ , $3$ with $20$ , and so on. Count every divisor of $60$ , then subtract the ones larger than $50$ .

Show the full solution

The marker

\frac{60}{n}

is a whole number exactly when

n

divides

60

, so hunt the divisors in pairs that multiply to

60

1 \times 60, 2 \times 30, 3 \times 20, 4 \times 15, 5 \times 12, 6 \times 10.

That gives

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

, twelve divisors. The machine only tries

n

up to

50

, and

60

is the single divisor above that ceiling, so

12 - 1 = 11

values light the lamp. Only the largest divisor can break the ceiling here, because its partner is

1

, the smallest divisor there is.

Practice these ideas

Practice

A spool is clamped at

9

evenly spaced points, cutting it into

9

equal lengths. The share-reading for one length is the answer to

1 \div 9

. Write it as a fraction

a / b

Show the solution

One coil cut into

9

equal pieces gives each piece

1 \div 9

, and that division is the fraction

1 / 9

. The top number is the one whole coil you started with, and the bottom number is how many equal pieces the clamps made.

Practice

Five sheets of gold leaf are shared equally among

8

picture frames, nothing wasted. Write how much gold leaf one frame receives as a fraction

\frac{a}{b}

Show the solution

Sharing

5

sheets equally among

8

frames is

5 \div 8

, so each frame receives

5 / 8

of a sheet. That is less than one full sheet, which fits, since

5

sheets cannot give

8

frames a whole sheet each.

Practice

A hose is cut into

7

equal sections. A sticker covers exactly

4

of them. Using the cut-and-take reading, write the fraction of the whole hose the sticker covers as

a / b

Show the solution

The hose is cut into

7

equal sections, so each section is

\frac{1}{7}

of it. The sticker covers

4

sections, which is

4

copies of

\frac{1}{7}

, or

4 / 7

of the hose. The bottom of the fraction is how many equal pieces make the whole, and the top is how many of them you take.

Practice

A beetle reaches

1

in exactly

12

equal steps, each of size

\frac{1}{12}

. How many of these steps does the beetle take to travel from $0$ to $1$ ?

Show the solution

A step of

\frac{1}{12}

is one piece out of

12

equal pieces that fill a whole, so twelve steps cover

12 \times \frac{1}{12} = \frac{12}{12} = 1

. The beetle takes

12

steps. The step count matches the denominator because the denominator was telling you how many equal pieces make one whole all along.

Practice

A token is

17

steps of

\frac{1}{5}

out from

0

, at address

\frac{17}{5}

. It falls between two whole numbers. Enter the smaller of those two whole numbers.

Show the solution

Five steps of

\frac{1}{5}

make one unit, so whole numbers sit at

0, 5, 10, 15, 20

steps out, which are the addresses

0, 1, 2, 3, 4

. The token is at

17

steps, between

15

and

20

, so

\frac{17}{5}

falls between

3

and

4

. The smaller is

3

. Checking directly,

3 = \frac{15}{5}

and

4 = \frac{20}{5}

, and

15 < 17 < 20

Practice

A hold sits at height

\frac{11}{3}

m, between two whole meter marks. Enter the larger of the two whole numbers that $\frac{11}{3}$ falls between.

Show the solution

Whole meters come in groups of three thirds, since

\frac{3}{3} = 1

. Three whole meters use up

\frac{9}{3}

, leaving

\frac{11}{3} - \frac{9}{3} = \frac{2}{3}

, so

\frac{11}{3} = 3 + \frac{2}{3}

. That is past the

3

mark and short of

4

, so the larger whole number is

4

Practice

A sensor hangs at

- \frac{9}{4}

m, between two whole meter marks. Which whole meter mark is the deeper (lower) of the two? Give it with its minus sign.

Show the solution

Since

\frac{9}{4} = 2 \frac{1}{4}

, the sensor sits at

- 2 \frac{1}{4}

, a quarter meter below the

- 2

mark and not yet down to

- 3

. The deeper of the two marks is

- 3

Deeper means more negative, so the lower mark is the one farther from the surface.

Practice

A log reads 84/7 bins.

\frac{84}{7}

lands on a whole number when

7

divides

84

exactly. Check divisibility. If it does, what whole number does

\frac{84}{7}

equal?

Show the solution

Counting up in sevens gives

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84

, and

84

lands right on the twelfth step. So

7

divides

84

with nothing left over, and

\frac{84}{7} = 12

bins.

Practice

Four ribbons laid end to end fill

0

4

. Split that stretch equally among

3

friends. Where does the first friend's piece end? Write that address as

\frac{a}{b}

Show the solution

The stretch from

0

4

measures

4

units, and sharing it among

3

friends means

4 \div 3

, so one piece is

\frac{4}{3}

long. The first piece starts at

0

, so it ends at address

4 / 3

That is a little past

1

, which makes sense, since

4

split three ways gives each person a bit more than one whole ribbon.

Practice

Three cards on a number line:

- \frac{8}{3}

\frac{8}{- 3}

- (\frac{8}{3})

. Trace where each points. How many distinct points do they name in total?

Show the solution

Read each card as one signed number. The first is already

- \frac{8}{3}

. A positive over a negative is negative, so

\frac{8}{- 3} = - \frac{8}{3}

. And

- (\frac{8}{3})

is the opposite of

\frac{8}{3}

, again

- \frac{8}{3}

. All three land on one spot, so they name

1

point. A minus sign on top, on the bottom, or out front all do the same job, they flip the sign once.

Practice

Five addresses:

\frac{80}{5}

\frac{80}{6}

\frac{80}{16}

\frac{80}{10}

\frac{80}{3}

. A locker is real when the fraction is whole. How many are real lockers?

Show the solution

The top is always

80

, so an address is a whole number exactly when its bottom divides

80

. Checking each one,

\frac{80}{5} = 16

\frac{80}{16} = 5

, and

\frac{80}{10} = 8

come out even, while

6

and

3

both leave a remainder. That is

3

real lockers.

Practice

Tiles

72

into

n

equal rows: needs

n ∣ 72

. For how many whole numbers $n \geq 1$ does $\frac{72}{n}$ land on a whole number?

Show the solution

\frac{72}{n}

is a whole number exactly when

n

divides

72

, so count the divisors of

72

. Hunting in pairs that multiply to

72

gives

1 \times 72

2 \times 36

3 \times 24

4 \times 18

6 \times 12

, and

8 \times 9

, which lists

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

. That is

12

values of

n

. Once the two partners in a pair meet in the middle, like

8

and

9

here, you have caught every divisor.

Practice

A loader accepts

n

trays when

\frac{n + 7}{n} = 1 + \frac{7}{n}

is a whole number, i.e., when

n

divides

7

. For how many whole numbers $n \geq 1$ does the loader accept the stack?

Show the solution

Split the top.

\frac{n + 7}{n} = \frac{n}{n} + \frac{7}{n} = 1 + \frac{7}{n} .

The

1

is already whole, so the whole expression is a whole number exactly when

\frac{7}{n}

is. Since

7

is prime, only

n = 1

and

n = 7

divide it, giving

\frac{8}{1} = 8

and

\frac{14}{7} = 2

. That is

2

values of

n

What a Fraction Means

Practice these ideas