Fractions — proper, improper, mixed numbers: Exercises

We lean on the parent Fractions topic and its engine, Division with remainder. Where we compare fractions we quietly use Equivalent fractions and the Number line.

Definition Ground rule for every problem below — the denominator

Throughout this page a fraction $\frac{a}{b}$ has a ==denominator $b$ that is a positive whole number== ( $b = 1, 2, 3, \dots$ ) and a whole-number numerator $a \geq 0$ .

Why not $b = 0$ ? Splitting a whole into $0$ equal pieces is meaningless — there is no piece-size to count, so $\frac{a}{0}$ is undefined. Picture trying to cut a pizza into "zero slices": you cannot even start.
Why do we keep $b$ positive here? A negative denominator like $\frac{3}{- 4}$ just re-writes as $- \frac{3}{4}$ ; it never gives a new amount, only a sign we could have put out front. So to keep the pictures clean (real slices of a real pizza), every $b$ on this page is a positive integer. Every answer inherits this restriction.
What about $a = 0$ ? That is allowed: $\frac{0}{b}$ means "take $0$ of the pieces" $= 0$ . And $b = 1$ is allowed too: $\frac{a}{1}$ means "pieces so big that one piece is the whole", so $\frac{a}{1} = a$ , a plain whole number. Both edge cases appear below.

Level 1 — Recognition

Goal: read a fraction and name its costume, no arithmetic tricks yet.

The fraction bars below show each classification before you read the algebra — a full row of cyan cells is one whole; count how far each bar spills past it.

Figure — Fractions — proper, improper, mixed numbers

Recall Solution 1.1

Compare the top (count of pieces) with the bottom (pieces per whole), and put each in one box:

$\frac{3}{8}$ : $3 < 8$ → proper (fewer pieces than a whole needs — the bar is only part-full).
$\frac{8}{3}$ : $8 > 3$ → improper (more than one whole — the bar overflows into a second row).
$\frac{5}{5}$ : $5 = 5$ → exactly one whole (5 out of 5 rebuilds the whole — one full row, nothing over). This is its own box: although the rule $a \geq b$ would technically call it improper, when top equals bottom it lands precisely on one whole, so we classify it as exactly one whole — not proper, not overflowing.
$\frac{12}{7}$ : $12 > 7$ → improper (one full row plus 5 more sevenths).

Recall Solution 1.2

Ask "how many whole fours fit in 11?" $2 \times 4 = 8$ and $3 \times 4 = 12$ . Since $8 \leq 11 < 12$ , we have $2 \leq \frac{11}{4} < 3$ . So $\frac{11}{4}$ sits between 2 and 3. Look at the figure below — the amber dot lands three quarter-steps past 2.

Recall Solution 1.3

$\frac{0}{5}$ : take $0$ of the five equal pieces → you have nothing, so $\frac{0}{5} = 0$ . Since $0 < 5$ , the top is smaller than the bottom, so it is a proper fraction, and it equals 0. (Every $\frac{0}{b}$ with $b \geq 1$ equals $0$ for the same reason — zero pieces is zero, whatever the piece-size.)
$\frac{7}{1}$ : the pieces are so big that one piece is the whole. Taking $7$ of them gives $7$ wholes, so $\frac{7}{1} = 7$ . Since $7 > 1$ it is improper, and it is exactly the whole number 7.

Level 2 — Application

Goal: run the conversion machinery both directions.

The picture first: shade 23 sixth-sized cells and watch them fill 3 full rows (each row = 6 sixths = one whole) with 5 cells left over — that is exactly $3 \frac{5}{6}$ .

Recall Solution 2.1

Use Division with remainder: write $23 = q \cdot 6 + r$ with $0 \leq r < 6$ .

How many whole sixes in 23? $3 \times 6 = 18$ , and $4 \times 6 = 24$ is too big. So $q = 3$ (the 3 full rows).
Leftover: $r = 23 - 18 = 5$ (the 5 stray cells).
Therefore $\frac{23}{6} = 3 + \frac{5}{6} = 3 \frac{5}{6}$ .
Check: $3 \times 6 + 5 = 23$ ✓.

Recall Solution 2.2

Rule MBAT (Multiply Bottom, Add Top): the 5 wholes must become ninths first.

Why re-slice the wholes first? A fraction can only add things that are the same size piece. Right now the "5" is measured in whole pizzas and the "7" is measured in ninths — different-sized units, so you cannot combine them yet, any more than you can add 5 metres to 7 centimetres without first converting. To turn 5 wholes into ninths, cut each whole into 9 slices: $5$ wholes $= 5 \times 9 = 45$ ninths. Now both counts are in ninths and can legally be added.
$5 \times 9 = 45$ (five wholes = $\frac{45}{9}$ ).
Add the extra $7$ ninths: $45 + 7 = 52$ .
So $5 \frac{7}{9} = \frac{52}{9}$ .
Check: $52 = 5 \times 9 + 7 = 45 + 7$ ✓, and $52 > 9$ so it is correctly improper.

Recall Solution 2.3

$\frac{9}{9}$ : $9 = 1 \times 9 + 0$ , so $q = 1, r = 0$ . That gives $1 + \frac{0}{9} = 1$ (exactly one whole, no fraction part).
$\frac{40}{8}$ : $40 = 5 \times 8 + 0$ , so $q = 5, r = 0$ . That gives $5$ (a whole number — the remainder rode on top as $\frac{0}{8} = 0$ ).

Level 3 — Analysis

Goal: compare, order, and reason about fractions without a calculator.

The two fraction bars below are drawn to the same total length so equal pieces mean equal widths; the amber bar ( $\frac{9}{5}$ ) clearly pokes one twentieth further right.

Recall Solution 3.1

To compare, the pieces must be the same size — this is Equivalent fractions. Common denominator of 4 and 5 is $20$ .

$\frac{7}{4} = \frac{7 \times 5}{20} = \frac{35}{20}$ .
$\frac{9}{5} = \frac{9 \times 4}{20} = \frac{36}{20}$ .
Now both count twentieths, so just compare tops: $36 > 35$ .
Therefore $\frac{9}{5} > \frac{7}{4}$ .
Sanity check as mixed numbers: $\frac{7}{4} = 1 \frac{3}{4} = 1.75$ , $\frac{9}{5} = 1 \frac{4}{5} = 1.8$ . Indeed $1.8 > 1.75$ ✓.

Recall Solution 3.2

Find the neighbouring multiples of 9: $11 \times 9 = 99$ and $12 \times 9 = 108$ .

$100$ is only $1$ above $99$ , but $8$ below $108$ .
So $\frac{100}{9} = 11 \frac{1}{9}$ — barely past 11, therefore closer to 11.
Check: $11 \times 9 + 1 = 100$ ✓.

Recall Solution 3.3

We need one common denominator so all three count the same-size piece.

Why 24 (and not, say, 48)? We want the smallest number that $6$ , $8$ and $12$ all divide evenly — the least common multiple. Build it from prime factors: $6 = 2 \times 3$ , $8 = 2^{3}$ , $12 = 2^{2} \times 3$ . Take the highest power of each prime that appears: $2^{3}$ (from the 8) and $3$ (from the 6 or 12). Multiply: $2^{3} \times 3 = 8 \times 3 = 24$ . So $24$ is the smallest common denominator. ( $48$ also works — every whole must divide it — but it makes bigger numbers for no benefit.)
$\frac{5}{6} = \frac{20}{24}$ (multiply by $\frac{4}{4}$ ).
$\frac{7}{8} = \frac{21}{24}$ (multiply by $\frac{3}{3}$ ).
$\frac{11}{12} = \frac{22}{24}$ (multiply by $\frac{2}{2}$ ).
Compare tops: $20 < 21 < 22$ .
So the order is $\frac{5}{6} < \frac{7}{8} < \frac{11}{12}$ .

Level 4 — Synthesis

Goal: combine conversion with other operations.

Recall Solution 4.1

Convert to improper first (cleaner for Adding and subtracting fractions).

$2 \frac{1}{3} = \frac{2 \times 3 + 1}{3} = \frac{7}{3}$ .
$1 \frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$ .
Common denominator 6: $\frac{7}{3} = \frac{14}{6}$ , $\frac{3}{2} = \frac{9}{6}$ .
Add: $\frac{14}{6} + \frac{9}{6} = \frac{23}{6}$ .
Back to mixed (from Problem 2.1!): $\frac{23}{6} = 3 \frac{5}{6}$ .
Check: $2 \frac{1}{3} + 1 \frac{1}{2} = 2.333 \dots + 1.5 = 3.833 \dots = 3 \frac{5}{6}$ ✓.

Recall Solution 4.2

"Per batch × number of batches" = multiply. Convert the mixed number to improper first.

$2 \frac{1}{2} = \frac{5}{2}$ .
Multiply: $\frac{3}{4} \times \frac{5}{2} = \frac{3 \times 5}{4 \times 2} = \frac{15}{8}$ .
Convert: $15 = 1 \times 8 + 7$ , so $\frac{15}{8} = 1 \frac{7}{8}$ .
Answer: $1 \frac{7}{8}$ cups. Check: $0.75 \times 2.5 = 1.875 = 1 \frac{7}{8}$ ✓.

Recall Solution 4.3

Link to Decimals. From Problem structure, $17 = 3 \times 5 + 2$ , so $\frac{17}{5} = 3 \frac{2}{5}$ .

The fraction part $\frac{2}{5} = \frac{4}{10} = 0.4$ .
So $\frac{17}{5} = 3.4$ .
Check: $3.4 \times 5 = 17$ ✓.

Level 5 — Mastery

Goal: reverse-engineer and construct fractions to meet constraints.

Recall Solution 5.1

Convert: $50 = 7 \times 7 + 1$ , so $q = 7, r = 1$ , giving $\frac{50}{7} = 7 \frac{1}{7}$ .
Now scale to denominator 21: multiply top and bottom by 3 (Equivalent fractions): $\frac{50 \times 3}{7 \times 3} = \frac{150}{21}$ .
Check: $\frac{150}{21}$ and $\frac{50}{7}$ are equal because $150 \times 7 = 1050 = 50 \times 21$ ✓. And $150 = 21 \times 7 + 3$ , so $\frac{150}{21} = 7 \frac{3}{21}$ .
Why does $7 \frac{3}{21}$ equal $7 \frac{1}{7}$ ? Look at the fraction part $\frac{3}{21}$ : both $3$ and $21$ are divisible by $3$ (their common factor). Cancel it — $\frac{3}{21} = \frac{3 \div 3}{21 \div 3} = \frac{1}{7}$ . This is Equivalent fractions run backwards (shrinking, not growing). So $7 \frac{3}{21} = 7 \frac{1}{7}$ — the very same value ✓.

Recall Solution 5.2

We need $4 < \frac{a}{6} < 5$ , i.e. $4 \times 6 < a < 5 \times 6$ , so $24 < a < 30$ .

The integers strictly between are $a = 25, 26, 27, 28, 29$ → that's 5 fractions: $\frac{25}{6}, \frac{26}{6}, \frac{27}{6}, \frac{28}{6}, \frac{29}{6}$ .
One example: $\frac{25}{6} = 4 \frac{1}{6}$ (since $25 = 4 \times 6 + 1$ ), which is indeed between 4 and 5.
General insight: between two consecutive wholes $n$ and $n + 1$ there are exactly $b - 1$ fractions with denominator $b$ (the numerators $nb + 1$ through $nb + (b - 1)$ ).

Recall Solution 5.3

$3 \frac{2}{5} = \frac{3 \times 5 + 2}{5} = \frac{17}{5}$ . So $\frac{a}{b} = \frac{17}{5}$ .

The smallest whole numbers giving this ratio are the ones already in lowest terms: $a = 17, b = 5$ (since 17 is prime and shares no factor with 5).
Check: $\frac{17}{5} = 3 \frac{2}{5}$ ✓, and no smaller pair works because $g cd (17, 5) = 1$ .

Flashcards

Convert 23/6 to a mixed number.

3 5/6 (since 23 = 3×6 + 5).

Convert 5 7/9 to improper.

52/9 (5×9 + 7 = 52).

Which is larger, 7/4 or 9/5?

9/5 (35/20 vs 36/20).

How many fractions with denominator 6 lie strictly between 4 and 5?

Five (numerators 25 to 29).

Add 2 1/3 + 1 1/2.

3 5/6 (14/6 + 9/6 = 23/6).

0.75 × 2 1/2 cups equals?

1 7/8 cups (15/8).

What is 0/5, and what type of fraction is it?

0 — a proper fraction (top smaller than bottom), value zero.

What is 7/1?

The whole number 7 (denominator 1 means each piece is a whole).

Why must a denominator be a positive whole number, never 0?

0 pieces gives no piece-size (undefined); a positive integer keeps real, countable slices.

Connections

Division with remainder — the engine for every improper→mixed step here.
Equivalent fractions — used in comparing, rescaling and cancelling (L3, L5).
Adding and subtracting fractions — common-denominator work in L4.
Decimals — cross-checking answers (Problem 4.3).
Number line — placing fractions between wholes (L1, L5).
Ratios and proportions — Problem 5.3 is a ratio in disguise.