Dividing Fractions

If you have half a pizza and share it equally among 3 people, each person gets one-sixth of the whole pizza. Splitting the half into 3 equal pieces gives \( \frac{1}{6} \) each. Alternatively, dividing by 3 is the same as multiplying by \( \frac{1}{3} \): \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \).

A check confirms it: \( \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \), so three lots of \( \frac{1}{6} \) really do make \( \frac{1}{2} \). Dividing a fraction by a whole number makes each piece smaller — one-third the size of what you started with.

The question asks “how many halves fit into 4?” Each whole contains 2 halves, so 4 wholes contain \( 4 \times 2 = 8 \) halves. A bar model makes this clear: draw 4 bars, split each in half, and count the pieces — you get 8. Dividing by \( \frac{1}{2} \) is equivalent to multiplying by 2.

Students often confuse “dividing by one half” with “dividing in half” (halving). Halving 4 gives 2, but that’s \( 4 \div 2 \), not \( 4 \div \frac{1}{2} \). The two operations are completely different: \( 4 \div 2 \) asks “what is half of 4?” while \( 4 \div \frac{1}{2} \) asks “how many halves are in 4?”

The question asks “how many quarters fit into two-thirds?” Converting to a common denominator: \( \frac{2}{3} = \frac{8}{12} \) and \( \frac{1}{4} = \frac{3}{12} \). Now: how many groups of 3 twelfths fit into 8 twelfths? That’s \( 8 \div 3 = \frac{8}{3} = 2\frac{2}{3} \), which is greater than 2.

This surprises students who expect division to make things smaller. That’s true when dividing by a number greater than 1, but when dividing by a proper fraction (less than 1), the answer is larger than what you started with. You’re asking “how many of this small piece fit?” — and small pieces fit lots of times.

Dividing by \( \frac{1}{5} \) asks “how many fifths fit in?” Each whole contains 5 fifths, so for any number, dividing by \( \frac{1}{5} \) counts how many fifths are in it — which is always 5 times the number. For example: \( 3 \div \frac{1}{5} = 15 = 3 \times 5 \). And \( \frac{1}{2} \div \frac{1}{5} = \frac{5}{2} = \frac{1}{2} \times 5 \).

This is why “keep, change, flip” works: the reciprocal of \( \frac{1}{5} \) is 5, so dividing by \( \frac{1}{5} \) is the same as multiplying by 5. The same logic extends to all fractions — dividing by \( \frac{a}{b} \) is the same as multiplying by \( \frac{b}{a} \), because the reciprocal “undoes” the fraction.

Example: \( 3 \div \frac{1}{2} = 6 \) — dividing 3 by one half gives 6, which is bigger than 3.

Another: \( \frac{1}{2} \div \frac{1}{4} = 2 \) — started with \( \frac{1}{2} \), answer is 2, which is bigger.

Creative: \( \frac{1}{10} \div \frac{1}{100} = 10 \) — dividing two tiny fractions gives a whole number much larger than either. Or \( \frac{1}{8} \div \frac{1}{12} = \frac{3}{2} \), since \( \frac{3}{2} > \frac{1}{8} \).

Trap: \( \frac{3}{4} \div 3 = \frac{1}{4} \) — this involves a fraction, so a student might think it qualifies. But \( \frac{1}{4} \) is smaller than \( \frac{3}{4} \). The condition requires dividing by a number less than 1 (a proper fraction), not just having a fraction somewhere in the calculation.

Example: \( \frac{3}{4} \div \frac{1}{4} = 3 \) — three quarters contain three lots of one quarter.

Another: \( \frac{1}{2} \div \frac{1}{6} = 3 \) — a half contains three sixths.

Creative: \( \frac{2}{3} \div \frac{1}{6} = 4 \) — the denominators aren’t the same, but two-thirds contains exactly four sixths. Or \( \frac{5}{8} \div \frac{5}{24} = 3 \), since \( \frac{5}{8} \times \frac{24}{5} = \frac{120}{40} = 3 \).

Trap: \( \frac{1}{3} \div \frac{1}{2} = \frac{2}{3} \) — a student might expect two “clean” unit fractions to produce a neat whole number, but \( \frac{2}{3} \) is not a whole number. The dividend must be an exact whole-number multiple of the divisor for the quotient to be an integer.

Example: \( \frac{1}{2} \div \frac{1}{3} = \frac{3}{2} \) but \( \frac{1}{3} \div \frac{1}{2} = \frac{2}{3} \) — swapping the fractions gives a completely different answer.

Another: \( \frac{3}{4} \div \frac{1}{4} = 3 \) but \( \frac{1}{4} \div \frac{3}{4} = \frac{1}{3} \) — one gives a whole number, the other a small fraction.

Creative: \( \frac{2}{5} \div \frac{3}{7} = \frac{14}{15} \) but \( \frac{3}{7} \div \frac{2}{5} = \frac{15}{14} \) — the two answers are reciprocals of each other, which is always true when you swap the order in a division.

Trap: \( \frac{1}{2} \div \frac{1}{2} = 1 \) — swapping gives the same answer. A student who learned that multiplication is commutative might assume division works the same way. But this only happens when both fractions are identical; for any two different fractions, order matters.

Example: \( 1\frac{1}{2} \div \frac{2}{3} = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4} = 2\frac{1}{4} \)

Another: \( 2\frac{1}{3} \div \frac{1}{2} = \frac{7}{3} \times 2 = \frac{14}{3} = 4\frac{2}{3} \)

Creative: \( 1\frac{1}{4} \div \frac{1}{3} = \frac{5}{4} \times 3 = \frac{15}{4} = 3\frac{3}{4} \) — the dividend, divisor, and answer all look completely different, making the relationship hard to spot.

Trap: \( 1\frac{1}{2} \div \frac{3}{4} = \frac{3}{2} \times \frac{4}{3} = \frac{12}{6} = 2 \) — this looks messy enough that you’d expect a mixed number answer, but it simplifies to exactly 2. Students assume that dividing mixed numbers by fractions always produces another mixed number, but the result can simplify to a whole number.

True when dividing a positive number by a proper fraction (less than 1): \( 3 \div \frac{1}{2} = 6 \), and \( 6 > 3 \). But false when dividing by an improper fraction (greater than 1): \( 3 \div \frac{3}{2} = 2 \), and \( 2 < 3 \).

It also fails when the starting number is 0: \( 0 \div \frac{1}{2} = 0 \), which is not larger. Students often learn “dividing by a fraction makes it bigger” without realising this only applies when the fraction is between 0 and 1 and the starting number is positive.

Stretch Challenge: What about negative numbers? \( -4 \div \frac{1}{2} = -8 \). Because \( -8 < -4 \), dividing a negative number by a proper fraction makes it smaller (further from zero). This creates a great moment of cognitive conflict!

If you divide any whole number \( n \) by a unit fraction \( \frac{1}{k} \), the result is \( n \times k \), which is always a whole number. For example: \( 3 \div \frac{1}{4} = 12 \), \( 7 \div \frac{1}{2} = 14 \), \( 1 \div \frac{1}{10} = 10 \). This is because each whole contains exactly \( k \) pieces of size \( \frac{1}{k} \), so \( n \) wholes contain \( nk \) pieces.

Students should recognise this as the foundation of why “keep, change, flip” works with unit fractions: dividing by \( \frac{1}{k} \) is the same as multiplying by \( k \). The result \( nk \) is always a whole number because it is the product of two whole numbers.

True case: \( \frac{1}{2} \div \frac{1}{3} = \frac{3}{2} \). Here \( \frac{3}{2} > \frac{1}{2} \) and \( \frac{3}{2} > \frac{1}{3} \), so the answer is larger than both.

False case: \( \frac{1}{4} \div \frac{3}{4} = \frac{1}{3} \). Here \( \frac{1}{3} > \frac{1}{4} \) (larger than the dividend) but \( \frac{1}{3} < \frac{3}{4} \) (smaller than the divisor). The answer is not larger than both. This happens when the dividend is much smaller than the divisor — the quotient ends up smaller than at least one of the original fractions.

Dividing any positive fraction by a whole number greater than 1 always makes it smaller. For example: \( \frac{3}{4} \div 2 = \frac{3}{8} \), and \( \frac{3}{8} < \frac{3}{4} \). Algebraically: \( \frac{a}{b} \div n = \frac{a}{bn} \), and since \( n > 1 \), the denominator increases, making the fraction smaller.

Students sometimes confuse this with dividing by a fraction (which can make things bigger). But dividing by a whole number greater than 1 always shrinks the result — just as \( 12 \div 3 = 4 \) is smaller than 12, \( \frac{1}{3} \div 5 = \frac{1}{15} \) is smaller than \( \frac{1}{3} \).

Students will often immediately say “Never” because they know division isn’t commutative. But it is true when \( a = b \) (for example, \( \frac{2}{3} \div \frac{2}{3} = 1 \), and swapping them still equals 1).

For any two different numbers, swapping the order gives the reciprocal of the original answer, not the same answer. This is a fantastic trap to test if students are actively considering identical values as a boundary case.

The student has flipped the wrong fraction. “Keep-change-flip” means keep the first fraction unchanged, change ÷ to ×, and flip the second fraction (the divisor). The student flipped the dividend (\( \frac{2}{3} \) became \( \frac{3}{2} \)) instead of the divisor.

The correct working is: \( \frac{2}{3} \div \frac{3}{4} = \frac{2}{3} \times \frac{4}{3} = \frac{8}{9} \). Notice the student’s answer (\( \frac{9}{8} \)) is the reciprocal of the correct answer (\( \frac{8}{9} \)) — this is always what happens when you flip the wrong fraction, which is a useful diagnostic clue.

The answer is correct, but the reasoning is based on “cancelling” matching numbers rather than understanding division. The student treats the operation like simplification — as if the matching 3s can just be crossed out. This works here only because the numerator happens to be divisible by the divisor.

Applying this logic to \( \frac{4}{5} \div 3 \), the student is stuck — there’s no “3 to cancel” in the numerator. The correct method is: \( \frac{3}{5} \div 3 = \frac{3}{5} \times \frac{1}{3} = \frac{3}{15} = \frac{1}{5} \). This method — multiplying by the reciprocal — works regardless of whether the numbers match. Relying on cancellation gives students no strategy when the numbers don’t cooperate.

The student has confused “dividing by one half” with “dividing in half” — one of the most common errors in fraction division. “Finding half” is \( 5 \div 2 = 2\frac{1}{2} \), but the question asks \( 5 \div \frac{1}{2} \): “how many halves fit into 5?”

The correct answer is \( 5 \div \frac{1}{2} = 5 \times 2 = 10 \). Each whole contains 2 halves, so 5 wholes contain 10 halves. The key distinction is: \( \div 2 \) means “split into 2 groups,” while \( \div \frac{1}{2} \) means “how many groups of size \( \frac{1}{2} \)?” These give very different answers, and mixing them up is the root cause of the error.

The student has multiplied the fractions directly instead of multiplying by the reciprocal — they’ve computed \( \frac{4}{5} \times \frac{2}{3} \) rather than \( \frac{4}{5} \div \frac{2}{3} \). This is the “forgetting to flip” misconception: the student remembers that division “becomes multiplication” but forgets the crucial step of taking the reciprocal of the divisor.

The correct working is: \( \frac{4}{5} \div \frac{2}{3} = \frac{4}{5} \times \frac{3}{2} = \frac{12}{10} = \frac{6}{5} = 1\frac{1}{5} \). A quick sense-check: \( \frac{4}{5} \) is larger than \( \frac{2}{3} \), so the answer should be greater than 1. The student’s answer of \( \frac{8}{15} \) is less than 1, which should signal something has gone wrong.

The student has wrongly assumed you can partition division like addition. While you can add \( 2\frac{1}{2} + 1\frac{1}{4} \) by adding the wholes and fractions separately, this does not work for division (or multiplication).

The correct method requires converting to improper fractions first: \( \frac{5}{2} \div \frac{5}{4} = \frac{5}{2} \times \frac{4}{5} = \frac{20}{10} = 2 \). A good sense-check for the student: \( 1\frac{1}{4} \) goes into \( 2\frac{1}{2} \) exactly twice, so the answer must be 2, not 4.