Adding and Subtracting Fractions
Questions designed to stretch thinking, reveal misconceptions, and spark mathematical reasoning.
Convince Me That…
Students must construct a mathematical argument for why each statement is true.
You can’t add halves and thirds directly because the pieces are different sizes. Converting to sixths: \( \frac{1}{2} = \frac{3}{6} \) and \( \frac{1}{3} = \frac{2}{6} \). Now the pieces are the same size, so \( \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \). Think of it as a pizza cut into 6 slices — eating 3 slices (half) plus 2 slices (a third) means you’ve eaten 5 of the 6 slices.
A common mistake is the “add the tops and add the bottoms” misconception: writing \( \frac{1}{2} + \frac{1}{3} = \frac{2}{5} \). But \( \frac{2}{5} \) is less than \( \frac{1}{2} \) (since \( \frac{2}{5} = 0.4 \) and \( \frac{1}{2} = 0.5 \)), so adding a positive fraction to \( \frac{1}{2} \) can’t possibly give something smaller than \( \frac{1}{2} \).
Since the denominators are the same, we subtract the numerators: \( \frac{3}{4} \; – \; \frac{1}{4} = \frac{2}{4} \). Then \( \frac{2}{4} \) simplifies to \( \frac{1}{2} \) because 2 and 4 share a common factor of 2. Visually, three-quarters of a shape minus one-quarter leaves two-quarters, which is exactly half.
The “not simplifying” misconception trips students up here — they stop at \( \frac{2}{4} \) and don’t recognise it as \( \frac{1}{2} \). Others may subtract incorrectly and get \( \frac{2}{0} \) (subtracting the denominators too) or simply write 2.
The denominators are 5 and 10. Since 10 is a multiple of 5, we only need to convert one fraction: \( \frac{2}{5} = \frac{4}{10} \). Now we can add: \( \frac{4}{10} + \frac{3}{10} = \frac{7}{10} \). We don’t need to use 50 as a common denominator — the lowest common denominator is 10 itself.
Converting to improper fractions: \( 2\frac{1}{3} = \frac{7}{3} \) and \( 1\frac{3}{4} = \frac{7}{4} \). The common denominator is 12: \( \frac{7}{3} = \frac{28}{12} \) and \( \frac{7}{4} = \frac{21}{12} \). So \( \frac{28}{12} \; – \; \frac{21}{12} = \frac{7}{12} \).
A common error is to subtract the whole numbers and fractions separately: \( 2 \; – \; 1 = 1 \), then attempt \( \frac{1}{3} \; – \; \frac{3}{4} \). Since \( \frac{1}{3} \) is smaller than \( \frac{3}{4} \), students often get stuck or make a mistake.
We can think of 3 as \( 2 + 1 \). We can rewrite the 1 as \( \frac{5}{5} \) (a whole pizza cut into 5 slices). So \( 3 = 2\frac{5}{5} \). Now the subtraction is straightforward: \( 2\frac{5}{5} \; – \; \frac{2}{5} = 2\frac{3}{5} \).
Many students struggle when there is no fraction part to subtract from, often just writing \( 2\frac{2}{5} \) (keeping the fraction) or getting stuck.
Give an Example Of…
Think carefully — the fourth box is a trap! Give a non-example that looks right but isn’t.
Example: \( \frac{1}{2} + \frac{3}{6} = \frac{1}{2} + \frac{1}{2} = 1 \) — using an equivalent fraction of \( \frac{1}{2} \).
Another: \( \frac{1}{3} + \frac{4}{6} = \frac{1}{3} + \frac{2}{3} = 1 \)
Creative: \( \frac{3}{8} + \frac{10}{16} \) — since \( \frac{10}{16} = \frac{5}{8} \), and \( \frac{3}{8} + \frac{5}{8} = 1 \).
Trap: \( \frac{1}{4} + \frac{2}{3} \) — a student might think “a quarter plus two-thirds covers most of the whole, so it must be exactly 1.” But \( \frac{1}{4} + \frac{2}{3} = \frac{3}{12} + \frac{8}{12} = \frac{11}{12} \), which falls short by \( \frac{1}{12} \).
Example: \( \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \), which simplifies to \( \frac{1}{3} \).
Another: \( \frac{1}{4} + \frac{1}{4} = \frac{2}{4} \), which simplifies to \( \frac{1}{2} \).
Creative: \( \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \) — the answer simplifies and the denominators were different to start with.
Trap: \( \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \) — a student might assume any answer with 12 as the denominator must simplify, but 7 and 12 share no common factors.
Example: \( \frac{1}{3} + \frac{1}{4} \) — LCD is 12, giving \( \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \).
Another: \( \frac{2}{3} + \frac{3}{5} \) — LCD is 15.
Creative: \( \frac{5}{6} + \frac{3}{8} \) — LCD is 24. Neither 6 nor 8 divides the other, and the LCD (24) is not their product (48), which makes this a trickier example.
Trap: \( \frac{1}{4} + \frac{3}{8} \) — the denominators are different, but 8 is a multiple of 4, so only \( \frac{1}{4} \) needs converting. You only convert both when neither denominator is a multiple of the other.
Example: \( \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} = 1\frac{1}{4} \)
Another: \( \frac{2}{3} + \frac{2}{3} = \frac{4}{3} = 1\frac{1}{3} \)
Creative: \( \frac{7}{8} + \frac{1}{5} = \frac{35}{40} + \frac{8}{40} = \frac{43}{40} = 1\frac{3}{40} \) — just barely over 1.
Trap: \( \frac{2}{3} + \frac{1}{6} = \frac{5}{6} \) — a student might reason “\( \frac{2}{3} \) is more than half and I’m adding more to it, so it must be over 1.” But \( \frac{5}{6} \) is still less than 1.
Example: \( 1\frac{1}{2} + 2\frac{1}{2} = 4 \) — the fraction parts sum to 1.
Another: \( 1\frac{1}{4} + 2\frac{3}{4} = 4 \).
Creative: \( 2\frac{3}{5} + 3\frac{4}{10} \) — converting \( \frac{4}{10} \) to \( \frac{2}{5} \) first to see the fractions sum to 1.
Trap: \( 1\frac{1}{3} + 1\frac{1}{3} = 2\frac{2}{3} \) — students might pick “nice” fractions but forget the parts must add to a whole (e.g. \( \frac{3}{3} \)) for the result to be an integer.
Always, Sometimes, Never
Is the statement always true, sometimes true, or never true? Students should justify their decision with examples.
Adding two positive fractions always gives a result larger than either fraction alone — just as adding two positive whole numbers gives something bigger than both. If \( \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \), then \( \frac{5}{6} > \frac{1}{2} \) and \( \frac{5}{6} > \frac{1}{3} \).
This addresses a critical sense-checking skill. Students who use the “add the tops and add the bottoms” method get answers like \( \frac{1}{2} + \frac{1}{3} = \frac{2}{5} \), but \( \frac{2}{5} \) is smaller than \( \frac{1}{2} \), which should immediately flag that something has gone wrong.
TRUE case: \( \frac{1}{4} + \frac{1}{3} = \frac{7}{12} \), which is proper (\( 7 < 12 \)). FALSE case: \( \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2} \), which is improper (\( 6 > 4 \)). It depends on how large the fractions are — two fractions that are each more than \( \frac{1}{2} \) will always sum to more than 1.
Fractions can only be added or subtracted when the denominators are the same — this is because the denominator tells you the size of each piece. When fractions already share a denominator (like \( \frac{3}{8} + \frac{1}{8} \)), you have a common denominator already.
This is fundamentally different from multiplication and division of fractions, which do not require a common denominator. Mixing up which operations need a common denominator is a very common error.
TRUE case: \( \frac{1}{2} \; – \; \frac{1}{2} = 0 \) (when fractions are equal, swapping doesn’t matter). FALSE case: \( \frac{3}{4} \; – \; \frac{1}{4} = \frac{1}{2} \), but \( \frac{1}{4} \; – \; \frac{3}{4} = -\frac{1}{2} \).
Students often assume subtraction is commutative because addition is. To challenge this, ask them: “Is \( 5 – 2 \) the same as \( 2 – 5 \)? If not, why would fractions be any different?”
Odd One Out
Which is the odd one out? Can you make a case for each one? There’s no single right answer!
Explain the Mistake
Each example contains a deliberate error targeting a common misconception. Can you find where and why the reasoning goes wrong?
Answer: \( \frac{2}{7} \)
Reasoning: “I added the numerators: 1 + 1 = 2. Then I added the denominators: 3 + 4 = 7. So the answer is \( \frac{2}{7} \).”
The student has added numerators and denominators separately. This method treats fractions like two separate whole-number additions.
The correct approach: \( \frac{1}{3} = \frac{4}{12} \) and \( \frac{1}{4} = \frac{3}{12} \), so \( \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \). A quick sense check confirms the student’s answer is wrong: adding a positive fraction to \( \frac{1}{3} \) can’t possibly give something smaller than \( \frac{1}{3} \).
(The answer is correct, but look at the reasoning)
Answer: \( \frac{4}{8} \) ✓
Reasoning: “When you add fractions, you just add the top numbers and the bottom number stays the same. So 3 + 1 = 4, keep the 8, answer is \( \frac{4}{8} \).”
The answer is correct, but the reasoning is dangerously over-generalised. The rule “add the numerators and keep the denominator” only works when the denominators are already the same. The student’s phrasing — “when you add fractions” — suggests they believe this works for all fraction addition.
Applying this faulty rule to \( \frac{1}{2} + \frac{1}{3} \) would lead to disaster (which denominator would they keep?). Getting the right answer with incomplete reasoning is often more dangerous than getting it wrong, as it builds false confidence.
Answer: \( \frac{1}{10} + \frac{2}{10} = \frac{3}{10} \)
Reasoning: “I found a common denominator of 10. Then I just put the fractions over 10, so \( \frac{1}{10} + \frac{2}{10} = \frac{3}{10} \).”
The student correctly identified 10 as the common denominator, but changed the denominators without adjusting the numerators. When converting \( \frac{1}{2} \) to tenths, both the numerator and denominator must be multiplied by the same number: \( \frac{1}{2} = \frac{5}{10} \).
The correct answer is \( \frac{5}{10} + \frac{4}{10} = \frac{9}{10} \). The student treated the conversion as just “rewriting the bottom number”.
Answer: \( 2\frac{2}{5} \)
Reasoning: “I did the whole numbers first: 3 − 1 = 2. Then the fractions: I couldn’t do \( \frac{1}{5} \; – \; \frac{3}{5} \) so I swapped them round and did \( \frac{3}{5} \; – \; \frac{1}{5} = \frac{2}{5} \). So the answer is \( 2\frac{2}{5} \).”
The student has subtracted the smaller fraction from the larger, ignoring the order. This is the fraction equivalent of the classic whole-number error where children compute 32 − 17 as 25.
Correct approach 1 (Improper Fractions): \( 3\frac{1}{5} = \frac{16}{5} \) and \( 1\frac{3}{5} = \frac{8}{5} \), so \( \frac{16}{5} \; – \; \frac{8}{5} = \frac{8}{5} = 1\frac{3}{5} \).
Correct approach 2 (Borrowing): Borrow 1 from the 3. Since \( 1 = \frac{5}{5} \), the first number becomes \( 2 + \frac{5}{5} + \frac{1}{5} = 2\frac{6}{5} \). Now subtract: \( 2\frac{6}{5} \; – \; 1\frac{3}{5} = 1\frac{3}{5} \).