Multiplying Fractions

Multiply the numerators: \( 3 \times 2 = 6 \). Multiply the denominators: \( 5 \times 3 = 15 \). This gives \( \frac{6}{15} \), which simplifies to \( \frac{2}{5} \) (dividing top and bottom by 3). Alternatively, you can cancel the common factor of 3 before multiplying: the 3 in the numerator of \( \frac{3}{5} \) cancels with the 3 in the denominator of \( \frac{2}{3} \), leaving \( \frac{1}{5} \times \frac{2}{1} = \frac{2}{5} \).

You can verify with a diagram: draw a rectangle, shade \( \frac{2}{3} \) of it horizontally, then shade \( \frac{3}{5} \) of it vertically. The overlap — the fraction that is shaded both ways — is 6 out of 15 small rectangles, which is \( \frac{2}{5} \) of the whole shape.

“\( \frac{3}{4} \) of 20” means: divide 20 into 4 equal parts (each part is 5), then take 3 of those parts: \( 3 \times 5 = 15 \). Now calculate \( \frac{3}{4} \times 20 = \frac{3 \times 20}{4} = \frac{60}{4} = 15 \). Both give 15 because they describe exactly the same process — divide by the denominator, multiply by the numerator.

Just like “dos” in Spanish and “deux” in French both translate to “two” in English, the word “of” in math word problems translates directly to the “\( \times \)” symbol. Understanding that “of” IS multiplication connects fraction arithmetic to real-world problems and helps students see that fraction multiplication isn’t a new operation — it formalises something they already understand.

\( 3 \times \frac{1}{4} \) means “3 lots of \( \frac{1}{4} \).” Adding three quarters: \( \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \). And \( 3 \times \frac{1}{4} = \frac{3}{4} \) (multiply the numerator by 3, keep the denominator). Both give \( \frac{3}{4} \), confirming they’re the same.

This connects to how multiplication works with whole numbers — \( 3 \times 7 \) means \( 7 + 7 + 7 \) — and shows students that the same principle applies with fractions. The multiplier tells you how many copies to add up. This is also why \( 3 \times \frac{1}{4} = \frac{3}{4} \): you’re just counting how many quarters you have.

Multiply across: \( \frac{2 \times 3}{3 \times 2} = \frac{6}{6} = 1 \). Alternatively, cancel before multiplying: the 2s cancel and the 3s cancel, leaving \( \frac{1}{1} \times \frac{1}{1} = 1 \). This works because \( \frac{3}{2} \) is the reciprocal of \( \frac{2}{3} \) — you’ve flipped the fraction upside down.

To see why reciprocals always multiply to 1: \( \frac{2}{3} \times \frac{3}{2} \) means “take \( \frac{2}{3} \), then multiply it by \( \frac{3}{2} \).” Multiplying by 3 gives you three times as much (undoing the ÷3 in the denominator), and dividing by 2 gives you half (undoing the ×2 in the numerator). Every step in making the fraction is reversed, bringing you back to 1.

It’s incredibly tempting to just multiply the whole numbers (\( 2 \times 3 = 6 \)) and multiply the fractions (\( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)). But this ignores the “cross-terms”!

To see the truth, convert them to improper fractions first: \( \frac{5}{2} \times \frac{7}{2} = \frac{35}{4} = 8\frac{3}{4} \). You can relate this to expanding double brackets in algebra or using a grid method: \( (2 + 0.5)(3 + 0.5) = 6 + 1 + 1.5 + 0.25 = 8.75 = 8\frac{3}{4} \).

Example: \( \frac{1}{4} \times 8 = \frac{8}{4} = 2 \)

Another: \( \frac{2}{3} \times 6 = \frac{12}{3} = 4 \)

Creative: \( \frac{7}{10} \times 30 = \frac{210}{10} = 21 \) — using a larger, less obvious pairing where the denominator (10) is a factor of the integer (30).

Trap: \( \frac{1}{3} \times 5 = \frac{5}{3} \) — a student might assume any integer multiplied by a fraction gives a whole number. But it only works when the denominator divides evenly into the product of the numerator and integer. Since 5 is not a multiple of 3, the answer is not a whole number.

Example: \( \frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2} \)

Another: \( \frac{5}{6} \times \frac{3}{5} = \frac{15}{30} = \frac{1}{2} \)

Creative: \( \frac{7}{8} \times \frac{4}{7} = \frac{28}{56} = \frac{1}{2} \) — using larger numbers and cross-cancellation (the 7s cancel and the 4 and 8 simplify) to produce the same result.

Trap: \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \), not \( \frac{1}{2} \) — a student might reason “half times half should be a half” because the word “half” appears twice. But multiplying two proper fractions always gives a result smaller than either fraction, so \( \frac{1}{2} \times \frac{1}{2} \) must be less than \( \frac{1}{2} \).

Example: \( \frac{3}{4} \times \frac{4}{3} = \frac{12}{12} = 1 \)

Another: \( \frac{5}{7} \times \frac{7}{5} = \frac{35}{35} = 1 \)

Creative: \( 8 \times \frac{1}{8} = \frac{8}{8} = 1 \) — using a whole number and its unit fraction reciprocal, which students may not recognise as the same pattern.

Trap: \( \frac{2}{5} \times \frac{5}{3} = \frac{10}{15} = \frac{2}{3} \) — a student might think “I just need to swap some numbers around” and swap only one part. The reciprocal of \( \frac{2}{5} \) is \( \frac{5}{2} \), not \( \frac{5}{3} \). You must flip the entire fraction, swapping the numerator and denominator exactly.

Example: \( 1\frac{1}{3} \times \frac{3}{4} = \frac{4}{3} \times \frac{3}{4} = \frac{12}{12} = 1 \)

Another: \( 2\frac{1}{2} \times \frac{4}{5} = \frac{5}{2} \times \frac{4}{5} = \frac{20}{10} = 2 \)

Creative: \( 3\frac{1}{3} \times \frac{9}{10} = \frac{10}{3} \times \frac{9}{10} = \frac{90}{30} = 3 \) — using larger values where the cancellation is less obvious.

Trap: \( 1\frac{1}{2} \times \frac{1}{2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \) — a student might try this because \( 1\frac{1}{2} \) contains a \( \frac{1}{2} \), making it feel like it should pair neatly with \( \frac{1}{2} \). But \( \frac{3}{4} \) is not a whole number. For the product to be whole, the improper fraction form \( \left(\frac{3}{2}\right) \) multiplied by the proper fraction must have a denominator that cancels completely — and \( 2 \times 2 = 4 \), which doesn’t divide into 3.

If both fractions are proper, their numerators are smaller than their denominators: \( a < b \) and \( c < d \). So \( ac < bd \), meaning the product \( \frac{ac}{bd} \) has a numerator smaller than its denominator — it’s proper. For example, \( \frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \frac{3}{10} \) (proper). Even fractions very close to 1, like \( \frac{99}{100} \times \frac{99}{100} = \frac{9801}{10000} \), still give a proper fraction.

Students might expect that multiplying two large proper fractions (like \( \frac{9}{10} \times \frac{9}{10} \)) could “push past 1,” but it can’t — you’d need at least one improper fraction to get a product ≥ 1.

TRUE when the fraction is between 0 and 1 (a proper fraction): \( 8 \times \frac{1}{2} = 4 \), which is smaller than 8. FALSE when the fraction is greater than 1 (an improper fraction): \( 8 \times \frac{3}{2} = 12 \), which is larger than 8. And if the fraction equals 1 (e.g. \( \frac{4}{4} \)), the number stays the same.

This is the fraction version of the “multiplication always makes bigger” misconception. Students need to recognise that the multiplier determines whether the result grows or shrinks: proper fractions shrink, improper fractions grow.

To multiply fractions, you multiply numerators together and denominators together: \( \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \). You never need to find a common denominator. For example, \( \frac{2}{3} \times \frac{4}{5} = \frac{8}{15} \) — just multiply straight across. Common denominators are needed for addition and subtraction, not multiplication.

This is one of the most widespread procedural mix-ups. Students who learn to find common denominators for adding fractions often carry this habit into multiplication, making the process unnecessarily complicated and sometimes producing wrong answers.

TRUE: \( \frac{2}{3} \times \frac{3}{4} = \frac{6}{12} \), and \( \frac{6}{12} \) simplifies to \( \frac{1}{2} \) (the 6 and 12 share a common factor of 6). FALSE: \( \frac{1}{3} \times \frac{1}{5} = \frac{1}{15} \), which is already in simplest form (GCD of 1 and 15 is 1).

Whether simplification is possible depends on whether the numerator and denominator of the product share any common factors. Cross-cancellation before multiplying can avoid large numbers — for example, in \( \frac{2}{3} \times \frac{3}{4} \), you can cancel the 3s first to get \( \frac{2}{4} = \frac{1}{2} \) directly.

Assuming we are dealing with positive values, this is always true. Mixed numbers (like \( 1\frac{1}{2} \) or \( 3\frac{1}{4} \)) are strictly greater than 1. When you multiply a number by something greater than 1, the result is scaled up.

For example, \( 1\frac{1}{2} \times 2\frac{1}{4} = \frac{3}{2} \times \frac{9}{4} = \frac{27}{8} = 3\frac{3}{8} \). The product (\( 3\frac{3}{8} \)) is larger than both \( 1\frac{1}{2} \) and \( 2\frac{1}{4} \). This combats the misconception addressed in Q2, reinforcing that fractions $> 1$ behave just like whole numbers when multiplying.

The student has correctly multiplied the numerators (\( 1 \times 1 = 1 \)) but has added the denominators instead of multiplying them. This is the “addition rule applied to multiplication” misconception — students confuse the procedure for adding fractions (which does involve working with denominators) with the procedure for multiplying them.

The correct method: multiply numerators AND multiply denominators. \( 1 \times 1 = 1 \) and \( 3 \times 4 = 12 \), giving \( \frac{1}{12} \).

The answer is correct, but the reasoning is dangerously wrong. The student’s method — “divide then add” — happens to work here only because \( 10 \div 5 = 2 \) and \( 2 + 2 = 4 \), which coincidentally matches the correct answer. The correct method is “divide then multiply”: \( 10 \div 5 = 2 \), then \( 2 \times 2 = 4 \).

The student’s method fails immediately with different numbers. Try \( \frac{3}{5} \times 10 \): the student would calculate \( 10 \div 5 = 2 \), then \( 3 + 2 = 5 \). But the correct answer is \( 3 \times 2 = 6 \). Getting the right answer with the wrong operation gives false confidence that collapses with harder questions.

The student has written 10 as \( \frac{10}{10} \), but \( \frac{10}{10} = 1 \), not 10. This is the “integer as \( \frac{n}{n} \) instead of \( \frac{n}{1} \)” misconception. To write a whole number as a fraction, the whole number goes in the numerator and 1 goes in the denominator: \( 10 = \frac{10}{1} \).

The correct calculation: \( \frac{3}{5} \times \frac{10}{1} = \frac{3 \times 10}{5 \times 1} = \frac{30}{5} = 6 \). The student has effectively multiplied by 1, not by 10.

The student has ignored the whole number part of the mixed number, treating \( 1\frac{1}{2} \) as if it were just \( \frac{1}{2} \). This is the “multiply parts separately” misconception — the student thinks the whole number and fraction parts of a mixed number can be handled independently.

The correct approach: convert the mixed number to an improper fraction first. \( 1\frac{1}{2} = \frac{3}{2} \). Then multiply: \( \frac{3}{2} \times \frac{2}{3} = \frac{6}{6} = 1 \).

The mistake: The student has confused the procedure for multiplying fractions with a shortcut sometimes used for comparing or dividing fractions. The term “cross-multiplying” is a frequent culprit here.

For multiplication, we multiply straight across: numerators together and denominators together. \( 2 \times 3 = 6 \) and \( 5 \times 4 = 20 \), giving \( \frac{6}{20} \), which simplifies to \( \frac{3}{10} \).