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Multiplying fractions

Numbers · level 4, lesson 2 of 7

A question to start

Ever since you learned your times tables, multiplying has made numbers bigger: \(3 \times 4 = 12\), which is more than 3 and more than 4. So what happens here?

\[\frac12 \times \frac12 = \frac14\]

The answer is smaller than both numbers you started with. Is multiplication broken for fractions? Not at all. By the end of this lesson, \(\tfrac12 \times \tfrac12 = \tfrac14\) will seem as obvious as \(2 \times 2 = 4\), and you will know exactly when multiplying makes things bigger and when it makes them smaller.

The idea

"Times" means "groups of", and "of" means "times"

\(3 \times 4\) means 3 groups of 4. Keep that meaning and see where it goes.

  • \(3 \times \tfrac25\) means 3 groups of \(\tfrac25\): \(\tfrac25 + \tfrac25 + \tfrac25 = \tfrac65\). Three lots of 2 fifths is 6 fifths. Only the numerator is multiplied, because the pieces are still fifths.
  • \(\tfrac12 \times 8\) means half a group of 8, which is half of 8: 4.
  • \(\tfrac12 \times \tfrac12\) means half of a half. Cut a half in two, and each part is a quarter. So it is \(\tfrac14\).

So when one of the numbers is a fraction, read the \(\times\) sign as "of". \(\tfrac34 \times \tfrac25\) means "three quarters of two fifths". That is why multiplying by a fraction smaller than 1 makes things smaller: you are taking only part of something.

A fraction of a whole number

What is \(\tfrac38\) of 16? Split 16 into 8 equal parts: each part is \(16 \div 8 = 2\). Take 3 of them: \(3 \times 2 = 6\).

So \(\tfrac38\) of \(16 = 6\). In general, to find a fraction of an amount, divide by the bottom and multiply by the top. (You can multiply first and divide after: \(3 \times 16 = 48\), and \(48 \div 8 = 6\). Same answer, bigger numbers in the middle.)

Gravity on Mars is very close to \(\tfrac38\) of Earth's. A backpack that feels like 16 kg on Earth would feel like about 6 kg on Mars.

A fraction of a fraction: the area model

What is \(\tfrac34\) of \(\tfrac25\)? Use a square as the whole.

  1. Cut the square into 5 equal columns and take 2 of them. That is \(\tfrac25\) of the square.
  2. Now take \(\tfrac34\) of that part. Cut the square into 4 equal rows and take 3 of them. The part that is in the chosen columns and the chosen rows is \(\tfrac34\) of \(\tfrac25\).
2/5 (2 columns) 3/4 (3 rows) green: in both = 6 of 20 pieces = 6/20 = 3/10

Count the pieces:

  • The whole square is cut into \(5 \times 4 = 20\) equal pieces. So each piece is \(\tfrac{1}{20}\).
  • The green part is 2 columns by 3 rows: \(2 \times 3 = 6\) pieces.

So \(\tfrac34 \times \tfrac25 = \tfrac{6}{20} = \tfrac{3}{10}\).

Why "multiply the tops and multiply the bottoms" works

The picture shows the rule, and it shows why:

\[\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}\]
  • The denominators multiply because cutting into \(d\) columns and \(b\) rows cuts the whole into \(b \times d\) equal pieces. That is the new size of piece.
  • The numerators multiply because the part we keep is \(c\) columns and \(a\) rows, which is \(a \times c\) pieces.

This is the same rectangle idea as multi-digit multiplication in level 3: area is rows times columns. Here the rectangle just sits inside a unit square.

(Notice the contrast with adding. For adding, the bottoms must match and are not added. For multiplying, nothing needs to match and both are multiplied. They are different operations doing different jobs.)

Simplify first: cancelling

\(\tfrac{5}{6} \times \tfrac{9}{10}\). Multiplying gives \(\tfrac{45}{60}\), and then we must simplify (divide by 15) to get \(\tfrac34\). It is easier to simplify before multiplying:

\[\frac{5}{6} \times \frac{9}{10} = \frac{5 \times 9}{6 \times 10}\]

5 is a factor on top and 10 on the bottom: divide both by 5, leaving 1 and 2. 9 on top and 6 on the bottom: divide both by 3, leaving 3 and 2. So it is \(\tfrac{1 \times 3}{2 \times 2} = \tfrac34\).

Why is this allowed? All four numbers are being multiplied together, top over bottom, so it does not matter which top number sits over which bottom number. Dividing any top factor and any bottom factor by the same number is just simplifying the final fraction early.

When does multiplying make a number smaller?

Compare multiplying 12 by different numbers:

\(12 \times\) 2 1 \(\tfrac12\) \(\tfrac14\) \(\tfrac32\)
answer 24 12 6 3 18
  • Multiply by a number bigger than 1, and the answer is bigger than 12.
  • Multiply by exactly 1, and nothing changes.
  • Multiply by a number between 0 and 1, and the answer is smaller, because you take only part of 12.

So the rule "multiplying makes bigger" was never really a rule. It was only true because you had only multiplied by whole numbers bigger than 1.

Mixed numbers

To multiply mixed numbers, change them to improper fractions first.

\[2\tfrac12 \times 1\tfrac13 = \frac52 \times \frac43 = \frac{20}{6} = \frac{10}{3} = 3\tfrac13\]

Check with the area model. A rectangle \(2\tfrac12\) wide and \(1\tfrac13\) tall splits into four parts: \(2 \times 1 = 2\), \(2 \times \tfrac13 = \tfrac23\), \(\tfrac12 \times 1 = \tfrac12\), \(\tfrac12 \times \tfrac13 = \tfrac16\). Total: \(2 + \tfrac46 + \tfrac36 + \tfrac16 = 2 + \tfrac86 = 3\tfrac26 = 3\tfrac13\). ✓

Worked examples

1. Whole number times a fraction. \(4 \times \tfrac23 = \tfrac83 = 2\tfrac23\) (4 lots of 2 thirds is 8 thirds).

2. Fraction of an amount. \(\tfrac56\) of 42. \(42 \div 6 = 7\), and \(5 \times 7 = 35\).

3. Fraction times fraction. \(\tfrac23 \times \tfrac45 = \tfrac{8}{15}\). Nothing cancels (2 and 4 on top share no factor with 3 and 5 on the bottom). Check the size: \(\tfrac23\) of something less than 1 should be less than \(\tfrac45\). And \(\tfrac{8}{15}\) is about half, which is less than \(\tfrac45\). ✓

4. Cancel first. \(\tfrac{7}{3} \times \tfrac{3}{2}\). The 3s cancel: \(\tfrac{7}{1} \times \tfrac{1}{2} = \tfrac72 = 3\tfrac12\).

5. Mixed numbers. A garden bed is \(2\tfrac34\) m long and \(3\tfrac13\) m wide. What is its area?

\(\tfrac{11}{4} \times \tfrac{10}{3} = \tfrac{110}{12} = \tfrac{55}{6} = 9\tfrac16\) square metres. (Or cancel 2 first: \(\tfrac{11}{2} \times \tfrac{5}{3} = \tfrac{55}{6}\).) Estimate: \(3 \times 3 = 9\). ✓

Try it

  1. \(5 \times \tfrac34\).
  2. \(\tfrac23\) of 27, and \(\tfrac{5}{8}\) of 40.
  3. Draw the area model for \(\tfrac12 \times \tfrac34\) and give the answer.
  4. \(\tfrac35 \times \tfrac{10}{21}\) (cancel first).
  5. Without working it out: is \(\tfrac{7}{8} \times 50\) more or less than 50? What about \(1\tfrac18 \times 50\)?
  6. \(1\tfrac12 \times 1\tfrac12\). Is it more or less than 2?
  7. On Earth a robot weighs as much as a 64 kg mass. On Mars it feels about \(\tfrac38\) as heavy. About how heavy does it feel on Mars?
  8. Harder: find two fractions whose product is \(\tfrac{1}{6}\) and whose sum is \(\tfrac56\).
Answers
  1. \(\tfrac{15}{4} = 3\tfrac34\).
  2. \(27 \div 3 = 9\), \(2 \times 9 = 18\). \(40 \div 8 = 5\), \(5 \times 5 = 25\).
  3. Cut a square into 4 columns and shade 3 (\(\tfrac34\)). Cut it into 2 rows and take 1. The overlap is 3 of the 8 pieces: \(\tfrac38\).
  4. 5 and 10 share 5: they become 1 and 2. 3 and 21 share 3: they become 1 and 7. So \(\tfrac{1 \times 2}{1 \times 7} = \tfrac27\).
  5. \(\tfrac78\) is less than 1, so \(\tfrac78 \times 50\) is less than 50 (it is 43.75). \(1\tfrac18\) is more than 1, so the answer is more than 50 (it is 56.25).
  6. \(\tfrac32 \times \tfrac32 = \tfrac94 = 2\tfrac14\). More than 2. (The area model shows why: \(1 + \tfrac12 + \tfrac12 + \tfrac14\). The little \(\tfrac14\) corner is what people forget when they guess "2".)
  7. \(64 \div 8 = 8\) and \(3 \times 8 = 24\). About 24 kg. (Its mass, the amount of stuff in it, is still 64 kg; only the pull of gravity is weaker.)
  8. Try simple pairs whose sum is \(\tfrac56\): \(\tfrac12 + \tfrac13 = \tfrac56\). Their product: \(\tfrac12 \times \tfrac13 = \tfrac16\). ✓ So \(\tfrac12\) and \(\tfrac13\).

Common mistake

Multiplying mixed numbers part by part. \(2\tfrac12 \times 1\tfrac13\) is not \(2 \times 1 + \tfrac12 \times \tfrac13 = 2\tfrac16\). The area model shows that this uses only two of the four pieces of the rectangle; the pieces \(2 \times \tfrac13\) and \(\tfrac12 \times 1\) are missing. Change to improper fractions first.

Finding a common denominator first. You do not need one to multiply. It is not wrong, just wasted work: \(\tfrac23 \times \tfrac34 = \tfrac{8}{12} \times \tfrac{9}{12} = \tfrac{72}{144} = \tfrac12\), but directly \(\tfrac{6}{12} = \tfrac12\) in one step.

Where it leads