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3r ESO · Rational numbers Notes 13 Oct 2026

Multiplying, dividing and combined operations

Multiply straight across, divide by cross-multiplying (or by the reciprocal), the sign rule and the order of combined operations.

Multiplication

To multiply two fractions, multiply straight across: numerator by numerator and denominator by denominator.

$$\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{a \cdot c}{b \cdot d}$$

Example.

$$\dfrac{2}{3} \cdot \dfrac{5}{7} = \dfrac{2 \cdot 5}{3 \cdot 7} = \dfrac{10}{21}$$

Example. When the numbers are large, it pays to break them into prime factors and simplify before multiplying:

$$\dfrac{30}{48} \cdot \dfrac{15}{27} \cdot \dfrac{6}{10} = \dfrac{2 \cdot 3 \cdot 5}{2^4 \cdot 3} \cdot \dfrac{3 \cdot 5}{3^3} \cdot \dfrac{2 \cdot 3}{2 \cdot 5} = \dfrac{2^2 \cdot 3^3 \cdot 5^2}{2^5 \cdot 3^4 \cdot 5} = \dfrac{5}{2^3 \cdot 3} = \dfrac{5}{24}$$
$$\dfrac{12}{28} \cdot \dfrac{49}{25} \cdot \dfrac{75}{9} = \dfrac{2^2 \cdot 3}{2^2 \cdot 7} \cdot \dfrac{7^2}{5^2} \cdot \dfrac{3 \cdot 5^2}{3^2} = \dfrac{2^2 \cdot 3^2 \cdot 5^2 \cdot 7^2}{2^2 \cdot 3^2 \cdot 5^2 \cdot 7} = \dfrac{7}{1} = 7$$

Division

To divide two fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ there are two ways, which give the same result:

→ Cross-multiply:

$$\dfrac{a}{b} : \dfrac{c}{d} = \dfrac{a \cdot d}{b \cdot c}$$

→ Multiply by the reciprocal of the second fraction:

$$\dfrac{a}{b} : \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c} = \dfrac{a \cdot d}{b \cdot c}$$

Example.

$$\dfrac{3}{4} : \dfrac{9}{8} = \dfrac{3 \cdot 8}{4 \cdot 9} = \dfrac{24}{36} = \dfrac{2}{3}$$

The sign rule

When you multiply two numbers, the sign of the result is:

$+$$\cdot$$+$$=$$+$
$+$$\cdot$$-$$=$$-$
$-$$\cdot$$+$$=$$-$
$-$$\cdot$$-$$=$$+$

The same goes for division.

Example. First decide the sign: there are three negative factors, and an odd number of minus signs gives a negative. Then divide as usual, multiplying by the reciprocal:

$$\left(-\dfrac{3}{4}\right) : \left(-\dfrac{5}{8}\right) : \left(-\dfrac{2}{5}\right) = -\dfrac{3}{4} : \dfrac{5}{8} : \dfrac{2}{5} = -\dfrac{3}{4} \cdot \dfrac{8}{5} \cdot \dfrac{5}{2} = -\dfrac{3 \cdot 2^3 \cdot 5}{2^2 \cdot 5 \cdot 2} = -3$$

Example. Here there are two negatives, so the result is positive. Operations of the same level are done from left to right:

$$\left(-\dfrac{1}{2}\right) \cdot \left(-\dfrac{1}{3}\right) : \dfrac{1}{4} \cdot \dfrac{1}{5} = \dfrac{1}{2} \cdot \dfrac{1}{3} \cdot \dfrac{4}{1} \cdot \dfrac{1}{5} = \dfrac{4}{30} = \dfrac{2}{15}$$

Combined operations

When there are several operations, the order is:

  1. Brackets, from the inside out: $(\;)$, $[\;]$
  2. Powers and roots: $a^n$, $\sqrt{\;}$
  3. Multiplication and division, from left to right: $\cdot$, $:$
  4. Addition and subtraction: $+$, $-$

Example. Multiplication comes before addition:

$$\dfrac{2}{3} + \dfrac{1}{3} \cdot \dfrac{7}{3} = \dfrac{2}{3} + \dfrac{7}{9} = \dfrac{6}{9} + \dfrac{7}{9} = \dfrac{13}{9}$$

Example. With brackets: first the division inside, then the subtraction, then the multiplication and finally the addition:

$$\dfrac{2}{3} + \dfrac{1}{3} \cdot \left(\dfrac{5}{2} : \dfrac{3}{4} - 1\right) = \dfrac{2}{3} + \dfrac{1}{3} \cdot \left(\dfrac{20}{6} - 1\right) = \dfrac{2}{3} + \dfrac{1}{3} \cdot \left(\dfrac{10}{3} - \dfrac{3}{3}\right) = \dfrac{2}{3} + \dfrac{1}{3} \cdot \dfrac{7}{3} = \dfrac{13}{9}$$