5. Homework solutions (15 Sep) — Fractions

5. Are they equivalent?

Say whether these pairs of fractions are equivalent.

a) $\displaystyle \dfrac{1}{3} \quad \text{and} \quad \dfrac{2}{5}$

Multiplying crosswise:

1 \cdot 5 = 5 \qquad 3 \cdot 2 = 6

The two products do not agree.

They are not equivalent.

b) $\displaystyle \dfrac{4}{5} \quad \text{and} \quad \dfrac{12}{15}$

Multiplying crosswise:

4 \cdot 15 = 60 \qquad 5 \cdot 12 = 60

They agree. Indeed, simplifying by $3$ gives $\dfrac{12}{15} = \dfrac{4}{5}$.

They are equivalent.

Work out the value of $R$ that makes the fractions below equivalent.

a) $\displaystyle \dfrac{42}{28} = \dfrac{18}{R}$

Cross product, then isolate $R$:

42 \cdot R = 28 \cdot 18 \;\Longrightarrow\; R = \dfrac{28 \cdot 18}{42}

We break the numbers into prime factors to avoid multiplying large numbers:

R = \dfrac{(2 \cdot 2 \cdot 7) \cdot (2 \cdot 3 \cdot 3)}{2 \cdot 3 \cdot 7} = 2 \cdot 2 \cdot 3

$$R = 12$$

Check: $\dfrac{42}{28} = \dfrac{3}{2}$ and $\dfrac{18}{12} = \dfrac{3}{2}$.

Order the fractions below from largest to smallest.

a) $\displaystyle \dfrac{27}{18}, \quad \dfrac{31}{36}, \quad \dfrac{3}{5}, \quad \dfrac{13}{9}$

We break the denominators into prime factors:

18 = 2 \cdot 3^2 \qquad 36 = 2^2 \cdot 3^2 \qquad 5 = 5 \qquad 9 = 3^2

The least common multiple takes each prime with its largest exponent:

\operatorname{lcm}(18, 36, 5, 9) = 2^2 \cdot 3^2 \cdot 5 = 180

We scale every fraction up to denominator $180$:

\dfrac{270}{180}, \quad \dfrac{155}{180}, \quad \dfrac{108}{180}, \quad \dfrac{260}{180}

With a shared denominator the numerators decide: $270 > 260 > 155 > 108$.

\dfrac{27}{18} > \dfrac{13}{9} > \dfrac{31}{36} > \dfrac{3}{5}