Rational numbers and their decimals

Once integers existed, dividing $7$ apples among $3$ children was no longer "impossible" , but the answer was no longer an integer either. The world of integers had to be enlarged to include fractions, like $\frac{7}{3}$. Together with their negatives and with $0$, these make the rational numbers.

Concept

A number is rational if it can be written in the form

$\dfrac{p}{q}, \quad p, q \in \mathbb{Z}, \quad q \ne 0.$

Every integer is rational (because $5 = \frac{5}{1}$). Every terminating decimal is rational ($0.75 = \frac{3}{4}$). Every recurring decimal is rational too , we will see why in a moment.

Equivalent fractions. Multiplying or dividing the top and bottom of a fraction by the same non-zero number gives an equal fraction: $\frac{3}{4} = \frac{6}{8} = \frac{15}{20}$. The simplest form (where $p$ and $q$ have no common factor) is unique.

On the number line. Every rational number has a definite home between two integers. To plot $\frac{7}{3}$, mark $2$ and $3$, divide the gap into $3$ equal parts, and the marker after the second part is $\frac{7}{3} = 2\frac{1}{3}$.

Decimal expansions of rationals. Convert by long division:

$\tfrac{3}{4} = 0.75, \quad \tfrac{1}{3} = 0.333\dots = 0.\overline{3}, \quad \tfrac{2}{7} = 0.\overline{285714}.$

There are exactly two possibilities:

1. The decimal terminates ($0.75$). This happens precisely when the denominator (in simplest form) has only the prime factors $2$ and $5$. So $\tfrac{3}{8}$ terminates because $8 = 2^3$. $\tfrac{7}{20}$ terminates because $20 = 2^2 \times 5$. But $\tfrac{1}{6}$ does not, because $6 = 2 \times 3$ contains a $3$.

2. The decimal recurs with a repeating block ($0.\overline{3}$, $0.\overline{142857}$). The length of the repeat is at most $q - 1$ where $q$ is the denominator.

So every rational gives a predictable decimal (terminating or recurring), and the rule for which type involves the prime factors of the denominator.

Going backwards. A terminating decimal becomes a fraction easily: $0.625 = \frac{625}{1000} = \frac{5}{8}$. Recurring decimals take a clever trick. To convert $x = 0.\overline{27}$:

$100x = 27.\overline{27}, \quad 100x - x = 27 \Rightarrow 99x = 27 \Rightarrow x = \tfrac{27}{99} = \tfrac{3}{11}.$

Density. Between any two rationals there are infinitely many other rationals. The simplest way to find one between $\tfrac{1}{3}$ and $\tfrac{1}{2}$ is to take their average: $\tfrac{1}{2}(\tfrac{1}{3} + \tfrac{1}{2}) = \tfrac{5}{12}$.

Worked examples

Example 1. Reduce $\dfrac{84}{120}$ to lowest terms.

• $\gcd(84, 120) = 12$.
• $\dfrac{84}{120} = \dfrac{7}{10}$.

Example 2. Without dividing, will $\dfrac{17}{40}$ have a terminating decimal?

• $40 = 2^3 \times 5$. Only $2$'s and $5$'s. So yes, it terminates.
• (In fact $\tfrac{17}{40} = 0.425$.)

Example 3. Convert $0.\overline{6}$ to a fraction.

• Let $x = 0.\overline{6}$. Then $10x = 6.\overline{6}$.
• Subtract: $9x = 6 \Rightarrow x = \tfrac{6}{9} = \tfrac{2}{3}$.

Example 4. Insert a rational number between $\dfrac{2}{5}$ and $\dfrac{3}{5}$.

• Average: $\tfrac{1}{2}\left(\tfrac{2}{5} + \tfrac{3}{5}\right) = \tfrac{1}{2} \cdot 1 = \tfrac{1}{2}$.

Try it yourself

1. Reduce $\dfrac{126}{210}$ to lowest terms.
2. Which of these have terminating decimals: $\dfrac{7}{8}, \dfrac{5}{12}, \dfrac{11}{125}, \dfrac{4}{15}$?
3. Convert $\dfrac{5}{6}$ to its decimal form.
4. Convert $0.\overline{3}$ and $0.\overline{18}$ to fractions.
5. Insert two rationals between $\dfrac{1}{4}$ and $\dfrac{1}{2}$.
6. Plot $\dfrac{11}{4}$ and $-\dfrac{7}{3}$ on a number line.
7. Find a rational between $-\dfrac{2}{3}$ and $-\dfrac{1}{2}$.
8. Is $0.121212\dots$ rational? If yes, write it as a fraction.

Activity

Decimal detective. Pick any prime $p \ne 2, 5$. Compute $\frac{1}{p}$ as a decimal by long division until it starts repeating. Try $p = 7$: $\frac{1}{7} = 0.\overline{142857}$. The repeating block has $6$ digits. Now try $\frac{2}{7}, \frac{3}{7}, \dots, \frac{6}{7}$. You will find the same block, just rotated. This is a beautiful theorem about cyclic numbers, and it works for many primes.