Between any two integers like $1$ and $2$, how many numbers exist? Not one, not ten, but an infinite universe of fractions! Rational numbers possess the property of denseness: between any two rational numbers, there exist infinitely many other rational numbers.
A number is Rational if it can be written as $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \ne 0$. Its decimal expansion is either:
- Terminating: Division ends with remainder 0 (e.g., $\frac{7}{8} = 0.875$). Occurs when prime factors of denominator $q$ are only of the form $2^m \times 5^n$.
- Non-Terminating Recurring (Repeating): Division never terminates, but a block of digits repeats infinitely (e.g., $\frac{1}{3} = 0.333\dots = 0.\bar{3}$).
Problem: Express $0.\bar{47} = 0.474747\dots$ in the form $\frac{p}{q}$.
Step 1: Let $x = 0.474747\dots$ — (Equation 1)
Step 2: Since 2 digits repeat, multiply both sides by $100$: $100x = 47.474747\dots$ — (Equation 2)
Step 3: Subtract Eq (1) from Eq (2): $100x - x = (47.4747\dots) - (0.4747\dots) \implies 99x = 47$
Result: $x = \frac{47}{99}$, where $p = 47, q = 99 \in \mathbb{Z}$ and $q \ne 0$.
In numbers like $0.2\bar{35}$, only the digits $35$ repeat! Students mistakenly multiply directly by $100$ without first shifting the non-repeating digit $2$.
Rule: First multiply by $10$ to get $10x = 2.\bar{35}$, then multiply by $1000$ to get $1000x = 235.\bar{35}$, and subtract: $990x = 233 \implies x = \frac{233}{990}$.
In financial banking systems, currency exchange calculations depend strictly on rational recurring decimal round-off rules to prevent micro-penny truncation fraud (salami slicing).