Chapter
Fractions, Decimals, and Periodicity
Terminating decimals, repeating decimals, decimal periods, and fraction-to-decimal structure.
Theory
1. Terminating Decimals
A reduced fraction \(a/b\) has a terminating decimal exactly when the prime factors of \(b\) are only \(2\) and \(5\).
2. Number of Decimal Places
If \(b=2^r5^s\), then \(a/b\) needs at most \(\max(r,s)\) digits after the decimal point.
3. Repeating Decimals
When another prime divides the denominator, long division eventually repeats a remainder.
Examples
Example 1. Terminating Criterion
This is the core theorem for terminating decimals.
Problem.
Explain why a reduced fraction whose denominator is \(2^a5^b\) has a terminating decimal.
Solution.
If \(m=\max(a,b)\), then \(2^a5^b\mid10^m\). So the fraction can be written with denominator \(10^m\).
Example 2. Decimal Places Needed
This turns the criterion into a quick computation.
Problem.
How many digits after the decimal point are needed to write \(\frac{7}{2^3\cdot5^5}\) as a terminating decimal?
Solution.
The denominator divides \(10^5\) but not a smaller power of \(10\). Therefore \(5\) decimal places are needed.
Example 3. Convert a Repeating Decimal
This is the standard algebraic conversion of a repeating decimal.
Problem.
Write \(0.\overline{27}\) as a reduced fraction.
Solution.
If \(x=0.\overline{27}\), then \(100x=27.\overline{27}\). Hence \(99x=27\), so \(x=3/11\).
Example 4. Fiftieth Digit of One Seventh
This is a period-indexing problem.
Problem.
Find the \(50\)-th digit after the decimal point in \(\frac17=0.\overline{142857}\).
Solution.
Since \(50\equiv2\pmod6\), the digit is the second digit of \(142857\), which is \(4\).
Problems
Problems
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Ladders
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