Decimal Expansions of Real Numbers

Every real number has a decimal expansion. Some look polite , $\dfrac{1}{4} = 0.25$ , end of story. Others wander forever but with a pattern, like $\dfrac{1}{7} = 0.\overline{142857}$ . A third kind wanders forever with no pattern at all, like $\pi$ or $\sqrt{2}$ . This lesson sorts decimals into three boxes and gives you tools to convert between fractions and decimals in either direction.

Definitions

A decimal terminates if it has a last non-zero digit, e.g. $0.5, 0.375, 12.6$ .

A decimal is non-terminating recurring if it goes on forever but, after some point, a fixed block of digits repeats. We write the repeating block under a bar: $\frac{1}{3} = 0.\overline{3}, \qquad \frac{1}{7} = 0.\overline{142857}, \qquad \frac{1}{6} = 0.1\overline{6}.$

A decimal is non-terminating non-recurring if it goes on forever with no eventually repeating block. Example: $0.1010010001\ldots$ , where the gaps grow one zero at a time.

Three boxes, one dictionary

Fundamental dictionary.

A real number is rational if and only if its decimal expansion either terminates or is eventually recurring. A real number is irrational if and only if its decimal expansion is non-terminating and non-recurring.

This single sentence is the most useful theorem in the chapter.

Which fractions terminate? A fraction $\dfrac{p}{q}$ in lowest terms terminates if and only if the denominator $q$ has no prime factors other than $2$ and $5$ . Examples:

$\dfrac{7}{8} = \dfrac{7}{2^3} = 0.875$ , terminates.
$\dfrac{17}{40} = \dfrac{17}{2^3 \cdot 5} = 0.425$ , terminates.
$\dfrac{1}{6} = \dfrac{1}{2 \cdot 3}$ , does not terminate (denominator has a $3$ ).
$\dfrac{1}{7}$ , does not terminate (denominator has a $7$ ).

Why $2$ s and $5$ s? A terminating decimal with $k$ digits after the point can be written as an integer over $10^k = 2^k \cdot 5^k$ . So in lowest terms the denominator can only use primes $2$ and $5$ .

Recurring $\to$ fraction. There is a beautiful trick to convert any recurring decimal to a fraction. Let the decimal be $x$ , multiply by a power of $10$ that shifts the repeat once to the left, subtract, and solve for $x$ .

For $x = 0.\overline{3}$ : $10x = 3.\overline{3}$ , so $10x - x = 3$ , $x = \dfrac{3}{9} = \dfrac{1}{3}$ .

For $x = 0.\overline{47}$ : $100x = 47.\overline{47}$ , so $99x = 47$ , $x = \dfrac{47}{99}$ .

For mixed cases like $x = 0.1\overline{27}$ : first shift past the non-repeating part. $10x = 1.\overline{27}$ and $1000x = 127.\overline{27}$ , so $1000x - 10x = 126$ , giving $x = \dfrac{126}{990} = \dfrac{7}{55}$ .

A famous oddity. $0.\overline{9} = 1$ . Apply the same trick: let $x = 0.\overline{9}$ . Then $10x = 9.\overline{9} = 9 + x$ , so $9x = 9$ and $x = 1$ . This is not a paradox , it tells you that some real numbers have two decimal names. The other name of $1$ is $0.\overline{9}$ ; the other name of $0.25$ is $0.24\overline{9}$ .

Irrational decimals. You cannot multiply an irrational decimal by a clean power of $10$ to repeat the pattern, because there is no pattern. That is the entire content of the second half of the dictionary.

Worked examples

Example 1. Without dividing, decide which of $\dfrac{13}{3125}, \dfrac{11}{12}, \dfrac{29}{343}, \dfrac{23}{50}$ terminate.

Factor denominators: $3125 = 5^5$ (only $5$ s; terminates); $12 = 2^2 \cdot 3$ (has $3$ ; non-terminating); $343 = 7^3$ (has $7$ ; non-terminating); $50 = 2 \cdot 5^2$ (only $2$ s and $5$ s; terminates).

Example 2. Express $0.\overline{6}$ as a fraction.

Let $x = 0.\overline{6}$ . Then $10x = 6.\overline{6} = 6 + x$ , so $9x = 6$ , $x = \dfrac{2}{3}$ .

Example 3. Express $0.5\overline{8}$ as a fraction.

Let $x = 0.5\overline{8}$ . Then $10x = 5.\overline{8}$ and $100x = 58.\overline{8}$ , so $100x - 10x = 53$ , giving $x = \dfrac{53}{90}$ .

Example 4. Find the decimal expansion of $\dfrac{1}{17}$ , at least the first $16$ digits, and explain why the expansion must repeat.

Long division gives $\dfrac{1}{17} = 0.\overline{0588235294117647}$ , a $16$ -digit cycle. The expansion must eventually repeat because at each step of long division the remainder lies between $1$ and $16$ . With only $16$ possible non-zero remainders, a remainder must recur within $16$ steps, and once it does the digits repeat.

Example 5. Write three numbers whose decimal expansions are non-terminating non-recurring.

Choices: $0.101001000100001\ldots$ , $0.123456789101112\ldots$ (Champernowne-style), and $\pi - 3 = 0.14159265\ldots$ . Each fails to repeat eventually, so each is irrational.

Try it yourself

Without dividing, decide whether each terminates: $\dfrac{8}{125}, \dfrac{7}{30}, \dfrac{14}{875}, \dfrac{9}{16}, \dfrac{2}{15}$ .
Convert each recurring decimal to a fraction: $0.\overline{5}, 0.\overline{45}, 0.4\overline{3}, 0.23\overline{47}$ .
Find the decimal expansion of $\dfrac{1}{13}$ and identify the repeating block.
Find one non-terminating non-recurring decimal between $0.4$ and $0.41$ .
Why is $\dfrac{1}{17}$ guaranteed to have a repeating block of length at most $16$ ?
Show that $0.\overline{9} = 1$ using a fraction argument.
Without using a calculator, decide whether $\dfrac{99}{800}$ terminates.
The decimal $0.142857142857\ldots$ equals which rational?
Is $0.1212212221\ldots$ rational or irrational? Justify.
If the decimal expansion of $\dfrac{p}{q}$ (lowest terms) terminates, what can you say about the prime factors of $q$ ?

Pitfalls / Insight

Lowest terms matters. $\dfrac{6}{30}$ looks bad but equals $\dfrac{1}{5}$ , which terminates.
Bars cover only the repeating block. $0.1\overline{6}$ means $0.16666\ldots$ , not $0.161616\ldots$ .
"Pattern" is not the same as "recurring". $0.123456789101112\ldots$ has a clear pattern but does not repeat a fixed block; it is irrational.

Insight. A long-division remainder remembers everything. If after some step the same remainder reappears, the digits from that step on must repeat , and remainders are bounded by the divisor. That is why every rational must terminate or recur.