Rational Numbers and Their Decimal Expansions

Two Kinds of Decimal Expansions

Think about what happens when you divide to convert a fraction into a decimal. Sometimes the division ends (terminates), and sometimes it goes on forever, repeating a block of digits.

• $\dfrac{1}{2} = 0.5$ — terminating (the division stops).
• $\dfrac{3}{4} = 0.75$ — terminating.
• $\dfrac{1}{3} = 0.333\dots = 0.\overline{3}$ — non-terminating recurring (the digit 3 repeats).
• $\dfrac{2}{11} = 0.1818\dots = 0.\overline{18}$ — non-terminating recurring (the block 18 repeats).

Key Point: Every rational number has a decimal expansion that is either terminating or non-terminating recurring. It can never be non-terminating non-recurring (that is the signature of an irrational number).

The beautiful result of this section: you can tell which type a fraction will give just by looking at its denominator — no long division needed.

[Board Important] The exam loves questions like 'Without actually dividing, state whether $\dfrac{13}{3125}$ has a terminating decimal expansion.' Master the denominator rule below.

The Terminating Decimal Rule

Here is the central theorem of this section.

Theorem: Let $x = \dfrac{p}{q}$ be a rational number in its simplest form (i.e. $p$ and $q$ are co-prime). Then $x$ has a terminating decimal expansion if and only if the denominator $q$ can be written in the form $$q = 2^n \times 5^m$$ where $n$ and $m$ are non-negative integers (0, 1, 2, …).

In plain words: after simplifying the fraction, look at the denominator's prime factors. If the only primes are 2's and 5's, the decimal terminates. If any other prime (3, 7, 11, …) appears, it does NOT terminate — it recurs.

Why $2$ and $5$?

Because our number system is base 10, and $10 = 2 \times 5$. A denominator made only of 2's and 5's can be turned into a power of 10, which gives a clean terminating decimal.

Common Mistake: You MUST simplify the fraction first. For example $\dfrac{6}{15}$ looks like it has a denominator 15 = 3 × 5 (non-terminating), but $\dfrac{6}{15} = \dfrac{2}{5}$, which terminates!

[Board Important] Always reduce to lowest terms before applying the rule. This single step trips up many students.

Converting to a Power of Ten

When a fraction does terminate, there is a neat way to find its decimal: make the denominator a power of 10.

Example: $\dfrac{13}{125}$

1. $125 = 5^3$ (only the prime 5 → terminates).
2. Multiply numerator and denominator by $2^3 = 8$ to balance the 5's: $\dfrac{13}{125} = \dfrac{13 \times 8}{125 \times 8} = \dfrac{104}{1000}$.
3. So $\dfrac{13}{125} = 0.104$.

Counting decimal places

If $q = 2^n \times 5^m$, the number of decimal places equals the larger of $n$ and $m$ (i.e. $\max(n, m)$). For $\dfrac{13}{125} = \dfrac{13}{2^0 \times 5^3}$, we have $\max(0, 3) = 3$ decimal places — and indeed 0.104 has 3 places.

Key Point: To convert, supply whichever of 2 or 5 is 'missing' so the denominator becomes a power of 10.

[Board Important] A quick-scoring question type: 'How many decimal places will $\dfrac{17}{8}$ have?' Since $8 = 2^3$, the answer is $\max(3,0) = 3$ places ($\dfrac{17}{8} = 2.125$).

The Non-Terminating Recurring Case

If, after simplifying, the denominator has any prime factor other than 2 or 5, the decimal expansion is non-terminating recurring.

Examples

• $\dfrac{1}{3}$: denominator 3 → non-terminating recurring ($0.\overline{3}$).
• $\dfrac{2}{7}$: denominator 7 → non-terminating recurring ($0.\overline{285714}$).
• $\dfrac{29}{343}$: $343 = 7^3$ → non-terminating recurring.

Key Point: Such a number is still rational — it just doesn't terminate. A repeating block always means rational. Don't confuse 'non-terminating' with 'irrational'; only non-terminating AND non-recurring is irrational.

Summary table

Denominator (in lowest terms)	Decimal type
Only 2's and 5's ($2^n 5^m$)	Terminating
Contains any other prime	Non-terminating recurring

[Board Important] A very common 1-mark MCQ: classify $\dfrac{64}{455}$. Since $455 = 5 \times 7 \times 13$ contains 7 and 13, it is non-terminating recurring.

Solved Examples

Example 1: Terminating or not?

Without dividing, state whether $\dfrac{13}{3125}$ has a terminating decimal expansion.

Solution:

1. Factorise the denominator: $3125 = 5^5$.
2. The only prime is 5, which is of the form $2^0 \times 5^5$.
3. So the fraction terminates.

Final Answer: Terminating.

Takeaway: Only 2's and 5's in the denominator ⇒ terminating.

Example 2: A non-terminating example

State whether $\dfrac{64}{455}$ has a terminating or non-terminating recurring expansion.

Solution:

1. Factorise the denominator: $455 = 5 \times 7 \times 13$.
2. Besides 5, the primes 7 and 13 are present.
3. Since primes other than 2 and 5 occur, it does not terminate.

Final Answer: Non-terminating recurring.

Takeaway: Any prime other than 2 or 5 in the denominator forces recurrence.

Example 3: Simplify first!

Does $\dfrac{6}{15}$ have a terminating decimal expansion?

Solution:

1. Simplify: $\dfrac{6}{15} = \dfrac{2}{5}$ (dividing top and bottom by 3).
2. Denominator is $5 = 2^0 \times 5^1$ — only the prime 5.
3. So it terminates: $\dfrac{2}{5} = 0.4$.

Final Answer: Terminating, equal to 0.4.

Takeaway: Always reduce to lowest terms before checking the denominator. The unsimplified 15 would have misled you.

Example 4: Convert by making a power of 10

Express $\dfrac{17}{8}$ as a decimal without long division.

Solution:

1. $8 = 2^3$, so multiply top and bottom by $5^3 = 125$.
2. $\dfrac{17}{8} = \dfrac{17 \times 125}{8 \times 125} = \dfrac{2125}{1000}$.
3. So $\dfrac{17}{8} = 2.125$.

Final Answer: $2.125$.

Takeaway: Supply the missing prime (here 5's) to make the denominator $10^3$.

Example 5: Number of decimal places

How many places of decimals will $\dfrac{23}{2^3 \times 5^2}$ have?

Solution:

1. The denominator is $2^3 \times 5^2$, so $n = 3$, $m = 2$.
2. Number of decimal places $= \max(n, m) = \max(3, 2) = 3$.

Final Answer: 3 decimal places. (Indeed $\dfrac{23}{200} = 0.115$.)

Takeaway: Decimal places $= \max(n, m)$ when $q = 2^n \times 5^m$.

Example 6: Classify several fractions

Classify as terminating (T) or non-terminating recurring (NT): $\dfrac{3}{8}$, $\dfrac{7}{12}$, $\dfrac{9}{125}$.

Solution:

1. $\dfrac{3}{8}$: $8 = 2^3$ → only 2's → T.
2. $\dfrac{7}{12}$: $12 = 2^2 \times 3$ → contains 3 → NT.
3. $\dfrac{9}{125}$: $125 = 5^3$ → only 5's → T.

Final Answer: T, NT, T.

Takeaway: Factorise each denominator; a stray 3 (or 7, 11, …) means non-terminating.

Example 7: Find the smallest multiplier to terminate

The fraction $\dfrac{1}{20}$ — express it as a terminating decimal.

Solution:

1. $20 = 2^2 \times 5$, so $n = 2$, $m = 1$; only 2's and 5's → terminates.
2. Make a power of 10: multiply top and bottom by $5^{1}$ to balance: $\dfrac{1}{20} = \dfrac{1 \times 5}{20 \times 5} = \dfrac{5}{100} = 0.05$.

Final Answer: $0.05$.

Takeaway: Multiply by whichever prime is in short supply (here one more 5) to reach $10^{\max(n,m)}$.

Example 8: Work backwards from a decimal

Write the terminating decimal $0.375$ as a fraction in lowest terms, and verify its denominator is of the form $2^n 5^m$.

Solution:

1. $0.375 = \dfrac{375}{1000}$.
2. Simplify: $\dfrac{375}{1000} = \dfrac{3}{8}$ (divide by 125).
3. Denominator $8 = 2^3 = 2^3 \times 5^0$ — only 2's. ✓

Final Answer: $0.375 = \dfrac{3}{8}$, denominator $2^3$.

Takeaway: Every terminating decimal, in lowest terms, has a denominator of the form $2^n 5^m$.

Example 9: A tricky disguised denominator

Does $\dfrac{35}{50}$ terminate?

Solution:

1. Simplify: $\dfrac{35}{50} = \dfrac{7}{10}$ (divide by 5).
2. $10 = 2 \times 5$ — only 2's and 5's → terminates.
3. $\dfrac{7}{10} = 0.7$.

Final Answer: Terminating, equal to 0.7.

Takeaway: Even though 50 = 2 × 5² already looks fine, simplifying makes the answer obvious and avoids errors.

Example 10: Recurring decimal to fraction (link back)

Express $0.\overline{36}$ as a fraction in lowest terms, and confirm its denominator is NOT of the form $2^n 5^m$.

Solution:

1. Let $x = 0.363636\dots$ Then $100x = 36.3636\dots$
2. Subtract: $100x - x = 36 \Rightarrow 99x = 36 \Rightarrow x = \dfrac{36}{99} = \dfrac{4}{11}$.
3. Denominator $11$ is a prime other than 2 or 5 — consistent with the number being non-terminating recurring.

Final Answer: $0.\overline{36} = \dfrac{4}{11}$.

Takeaway: Recurring decimals are rational, and their reduced denominators always contain a prime other than 2 or 5.