Revisiting rational and irrational number properties

Take any fraction in lowest terms and, without doing a single division, you can already say whether its decimal expansion stops dead or runs forever in a repeating loop. The secret is not in the numerator at all — it is hiding entirely in the prime factors of the denominator. This closing topic of the Class 10 chapter Real Numbers turns that observation into a precise, provable rule: a rational number $\dfrac{p}{q}$ in lowest terms terminates exactly when $q$ 's only prime factors are $2$ and $5$ , and repeats forever otherwise. The same idea then completes the picture of the entire real number line, sorting every real number into one of three decimal types and showing exactly where irrational numbers such as $\sqrt2$ and $\pi$ fit in.

The core insight

A rational number $p/q$ in lowest terms has a terminating decimal expansion if and only if $q$ 's prime factorisation contains no primes other than $2$ and $5$ ; otherwise its decimal expansion is non-terminating repeating.

The core idea

Every real number falls into exactly one of three decimal categories, and rational numbers split cleanly between the first two:

Terminating — the digits stop after finitely many places, like $0.375$ or $0.00416$ .
Non-terminating repeating (recurring) — the digits continue forever but settle into a block that repeats exactly, like $0.3\overline{8} = 0.38888\ldots$
Non-terminating non-repeating — the digits continue forever with no repeating block at all. These are precisely the irrational numbers, such as $\sqrt2 = 1.41421356\ldots$ and $\pi = 3.14159265\ldots$

For a rational number, which of the first two categories it falls into is decided entirely by the denominator once the fraction is written in lowest terms — that is, once $p$ and $q$ share no common factor other than $1$ .

The denominator test. Let $x = \dfrac{p}{q}$ be in lowest terms, with $q \ne 0$ . Then $x$ terminates exactly when

q = 2^m \times 5^n

where: $m$ and $n$ are non-negative integers (either can be zero, and $q = 1$ is allowed as $2^05^0$ ). If $q$ carries any prime factor other than $2$ or $5$ — such as $3$ , $7$ , $11$ , or $13$ — then $x$ is non-terminating repeating instead. The size of the exponents $m$ and $n$ never matters for this classification; only whether $2$ and $5$ are the only primes present in $q$ decides the outcome. The reason $2$ and $5$ are singled out traces back to how decimals are built. Every terminating decimal can be written as a fraction over a power of $10$ — for instance $0.375 = \dfrac{375}{1000}$ — and because $10 = 2 \times 5$ , every power of $10$ factorises as $10^k = 2^k \times 5^k$ . So a denominator that already consists only of $2$ s and $5$ s can always be "completed" into a power of $10$ by multiplying by the missing $2$ s or $5$ s, which is exactly what produces a finite string of decimal digits. Any other prime sitting in the denominator can never be cleared this way, no matter how many $2$ s and $5$ s you throw at it — the division never lands on an exact power of $10$ , so it must go on forever, and it is forced to repeat because only finitely many remainders are possible at each step of long division.

Figure

A branching diagram sorting a rational number into terminating or non-terminating repeating based on whether the lowest-terms denominator has only 2 and 5 as prime factors

Why it works

Derivation AWhy a denominator of only 2s and 5s gives a terminating decimal

The goal is to show, using nothing beyond exponent rules, that whenever a fraction in lowest terms has denominator $q = 2^m5^n$ , it can always be rewritten with a denominator that is a pure power of $10$ , and that this rewriting is exactly what forces the decimal expansion to stop after a finite number of digits.

Step 1. Start from $x = \dfrac{p}{2^m5^n}$ with $p, m, n$ as before, and suppose without loss of generality that $m \ge n$ (the argument is symmetric if $n \ge m$ instead). Let $k = m$ , the larger of the two exponents.

Step 2. Multiply numerator and denominator by $5^{k-n} = 5^{m-n}$ , which is a whole number since $m \ge n$ . This gives $x = \dfrac{p \times 5^{m-n}}{2^m5^n \times 5^{m-n}} = \dfrac{p \times 5^{m-n}}{2^m5^m}$ .

Step 3. Combine the denominator using $2^m5^m = (2 \times 5)^m = 10^m$ , so $x = \dfrac{p \times 5^{m-n}}{10^m}$ . The numerator $p \times 5^{m-n}$ is a whole number, call it $N$ , so $x = \dfrac{N}{10^m}$ .

Step 4. A fraction with denominator $10^m$ is, by definition, obtained from $N$ by shifting the decimal point $m$ places to the left — this always produces a finite string of digits, at most $m$ places long. So $x$ terminates, and the number of decimal places is exactly the larger of $m$ and $n$ (or fewer, if trailing zeros appear). Verified: for $q = 2^0 \times 5^5 = 3125$ , $k = m$ becomes $n = 5$ by symmetry and $x = N/10^5$ with at most $5$ decimal places, matching the worked example below.

Worked example

Without actual division, state whether $\dfrac{13}{3125}$ has a terminating or non-terminating repeating decimal expansion, and if it terminates, find it.

Step 1. Check lowest terms. $13$ is prime and $3125$ is not a multiple of $13$ , so the fraction is already in lowest terms — the denominator test can be applied directly.

Step 2. Factorise the denominator: $3125 = 5^5$ , which is $2^0 \times 5^5$ . Only the primes $2$ and $5$ appear (with $m = 0$ ), so by the rule, $\dfrac{13}{3125}$ has a terminating decimal expansion.

Step 3. Apply the derivation to actually find it. Since $n = 5 > m = 0$ , multiply top and bottom by $2^5 = 32$ : $\dfrac{13}{3125} = \dfrac{13 \times 32}{3125 \times 32} = \dfrac{416}{100000}$ .

Step 4. Read off the decimal by shifting the point five places: $\dfrac{416}{100000} = \mathbf{0.00416}$ , confirming both the prediction and the exact value.

Competitive stretch

Without division, classify $\dfrac{7}{18}$ , and if it repeats, bound the length of the repeating block. Check lowest terms first: $7$ is prime and does not divide $18$ , so the fraction is reduced. Factorise: $18 = 2^1 \times 3^2$ . The factor of $3$ is neither $2$ nor $5$ , so $\dfrac{7}{18}$ is non-terminating repeating. A useful fact beyond the core rule: for a denominator $q$ in lowest terms with no factors of $2$ or $5$ , the repeating block has length at most $q - 1$ digits (a consequence of long division having only $q-1$ possible nonzero remainders); here that bounds the block at $17$ digits, though in practice it is much shorter — indeed $\dfrac{7}{18} = 0.3\overline{8} = 0.38888\ldots$ , a block of length one.

Common traps

Applying the rule to a fraction that is not yet in lowest terms. $\dfrac{6}{15}$ has denominator $15 = 3 \times 5$ , which looks non-terminating, but in lowest terms it is $\dfrac{2}{5}$ , which terminates. Always simplify first.
Thinking the values of $m$ or $n$ affect whether the decimal terminates. They do not — only whether $2$ and $5$ are the sole primes in $q$ matters, however large those exponents are.
Confusing "terminates" with "short." $\dfrac{1}{2^{10}} = \dfrac{1}{1024}$ terminates but runs to ten decimal places; terminating is about the type of expansion, not its length.
Forgetting that an integer denominator like $q = 1$ trivially counts as $2^05^0$ and terminates with zero decimal digits, and that the rule applies only to rational numbers — irrational numbers never terminate or repeat at all.

Ask the tutor: Want to test the denominator rule on a fraction of your own, or see why every power of 10 factors into only 2s and 5s? Ask the AI tutor to work through it with you.

Why does only the prime pair 2 and 5 matter, and not say 2 and 3

Because our number system is base ten, and $10 = 2 \times 5$ . A terminating decimal is always a fraction over some power of $10$ , and $10^k$ factorises only into $2^k$ and $5^k$ — no other primes ever appear in a power of ten. So a denominator built purely from $2$ s and $5$ s can always be topped up with the missing factors to become a power of $10$ , while any other prime, like $3$ , can never be cleared this way and forces the division to run forever.

Does this rule tell me what the repeating block actually looks like, or just that one exists

The rule as stated only predicts termination versus repetition from the denominator's prime factors — it does not hand you the digits or the block length directly. To find the actual repeating digits you still need to carry out the long division, watching the remainders cycle. What the theory does guarantee is an upper bound: for denominator $q$ (with no factors of $2$ or $5$ ) the block length is at most $q - 1$ digits, since only that many distinct nonzero remainders exist.

Is every non-terminating decimal automatically irrational

No, and this is a common slip. A non-terminating decimal is irrational only if it also never settles into a repeating block. $0.38888\ldots$ is non-terminating but it does repeat (the digit $8$ forever), so it is rational — equal to $\dfrac{7}{18}$ . True irrational numbers like $\sqrt2$ or $\pi$ are non-terminating and non-repeating: no block of digits ever recurs exactly, no matter how far you compute. Always check both conditions before calling a decimal irrational.