Real numbers and decimal expansions

Two and only two decimal behaviours for rationals. Take any rational number $\dfrac{p}{q}$, $q \neq 0$, and reduce it to lowest terms (HCF of $p$ and $q$ equal to $1$). Long division of $p$ by $q$ can only ever behave in one of two ways: it terminates, meaning the remainder becomes $0$ after finitely many steps, or it recurs, meaning it never hits remainder $0$ but the sequence of remainders starts repeating, forcing a block of digits to repeat forever. There is no third option, because at every division step the remainder is one of the $q$ possible values $0, 1, 2, \ldots, q-1$; if $0$ never appears, some non-zero remainder must eventually reappear (there are only finitely many choices), and from that point on the division repeats exactly as before.

The denominator test. Whether the decimal terminates depends only on the prime factorisation of the reduced denominator $q$:

where: $p, q$ are integers in lowest terms, $q \neq 0$, and $m, n$ are the powers of the primes $2$ and $5$ in $q$'s factorisation. The reasoning is that a terminating decimal with $k$ digits after the point is really $\dfrac{p}{q} = \dfrac{N}{10^k}$ for some integer $N$, and $10^k = 2^k 5^k$ — so the denominator, once fully reduced, can only ever carry the primes $2$ and $5$. If any other prime (3, 7, 11, ...) survives in $q$ after reduction, no power of $10$ can ever absorb it, and the division is forced to recur.

Converting a recurring decimal back to $\dfrac{p}{q}$. Every non-terminating recurring decimal is rational, and the standard technique is "multiply and subtract": let $x$ equal the decimal, multiply by the power of $10$ that shifts the repeating block exactly one full cycle, then subtract the original equation so the infinite repeating tail cancels, leaving a simple linear equation in $x$.

Successive magnification, extended to irrationals. For a terminating or recurring decimal, zooming into successive intervals of width $10^{-1}, 10^{-2}, 10^{-3}, \ldots$ around the true value eventually nails it down exactly (or the digits settle into their repeating cycle). A non-terminating, non-recurring decimal — an irrational number such as $\sqrt2 = 1.41421356\ldots$ — never lets the digits settle, so the zooming process never "finishes." But at every stage the interval still shrinks by a factor of $10$ and still contains the number, so the process still identifies one and only one point on the number line, just one that requires infinitely many stages rather than finitely many. This is precisely why the real number line — rationals and irrationals together — has no gaps: every possible decimal string, finite or infinite, periodic or not, lands on exactly one point.

Determine, without doing the division, whether $\dfrac{7}{8}$ and $\dfrac{6}{15}$ have terminating or recurring decimal expansions, then find the exact decimal for whichever one terminates.

Step 1. Check $\dfrac{7}{8}$: is it already in lowest terms? HCF$(7,8) = 1$, yes. Factorise the denominator: $8 = 2^3$. Only the prime $2$ appears, so by the denominator test this decimal must terminate.

Step 2. Check $\dfrac{6}{15}$: is it in lowest terms? HCF$(6, 15) = 3$, so it is not reduced. Reduce first: $\dfrac{6}{15} = \dfrac{6 \div 3}{15 \div 3} = \dfrac{2}{5}$. Now factorise the reduced denominator: $5 = 5^1$. Only the prime $5$ appears, so this decimal must also terminate. (Skipping the reduction step and testing $15 = 3 \times 5$ directly would have wrongly suggested a recurring decimal — the $3$ only appeared because the fraction was unreduced.)

Step 3. Find the exact decimal for $\dfrac{7}{8}$ by converting to a power-of-10 denominator: multiply numerator and denominator by $125$ (since $8 \times 125 = 1000$): $\dfrac{7}{8} = \dfrac{7 \times 125}{8 \times 125} = \dfrac{875}{1000} = 0.875$.

Step 4. Find the exact decimal for $\dfrac{2}{5}$ similarly: multiply by $2$ so the denominator becomes $10$: $\dfrac{2}{5} = \dfrac{4}{10} = 0.4$.

Both terminate: $\dfrac{7}{8} = 0.875$ and $\dfrac{6}{15} = \dfrac{2}{5} = 0.4$.

Competitive stretch

A slightly trickier check: does $\dfrac{9}{2^2 \times 5 \times 3}$ terminate? First reduce — $9 = 3^2$ cancels one factor of $3$ in the denominator, leaving $\dfrac{3}{2^2 \times 5} = \dfrac{3}{20}$, whose denominator $20 = 2^2 \times 5$ has only primes $2$ and $5$. So it terminates: $\dfrac{3}{20} = \dfrac{15}{100} = 0.15$. The lesson repeats — always reduce to lowest terms before reading off the denominator's primes.

• Reading the denominator's prime factors before reducing the fraction to lowest terms — an unreduced denominator can carry a stray prime (like the $3$ in $6/15$) that cancels away and never actually affects the decimal.
• Choosing the wrong power of $10$ when converting a recurring decimal to a fraction — the multiplier in each step must match exactly how many digits need to shift past the non-repeating part or past one full repeating cycle, not a guessed value.
• Believing a decimal that merely looks repetitive for a few digits (like $0.454545454\ldots$ truncated on a calculator screen) is automatically recurring — genuine recurring decimals repeat the identical block forever, which calculators can only display, not prove.
• Thinking successive magnification only works for terminating decimals — the same interval-halving-by-ten idea, continued without end, is exactly what gives irrational numbers their unique point on the number line too.

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