How this instrument works
A fraction terminates in decimal form exactly when its REDUCED denominator's only prime factors are 2 and 5 — the same two prime factors that build 10 itself. 1⁄8 terminates (0.125) because 8 is 2³, built entirely from 2s; 1⁄3 repeats forever (0.333…) because 3 shares no factor with 10 at all, however far the division is carried out.
This page tests that rule with a single closed-form calculation rather than a long division: it computes 10 raised to the 15th power, then finds the remainder when that huge number is divided by the fraction's own reduced denominator. Because 10¹⁵ is itself built entirely from 2s and 5s (fifteen of each), it divides evenly ONLY by denominators also built entirely from 2s and 5s — a remainder of exactly 0 means the fraction terminates, and any nonzero remainder means it repeats.
The reduction step matters: a fraction like 5⁄10 first simplifies to 1⁄2 before the test runs, since it's the REDUCED denominator's prime factors that decide the outcome, not the original, possibly un-simplified one.
- Enter the fraction's numerator into the Numerator field.
- Enter its denominator into the Denominator field.
- Read Test value: exactly 0 means the decimal terminates; any nonzero value means it repeats forever.
- Try a denominator built only from 2s and 5s (like 8, 20, or 25) alongside one that isn't (like 3, 6, or 7) to see both outcomes side by side.
Worked example — testing 1⁄8 and 1⁄3
1⁄8 is already in lowest terms, with denominator 8=2³ — built purely from factors of 2. Dividing 10¹⁵ by 8 leaves a remainder of exactly 0, confirming 1⁄8=0.125 terminates cleanly with no repeating digits.
1⁄3, by contrast, has a denominator of 3, which shares no factor with 10 at all — dividing 10¹⁵ by 3 leaves a nonzero remainder (1), confirming 1⁄3=0.333… repeats forever. 5⁄6 lands in between: reduced, its denominator is 6=2×3, and that stray factor of 3 leaves a nonzero remainder (4), confirming 5⁄6=0.8333… also repeats, despite carrying one valid factor of 2 alongside the problem factor of 3.
Questions
How can you tell if a fraction terminates without doing long division?
Reduce it to lowest terms, then check its denominator's prime factors — if the only prime factors present are 2 and 5 (the prime factors of 10 itself), the decimal terminates; if any other prime factor is present, it repeats forever.
Why does this page use 10 to the 15th power?
10¹⁵ is built entirely from fifteen factors of 2 and fifteen factors of 5, so it divides evenly by any denominator that's ALSO built purely from 2s and 5s (up to that many of each) — turning the terminating-decimal test into one modulo calculation instead of a manual factorization.
Why must the fraction be reduced first?
An un-simplified denominator can carry factors that actually cancel with the numerator — 5⁄10 looks like it has a problematic factor of 5 alone, but once reduced to 1⁄2, its true denominator (2) is perfectly fine, so the test only makes sense applied after reduction.
What's an example of a fraction that repeats?
1⁄3, 1⁄6, 1⁄7, and 1⁄9 all repeat forever in decimal form, since 3, 6, 7, and 9 each carry a prime factor (3 or 7) that no power of 10 can ever absorb.
What's an example of a fraction that terminates?
1⁄2, 1⁄4, 1⁄5, 1⁄8, 1⁄10, and 1⁄20 all terminate cleanly, since 2, 4, 5, 8, 10, and 20 are each built entirely from factors of 2 and 5 alone.