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For a reduced fraction $p/q$ with $0<p<q$, $q>1$, and $q,10$ coprime integers, its decimal representation repeats immediately after the decimal point. The shortest repeating block has lengthHere $\mathrm{ord}$ denotes multiplicative order, and $\phi$ denotes Euler's totient function. Changing the numerator preserves the decimal period as long as the fraction remains reduced.^[ref]

For example, long division of $1/7$ gives the remainders $1,3,2,6,4,5,1$, with emitted digits $1,4,2,8,5,7$. ConsequentlyEach remainder determines both the next digit and the next remainder; returning to one restarts the same six steps.

More generally, initialize long division with $r_0=p$. At each step, division with remainder givesso $r_j\equiv p10^j(\mathrm{mod}q)$. Since $p,q$ are coprime integers, the modular inverse of $p$ shows that $r_L=r_0$ exactly when $10^L\equiv 1(\mathrm{mod}q)$. To see why the digit block cannot repeat sooner than the remainders, a purely repeating decimal with a block of $L$ digits represents $D/(10^L-1)$ for an integer $D$. A reduced fraction $p/q$ with that value forces $q\mid 10^L-1$. Hence the shortest digit repeat and the multiplicative order coincide.

Knowing the decimal period predicts how many digits must be computed and stored: one block determines the whole repeating decimal. It also constructs the block directly. With $L={\mathrm{ord}}_q(10)$, the integeris the block, padded on the left with zeros to $L$ digits. For $1/11$, $L=2$ and $D=9$, so the block is $09$ and $1/11=0.\overline{09}$. Reduction matters: $7/21=1/3=0.\overline{3}$ has decimal period one, whereas ${\mathrm{ord}}_{21}(10)=6$.

If the reduced denominator contains prime factors two or five, ten has no multiplicative order modulo that whole denominator. WriteThe decimal preperiod has exactly $k_0$ digits. If $m>1$, the remaining repeating decimal has decimal period ${\mathrm{ord}}_m(10)$. For instance, $28=2^2\cdot 7$ gives $1/28=0.03\overline{571428}$: two prefix digits, then six repeating digits. If $m=1$, the decimal representation is a terminating decimal with exactly $k_0$ fractional places. We use the representation ending in zeros, rather than its alternative ending in nines; for example $1/40=0.025$ terminates after three places.

A short derivation explains both cases. After $k$ digits, multiplying the reduced fraction by $10^k$ and discarding its integer part leaves a reduced fraction whose denominator is $q/\gcd (q,10^k)$. This denominator first becomes coprime to ten at $k=k_0$. It then equals $m$, giving an immediate repeat when $m>1$, or an integer when $m=1$. The same long division argument works for positional notation in any base $b\ge 2$: for a reduced denominator $q>1$ coprime to $b$, the shortest repeating block has length ${\mathrm{ord}}_q(b)$.

A decimal representation is positional notation with radix ten and numeral digits zero through nine. For an integer, the nonnegative part is a finite decimal expansion; a negative integer has a separate minus sign. For example, $205=2\cdot 100+0\cdot 10+5$, whereas $250=2\cdot 100+5\cdot 10+0$. The numeral digits alone do not determine the value: their positions do.

Fractional positions extend the same place value rule to the right of the decimal point, with successive weights $1/10,1/100,1/1000,\dots$. ^[ref] For example,The matching binary representation is $11001101.011_2$, since its fractional part is $0\cdot 2^{-1}+1\cdot 2^{-2}+1\cdot 2^{-3}=3/8$. These representations terminate in both bases; finiteness in one radix does not imply finiteness in another.

Leading zeros in the integer part and trailing zeros in the fractional part leave the value unchanged: $00205.3750=205.375$. A zero appended to the right of an integer string has a different effect: $2050=10\cdot 205$. Keeping the decimal point fixed prevents confusing these two operations.

Every integer has a finite decimal expansion, but a noninteger rational number may need infinitely many fractional numeral digits, as in $1/3=0.333\dots$. A terminating decimal has a zero tail by convention. Infinite decimal strings have a further ambiguity that finite canonical integer strings do not: $0.999\dots =1$, because the geometric series ${\sum }_{j=1}^{\infty }9\cdot 10^{-j}$ sums to one. An infinite decimal tail denotes a limit, not a sufficiently long rounded approximation.