Exercises — Rational numbers — definition, decimal expansion (terminating - repeating)

$\varphi (q)$? (Euler's theorem sketch) Take $\frac{1}{q}$ with $\gcd (q,10)=1$ (no 2s or 5s in $q$). In long division the remainders follow the rule $r\mapsto 10r\mathrm{mod}q$ (see Modular arithmetic). Starting from $r=1$, after $k$ steps the remainder is $10^k\mathrm{mod}q$. The decimal repeats exactly when a remainder returns to $1$ — that is, when $10^k\equiv 1(\mathrm{mod}q)$. The smallest such $k$ is called the order of 10 mod $q$, and it equals the repetend length.

Now Euler's theorem says: for any $a$ coprime to $q$, $a^{\varphi (q)}\equiv 1(\mathrm{mod}q).$ In particular $10^{\varphi (q)}\equiv 1(\mathrm{mod}q)$, so $k=\varphi (q)$ is one exponent that works. The smallest working exponent (the order) must divide every working exponent — because if the order is $k$ and some other exponent $m$ also gives $10^m\equiv 1$, write $m=k\cdot t+s$ with remainder $0\le s<k$; then $10^s\equiv 1$ too, and since $k$ is smallest we're forced to $s=0$, i.e. $k\mid m$. Taking $m=\varphi (q)$ gives $k\mid \varphi (q)$. That is precisely "repetend length divides $\varphi (q)$." ∎

The rule (from the parent note): write the denominator in lowest terms, then factor it. If only the primes 2 and 5 appear → terminates. Any other prime → repeats.

• $\frac{3}{20}$: $\gcd (3,20)=1$ (already lowest terms), and $20=2^2\cdot 5$. Only 2 and 5 → terminates.
• $\frac{7}{15}$: $\gcd (7,15)=1$, and $15=3\cdot 5$. The prime $3$ appears → repeats.
• $\frac{9}{40}$: $\gcd (9,40)=1$, and $40=2^3\cdot 5$. Only 2 and 5 → terminates.
• $\frac{11}{6}$: $\gcd (11,6)=1$, and $6=2\cdot 3$. The prime $3$ appears → repeats.

Trap-avoidance first: reduce before factoring. As given, $48=2^4\cdot 3$, which looks like it repeats. But the fraction is not in lowest terms — here $\gcd (15,48)=3$, so we can cancel: $\frac{15}{48}=\frac{15÷3}{48÷3}=\frac{5}{16},16=2^4.$ Now only the prime 2 appears → terminates (indeed $\frac{5}{16}=0.3125$).

$16=2^4$, so it terminates. To reach a power of 10 we need matching factors of 5. We have $2^4$ and no $5$s, so multiply top and bottom by $5^4=625$:

$11$ is a prime other than 2 or 5 → repeats. Track the remainders:

2 ÷ 11:
 remainder 2 → 20 ÷ 11 = 1 remainder 9
 remainder 9 → 90 ÷ 11 = 8 remainder 2  ← remainder 2 again!

Once remainder 2 returns, the block $18$ repeats. The repetend (repeating block) is "$18$": $\frac{2}{11}=0.\overline{18}.$ Repetend length $=2$. (Consistent with theory: $\varphi (11)=10$ and $2\mid 10$.)

Let $x=0.\overline{45}$. Repetend length is 2, so multiply by $10^2=100$: $100x=45.\overline{45}.$ Subtract $x=0.\overline{45}$ so the infinite tails cancel exactly: $100x-x=45\Rightarrow 99x=45\Rightarrow x=\frac{45}{99}=\frac{5}{11}.$

Non-repeating part = "16" (2 digits); repeating part = "$6$" (1 digit). We use two multipliers, and here is exactly why those two: one multiplier must slide the decimal point to just past the non-repeating prefix, the other must slide it one full repetend further, so that when we subtract, the two infinite tails sit digit-for-digit on top of each other and vanish.

Multiplier A ($10^2=100$, because the prefix "16" is 2 digits) moves the point past the prefix: $100x=16.\underline{6666\dots }$ Multiplier B ($10^{2+1}=1000$, prefix 2 digits plus repetend 1 digit) moves it one repetend further: $1000x=166.\underline{6666\dots }$ Line them up and watch the tails:

The two underlined tails are identical — both are $0.6666\dots$ starting at the same place — so subtracting them gives exactly $0$. Only the finite whole-number parts survive: $166-16=150$. Hence $900x=150\Rightarrow x=\frac{150}{900}=\frac{1}{6}.$ Check: $\frac{1}{6}=0.1\overline{6}$, which is the same as $0.16\overline{6}$. ✓ The rule this demonstrates: prefix length $m$ → subtract $10^{m}x$ from $10^{m+n}x$, where $n$ is the repetend length. Here $m=2,n=1$, giving $100x$ and $1000x$.

First, recall exactly how long division turns a remainder into a decimal digit. At each step you take the current remainder $r$, append a zero (i.e. multiply by 10), then ask "how many $7$s fit?"

• The digit you write down is the whole-number quotient: $\text{digit}=\lfloor 10r/7\rfloor$ (the $\lfloor \rfloor$ means "round down to a whole number").
• The next remainder is what's left over: $r\mapsto 10r\mathrm{mod}7$ (the leftover after dividing $10r$ by 7 — see Modular arithmetic).

Run it for $\frac{1}{7}$ starting at $r=1$:

Reading the middle column top-to-bottom spells $142857$ — exactly $0.\overline{142857}$.

Key insight: the same remainder wheel $1\to 3\to 2\to 6\to 4\to 5\to 1$ drives $\frac{k}{7}$ for every $k$; a different numerator just starts the division at a different remainder. For $\frac{3}{7}$ the first remainder is $3$, which sits at the second slot of the wheel. Reading the digit column starting from that slot ($4,2,8,5,7,1$) gives: $\frac{3}{7}=0.\overline{428571}.$ Verify quickly: $\frac{3}{7}\approx 0.42857\dots$ ✓

Predict: $13$ is prime and $\ne 2,5$, so $\frac{1}{13}$ repeats. The length must divide $\varphi (13)=12$ (see the sketch in the intro). Candidate lengths: $1,2,3,4,6,12$. So it is at most 12; we must divide to find which divisor.

Confirm with the same two-column table as before (digit $=\lfloor 10r/13\rfloor$, next $r=10r\mathrm{mod}13$), starting at $r=1$:

The remainder $1$ returns after 6 steps, and the digit column spells $076923$: $\frac{1}{13}=0.\overline{076923},\text{length }6.$ And indeed $6\mid 12$. ✓ The length is a divisor of $\varphi (13)$, not necessarily $\varphi (13)$ itself.

Split off the whole number: $2.\overline{3}=2+0.\overline{3}$. For $y=0.\overline{3}$: $10y=3.\overline{3}$, so $9y=3$, giving $y=\frac{1}{3}$. Therefore $2.\overline{3}=2+\frac{1}{3}=\frac{7}{3}.$ Sanity check: $\frac{7}{3}=2.333\dots$ ✓

Let $x=0.\overline{9}$. Repetend length 1, so $10x=9.\overline{9}$. Subtract: $10x-x=9.\overline{9}-0.\overline{9}=9\Rightarrow 9x=9\Rightarrow x=1.$ So $0.\overline{9}$ is not "just below 1" — it is $1$, written a different way. This is the same subtraction trick used everywhere on this page, applied to a surprising case.

A fraction $\frac{p}{q}$ with $q=2^a{5}^b$ terminates after exactly $\max (a,b)$ digits, because you must multiply up to $10^{\max (a,b)}$ to clear both prime powers. To force exactly 3 digits, pick $\max (a,b)=3$ with $p$ sharing no factor with $q$ (i.e. $\gcd (p,q)=1$).

A valid answer: $\frac{7}{8}=\frac{7}{2^3}$. Here $\max (3,0)=3$, so: $\frac{7}{8}=\frac{7\cdot 5^3}{10^3}=\frac{875}{1000}=\mathrm{0.875.}$ It cannot stop at 2 digits: that would require $q\mid 100=2^2{5}^2$, but $2^3\nmid 2^2{5}^2$. (Another valid answer: $\frac{1}{40}=0.025$, since $40=2^3\cdot 5$, again $\max =3$.)

Length 2 means the smallest $k$ with $10^k\equiv 1(\mathrm{mod}q)$ is $k=2$. Test small $q$ coprime to 10: for $q=11$, $10^1=10\equiv 10\not\equiv 1$, and $10^2=100\equiv 1(\mathrm{mod}11)$ (since $100=9\cdot 11+1$). The order is exactly 2, so: $\frac{1}{11}=0.\overline{09},\text{length }2.✓$ (We reject $q=3$: $10\equiv 1(\mathrm{mod}3)$ gives length 1. We reject $q=7$: length 6.)

Every rational number's decimal either terminates or eventually settles into a fixed repeating block of some finite length $L$ (parent note). We argue by contradiction: suppose our number were rational. Then there is a finite block length $L$ and a starting position $N$ such that from digit $N$ onward, the digits repeat with period $L$ — every digit equals the one $L$ places earlier.

Now use the structure of our number. Its digits are: a $1$, then one $0$, a $1$, then two $0$s, a $1$, then three $0$s, and so on — the runs of consecutive zeros are $1,2,3,4,\dots$ long and grow without bound. So somewhere past position $N$ there is a run of zeros longer than $L$.

Look at one full period-window of $L$ digits sitting inside that long run of zeros: it is entirely zeros. By our assumed period-$L$ repetition, every window from that point on must be a copy of this all-zero window — meaning from there on the number is all zeros forever.

But that is false: further along, another $1$ always appears (the $1$s never stop, since between consecutive $1$s we only ever have a finite run of zeros). This directly contradicts "all zeros forever." Our assumption of rationality collapses. Therefore no finite repeating block exists → the number is irrational. ∎

This is exactly the boundary between rationals and Irrational numbers living inside the Real numbers — and it's why the rationals, though dense, still leave gaps.

Rational ⟺ decimal terminates or repeats ::: and it terminates ⟺ its reduced denominator is $2^a{5}^b$; otherwise it repeats with length dividing $\varphi (q)$.