Rational numbers — definition, decimal expansion (terminating - repeating): Question bank

Two words we lean on constantly. A repetend is the block of digits that repeats forever, written with a bar, like the $\overline{3}$ in $0.8 \overline{3}$ . And lowest terms means the fraction $\frac{p}{q}$ has no common factor left to cancel — $g cd (p, q) = 1$ .

The figure below shows this split as a prime-factor picture — the whole game of "terminate vs repeat" lives in which coloured primes appear.

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

True or false — justify

True or false:

0.4999 \dots

(nines forever) is a different number from

0.5

False — they are the same number.

0.4 \overline{9} = 0.5

exactly, because

0. \overline{9} = 1

(multiply

x = 0. \overline{9}

by 10, subtract:

9 x = 9

). Two decimal strings can name one rational.

The number-line picture below makes this concrete: $0.9, 0.99, 0.999, \dots$ crowd up to $1$ with the gap shrinking to zero, so their infinite limit is $1$ .

True or false: every terminating decimal is also a repeating decimal.

True, in a trivial sense — you can append repeating zeros:

0.375 = 0.375 \overline{0}

. Terminating is just the special case where the repetend is

\overline{0}

True or false: if a decimal repeats, the number must be rational.

True. A repeating pattern lets you subtract two shifted copies and cancel the tail, always leaving an integer over

1 0^{n} - 1

(times a power of 10) — a fraction. Repetition guarantees rationality.

True or false:

\frac{1}{3} + \frac{1}{6} = 0.333 \dots + 0.1666 \dots

is irrational because both addends have infinite decimals.

False. The sum is

\frac{1}{2} = 0.5

, which terminates. "Infinite decimal" does not mean irrational — rationals with a forbidden prime also run forever (but repeat).

True or false: a fraction

\frac{p}{q}

in lowest terms with

q = 40

terminates.

True.

40 = 2^{3} \cdot 5

, only allowed primes, so it terminates regardless of

p

. Example:

\frac{7}{40} = 0.175

True or false: whether

\frac{p}{q}

terminates can depend on the numerator

p

False, once in lowest terms. Terminating depends only on

q

's prime factors. But

p

matters for reducing:

\frac{6}{15}

looks like it has factor 3, yet reduces to

\frac{2}{5} = 0.4

, which terminates.

True or false: putting a minus sign in front, like

- \frac{5}{6}

, can change whether the decimal terminates or repeats.

False. The sign only flips the number to the other side of zero; it leaves every digit and the whole terminate/repeat pattern untouched.

- \frac{5}{6} = - 0.8 \overline{3}

repeats exactly as

\frac{5}{6} = 0.8 \overline{3}

does.

True or false: the number

0.123456789101112 \dots

(concatenating the counting numbers) is rational.

False. Its digit blocks keep growing and never settle into a fixed repeating block, so it cannot terminate or repeat — it is irrational.

True or false: between

\frac{1}{3}

and

\frac{1}{2}

there are only finitely many rationals.

False. The rationals are a dense set — between any two of them lies another (e.g. their average), so there are infinitely many.

Spot the error

\frac{1}{6} = 0.1 \overline{6}

, so 6 has a prime factor other than 2 and 5 — but 6 also has factor 2, contradiction!" Where's the flaw?

No contradiction. Having some forbidden prime (the 3) is enough to force repeating. The 2 alone would terminate; the 3 ruins it. Termination needs the denominator to have only allowed primes — one forbidden prime spoils it.

\frac{1}{7}

has repetend length 6, so

\frac{1}{n}

always repeats with

n - 1

digits." Find the error.

The rule is that repetend length divides

ϕ (n)

, it does not equal

n - 1

. For

\frac{1}{7}

ϕ (7) = 6 = 7 - 1

only because 7 is prime;

\frac{1}{13}

has length 6 while

ϕ (13) = 12

"To convert

0.58 \overline{3}

I multiply by 10 once and subtract." Why does this fail?

One multiply-and-subtract only cancels the tail if the repetend and the whole number are already aligned. With a non-repeating head (

58

), you need two shifts — one to jump past the head, one more to line up the repetend — then subtract.

q = 9 = 3^{2}

has no 2 or 5, so

\frac{1}{9}

never gives a whole-number remainder in long division." Where's the slip?

The statement about remainders is fine, but its phrasing hides the real point: remainders never hit 0 only because they must recycle among

{1, \dots, 8}

, forcing repetition, not termination.

\frac{1}{9} = 0. \overline{1}

0. \overline{9} = 1

must be wrong, because

0. \overline{9}

is clearly less than 1." Spot the mistaken intuition.

The intuition treats

0. \overline{9}

as "stopping just short." It doesn't stop — with infinitely many nines the gap to 1 is exactly zero. Algebra (

9 x = 9

) and limits both confirm

0. \overline{9} = 1

"Since

π = \frac{22}{7}

π

is rational." What's wrong?

\frac{22}{7} = 3. \overline{142857}

is a rational approximation, not

π

. The true

π = 3.14159 \dots

never repeats, so it is irrational;

\frac{22}{7}

merely agrees for a couple of digits.

"The repetend of

\frac{p}{q}

can start with a non-repeating part only if

q

is even." Correct the claim.

The non-repeating "head" appears whenever

q

has any allowed prime alongside a forbidden prime — so it can happen for

q = 15 = 3 \cdot 5

(odd) too:

\frac{1}{15} = 0.0 \overline{6}

Why questions

Why can a long division of

\frac{p}{q}

never repeat for more than

q - 1

digits before a remainder recurs?

The nonzero remainders live in

{1, 2, \dots, q - 1}

— only

q - 1

possible values. By the pigeonhole principle a remainder must repeat within

q

steps, and once a remainder returns the whole cycle returns.

The remainder-cycle figure below traces this for $\frac{1}{7}$ : the remainders $1 \to 3 \to 2 \to 6 \to 4 \to 5$ loop back to $1$ , and the digits ride along that loop.

Why does the denominator being

2^{a} \cdot 5^{b}

force termination, in one sentence?

You can multiply top and bottom to make the denominator a power of ten (

1 0^{n} = 2^{n} 5^{n}

), and any integer over

1 0^{n}

is just that integer with the decimal point moved

n

places — a finite string.

Why do we insist on "lowest terms" before reading off the terminate/repeat rule?

Because hidden common factors can cancel a forbidden prime:

\frac{3}{6}

looks like it has a 3, but reduces to

\frac{1}{2}

. Only the reduced denominator's primes decide the behaviour.

Why is

ϕ (q)

(from Euler's totient function) the natural bound on repetend length for

q

coprime to 10?

The repetend length is the smallest power

k

with

1 0^{k} \equiv 1 (mod q)

(the order of 10 mod

q

in Modular arithmetic), and by Euler's theorem that order divides

ϕ (q)

Why does multiplying

x = 0. \overline{ab c}

1 0^{n}

(with

n

= repetend length) make subtraction cancel the infinite tail?

Shifting by exactly one full period lines the tails up digit-for-digit, so

1 0^{n} x

and

x

have identical fractional parts — subtracting kills them and leaves an integer.

Why do rationals leave "gaps" in the number line even though they are dense?

Dense means no two rationals sit with empty space of rationals between them, but limits of rational sequences (like

2

) can fall between them — those limit points are the irrationals filling the gaps to form the Real numbers.

Edge cases

What is the decimal expansion of

\frac{0}{7}

, and does it terminate or repeat?

0

exactly —

\frac{0}{q} = 0

for any nonzero

q

. It terminates trivially; the "repeat vs terminate" rule is about nonzero numerators.

Does

\frac{p}{q}

with

q = 1

(an integer) terminate? Which prime-factor rule applies?

Yes —

q = 1 = 2^{0} \cdot 5^{0}

fits "only allowed primes" vacuously, so integers terminate:

5 = 5.0

. No forbidden primes exist to cause repetition.

0. \overline{0}

a repeating decimal or a terminating one?

We classify it as terminating, using the sharp rule "a decimal terminates if from some point on every digit is

0

." Since

0. \overline{0}

is all zeros, it satisfies that rule; a decimal is called repeating only when its shortest repetend is a non-zero block — so

0. \overline{0} = 0

sits firmly on the terminating side, not in between.

How does a leading minus sign, as in

- \frac{7}{8}

, interact with the terminate/repeat test?

It doesn't — the test reads only the reduced denominator's primes, and the sign lives with the numerator.

- \frac{7}{8} = - 0.875

terminates just like

\frac{7}{8} = 0.875

; reflect any expansion through zero and the pattern is unchanged.

For

\frac{1}{q}

with

q

having both an allowed prime and a forbidden prime (say

q = 6

), what does the decimal look like structurally?

A finite non-repeating head followed by a repeating tail, e.g.

\frac{1}{6} = 0.1 \overline{6}

: the 2 produces the one-digit head, the 3 produces the repeating part.

Can a rational number have a repetend longer than the number of digits in its denominator?

Yes. Length is bounded by

q - 1

, not the digit-count of

q

. For example

\frac{1}{97}

has a 96-digit repetend, far longer than the two digits of "97."

What happens to the terminate/repeat classification if you write the number in base 6 instead of base 10?

It can change, because the "allowed primes" are now the prime factors of the new base. In general, in base

b

a reduced

\frac{p}{q}

terminates exactly when every prime factor of

q

also divides

b

— otherwise it repeats. Base

6 = 2 \cdot 3

makes

2

and

3

allowed, so

\frac{1}{3}

terminates as

0. 2_{6}

even though it repeats in base 10.

The table figure below lines up $\frac{1}{3}, \frac{1}{5}, \frac{1}{7}$ across bases 10, 6 and 2 so you can watch "allowed primes" swap as the base changes.

Recall One-line summary of every trap

Terminating vs repeating depends only on the reduced denominator's primes relative to the base (allowed = factors of the base); repetition (not "infinite length") is what guarantees rationality; a sign never changes the pattern; and $ϕ (q)$ bounds — never equals — the repetend length.