The definition of a Irrational number is "Irrational numbers don't include integers OR fractions. However, irrational numbers can have a decimal value that continues forever WITHOUT a pattern."

So my question is why is a base 10 decimal "fraction" okay but a non base 10 fraction not okay?

Thanks for your expertise!

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    $\begingroup$ Irrational numbers fail to repeat in any base, and rationals repeat in all bases. $\endgroup$ Jul 16, 2014 at 16:05
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    $\begingroup$ Actually I think it's more natural to define what a rational number is, then say irrational numbers are those which are not rational. A rational number is something of the form $p/q$ where $p$ and $q$ are integers (and $q\neq 0$). This should be fine base anything. $\endgroup$ Jul 16, 2014 at 16:05
  • $\begingroup$ en.wikipedia.org/wiki/Repeating_decimal even have proof regarding why it must be the case that a number is rational if and only if all, or any, of its fraction are repeating in any bases. Not very rigorous, but can be easily turn into one. $\endgroup$
    – Gina
    Jul 16, 2014 at 16:52
  • $\begingroup$ @Thomas. Technically, irrational numbers only fail to repeat in all integral bases. There are bases (e.g. golden ratio base, base $\pi$, base $\sqrt 2$) where some irrationals do not repeat, and all rationals (possibly excluding integers) do repeat. $\endgroup$
    – hatch22
    Jul 16, 2014 at 17:17

2 Answers 2


First, some thoughts on expanding a rational number in some base, such $s=10/27$ in bases 3 and 10. (Decimal representation is assumed unless otherwise indicated.) If we multiply by a sufficiently large power of 3 then we can turn this into an integer---in this case, $27s=10$---and then we can express the result in base-3 as $9+1=11_3.$ Dividing by $3^3$ then gives $s=0.011_3$.

In base ten, no power of 10 turns $s$ into an integer. But observe what happens if use $10^3$: then $$1000 s= \frac{10000}{27}=\frac{370*27+10}{27}=370+\frac{10}{27}=370+s,$$ and so $$s=0.370+0.001s=0.370+0.001\cdot (0.370+0.001s)=\cdots=0.370370\ldots=0.\overline{370}.$$ So we can get a repeating expansion.

The key point is that we were able to write down some equation with integer coefficients which this rational number satisfied, chosen appropriately for whichever base $b$ we were interested in. This characterizes the rational numbers, and so tells us that their digits must either terminate or repeat for a given base. But this also tells us that if a number is irrational (i.e. it satisfies no equation of this kind) then its digits cannot repeat or terminate in any base.


Any base-$n$-fraction $\frac pq$ will continue WITH a pattern, repeating with a length of at most $q-1$ repeating digits where $q$ is its denominator - no matter what base you are looking at.

For example, while $\frac17=0.\overline{142857}_{10}=0.1_7$, $\frac1{10}=0.1_{10}=0.\overline{0462}_7$.

If a number repeats or stops in base $10$, it will continue or stop in any other base $n$ as well.

On the other hand, irrational numbers are those where you cannot find any repetitions in the digits in any base.

  • $\begingroup$ the bound on the length of the repeating part is the denominator of the fraction, not the base. $\endgroup$
    – Ned
    Jul 16, 2014 at 17:15
  • $\begingroup$ @Ned: thank you for noticing, i fixed it in the answer $\endgroup$
    – pascalhein
    Jul 16, 2014 at 17:27

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