# If a number is irrational, then does its mantissa contain every possible digit sequence of finite length?

For a number to be irrational, it must be impossible to express the value as a ratio of integers. So, if I look at the infinite string of digits to the right of the number's decimal point, can I find any given integer sequence of length n (n natural), or is that not guaranteed?

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You're not even guaranteed to find any given digit. For example, $0.110100100010000...$. –  Qiaochu Yuan Aug 4 '11 at 1:58

## 2 Answers

It is not guaranteed. For example, $0.101001000100001...$ is irrational. A number is rational if and only if its decimal expansion either terminates or from some point on repeats, so any non-terminating decimal that doesn’t end in an infinite string of repetitions of a single finite block of digits must represent an irrational number.

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Thanks, I was confusing irrational numbers with normal numbers. –  Zach Rattner Aug 4 '11 at 14:08
@Rattner: Normal numbers satisfy a very strong version of "every possible string of digits appears" (sometimes called "a lexicon"). Readers with the appropriate backgound may be interested in the fact that almost all real numbers in both the Baire category sense and the Lebesgue measure sense (and more, since the non-lexicon numbers are a countable union of uniform Cantor sets) are lexicons. On the other hand, Lebesgue-almost-all real numbers are normal and Baire-almost-all real numbers are NOT normal (even non-normal in extremely strong ways). –  Dave L. Renfro Aug 4 '11 at 14:32

No. Consider e.g. $.1101001000100001\ldots$.

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