I understand that an irrational number has no periodic numerical pattern. I was wondering, however, if we could find logical patterns instead and if they would even be useful.

For example, let's say we have a number g, approximately: g=5.123768322988 and let's assume this is irrational. Let's say that I was able to prove that the numbers following the decimal in g follow a particular logical pattern: "three numbers less than 4 followed by 3 numbers greater than 5". Is this something that could be proven for irrational numbers like pi?

If so, would it even be useful for example in computing precise approximations in a cheaper way?

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    **an irrational number has no numerical patterns.** wrong. The decimal expansion of an irrational number is not eventually periodic. Any other patterns are OK. – GEdgar Jul 08 '21 at 14:51
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    For instance, $\sum 10^{-n^2}=.100100001\cdots$ is irrational (clearly) though the digits in the decimal expansion follow an obvious pattern. – lulu Jul 08 '21 at 14:57
  • [Related](https://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations) – Arctic Char Jul 08 '21 at 15:01
  • You haven’t defined $g.$ You’ve only give an approximation. The possible values of $g$ are irrational and rational, and some of the irrationals have some patterns in their digits, other than simple repeating. A common error when thinking about this is thinking numbers are the digits. The base $10$ representation is just one way to represent numbers, and no representation is the “primary” one. And I’m not just talking about other bases. Continued fractions are another way. – Thomas Andrews Jul 08 '21 at 15:13
  • @ThomasAndrews that is true, it was a poor choice of words, from the fact that I was thinking about the logical pattern for it as the definition. I've corrected my text to reflect what I meant more clearly. – Reacf Jul 08 '21 at 15:18
  • @GEdgar i hadn't thought of that, that's a great correction thanks! – Reacf Jul 08 '21 at 15:20
  • The clarification is good, but misses my point. You will rarely find such a $g,$ unless you define it specifically to satisfy this sort of condition. This is because if $g$ satisfies this rule in base $10,$ it probably won’t satisfy any similar rule in any other base, other than bases $10^k.$ So, it is a property of the number in the base $10.$ Unless you specifically defined $g$ to have properties, base $10,$ the odds that some number you’ve defined has properties base $10$ is highly unlikely. There are better ways to estimate numbers than base $10.$ – Thomas Andrews Jul 08 '21 at 15:40
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    We fetishize base $10$ digits (of $\pi$ in particular) for historical and psychological reasons, not out of any other need. But expecting patterns like this in base $10$ misconstrues how real numbers actually work. – Thomas Andrews Jul 08 '21 at 15:41
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    You can have numbers with pretty predictable (and even algorithmic) patterns, which are not only **irrational** but also **transcendental**, For an example see https://en.wikipedia.org/wiki/Champernowne_constant – CiaPan Jul 08 '21 at 16:36
  • You can predict the digits of any number using its series expansion and the nth digit would be a consequence of adding the series terms and “carrying over the 1s”. Would this work? – Tyma Gaidash Jul 08 '21 at 19:51

3 Answers3


Is this something that could be proven for irrational numbers like pi?

If this were proved for pi it would be a major accomplishment. To date, the majority view is that the decimal for pi is normal (in all bases), and therefore does not have the pattern you suggest.

Even if you proved it for $\sqrt{2}$ it would be major news (to mathematicians).

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I was wondering, however, if we could find logical patterns instead and if they would even be useful.

Sorry but no there are no proved logical pattern for many irrational numbers.

Any irrational number has a non-terminating, non-repeating sequence of digits in its decimal representation (or the representation in any whole number base). This is easy to prove by contradiction. Any terminating decimal is obviously rational. Any repeating sequence with period n can be converted to a terminating decimal by multiplying by $(10^n−1)$ .

On a side note let me tell you that irrational numbers like $0.1010010001...$ are irrational having pattern.

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    @AndrésE.Caicedo Thanks for guiding me in making this answer more accurate –  Jul 08 '21 at 16:22

I think what you may be looking for is a continued fraction expansion for irrational numbers. This does help you find the pattern that you are suggesting for many irrational numbers (such as sqrt 2, the golden ratio, e, etc). However, to my knowledge, there is no pattern in the continued fraction for pi.

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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Feb 11 '22 at 15:09
  • There are (non-simple) continued fractions with a simple pattern known for pi, though. – Martin Väth Feb 11 '22 at 15:09