In the decimal system, circumference / diameter = 3.14159265359...

But, for example, in binary, that ratio is 11.00100100001111110111... Or in hexadecimal, 3.243F6A8885A308D313198A2E037073...

We also obtain an irrational number, independent of the numeric system that we use.

Has something to do with the fact that we use decimal system, or with the dimensions of the given space? Could it be otherwise, with another set of rules or number of spatial dimensions? (Obtain an exact number)

  • 4
    $\begingroup$ The rationality of a number does neither depend on its representation nor on the base-system it is displayed. "why" $\pi$ turned out to be irrational , even transcendental , is a philosophical question. One might argue that "almost every real number" is transcendental, but this is not a satisfying answer. $\endgroup$
    – Peter
    Sep 21, 2022 at 16:24
  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Sep 21, 2022 at 16:24
  • $\begingroup$ @Peter Thanks! But is it a philosophical question? Has it in fact been proven that is not a geometrical result given the properties of the given space? $\endgroup$
    – KPrv
    Sep 21, 2022 at 16:29
  • 1
    $\begingroup$ Well, if we take the ratio of the length of a diagonal in a square to the length of this square, we know that the result is algebraic irrational , namely $\sqrt{2}$. This follows from the Pythagorean theorem, but "why" does this theorem hold ? I have no clue, maybe someone knows the "reason". $\endgroup$
    – Peter
    Sep 21, 2022 at 16:33
  • $\begingroup$ @Peter I agree. A proof is not a reason, however much people like to think it is. Yes, it is proven that $\pi$ is irrational, but there isn't any particular, fundamental, underlying reason for it. There are just ways of demonstrating that it is true. (There do exist reasons why one might expect a certain statement to be true, and some of those could be expanded into a full proof, but that's different.) $\endgroup$
    – Arthur
    Sep 21, 2022 at 16:37

1 Answer 1


The representation is irrelevant since the repeating pattern forces the number to be rational. Say we have a number that repeats eventually in base $n$ called $r$ and say the digits repeat after $k$ digits. In general there will be an integer portion, a non-repeating portion after the decimal, and then the repeating portion. The first step is to find some appropriate power of $n$ so that the repeating pattern starts at the decimal point. So say I have $r=0.124737373...$ in base $8$ I would multiply by $8^3$ so that $8^3s=124.737373....$ and from here we then multiply by $n^k$ to move the pattern to match up with itself again. So in our running example we would have that $8^28^3s=8^5s = 12473.737373...$ and now we subtract to remove the portion after the decimal place so in our example we take $8^5s -8^3s = 12473.737373... - 124.737373... = 12349$ and so we have that $(8^5-8^3)s = 12349$ Simply noticing that $8 = 10$ base $8$ this gives us $(10000-1000)s = 1700s=12349$ and so solving for $s$ we recover $s=12349/1700$. This works in every base just like it does in base 10. It's a property of the notation itself rather than the base.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .