Is the following (decimal) number irrational?

0.10100100010000100000100000010000000100... etc.

My intuition tells me it is irrational. My informal "proof" is simply that it doesn't contain a repeating set of digits.

  1. For irrationality, is it both a necessary and sufficient condition that the digits never revert to a repeating sequence?

  2. Is there a more formal proof for this case?

  • 2
    $\begingroup$ Of course! Every rational has the property that the digits revert to a repeating sequence, and conversely, so 1) is true. Now, for 2) note that the number of zeros that occur between two ones never repeat.. $\endgroup$ – Rijul Saini Sep 5 '12 at 14:34
  • $\begingroup$ number is irrational if it can't be expresses as quotient of two rational numbers $\endgroup$ – dato datuashvili Sep 5 '12 at 14:34

A formal justification of your informal proof can be achieved by noting that in the process of long division, the fact that you have a finite number of possible remainders guarantees that eventually a remainder will be repeated. That is, for any rational number its decimal expansion becomes periodic.

  • $\begingroup$ Since the OP asked about a formal proof, maybe you could do more than just comment saying that it can be made formal. $\endgroup$ – Thomas Sep 5 '12 at 14:46
  • $\begingroup$ Thanks, that's what I guessed. I wondered if a more formal proof would be required to demonstrate that the decimal cannot be represented as the quotient of two integers - similar to the proof that sqrt(2) is irrational $\endgroup$ – njr101 Sep 5 '12 at 14:48
  • $\begingroup$ @Curious: Martin Argerami indicated how "rational $\implies$ periodic decimal" can be proved. The converse is "periodic decimal $\implies$ rational", and the contrapositive of the converse is what you asked. (This is mostly for other readers, as the context of your question suggests that you're aware of this.) The converse can be proved using the high school geometric series formula. (continued) $\endgroup$ – Dave L. Renfro Sep 5 '12 at 15:12
  • $\begingroup$ @Curious: continued) Consider, for example, $342.027665696565...$ This is the sum of two rational numbers, $342.0276$ and the sum of a certain geometric series with rational first term and rational common ratio. Note that $a$ rational and $r$ rational implies that $\frac{a}{1-r}$ is rational (when defined). $\endgroup$ – Dave L. Renfro Sep 5 '12 at 15:12
  • 1
    $\begingroup$ There's a typo (if, not of) which I'm not allowed to fix $\endgroup$ – SwiftsNamesake Jan 11 '17 at 8:31

Yes, a number is rational if and only if its decimal representation is eventually periodic (including the possibility of a period $\overline 0$)

A formal proof for your specific number requires a formal definition. I assume that your number is $$\sum_{n=1}^\infty 10^{-\frac{n^2+n}2}.$$ Any decimal representation that has infinitely non-zero digits (which is the case for your number) and has blocks of zeroes of arbitrary size (which is also the case for your number) cannot be eventually periodic: Some late period must lie completely in a sufficiently big block of zeroes, hence the period must be all zeroes, contradicting the fact that some non-zero digit occurs further to the right.

A number with a similar expression can even be shown to be not only irrational, but in fact transcendent: $$\sum_{n=1}^\infty 10^{-n!}.$$


What you have there is a trancedental number; a number for which there is no variable polynomial equation with rational coefficients that has this number as a root. Trancedental numbers are always irrational (but not all irrational numbers are trancedental). Therefore, yes, your number is irrational.

The proof is in the construction; like the well-known trancedental numbers $\pi$ and $e$, your number is the asymptotic limit of the sum of an infinite series; in this case, the sum of a reducing fraction:

$$\sum_{n=1}^{\infty} \dfrac{1}{10^{\dfrac{n(n+1)}{2}}}$$

This is similar to the construction of the Champernowne constant which is proven transcedental. The sum is constructed such that no 2 terms ever modify the value of a decimal place of the same order of magnitude, very much like $C_{10}$, and so the number constantly increases but can never reach a rational sum, unlike the infinite sum of $\frac{1}{2^n}$.

  • $\begingroup$ Thanks for the extra info. But by this logic, then surely the sum to infinty of (2^-n) also "always has one more term to be added". But this series sums to 1 exactly, which is obviously an integer. $\endgroup$ – njr101 Sep 5 '12 at 15:22
  • $\begingroup$ This answer is a double post. Please delete. $\endgroup$ – njr101 Sep 5 '12 at 15:32
  • $\begingroup$ Edited and removed the double post. The difference between this and $1/2^n$ is that, like $C_{10}$, each term never modifies the value of any decimal place already known by the sum of all previous terms. As such, there is no possible "rollover" to a rational value as n grows large. $\endgroup$ – KeithS Sep 5 '12 at 15:34
  • $\begingroup$ I see no reason to believe that the given number is transcendental. $\endgroup$ – Sean Eberhard Sep 5 '12 at 16:04

The number you gave when you asked the question is indeed an irrational number because it goes on without repeating or ending. I've used to learn that. It keeps adding zeroes on and on and on beside the ones. This never goes into a repeating sequence because it keeps adding on the zeroes. I don't know if there's a more formal proof for this kind of case, but remember that an irrational number can never be written as a fraction.

  • $\begingroup$ This answer is not needed. The question was asking for a formal proof of the statement which none of the answers gave, not for an answer that convinces other people of the truth of the statement. If you can figure out how to write a formal proof, it's fine to include it in your answer but when it's easy for experts to find one, saying there might not be one probably belongs in a comment rather than an answer. Just do your best and don't worry if your best isn't that great and you end up getting blocked. The purpose of blocking is probably not to punish. $\endgroup$ – Timothy Jul 28 '18 at 17:32

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