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How Do We Know If Irrational or Transcendental Numbers Repeat?

  1. Mar 25, 2014 #1
    Okay, so this is a problem I've been pondering for a while. I've heard from many people that pi doesn't repeat. Nor does e, or √2, or any other irrational or transcendental number. But what I'm wondering is, how do we know? If there truly is an infinite amount of digits, isn't it bound to repeat? I guess it's similar to Zeno's Paradox in a way, theoretically, it should never reach, but a proof says otherwise. Speaking of proofs, are there any to see if a number repeats? An infinite spigot algorithm? I'm sure that a proof for something like this would be easier with an algebraic number, but how would you do so for a transcendental number like e or pi or tau?

    So my main point is, theoretically, shouldn't all numbers, including irrational and transcendental, repeat? Or is it just an absence of a proof that causes this to be false?
  2. jcsd
  3. Mar 25, 2014 #2
    It is very easy to construct numbers that don't repeat. For example


    this doesn't repeat because I always keep adding more zeroes between the ones. Another example is the Champernowne's constant, which is just


    A moment's thought will convince you it doesn't repeat.

    It can be proven (but this is much harder) that ##\pi##, ##e## or ##\sqrt{2}## don't repeat. So we don't claim it because of absence of proof, we can actually prove that they don't repeat.

    Also, any repeating number must be a rational number. This is easy to see by the following method. For example, take

    [tex]x = 0.213131313...[/tex]


    [tex]10x = 2.131313131313....~\text{and}~1000x = 213.1313131313....[/tex]


    [tex]1000x - 10x = 211[/tex]

    thus ##x = \frac{211}{990}##. Thus method works for all repeating numbers. So all repeating numbers are fractions of integers. On the other hand, using long division, we can show that fractions of integers necessarily repeat.
  4. Mar 25, 2014 #3


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    If the numbers "repeat", and by this you mean a fixed sequence of numbers repeats indefinitely, then the number is rational.

    To see this, you can consider the repeating sequence as a geometric progression, which will have a rational sum.

    E.g. if the numbers 3456 repeat indefinitely, then you have a geometric progression with a common ratio of 10000.
  5. Mar 25, 2014 #4


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    If a number repeats, meaning that its fractional part consists of a finite sequence of digits ##a_1 a_2 \ldots a_N## which is repeated forever, then the number is rational. Let's prove this for the case of decimal digits, for example. We can ignore the integer part of the number without changing whether it is rational or irrational, so let's just focus on the digits to the right of the decimal. If they are ##a_1 a_2 \ldots a_N a_1 a_2 \ldots a_N \ldots## then this is equal to
    $$\sum_{k=0}^{\infty}\left(\sum_{n=1}^{N} a_{n} 10^{-n - Nk}\right)$$
    We can slide the ##10^{-Nk}## factor out of the first sum because it does not depend on ##n##, and we get
    $$\sum_{k=0}^{\infty} 10^{-Nk} \left(\sum_{n=1}^{N}a_n 10^{-n}\right)$$
    The expression inside the parentheses is not a function of ##k##. It is just a constant with respect to the sum. Let us give this constant a name:
    $$C = \sum_{n=1}^{N} a_n 10^{-n}$$
    Note that ##C## is rational because it is the sum of finitely many rationals. Now the previous expression reduces to
    $$C\sum_{k=0}^{\infty}10^{-Nk} = C \sum_{k=0}^{\infty}(10^{-N})^k$$
    This is a geometric series of the form ##\sum_{k=0}^\infty x^k##, where ##x = 10^{-N}##. Since ##|x|< 1##, the sum converges to ##1/(1-x) = 1/(1 - 10^{-N})##, which is a rational number because it is the quotient of rationals. Thus the original sum is also rational.

    So the above shows that any real number whose fractional part repeats (base 10, but the same proof works with any base) must be rational. The converse of this is that an irrational number CANNOT repeat. Since ##e## and ##\pi## have been shown to be irrational - see any good real analysis book, or (I thnk) Spivak's Calculus for proofs - this means that they do not repeat.
  6. Mar 25, 2014 #5
    Thank you, that helped a lot. The one thing I just don't really understand is about the proof you showed for a repeating number. Lets say you apply that proof to pi, and you would get about 3110/990 or so, but the numerator still goes on forever. It seems to me to be similar to the effect of circle squaring, you can get so close, but never really there. I feel like it has to do with the special properties of infinity, like 1+2+3... =-1/12. If you have an infinite string of numbers, you can't use finite numbers like 10 or 1000 if you have a repeating string that could be 10^googol^googol^googol long or something like that.
    Last edited: Mar 25, 2014
  7. Mar 26, 2014 #6
    Rest assured, 1+2+3+...≠-1/12.
  8. Mar 26, 2014 #7
    In case you're not aware, there was a popular video circulating by a respectable physicist who performs a number of invalid operations on divergent series to arrive at this answer as an analytic continuation of the Riemann zeta function.

    Sadly, he didn't provide any context or justification for his manipulations and was portrayed as defending this result by arguing that its validity lies in the fact that you can't actually sum indefinitely.

    I think it left millions of viewers with the wrong impression.
    Last edited: Mar 26, 2014
  9. Mar 26, 2014 #8
    I was aware. It is painfully obvious, however, given that the left hand side of the equation is a sum of increasing positive terms, their total (whatever it may be) cannot be negative. Just because we often meet counter intuitive results in mathematics it does not mean we throw good sense out of the window.
  10. Mar 26, 2014 #9
    What do you mean by this? There is no reason to apply the technique that micromass presented because ##\pi## is not rational. Since it is not rational, it the decimal expansion does not repeat. The proof that ##\pi## is not rational is separate from what micromass wrote. That just shows that one need not show that ##\pi## never repeats directly, but one only needs to show that ##\pi## is irrational.
  11. Mar 28, 2014 #10

    What I mean is that since you cannot apply that technique to pi, and it results in such an answer as I described, it's more of a lack of a result rather than a result that proves pi does not repeat. It's indeterminate.
  12. Mar 28, 2014 #11


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    You may have missed a step in the argument here.

    1) We know, by micromass's argument, that if a number does repeat, it is rational.
    2) From #1, we take the contrapositive ("If A then B" implies "If not B then not A") to show if a number is not rational then it does not repeat.
    3) We know that pi is not rational (several proofs available).

    By #2 and #3, we arrive at pi does not repeat.
    Note that we never have to apply micromass's technique to pi.
    Last edited: Mar 28, 2014
  13. Mar 28, 2014 #12
    And this is what someone has yet to give me. A true proof that pi does not repeat. Unless, by some strange theory of higher level math, a non repeating decimal can be rational. Or if there's a different definition of a rational number that I don't know about.

    But a proof would be nice, please. I almost feel like the question is being avoided in a way. By the way, I'm not trying to step on anyone's toes or get on anyone's nerves, I'm just an aspiring high school student who loves number theory and other kinds of theoretical math :) (interesting term, isnt it? One would think math is set in stone)
  14. Mar 28, 2014 #13
  15. Mar 28, 2014 #14
    I think that we all read your question as "how do we know that irrational functions have non-repeating decimal expansions" which micromass addressed. Gopher's link(s) should fill in the rest, but there is no simple proof. ##\sqrt{2}## would be easier. That is a proof that a high school student should be able to follow.
  16. Mar 28, 2014 #15


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    "Not repeating" and "not rational" are different things.

    "Not rational" means that the number cannot be written in the form ##a/b## where ##a## and ##b## are both integers. That pi cannot be written in this form was proven in 1761 by Johaan Lambert and there are a number of other proofs out there as well. The wikipedia article at http://en.wikipedia.org/wiki/Proof_that_π_is_irrational is a good start.

    Given these proofs that pi is not rational (in fact it's even transcendental, but that's overkill here), we can use the also proven fact that no irrational number can repeat to conclude that pi doesn't repeat.
  17. Mar 28, 2014 #16


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    geometric series.
  18. Mar 30, 2014 #17
    It is not as if we just witness no repetition for the first billion numbers and then conclude that it doesn't repeat.

    We conclude that it does not repeat because if it did, it would be rational. In other words, if a repetition existed (anywhere) it would be impossible to show that root 2 was irrational. We can show that, so a repetition does not exist, ever.
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