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My question is referring to the exact definition mathematicians use when describing the decimal expansions of irrational numbers as "nonterminating and nonrepeating." Now, I understand, at least ostensibly, what is meant by "nonterminating" and the phrase "nonrepeating" seems simple enough to understand, but I've always wondered what is meant by the exact definition:

It was always my understanding that the term "nonrepeating" was referring to a specific sequence of numbers showing up no more than once in the decimal expansion. I'm confused as to the exact criteria for fulfilling this requirement.

  • Surely it can't be just a sequence of $1$ number. In the sense that $\pi$ starts with the number $3$ and then the number $3$ shows up again, and again and again an infinite number of times.

  • Is it a sequence of $2$ numbers repeating then? For instance in the golden ratio $\phi = 1.61803398874989$ we could take any $2$ number sequence, say $61$ or $98$ or $33$ and would it be sufficient to say that that particular sequence never shows up again? That seems highly unlikely given the "nonterminating" nature of the decimal expansions for irrational numbers.

  • If not, then what sequence of $n$ numbers is sufficient to declare a number "nonrepeating?"

  • Moreover, philosophically, how does it make sense that any sequence of numbers doesn't show up more than once? When, necessarily, an irrational number has an infinitely long decimal expansion and a sequence of numbers (at least for practical determination) would be finite up to some $n \in \mathbb{N}$

    • I mean, the idea that the sequence length be infinitely long just seems like a convenient workaround that dilutes the significance of the "nonrepeating" quality of irrational numbers in the first place. Since, if you ever came upon a sequence that repeated for whatever $n$-digit sequence you had you could always just say "oh actually I meant this $(n+1)$-digit sequence instead!" and keep adding digits to the sequence ad infinitum.

Perhaps the term isn't referring to repeating sequences of numbers but rather the same numbers repeating one after another.

  • But this cannot be the case as we saw above with the Golden Ratio, in the short approximation written out we have $2$ cases where the same number is repeated immediately (i.e., $33$ and $88$) we also see this in this approximation for $\pi = 3.1415926535897932384626433$

So, if the term "nonrepeating" doesn't refer to repetition of sequences of numbers $n$ digits long, nor does it the consecutive repetition of the same number, then what else would it refer to?

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    I think "non-periodic" would be a better term. For example, the decimal expansion of 1/11 is 0.09090909, i.e. it has a repeating sub-sequence '09', and thus the sequence has a period of 2. BTW, the sequence can be non-periodic for a while, but it is now allowed to have a periodic tail. For example 691/22000 = 0.31409090909..., where the periodic tail starts after 0.314. – blamocur Apr 24 '22 at 22:14
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    @blamocur And "non-eventually-periodic" is even more accurate. – Noah Schweber Apr 24 '22 at 22:18
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    Is math-history really a relevant topic here? – George Lee Apr 25 '22 at 15:53
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    "Non-repeating" is in fact a bad terminology. In a normal number , every finite digit string appears infinite many often, but still such a number is irrational. – Peter Apr 26 '22 at 16:57
  • $e$ is an even better example, very soon a $4$ digit-block repeats, but $e$ is known to be not only irrational, but even transcendental. – Peter Apr 26 '22 at 17:01
  • Have the answers below resolved your question? – Noah Schweber Jun 17 '22 at 05:24
  • @NoahSchweber Yes, they have. My apologies, I will select a best answer. – Numerical Disintegration Jun 17 '22 at 19:27

6 Answers6

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The phrase "non-repeating" can be a bit confusing when first introduced. A more precise, if less snappy, term is "not eventually periodic" (and this is what mathematicians mean when they say "non-repeating" in the context in question).

A sequence of numbers $(a_i)_{i\in\mathbb{N}}$ is eventually periodic iff there are $m,k$ such that for all $n>m$ we have $a_n=a_{n+k}$. The "eventually" here is connected to the "$m$" - the sequence $$0,1,2,3,4,5,6,4,5,6,4,5,6,...$$ is not periodic but it is eventually periodic (take $m=4$ and $k=3$). On the other hand, the sequence $$0,1,0,0,1,0,0,0,1,0,0,0,0,1,...$$ is not even eventually periodic (although of course it does have lots of repetition in it).

The connection with irrationality is this:

For a real number $r$, the following are equivalent:

  • $r$ is irrational.

  • Some decimal expansion of $r$ is not eventually periodic.

  • No decimal expansion of $r$ is eventually periodic.

(The issue re: these last two bulletpoints is that a few numbers have multiple decimal expansions. But this isn't a big deal to focus on at first.) In particular, the number $$0.01001000100001000001...$$ is irrational.

And base $10$, unsurprisingly, plays no role here: the above characterization works with "decimal expansion" replaced with "base-$b$ expansion" for any $b$.

Noah Schweber
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    Worth noting that even integers have periodic decimal expansion - it just happens to all be 0's. This can help when understanding that the same number may appear to have repeating digits in one base and not in another, e.g. 1/3 in base 10 is 0.33333... but in base 12, it's just 0.4. But if you think about those trailing 0's, it's really 0.40000..., so it has repeating digits regardless of base. – Darrel Hoffman Apr 25 '22 at 13:18
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    And, in base 12, 0.400000... is also 0.3BBBBBB... (the multiple decimal expansion clause above). – Yakk Apr 25 '22 at 13:26
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    Any integer base "b" (or possibly any rational base "b"). Pi, for example, is eventually periodic when written in base pi. – Mark Apr 25 '22 at 23:44
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    @Mark That was a deliberate omission: I think it best to only treat base $b$ for integral $b>1$, given the OP's background. (But yes.) – Noah Schweber Apr 25 '22 at 23:47
  • The second and third bullet points don't seem to be equivalent just based on logical reasoning. Should it be "any" instead of "some"? Unless maybe "not eventually periodic" is not the opposite of "eventually periodic"... – silvado Apr 27 '22 at 11:05
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    @silvado They are not trivially equivalent, but they do indeed happen to be equivalent: if one decimal expansion of a number is not eventually periodic, then all decimal expansions of that number will be not eventually periodic. This is a good exercise. – Noah Schweber Apr 27 '22 at 13:46
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Others have provided a solid, formal definition, which is great. I'll post an answer that focuses more on the conceptual understanding of the thing.

It was always my understanding that the term "nonrepeating" was referring to a specific sequence of numbers showing up no more than once in the decimal expansion.

No. Repeating means that the decimal expansion at some point has a specific block of digits that repeats, one after the other, and nothing else ever happens. Nonrepeating means anything other than that.

And perhaps it bears highlighting that, logically, "not always repeating" (nonrepeating) is not the same thing as "always not repeating". It is, rather, equivalent to "sometimes not repeating". (Related question.)

I think some would say that the quality is rather obvious if you understand the mechanism that generates repeating decimal sequences. Take any fraction a/b. Start doing the long division: at each step there are only b possible remainders (specifically, the integers 0 to b-1). Because of this finite set of possible remainders, if the division doesn't terminate, then at some point it will hit one of the previously-seen remainders. And once that happens then you're literally repeating a step of division seen earlier in the process, which will start reproducing the exact same sequence of quotients and divisors after that point, forever.

A great exercise would be to actually do several of these long-divisions by hand and witness the repetition of digits as it occurs. Hopefully that experience will make the meaning of "repeating" very concrete.

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    While the formal definitions are important, your division example is exactly the sort of thing needed to make it obvious. – Auspex Apr 26 '22 at 13:38
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    As a converse, any eventually-repeating decimal sequence can be written as a fraction $a/b$ (try this, choose an arbitrary sequence, then see what happens if you divide the repeating string by an equal-length string of 9s, times a power of 10 (e.g. $0.125363636...=125/1000+36/99000$). – 1Rock Apr 26 '22 at 21:59
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A decimal $x$ number is repeating if there exists $m,n \in \mathbb{N}$ with $m > n$ such that $10^mx-10^nx \in \mathbb{Z}$. This means that $10^mx$ and $10^nx$ have the same periodic pattern after the decimal point. Note that such a number must be rational because if $10^mx-10^nx=k$ for some integer $k$ then $x = k/(10^m-10^n)$ which is the ratio of two rational numbers, and so is rational. Therefore for a number to be irrational the decimal expansion cannot be eventually periodic.

CyclotomicField
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  • Couldn't this be simplified to "there exists $m \in \mathbb{N}$ such that $10^mx-x \in \mathbb{Z}$? It's the difference between $m$ and $n$ that matters. – chepner Apr 25 '22 at 15:29
  • @chepner it's similar but not exact. E.g. $x=0.1$ then there is no natural $m>0$ s.t. $10^mx-x\in\mathbb{Z}$. – David Pement Apr 25 '22 at 17:09
  • Why does $m$ have to be greater than 0? (Can we change to $m >=n $, or are we in "is zero a natural number or not" territory?) – chepner Apr 25 '22 at 17:16
  • @chepner Note that if $m=n$ we have $10^mx-10^mx=0$ so it would always be an integer. If such an $m$ and $n$ exist then some minimal $m$ and $n$ exist, and if they're minimal then the length of the period will be $m-n$. In this interpretation the reason $m \neq n$ is to ensure it has some non-trivial period, otherwise the period would be of length zero, meaning no pattern. – CyclotomicField Apr 25 '22 at 20:08
  • OK, yeah, I think I've been stuck thinking about values that have a (non-zero) repeating portion. – chepner Apr 25 '22 at 20:27
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The following is a typical example of a decimal number $x$. There must exists a natural number $n$, and a finite sequence of digits $d_1, d_2,d_3,\ldots d_n$, such that the following equation holds.

$$x = 0.d_1d_2d_3\ldots d_nd_1d_2\ldots$$

repeating in the obvious way. If you prefer, you could say that the $N^{\text{th}}$ digit of $x$ is $d_{N \% n}$, where $\%$ refers to the modulus.

Repeating decimals are precisely those which end in a repeating sequence of digits. In other words, it starts with any (finite!) sequence of digits you like whatsoever, and then eventually repeats as above. The repeating need not start at the decimal point. These are precisely the decimal numbers that can be expressed as fractions.

$\phi$ does not have this property for $n=2$ and $(d_1,d_2) = (6,1)$ because the first $61$ is followed by $80$, not $61$. In fact, $\phi$ has no finite subsequence which repeats forever. However, it probably does contain the subsequences $6161$, $616161$, $61616161$, and so on. The subsequence $61$ may occur infinitely often. But that's not enough for it to be a repeating decimal.

If not, then what sequence of n numbers is sufficient to declare a number "nonrepeating?"

You cannot tell whether a number is irrational (i.e. not eventually repeating) by looking at finitely many digits. You need to use other techniques. In general it can be very hard to prove that a given real number (defined other than by its digits) is irrational. But it can be done sometimes.

if you ever came upon a sequence that repeated for whatever n-digit sequence you had you could always just say "oh actually I meant this (n+1)-digit sequence instead!" and keep adding digits to the sequence ad infinitum.

This game will not work if you are allowed to change digits after observing them. The whole point of calculating a decimal expansion is that you pick the number first. How do you know what number if you don't know its decimals? I hear you cry. There are lots of ways. For example: $\tfrac{7}{3}$. Or I might say "the unique positive solution to the equation $x^2=2$". That's a perfectly good definition (at least it can be shown to be a perfectly good definition) and it allows you to compute all the decimals. But you don't get to count some of them and then change it.

If you mean instead that the sequence that you think is the repeating part might keep changing - i.e. you might allow a wider and wider window to represent the repeating part - then your trick doesn't break anything. Take the golden ratio, for example. Is the repeating part 6? No, the next digit is 1. Is the repeating part 61? no, the next digit is 8. Is it 618? No... And we continue forever. You'll never ever find a sequence that keeps repeating. That's what it means for it to be irrational.

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    This is not the usual definition of a repeating decimal $-$ you are allowed a finite number of non-repeating digits before the repetition kicks in. – TonyK Apr 25 '22 at 12:38
  • @TonyK Of course you are correct. I managed the correct statement in the next paragraph "Rational numbers are precisely those which end in a repeating sequence of digits" but will move the word definition to make it correct. – preferred_anon Apr 28 '22 at 08:11
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Repeating decimals can be written as $a.b\bar c$ where $a$, $b$, and $c$ are some (possibly null/empty) sequence of digits. $\overline c$ means "repeat this sequence of digits". For instance, $571.45\overline{2973}$ is a repeating decimal, with $a$ being the sequence $571$, $b$ being the sequence $45$, $c$ being the sequence $2973$, and $\overline{2973}$ means "repeat the sequence $2973$ an infinite number of times". I.e. $571.45\overline{2973} = 571.4529732973297329732973297329732973...$, where the meaning of an infinite decimal representation is defined in terms of the limit of the sequence of finite truncations of the representation as the length of the truncation goes to infinity.

All rational numbers can be written as a (not necessarily unique) repeating decimal (this is true of all number bases). Irrational numbers cannot.

Acccumulation
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A repeating decimal may be represented by:

$$ \sum_n^\infty \frac{d}{10^{kn}} $$

For example, $3.141414...$ would be $3 + \sum\limits_{n=1}^\infty \frac{14}{10^{2n}}$. This sum is an example of the geometric series which may be converted to a fraction (exercise left to the reader) which ends up being $\frac{14}{99}$. Similarly, $3.1415926 \,1415926 \,1415926...$ would be $3 + \sum\limits_{n=1}^\infty \frac{1415926}{10^{7n}}$.

In base $b$, a repeating decimal would be represented by:

$$ \sum_n^\infty \frac{d}{b^{kn}} $$

A second exercise left to the reader is to show that this infinite series may be represented by a single fraction.

Irrational numbers, by definition, may not be represented by a finite-length sum of ratios. Since the geometric series conveniently turns every repeating decimal into a fraction of finite length, we actually say that the repeating decimal prohibition is a consequence of irrationality instead of a condition.

  • "the repeating decimal prohibition is a consequence of irrationality instead of a condition" is a good observation. – Erick G. Hagstrom Apr 27 '22 at 13:29
  • You are making the same mistake as preferred_anon $-$ a finite number of non-repeating digits (before or after the dedimal point) are allowed before the repetition kicks in. – TonyK Apr 27 '22 at 15:36
  • @TonyK Clearly I'm not. Both examples have an added constant. Would dividing either constant by some power of 10 make it clearer? – user121330 Apr 27 '22 at 18:16
  • Clearly you are. Both your examples have an added constant before the decimal point. So your answer looks like the answer of somebody who thinks that the added constant has to be an integer. – TonyK Apr 27 '22 at 19:44
  • @TonyK The OP asked what "repeating decimals" means and I defined them. They asked in the context of irrational numbers, so I half-heartedly defined those too. Would it be clearer if I re-worded that?: "rational numbers, by definition must be representable by a finite-length sum of ratios." Ratio does mean fraction which is often less than 1, right? – user121330 Apr 28 '22 at 05:14