What is meant, exactly, by nonrepeating when talking about irrational numbers?

Question

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?

Answer 1

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 $b$.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.