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As $\pi$ has infinite digits in its decimal expansion, one could argue that its digits will repeat after a finite number of digits. If so, it is a rational number. What's wrong with this argument?

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No, one could not argue that, as the digits do not repeat. For example, .10100100010000... with $n$ zeros after the $n^{th}$ one never repeats. –  Alex Becker Jan 17 '12 at 3:28
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Technical note: $\pi$ does not have infinite digits (digits that are larger than any natural number), but rather infinitely many digits. –  Charles Jan 17 '12 at 3:31
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$\pi$ does not have infinite digits in its decimal expansion; rather it has infinitely many digits in its decimal expansion. But there is not one of them that is infinite. –  Michael Hardy Jan 17 '12 at 3:31
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Why does the question have 3 unexplained downvotes? The question appears to be the result of a sincere confusion by an amateur and cannot be resolved by simply (e.g.) looking up on Wikipedia. –  Srivatsan Jan 17 '12 at 8:08
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@Srivatsan: I didn't downvote, but I could see why a person would -- the question is quite imprecise. Rather than explaining the (faulty) argument, one is merely postulated. –  Charles Jan 17 '12 at 19:35

5 Answers 5

up vote 3 down vote accepted

There's nothing wrong with the argument

If pi's digits repeat after some finite segment, then it is rational.

in the same way that there's nothing wrong with the argument

If 1+1 = 3, then 4 = 6.

But in both cases the left side is false so the implication is true only trivially. $\pi$ is known to be irrational (in fact, transcendental).

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But his first point "as π has infinite digits in its decimal expansion, one could argue that its digits will repeat" does not take repetition as a premise, so is invalid. –  Alex Becker Jan 17 '12 at 3:31
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@AlexBecker: I guess that depends on your interpretation of "one could argue". –  Charles Jan 17 '12 at 3:33
    
Fair point. I interpreted it as "one could correctly argue". If someone meant "one could say the following words" I don't see why anyone would bother to say it. –  Alex Becker Jan 17 '12 at 3:35
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People who are downvoting this should at least explain their vote. –  Alex Becker Jan 17 '12 at 4:02
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@AlexBecker - I didn't downvote, but I considered it. The OP makes it clear they know $\pi$ is irrational (from "What's wrong with this argument?") so the question appears aimed at at helping the OP resolve the cognitive dissonance between this and the idea that in all of the infinitude of $\pi$'s digits, somewhere there must eventually be some repeating ones. This answer seems to ignore the intent of the question in favor of the opportunity to be smart-alecky. I didn't downvote because I didn't feel sure Charles intended to be smart-alecky. –  Ben Blum-Smith Jan 17 '12 at 4:24

I am guessing that you are mixing up the fact that some digit will have to reappear infinitely many times in the expansion (since there are infinitely many decimal places to fill and only 10 choices for each) - this is correct - with the (incorrect) idea that this means the expansion will be repeating.

Since the argument does not make use of any special features of $\pi$, to see what's going on we could consider any irrational number. Here is one manufactured to make it clear what's going on with the digits in the long run:

$0.101100111000111100001111100000...$

This number was designed so that I could make sure the decimal expansion is not eventually repeating. There is no segment of the decimals such that eventually the expansion consists of this segment over and over again, because the sequences of 1's and 0's get longer and longer.

Now, what I took you to be saying is that because there are infinitely many places in the expansion, some digits have to happen infinitely often. This is correct. (It's a consequence of the pigeonhole principle.) However, they don't happen in a repeating pattern. In the case at hand, the digits 0 and 1 both occur infinitely often, but not in a repeating way. Similarly, in $\pi$, some digit must occur infinitely often, but they never settle into a cycle that repeats itself.

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I was under the impression that whether any fixed digit appears in the decimal expansion of $\pi$ infinitely many times is still open, even more so whether they all do. –  Alex Becker Jan 17 '12 at 3:39
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@RaviKulkarni - The point is that there's no reason to expect this. The example shows it's possible for an infinite sequence of digits to never repeat. We also have the theorem that if the digits eventually settle into a repeating pattern the number is rational (this theorem is not hard) and the deep theorem of Lindemann that $\pi$ is irrational. Together these imply the digits of $\pi$ never repeat. They may look random, which makes them different from my example, but they can't repeat by the theorem of Lindemann. –  Ben Blum-Smith Jan 17 '12 at 3:54
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@RaviKulkarni - In the hopes of further illustrating: actually, a truly random sequence of digits is also very unlikely to settle into a repeating pattern. A repeating pattern (as in a rational number) is an extremely _un_random thing: it has to be that, except for a finite number of digits at the beginning, THE WHOLE THING looks like the same finite fragment over and over again. Example $0.223409844535445354453544535445354453544535...$. The vast majority of the sequence of digits has to have a very peculiar structure. –  Ben Blum-Smith Jan 17 '12 at 3:57
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@BenBlum-Smith From Wikipedia's article on $\pi$. –  Alex Becker Jan 17 '12 at 3:58
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@AlexBecker - Thanks, good catch! I guess in the vague recollection of what I read or heard I was confusing the answer ("pi is normal") with the question ("is pi normal?"). –  Ben Blum-Smith Jan 17 '12 at 4:10

$$ 3.1415926535\ldots $$ The digit $1$ "repeats" since it appears in the third place after the decimal point after appearing earlier in the first place after the decimal point; likewise $3$ repeats since it appears in the ninth place after the decimal point after appearing earlier before the decimal point, and $5$ similarly repeats.

The pigeonhole principle says that kind of "repetition" must happen no later than the 11th digit, since only $10$ digits can be distinct.

But that's not the sort of "repetition" from which one can infer that a number is rational. That involves periodicity.

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Consider the Champernowne constant: $$ 0\ .\ 1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12\ 13\ \dots $$

It is made by counting out loud and writing down whatever number you are saying. (Notice I don't intend for the number "13" to occupy a single place value, but rather I write down a "1" and then a "3".)

This number has infinitely-many digits, since there are infinitely-many counting numbers. The number cannot be expressed as some finite string of digits repeating infinitely-often, since you won't find a repeat in the set of counting numbers.

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By the way, this is known as the Champernowne constant. –  Rahul Jan 17 '12 at 3:46
    
@RahulNarain Thank you for the reference. –  Austin Mohr Jan 17 '12 at 3:53

I struggled with this theory until I read an article similar to the one listed below outlining the reality of infinity.

http://www.scientificamerican.com/article.cfm?id=strange-but-true-infinity-comes-in-different-sizes

"Take, for instance, the so-called natural numbers: 1, 2, 3 and so on. These numbers are unbounded, and so the collection, or set, of all the natural numbers is infinite in size. But just how infinite is it? Cantor used an elegant argument to show that the naturals, although infinitely numerous, are actually less numerous than another common family of numbers, the "reals." (This set comprises all numbers that can be represented as a decimal, even if that decimal representation is infinite in length. Hence, 27 is a real number, as is π, or 3.14159….)

In fact, Cantor showed, there are more real numbers packed in between zero and one than there are numbers in the entire range of naturals. He did this by contradiction, logically: He assumes that these infinite sets are the same size, then follows a series of logical steps to find a flaw that undermines that assumption. He reasons that the naturals and this zero-to-one subset of the reals having equally many members implies that the two sets can be put into a one-to-one correspondence. That is, the two sets can be paired so that every element in each set has one—and only one—"partner" in the other set."

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