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?
Repeating digits in $\pi$
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5No, one could not argue that, as the digits *do not* repeat. For example, .10100100010000... with $n$ zeros after the $n^{th}$ one never repeats. – 2012-01-17
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9Technical note: $\pi$ does not have _infinite_ digits (digits that are larger than any natural number), but rather _infinitely many_ digits. – 2012-01-17
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2$\pi$ 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. – 2012-01-17
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5Why 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. – 2012-01-17
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1@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. – 2012-01-17
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2@Charles: (My 2 cents.) I agree that the question is imprecise, but I feel that is where this site comes in. Amateurs often find it hard to express their thoughts precisely; in fact, perhaps many of them don't even find the need for being precise. As experts, we are in a position to help the OP clarify the question; and unexplained downvotes seem to be counter to that goal. I myself upvoted the post because I think that the post hints at an interesting confusion, and has attracted a few good answers. – 2012-01-17
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0@Charles: "Rather than explaining the (faulty) argument, one is merely postulated." -- Isn't the question essentially asking for an explanation for why the argument is faulty? If the OP could resolve the issue on their own, then why would they have to post it in this forum? – 2012-01-17
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0@Srivatsan: I'm not saying that the question should have explained why the argument is faulty, I'm saying that the question _should have said what the argument was_. Ben and I both guessed what the argument might have been, and it's entirely possible that yet a third argument was intended. Michael made the same guess as Ben. Austin seems to have no idea what it might have been, so he explains the general concept of irrationality. So yes, I think more precision would have been useful. – 2012-01-17
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0@Charles: I see; thanks for the clarification. – 2012-01-17
5 Answers
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."
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.
(Addendum: actually it occurs to me you may have been thinking about the idea that some block of digits will reappear infinitely often. This is also correct. In fact, for the same reason that at least one digit will appear infinitely often, there will be a block of digits of length $n$, for any $n$, that repeats infinitely many times. But as I hope the comments below show, this is still different from settling into a repeating pattern.)
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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0I 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. – 2012-01-17
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0Being an amateur, I tend to make many such inaccurate statements, so please indulge me. What I meant was that, as there are infinitely many digits in the decimal expansion of $\pi$, there must be a finite sequence of 141592..... that may repeat eventually. The digits are seemingly random and the $\pi$ is not artificially constructed as you did in your example. – 2012-01-17
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0@AlexBecker-For some reason I recall being told or reading that it was a theorem that they are actually asymptotically uniformly distributed; but since I have no idea where I got this, so have no idea if it's right, I'll edit the answer to take that statement out. – 2012-01-17
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2@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. – 2012-01-17
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2@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. – 2012-01-17
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1@BenBlum-Smith From [Wikipedia](http://en.wikipedia.org/wiki/Pi#Open_questions)'s article on $\pi$. – 2012-01-17
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1@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?"). – 2012-01-17
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0@RaviKulkarni - A minor correction to my earlier comment - the theorem of Lindemann actually states that $\pi$ is not only irrational but transcendental. The theorem that $\pi$ is irrational is due to Lambert. – 2015-04-03
$$ 3.1415926535\ldots $$ 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.
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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1By the way, this is known as the [Champernowne constant](http://en.wikipedia.org/wiki/Champernowne_constant). – 2012-01-17
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0@RahulNarain Thank you for the reference. – 2012-01-17
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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4But 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. – 2012-01-17
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1@AlexBecker: I guess that depends on your interpretation of "one could argue". – 2012-01-17
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0Fair 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. – 2012-01-17
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3People who are downvoting this should at least explain their vote. – 2012-01-17
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3@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. – 2012-01-17
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0@BenBlum-Smith: I didn't even consider that interpretation of the question. I was just answering the question literally. – 2012-01-17