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How Do We Know If Irrational or Transcendental Numbers Repeat? 
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#1
Mar2514, 04:07 PM

P: 12

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
Mar2514, 04:17 PM

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P: 18,346

It is very easy to construct numbers that don't repeat. For example
[tex]0.101001000100001000001000000100000001...[/tex] this doesn't repeat because I always keep adding more zeroes between the ones. Another example is the Champernowne's constant, which is just [tex]0.1234567891011121314151617181920...[/tex] 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] then [tex]10x = 2.131313131313....~\text{and}~1000x = 213.1313131313....[/tex] thus [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. 


#3
Mar2514, 04:20 PM

P: 446

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. 


#4
Mar2514, 04:21 PM

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P: 3,289

How Do We Know If Irrational or Transcendental Numbers Repeat?
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/(1x) = 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. 


#5
Mar2514, 04:44 PM

P: 12




#6
Mar2614, 04:00 AM

P: 25

Rest assured, 1+2+3+...≠1/12.



#7
Mar2614, 04:59 AM

P: 428

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. 


#8
Mar2614, 05:28 AM

P: 25

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.



#9
Mar2614, 09:13 PM

P: 460




#10
Mar2814, 02:44 PM

P: 12

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. 


#11
Mar2814, 03:27 PM

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P: 3,949

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. 


#12
Mar2814, 06:08 PM

P: 12

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) 


#13
Mar2814, 06:17 PM

P: 507



#14
Mar2814, 07:12 PM

P: 460




#15
Mar2814, 07:48 PM

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P: 3,949

"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_t..._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
Mar3014, 05:29 PM

P: 1,304

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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