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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?

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- #1

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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

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[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

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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

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$$\sum_{k=0}^{\infty}\left(\sum_{n=1}^{N} a_{n} 10^{-n - Nk}\right)$$

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:

$$C = \sum_{n=1}^{N} a_n 10^{-n}$$

$$C\sum_{k=0}^{\infty}10^{-Nk} = C \sum_{k=0}^{\infty}(10^{-N})^k$$

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/(1-x) = 1/(1 - 10^{-N})##,

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

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Thank you, that helped a lot. The one thing I just don't really understand is about the proof you showed for a repeating number. Lets say you apply that proof to pi, and you would get about 3110/990 or so, but the numerator still goes on forever. It seems to me to be similar to the effect of circle squaring, you can get so close, but never really there. I feel like it has to do with the special properties of infinity, like 1+2+3... =-1/12. If you have an infinite string of numbers, you can't use finite numbers like 10 or 1000 if you have a repeating string that could be 10^googol^googol^googol long or something like that.

[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.

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- #6

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Rest assured, 1+2+3+...≠-1/12.

- #7

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In case you're not aware, there was a popular video circulating by a respectable physicist who performs a number of invalid operations on divergent series to arrive at this answer as an analytic continuation of the Riemann zeta function.Rest assured, 1+2+3+...≠-1/12.

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.

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- #8

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- #9

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What do you mean by this? There is no reason to apply the technique that micromass presented because ##\pi## is not rational. Since it is not rational, it the decimal expansion does not repeat. The proof that ##\pi## is not rational is separate from what micromass wrote. That just shows that one need not show that ##\pi## never repeats directly, but one only needs to show that ##\pi## is irrational.Lets say you apply that proof to pi, and you would get about 3110/990 or so, but the numerator still goes on forever.

- #10

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What do you mean by this? There is no reason to apply the technique that micromass presented because ##\pi## is not rational. Since it is not rational, it the decimal expansion does not repeat. The proof that ##\pi## is not rational is separate from what micromass wrote. That just shows that one need not show that ##\pi## never repeats directly, but one only needs to show that ##\pi## is irrational.

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

Nugatory

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You may have missed a step in the argument here.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.

1) We know, by micromass's argument, that

2) From #1, we take the contrapositive ("If A then B" implies "If not B then not A") to show

3) We know that

By #2 and #3, we arrive at

Note that we never have to apply micromass's technique to pi.

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- #12

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And this is what someone has yet to give me. A true proof that pi does not repeat. Unless, by some strange theory of higher level math, a non repeating decimal can be rational. Or if there's a different definition of a rational number that I don't know about.We know thatpi is not rational(several proofs available).

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

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I think that we all read your question as "how do we know that irrational functions have non-repeating decimal expansions" which micromass addressed. Gopher's link(s) should fill in the rest, but there is no simple proof. ##\sqrt{2}## would be easier. That is a proof that a high school student should be able to follow.And this is what someone has yet to give me. A true proof that pi does not repeat.

- #15

Nugatory

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We know that pi is not rational (several proofs available).

"Not repeating" and "not rational" are different things.And this is what someone has yet to give me. A true proof that pi does not repeat.

"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_that_π_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.

- #16

mathwonk

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geometric series.

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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

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