[Math] Maximum length of repeating digits in a decimal

decimal-expansion

Total math novice here. I'm wondering if (as I think should be the case) there is no maximum length of a repeating decimal period, for example:

- 0.33333... (period 1)
- 0.252525... (period 2)
- 0.142857142857... (period 6) 
- etc.

The reason I am curious about this is that I have learned that the decimal expansions of irrational numbers don't terminate or repeat. But if there is no upper limit to the period that a repeating decimal can have, then where is the difference between the two? I am guessing that although the repeting pattern can grow without bound it is always finite (I think?) whereas with an irrational number the pattern never restarts again after a finite period. But it still makes my head hurt 🙂 so I am wondering if there are any explanations for it that I might be able to understand as a non-practitioner.

Thanks for any help!

Best Answer

To answer the first part, yes these period can get arbitrarily large. Take a look at the number: $$ \frac{1}{10^n-1}$$

For $n=1$, this gives $\frac{1}{9}=0.1111...$

For $n=2$, this gives $\frac {1}{99}=0.010101...$

For $n=3$, this gives $\frac {1}{999}=0.001001001...$

The decimal expansion of each number has period $n$ (the decimal expasion is just the repeating pattern of $n-1$ zeros followed by a $1$), and we can make $n$ as big as we would like.

For the second part of your question, the difference is that irrational numbers don't have a period. It's not that they have an infinitely large period after the decimal place; they just don't have a period whatsoever. They never get to a point where they start repeating. If the number ever continues to infinitely repeat itself, then that number is a rational number.

The difference is that every single individual rational number has a finite period. There are rational number with whatever finite period size we want, but there is never a rational number with an infinite period size.

Related Solutions

[Math] How to evalulate a fraction (example 1/998001) accurately to 100 decimal places and then display it in maxima/wxmaxima?

$$ \frac{1}{(1000-1)^2} =\frac{10^{-6}}{(1-\frac{1}{1000})^2} =\frac{10^{-6}}{(1-x)^2} =10^{-6}\left(\sum_{n=0}^{\infty}x^n\right)^2 =10^{-6}\sum_{n=1}^{\infty}nx^{n-1} $$ for $x=.001$, which has no carries in its decimal expansion for $n<1000$. Thus the first $1000$ triplets ($3000$ digits after the decimal place) will be $0.000\;001\;002\;003\;\dots\;996\;997\;999\;000$, which brings us up to the first digit which receives a carry. If you want to see it without the scientific notation, add one before formatting. In sage:

(1+(10^3-1)^(-2)).n(digits=3001) # ignore the leading one in the output!

or in scientific notation:

((10^3-1)^(-2)).n(digits=3001) # note the e-6 at the end, meaning times 10^{-6}

As to the formula above, it can be derived by differentiation: $$ \left(\sum_{n=0}^{\infty}x^n\right)^2 =(1-x)^{-2} =\frac{d}{dx}(1-x)^{-1} =\frac{d}{dx}\left(\sum_{n=0}^{\infty}x^n\right) =\sum_{n=1}^{\infty}nx^{n-1} \qquad\text{for} \qquad|x|<1 $$

[Math] Are there any irrational/transcendental numbers for which the distribution of decimal digits is not uniform

What you mention is not true for all irrational numbers, but for a special subset of them called Normal numbers.

From the wiki article:

While a general proof can be given that almost all real numbers are normal (in the sense that the set of exceptions has Lebesgue measure zero), this proof is not constructive and only very few specific numbers have been shown to be normal.

And

It is widely believed that the (computable) numbers $\sqrt{2}$, $\pi$, and $e$ are normal, but a proof remains elusive.

Note that there are infinitely many irrational numbers that are not normal. In 1909, Borel introduced the concept of a Normal number and proved (with a few gaps resolved later) the following theorem:

Almost all real numbers are normal, in the sense that the set of non-normal irrational numbers has Lebesgue measure zero.

Some additional points of interest: (added with thanks to @leonbloy)

The number of non-normal irrational numbers is uncountable Theorem 4 of this reference.
There is a subset of normal numbers called Abnormal numbers and Absolutely Abnormal numbers which are uncountable. Abnormal numbers are not normal to a given base $b$ while Absolutely Abnormal numbers are not normal to any base $b \ge 2$.

Best Answer

Related Solutions

[Math] How to evalulate a fraction (example 1/998001) accurately to 100 decimal places and then display it in maxima/wxmaxima?

[Math] Are there any irrational/transcendental numbers for which the distribution of decimal digits is not uniform

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