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This question came up because I was wondering the following: If the digits of PI are placed in ascending order, what is the <insert-large-finite-number-here>th digit?

I believe that the answer is 0, but I am not sure. The reason why I'm not sure is dependent on the answer to whether an irrational number can have a finite number of a certain digit. If this is the case, then it is conceivable that there could be a finite number of zeros. Taking an example, let us say that we are going to encode every digit of PI through the function $$ f(x) = \left\{ \begin{array}{c} 10+x,\quad when \quad x≠0\\ 99,\quad when \quad x=0 \end{array} \right. $$ That is, for example if I am encoding the digits of $20$, then I'd replace the $2$ with $f(2)$ and the $0$ with $f(0)$ I'd get, as my end result, $1299$. This mapping function seems to totally eliminate zeros from the answer. By doing the same thing to PI, I can totally eliminate its zeros. It also seems conceivable that I can still generate a unique mapping if I apply the function to digits only after the first zero.

So: does my argument hold water? Is it possible to turn an irrational number into one with a finite number of a certain digit? Additionally, is it possible to prove whether an arbitrary irrational number has a finite number of a certain digit?

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So You are in fact asking: does every irrational number...? –  leo Feb 14 '13 at 5:01

3 Answers 3

up vote 4 down vote accepted

The number $0.010010001000010000010000001\ldots$ is irrational and has no instances of any digit other than $0$ and $1$. More generally, pick any two digits $d$ and $e$, and form the number

$$0.\mathrm{dedeedeeedeeeedeeeeed}\dots\;;$$

it never becomes periodic, so it’s irrational. Of course one can insert any finite string of digits between the decimal point and the expansion of an irrational number and still have an irrational number, e.g., $0.345345010010001000010000010000001\ldots$.

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I think the OP meant something like $0.8787100100010000\cdots$. There the $8$ and $7$ occur only twice. Basically we want a sum of a rational plus a "lacunary" irrational, I guess. –  Pedro Tamaroff Feb 14 '13 at 3:28
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@Peter: I thought it obvious that one could prefix any finite string, but I’ll add that. –  Brian M. Scott Feb 14 '13 at 3:29
    
But is it possible to prove whether or not an arbitrary irrational number has a fixed number of a certain digit? –  Maz Feb 16 '13 at 16:43
    
@Maz: Not in general, no. –  Brian M. Scott Feb 16 '13 at 23:08

Yes. For an extreme example, let $a$ have decimal expansion $$0.101001000100001000001\dots.$$ The number of $0$'s between consecutive $1$'s is $1$, $2$, $3$, $4$, $5$, and so on. Thus the decimal expansion of $a$ cannot be ultimately periodic, and therefore $a$ is irrational.

By the way, it is not known whether the decimal expansion of $\pi$ has infinitely many $0$'s. The same applies to any other digit.

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Of course one can add any finite string of numbers before the "binary" sequence to fulfill the OP's request. –  Pedro Tamaroff Feb 14 '13 at 3:27

Yes, and without giving you another example let me instead give you the vocabulary. An irrational number is said to be a "normal" number if each digit, 0 through 9 in base-10, and in each base numbering system, occurs in equal quantities. All irrational numbers must contain at least two unique digits in infinite quantities, but not necessarily in equal proportion.

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Doesn't normality require that each string, not just each digit, be uniformly distributed? Does requiring that each digit be uniformly distributed in EVERY base imply that every srtring is also uniformly distributed? –  Quinn Culver Feb 14 '13 at 13:04
    
Actually at least two must be in the same quantity (in the same way that there are as many even numbers as natural numbers.) Assume that all the digits occur a different number of times. Than one will occur the most. Since the number of digits in an irrational number is $\aleph_0$ the sum of the occurrences will be that. For this to be true, the greatest one must be $\aleph_0$. Then the others must be less, and therefore finite in occurrence. They will run out eventually, and then it will just the greatest occurring digit, which makes it rational, which is a contradiction! –  PyRulez Feb 16 '13 at 17:12
    
Therefore, two must occur the same number of times, $\aleph_0$. –  PyRulez Feb 16 '13 at 17:13
    
@QuinnCulver It turns out that the answer is yes, but this is a non-trivial result - I suggest looking at section 3.5 in volume 2 of Donald Knuth's The Art of Computer Programming, specifically Theorem C there. –  Steven Stadnicki Feb 27 '13 at 23:09

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