# Finite string of numbers

• Nov 17th 2009, 12:35 AM
redsoxfan325
Finite string of numbers
Question: For any pair $(x,n)$, where $x$ is an irrational number such that $0, so $x=\sum_{i=1}^{\infty}\frac{x_i}{10^i}$, and $n$ is a natural number, so $n=\sum_{i=0}^m 10^i n_i$, what's the probability that the finite string of digits in $n$ will be contained in the decimal expansion of $x$? In other words, what's the probability that $\exists N$ such that $x_{N+k}=n_k$ for all $k\in[0,m]$?

Extra Questions: 1.) What's the probability that the string of digits in $n$ occurs in $x$ twice? $r$ times? Infinitely many times?
2.) How does the answer change (if at all) if we allow $x$ to be any real number?

Example: Say $x=\sqrt{2}-1$ and $n=1356237$. Then you have $\sqrt{2}-1=0.4142\textbf{1356237}31...$ so the string of digits of $n$ does occur in $x$.

Any ideas?
• Nov 17th 2009, 12:53 AM
aman_cc
Quote:

Originally Posted by redsoxfan325
If you take the decimal expansion of any irrational number less than one, i.e. $0.a_1a_2a_3a_4...$ and any finite string of digits $n_0n_1n_2...n_m$ (i.e. a natural number), will you always find that exact string somewhere in the decimal expansion of the irrational number. In other words, does there exist $N$ such that $a_{N+k}=n_k$ for all $k\in[0,m]$?

For instance, say you are given the irrational number $\sqrt{2}-1$ and a string of numbers $1356237$. Then you have $\sqrt{2}-1=0.4142\textbf{1356237}31...$

Intuitively, it seems like you should be able to, but I have no idea how to justify that claim.

Any ideas?

Hi - A very interesting question indeed.
My thoughts:
We know that any rational has a repeating decimal representation.
So consider
x=0.121221222122221222221222222122222221......
This can never be repeating hence x is an irrational number.
Now consider, say, 3. You will never find it in the decimal expansion of x. So I would say claim is false.

Does it sound good?
• Nov 17th 2009, 12:57 AM
redsoxfan325
Quote:

Originally Posted by aman_cc
Hi - A very interesting question indeed.
My thoughts:
We know that any rational has a repeating decimal representation.
So consider
x=0.121221222122221222221222222122222221......
This can never be repeating hence x is an irrational number.
Now consider, say, 3. You will never find it in the decimal expansion of x.

Does it sound good?

Ah yes, true. I should refine my question then.

See above for refined question.

I feel like this is one of those questions where the answer is either zero or one hundred percent.
• Nov 17th 2009, 01:05 AM
aman_cc
Quote:

Originally Posted by redsoxfan325
Ah yes, true. I should refine my question then.

If you pick an irrational number (less than one) out of a (large) hat and a natural number out of a hat, what's the probability that the finite string will be contained in the decimal expansion?

I feel like this is one of those questions where the answer is either zero or one hundred percent.

Sorry but I guess your question is not defined well. What is 'large' hat? And I feel this is going to be one difficult question.

An answer based on pure/raw/unsupported gut feel is 0.
• Nov 17th 2009, 01:08 AM
redsoxfan325
Quote:

Originally Posted by aman_cc
Sorry but I guess your question is not defined well. What is 'large' hat? And I feel this is going to be one difficult question.

A answer based on pure/raw/unsupported gut feel is 0.

The 'large' part was a joke.
• Nov 17th 2009, 02:38 AM
gdmath
Hi redsoxfan325,

I work on arbitary arithmetic patterns.

One of my concerns is if an irrational number, defined in the following form
$
r=\sum_{i=1}^{\propto}a_i
$

Does have "random distribution" (or more proper "normal distribution") on its decimal digits.

I think that your question does have deeper relation to this condition

However, as aman_cc noticed, we must be very carefull to the mathematic definitions that we use to define the problem. In fact this is one of my biggest concerns in my work.

We keep in touch.
• Nov 17th 2009, 03:05 AM
redsoxfan325
I wasn't trying to be mathematically precise when I worded the second question, but I have edited so that I think it is.
• Nov 17th 2009, 05:26 AM
gdmath
• Nov 17th 2009, 05:34 AM
aman_cc
Quote:

Originally Posted by gdmath

The question posed appears pretty tough. I don't even know how to start. :)
• Nov 17th 2009, 06:48 AM
Drexel28
Isn't this a very tough question? It depends on whether or not the digits of the number are random. Suppose that $x$ has random digits. Define $\sigma_n=\frac{\text{number of }\ell\text{s in the first n digits}}{n}$ where $0\le \ell \le 9$. Then $\lim_{n\to\infty}\sigma_n=\frac{1}{10}$. So we must theoretically only look at the first ten numbers to find an occurence of $\ell$. If $\ell$ were a string of $k$ digits we would theoretically have to only look ath the first $10^k$ digits to find an occurence of that string.
• Nov 17th 2009, 08:35 AM
chiph588@
It's unfortunate, but mathematicians don't know that much about normal numbers and hence I believe your question can't be answered...
• Nov 17th 2009, 09:05 AM
redsoxfan325
It was just something I was thinking about last night, so I figured I'd post it and see what people thought. It's certainly possible that it's too difficult to solve.

Though the fact that almost all real numbers are normal seems to indicate (intuitively, not mathematically) that there's a high chance of a string of digits appearing in the expansion.