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Problem. Let $\left \{ a_{n} \right \}_{n\in \mathbb{N}}$ be a sequence where $a_{n}\in \mathbb{N}$, $a_{n}<a_{n+1}$, $a_{n+1}\leqslant 10\cdot a_{n}$ for $\forall n\in \mathbb{N}$. What type of number is $x=0.a_{1}a_{2}a_{3}...a_{n}...$ (concatenation)? (I.e. rational or irrational)

Example: $a_{1}=1, a_{2}=10, a_{3}=100, ... a_{n}=10^{n-1}$ and $x=0.110100...10000...$

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What does $x=0.a_{1}a_{2}a_{3}...a_{n}...$ mean when $a_n$ can be greater than 9, and will certainly be for $n>10$? –  Henry Jan 10 '12 at 14:33
    
Example $a_{1} = 1, a_{2}=10, a_{3}=100 ... a_{n}=10^{n-1}$ and $x=0.1101001000...100000...$ –  rtybase Jan 10 '12 at 14:38
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So concatenation –  Henry Jan 10 '12 at 14:51
    
What do you know about the decimal expansion of a rational number? Can you find a sequence that produces such an expansion? –  Marc van Leeuwen Jan 10 '12 at 15:00
    
One way possible way to address the problem is using a result from en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem which says: α - irrational <=> for ∀ε>0, there ∃ infinite many integers m,n so that 0 < |mα - n| < ε. –  rtybase Jan 10 '12 at 15:05

1 Answer 1

I assume $x=0.a_1 a_2 a_3...a_n...$ just means to concatenate all the $\{a_{n}\}$ to form the decimal expansion of $x$. So the number could be $$0.123456789101112...\qquad(a_{n+1}=a_{n}+1),$$ or $$0.110100100010000...\qquad(a_{n+1}=10a_{n}),$$or whatever. What it can't be is ultimately periodic.

Let $l_{i}$ be the length (in digits) of $a_{i}$. Since $a_{n+1}>a_{n}$, we have $a_{n}\rightarrow\infty$ and hence $l_{n}\rightarrow\infty$; and since $a_{n+1}\le 10a_{n}$, we have $l_{n+1}\le l_{n}+1$. In other words, the sequence of lengths $\{l_n\}$ grows indefinitely without skipping any values. Now suppose $x$ is periodic with period $P$ after the $N$-th digit. Then there must be some $M\ge N+1$ for which $l_{M}$ is a multiple of (say, $k$ times) the periodicity $P$. Hence, $a_M$ falls in the periodic portion of $x$ (since it starts at digit $(\sum_{i=1}^{M-1}l_i) \ge M-1 \ge N$ after the decimal point), and it covers exactly $k$ repeats of the periodic substring. Now, the next $l_{M+1}$ digits of $x$, either $kP$ or $kP+1$, must coincide with $a_{M+1}$. But neither of these cases is possible: the next $kP$ digits are just $a_{M}$ again, and we know that $a_{M+1}>a_{M}$; while the next $kP+1$ digits are $a_{M}$ with a single nonzero digit (the first digit of $a_{M}$) appended, i.e., a number that is strictly greater than $10a_{M}$, and we know that $a_{M+1}\le 10a_{M}$.

We've shown that $x$ can't be ultimately periodic, and hence it must be irrational.

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mjqxxxx's first example is called the (base-10) Champernowne number. –  J. M. Jan 10 '12 at 15:37
    
I didn't understood the proof completely, but I got the idea that to prove a number rational, we need to prove some Periodicity into the decimal. And if we prove it is not there, the number is irrational. I know these are very basics, and themathpage.com/aprecalc/… link also helped. Thank you :) –  user840077 Jan 11 '12 at 5:09

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