How prove $0.a_{1}a_{2}a_{3}a_{4}\cdots $ is irrational. Question:

Define sequence $\{a_{n}\}$,and such $$a_{n}\in \{0,1,2\},a_{n}\equiv \binom{2n}{n}\pmod 3$$
show that:
$0.a_{1}a_{2}a_{3}a_{4}\cdots $ is irrational.

My idea: I found
$$a_{1}=2,a_{2}=0,a_{3}=2,a_{4}=1,a_{5}=0,a_{6}=0,a_{7}=0,a_{8}=0,a_{9}=1$$
then I can't prove it.Thank you
 A: Let us prove that there are arbitrarily long sequences of consecutive $0$s in the decimal expansion. This will prove that the expansion is aperiodic and that the number is irrational.
Consider Kummer's theorem which says that the largest power of $p$ divides the binomial coefficient $\binom{n}{m}$ is given by the number of carries when $m$ is added to $n-m$ in base $p$. An immediate corollary of Kummer's theorem for our purposes is this:
Lemma: We have
$$\binom{2n}{n}\equiv 0\pmod{3}$$
if and only if the base-$3$ expansion of $n$ contains a $2$.
Proof: By Kummer's theorem, the highest power of $3$ dividing the central binomial coefficient is equal to the number of carries when $n$ is added to itself in base-$3$. A $2$ in the expansion necessitates a carry and so $3$ must divide $\binom{2n}{n}$. 
Conversely, if the base-$3$ expansion of $n$ contains no $2$ then there will be no carries. Therefore $3$ does not divide $\binom{2n}{n}$. $\square$
Now the sequence of numbers from $2\cdot 3^k$ to $3^{k+1}-1$ all have base $3$ expansions beginning with a $2$ and so form a block of consecutive $0$s of length $3^k - 1$ in the expansion of our number. This means that our number is necessarily irrational. 
A: I would suggest using Lucas Theorem http://en.wikipedia.org/wiki/Lucas%27_theorem to show that there are arbitrary number of consecutive terms in the sequence $a_i$ that will be equal to $0$ (or maybe $1$ or $2$) this way you can show that the number $0.a_1a_2a_3 \ldots$ will not have a recurring decimal and hence irrational.
Note: when I say arbitrary number of consecutive terms: I mean a number like $0.1001000100001 \ldots$ have arbitrary zeros in between $1'$s hence the decimal digits cannot be periodic.
