$0.101001000100001000001$... is irrational. How do I show that $0.101001000100001000001...$ is irrational? How do I generalize this to decimal expansions of a similar type, i.e. what is a criterion for decimal expansions with $0$'s and $1$'s involving long gaps of zeros and which guarantees irrationality?
 A: A well-known, general criterion to decide if a number is irrational or not from its decimal expression is that a number is rational if and only if its decimal expression is finite or periodic.
The decimal expression of the number mentioned in your question has arbitrarily long gaps of zeroes. So it can not be periodic, or much less finite.
A: If $0 < p < q$, where $p,\, q \in \mathbb{Z}$, the decimal expansion for $p/q$ is eventually periodic. Indeed, the first $k$ digits after the decimal point are given by $\lfloor 10^kp/q\rfloor$. Write $m \Delta n$ if $10^m$ and $10^n$ have the same remainder when we divide by $q$; the pigeonhole principle implies$$m \Delta n, \text{ for some }1 \le m < n \le q + 1.$$
If $m \Delta n$, then
\begin{align} q\,|\,(10^mp - 10^np)
&\implies q\,|\,(10(10^mp - 10^np))\\
&\implies q\,|\,(10^{m+1}p - 10^{n+1}p)\implies (m+1)\Delta(n+1),\end{align}
and by induction, $10^{m+k}\Delta10^{m+k}$ for all $k$. Next, note that if $m \Delta n$, and $r$ is the remainder when $10^mp$ and $10^np$ are divided by $q$, then $\lfloor10r/q\rfloor$ is the $(m+1)$st digit of the decimal expansion of $p/q$, and the $(n+1)$st. Thus, for all $k \ge 1$, we conclude that the $(m+k)$th digit and the $(n+k)$th digit of the expansion of $p/q$ are equal. Thus, $p/q$ has eventually periodic expansion with period $n-m$.
Now, the given decimal does not have period $P$ for any $P$ because for arbitrarily large $N$, there is a $1$ in the $N$th decimal place followed by $P$ zeros. More generally, if the decimal expansion of any $x \in \mathbb{Q}$ has arbitrarily long strings of zeros, and repeats with period $K$, then for any $N$, there exists $K$ consecutive zeros after the $N$th spot, so that the $K$ repeating digits are zeros.
A: If $a=0.1010010001\dots$ is rational, then it either has a finite decimal representation, or it's decimal representation repeats with some period $p$.
This decimal is obviously not finite, so if it is rational, the decimal repeats with period $p$.
However, somewhere after the repeating starts, we get a string of $0$'s followed by a $1$ of length greater than $p$. Hence the number is irrational.
A: The given number $\displaystyle\;x = \sum_{n=1}^\infty 10^{-n(n+1)/2}\;$ is irrational.
Given any rational number $r > 0$, we can rewrite it into the form
$$r = \frac{1}{10^a}\left(b + \frac{p}{q}\right)$$
 where $a, b, p, q \in \mathbb{N}$, $0 \le p < q$ and $\gcd(10,q) = 1$. If $p =0$ or $q = 1$, everything is obvious.
Consider the case $0 < p < q$ and let $\varphi(q)$ be the
Euler's totient function.
By Fermat-Euler theorem, we have
$$10^{\varphi(q)} \equiv 1 \pmod q 
\quad\implies\quad
10^{\varphi(q)} - 1 = qs \quad\text{ for some integer } s$$
This leads to a decimal expansion of $\displaystyle\;\frac{1}{q}$, 
$$\frac{1}{q} = \frac{s}{10^{\varphi(q)} - 1} = \sum_{k=1}^\infty s \times 10^{-k\varphi(q)}$$
which repeats every $\varphi(q)$ digits.
Multiply this expansion by $p$ gives us a decimal expansion of $\frac{p}{q}$ which again repeat every $\varphi(q)$ digits. Adding $b$ and then divide by
$10^a$ will introduce some garbage in the front and shift the whole expansion
to the right for $a$ digits. At the end, we continue to have an expansion which is
ultimately periodic with a period which is a divisor of $\varphi(q)$.
Back to our number $x$, one find the sequence of '$0$' in its decimal expansion can be arbitrary long. Given any $q$, one can always a find a sequence of '$0$' longer than $\varphi(q)$. This rules out the possibilities that $x$ is 
a rational number with a corresponding $q$. As a result, $x$ cannot be rational.
In fact, the number $x$ is known to be transcendental. See the refs in my answer to a related question.
A: 
what is a criterion for decimal expansions with 0's and 1's involving long gaps of zeros and which guarantees irrationality?

To give one answer to this more general part of your question, a sufficient criterion for irrationality is the existence of arbitrarily long sequences of $0$s and $1$s in the expansion. From this, you can show aperiodicity, and in turn irrationality. To show that such sequences must be aperiodic, consider such a sequence.
Now, assume towards a contradiction that it eventually has a periodic repeating element $p$. First, you can easily show that $p$ must contain at least one $1$ and one $0$, otherwise the sequence would end in all $0$s or all $1$s, contradicting the existence of arbitrary length $0$s and $1$s. But then, by similar reasoning to the other answers, we use the arbitrary-length $0$s condition to show that $p$ must be contained within some sequence of all $0$s, contradicting the fact that it contains a $1$. So such a sequence is aperiodic, and hence irrational.
Note that this is a sufficient condition, not a necessary one. It can be strengthened. In fact, your sequence doesn't meet this condition (it only has arbitrary length $0$s, not $1$s).
