Prove that 2.101001000100001... is an irrational number. My try:
This number is non-terminating and non-repeating, so this is an irrational number.
But how do I prove it more formally in a more mathematically rigorous way?
 A: Hint:
2 is rational
$0.1010010001\ldots$ is easy to show irrational (proof in first chapter of Rudin IIRC)
rational + irrational = irrational
A: You have $$x=\sum_{n=0}^{\infty}10^{-\binom{n}{2}}$$
If you have a repeating decimal then you have finitely many digits appearing first, followed by strings of any length that repeat (any number of times) later in the representation. This decimal representation is not repeating, since for any $k$, we can identify a string of digits $1\overbrace{00\cdots00}^k1$ that is present in the decimal representation and never appears again. If all such strings were present in the nonrepeating part, we have a contradiction, since the nonrepeating part is only finitely many digits long.
A: Assume that $2.101001000100001…$ is a rational number. Since 2 is rational we have that $(2.101001000100001…-2)=0.101001000100001…$ is also rational.Then there exists $ p,q\in \mathbb{Z} $ (with $ q\neq 0 $) such that $ 0.101001000100001…=\frac{p}{q} $.
That is$$ \sum_{k=1}^{\infty}\frac{1}{10^{\frac{k(k+1)}{2}}}=\frac{p}{q} .$$
Now choose $ n\in \mathbb{N} $ such that $ n=\frac{k_{0}(k_{0}+1)}{2} $ for some $ k_{0}\in \mathbb{N} $ with $ 10^{k_{0}}>q $.
Put $$ x=q10^{n}\left( \frac{p}{q}-\sum_{k=1}^{k_{0}}\frac{1}{10^{\frac{k(k+1)}{2}}}\right). $$
Then $$x=10^{n}p-q10^{n}\sum_{k=1}^{k_{0}}\frac{1}{10^{\frac{k(k+1)}{2}}}$$ and hence $x$ is an integer.
Observe that $$x=q10^{n}\sum_{k=k_{0}+1}^{\infty}\frac{1}{10^{\frac{k(k+1)}{2}}}>0$$and
$$x=q10^{n}\sum_{k=k_{0}+1}^{\infty}\frac{1}{10^{\frac{k(k+1)}{2}}}<q10^{n}\sum_{k=k_{0}+1}^{\infty}\frac{1}{10^{n+k}}=q\sum_{k=k_{0}+1}^{\infty}\frac{1}{10^{k}}<q\sum_{k=k_{0}+1}^{\infty}\frac{9}{10^{k}}=\frac{q}{10^{k_{0}}}<1.$$
This is a contradiction since $x$ is an integer and $0<x<1$. Therefore our assumption is false. Hence $2.101001000100001…$ is irrational.$\square$
A: Let $x = 2 + \sum\limits_{k=1}^\infty10^{-k(k+1)/2}$ be the number at hand.
If $x$ is rational, say $x = \frac{p}{q}$ for some positive integers $p,q$, we can pick a $n > 1$ such that $10^n > q + 1$. It is clear 
$$qx \times 10^{n(n-1)/2} = p \times 10^{n(n-1)/2}$$
is also an integer. However, the fractional part of this number is equal to
$$ \left\{ q \times 10^{n(n-1)/2} \left(2 + \sum_{k=1}^\infty 10^{-k(k+1)/2}\right)\right\}
= \left\{ q \times \sum_{k=1}^\infty 10^{-k(k+2n-1)/2} \right\}
$$
which belongs to $(q \times 10^{-n}, (q+1)\times 10^{-n} ) \subset (0,1)$.
Since $(0,1)$ doesn't contain any integer, this leads to a contradiction and hence $x$ is irrational.
A: It is enough to show this number is not periodic, as a number is periodic if and only if the number is rational. Here is a proof by contradiction:
Assume $x=2.1010010001...$ is periodic, then there exists some numbers $b_1, b_2,...., b_p$ which repeat forever at the end, where $p \ge 1$ is the period of $x$. 
One of these numbers must be 1 as $x$ cannot have recurring zeros at the end. 
If one of these numbers is 1, say $b_i=1$ for some integer i where $1 \le i \le p$ then the number of zeros until the next 1 is at most $p-1$, which remains true for every subsequent 1, which is clearly not possible.
It follows that there are no cases where $x$ has a repeating sequence at the end, which is a contradiction. The number $x$ is therefore not periodic and therefore not rational. 
