# 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?

-
Try to represent it as a fraction $p/q$ with $p,q\in\mathbb Z$ and then reach a contradiction. –  Ian Coley Feb 6 '14 at 17:55
A number is rational if and only if it's decimal expansion is eventually repeating by means of a geometric series. –  nayrb Feb 6 '14 at 17:56
That sounds pretty rigorous to me. What more exactly do you think you need to do? Perhaps you could explain why it's not repeating. –  Nate Feb 6 '14 at 17:58
@Nate And why is it not repeating? –  user2369284 Feb 6 '14 at 17:59
–  MJD May 2 '14 at 18:50

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.

-

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.$\square$

-

Hint:

2 is rational
$0.1010010001\ldots$ is easy to show irrational (proof in first chapter of Rudin IIRC)

rational + irrational = irrational

-
This is a good hint. +1 –  Jeel Shah Feb 6 '14 at 18:05
@Newb How can I show 0.1010010001… is irrational(I don't even know that). –  user2369284 Feb 6 '14 at 18:08
@tgymasb iirc means "if i recall correctly" –  Newb Feb 10 at 17:26
@user2369284 Try supposing that the decimal expansion is periodic with some period $p$. You know that the length of the strings of $0$'s grows from each string to string. So what happens when the length is larger than that period? Can this happen for any period? Try to use this to conclude it isn't rational. –  Alfred Yerger Feb 16 at 15:46

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.

-