Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
Try to represent it as a fraction $p/q$ with $p,q\in\mathbb Z$ and then reach a contradiction. –  Ian Coley Feb 6 at 17:55
1  
A number is rational if and only if it's decimal expansion is eventually repeating by means of a geometric series. –  nayrb Feb 6 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 at 17:58
    
@Nate And why is it not repeating? –  user2369284 Feb 6 at 17:59
    
If it were repeating with "length" $l$, show that the repeating part would be all zeroes. Contradiction. –  David Mitra Feb 6 at 18:00
show 4 more comments

3 Answers 3

up vote 5 down vote accepted

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.

share|improve this answer
add comment

Hint:

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

rational + irrational = irrational

share|improve this answer
    
This is a good hint. +1 –  gekkostate Feb 6 at 18:05
    
@Newb How can I show 0.1010010001… is irrational(I don't even know that). –  user2369284 Feb 6 at 18:08
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.