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I am confused in my thoughts about the irrational numbers in real line. My confusion is: If $x\in$$\mathbb R$$-\mathbb Q$ then for $\epsilon>0$ as small as you please, the element ($x+\epsilon$) belongs to $\mathbb Q$ or $ \mathbb R$$-\mathbb Q$ ?

  • My thought(which confused me): (1)As any interval of $\mathbb R$ doesn't contain only irrationals (or rationals), it suggests that irrationals are disconnected and hence the nearest neighbor of an irrational number must be rational thereby making me think that ($x+\epsilon$) is rational.
  • (2) If I am correct in (1), then what are the unique $p$ and $q$ s.t. $x+\epsilon=\frac{p}{q}$ where $gcd(p,q)=1$? I took $x=√2$ but unable to find such $p$ and $q$. Kindly help.
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    $\begingroup$ Both the rational numbers and the irrational numbers are dense. It is possible then that $x+\epsilon$ will be rational. It is also possible that $x+\epsilon$ will be irrational. It depends on the specific value of $\epsilon$ in relation to $x$. $\endgroup$ – JMoravitz Jun 25 '16 at 15:22
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    $\begingroup$ The short answer is: It depends on $\epsilon$. It could be a rational number, it could be an irrational number, and there is no way to tell in advance which is which. You just have to calculate $x + \epsilon$ and see. $\endgroup$ – Arthur Jun 25 '16 at 15:23
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    $\begingroup$ There are both rationals and irrationals as close to $x$ as desired. Think of the decimal expansion of $x$...you can keep the first $N$ the same and then have all $0's$ after (to make a rational extremely close to $x$) or , after the first $N$, you can replace the digits of $x$ with those of your favorite irrational to get an irrational extremely close to $x$. $\endgroup$ – lulu Jun 25 '16 at 15:24
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    $\begingroup$ Also, you say "the nearest neighbor of an irrational number"... but there is no nearest... If you suggest a number $y$ to be the nearest number to $x$ while being unequal, then I will show you $\frac{x+y}{2}$ which is nearer. $\endgroup$ – JMoravitz Jun 25 '16 at 15:24
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You say "the nearest neighbor of an irrational number". But there is no such thing. If you suggest a number $y\ne x$ as the nearest number to $x$, then you must be wrong because $\frac{x+y}{2}$ is nearer. [Thanks to @JMoravitz ]

You can get both rationals and irrationals as close to $x$ as you wish. Think of the decimal expansion of $x$ ... you can keep the first $N$ places the same and then all $0$s to get a close rational or you can replace the remaining digits with those from $\pi$ to get a close irrational. [Thanks to @lulu ]

Another way of looking at this is that both the rational numbers and the irrational numbers are dense in the reals.

So the short answer to your question is it could be either: $x+\epsilon$ could be rational or irrational depending on the value of $\epsilon$. [Thanks to @Arthur ]

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  • $\begingroup$ Bother, I still haven't got the hang of displaying hyperlinks properly. $\endgroup$ – almagest Jun 25 '16 at 15:44
  • $\begingroup$ very smart though...@almagest. $\endgroup$ – Nitin Uniyal Jun 25 '16 at 15:45
  • $\begingroup$ @almagest no spaces between the square braces and parenthesis. Also http:// is a must. Adding a bit more words...[google](http://www.google.com) displays as google. $\endgroup$ – JMoravitz Jun 25 '16 at 15:46
  • $\begingroup$ Many thanks! Maybe next time it will work! $\endgroup$ – almagest Jun 25 '16 at 15:47

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