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.
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 ]