Let $ \alpha $ is fixed irrational number. Let $ [-h,h] $ be an interval. Is it true that for any sequence of irrational numbers $ \{h_n\} $ converges to zero in $ [-h,h] \, $ both $ \, \alpha + h_n \, \mbox{and} \, \alpha - h_n \, $ are always irrationals ?
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You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it. – 5xum Jan 28 '19 at 12:55
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Most certainly not.
In fact, take any sequence $q_n$ of rational numbers in $[\alpha - h, \alpha + h]$ that converges to $\alpha$ [1]. Now, define $h_n=q_n-\alpha$. Then,
- $h_n$ is in $[-h,h]$.
- $h_n$ converges to $0$.
- for all $n\in\mathbb N$, $h_n$ is irrational
However, $\alpha + h_n = q_n$, which means that $\alpha+h_n$ is always rational.
[1] There are plenty such sequences, for example, $q_n=\alpha + \frac{1}{n+M}$ where $M$ is large enough so that $q_1\in[\alpha-h,\alpha+h]$).
5xum
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