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The original question: If we have a sequence of real numbers and on that, we define a sequence of $n^{th}$ rational numbers and another of irrational numbers, then they are both subsequences of the original sequence of real numbers?

In other words, can a rational sequence be a complementary subsequence of an irrational sequence?

That is can there exist a sequence of real numbers which has both a rational subsequence and an irrational subsequence?

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Can a rational sequence and an irrational sequence have same limit?

Yes. Take the sequence of irrational numbers $\frac{1}{\sqrt{p}}$ as $p\rightarrow \infty$, which converges to zero. Now consider the sequence of rational numbers $\frac{1}{p}$, as $p\rightarrow \infty$, which converges to zero.

Can there exist a sequence of real numbers which has both a rational subsequence and an irrational subsequence?

Sure. When constructing sequences, you have great freedom. As an exercise, do a $\delta - \epsilon$ proof that the sequence ${\frac{1}{p_1}, \frac{1}{\sqrt{p_1}}, \frac{1}{p_2}, \frac{1}{\sqrt{p_2}},...}$ converges to zero. Then show each "compliment" sub-sequence converges to zero.

Here's another thing you could maybe think about playing with, the AM-GM inequality: $$\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} \geq \sqrt[n]{a_1\cdot a_2 \cdot a_3 \cdot \cdots \cdot a_n} $$

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  • $\begingroup$ Yeah I did that and put that as an example, but option was given both can't exist together. I was worried I misunderstood it Thanks a lot. I was so scared I'll lose marks. Thank you. $\endgroup$ – Shreya Singh Nov 12 at 12:07
  • $\begingroup$ @ShreyaSingh, if "both can't exist together" means that the two subsequences must be disjoint, then it is the case in the example above. Barry's answer explicitly shows this disjointness, since each rational number is zero. $\endgroup$ – Lmidh Nov 12 at 13:05
  • $\begingroup$ Yeah yeah saw that $\endgroup$ – Shreya Singh Nov 12 at 13:05
  • $\begingroup$ I have a good understanding of the things I understand in those topics; unfortunately, it isn't much by any measure. $\endgroup$ – Lmidh Nov 12 at 13:08
  • $\begingroup$ That isn't what the comment sections are for. If you have more questions, ask them here to the community, and hopefully we can be of help. If your previous questions are getting deleted, then you need experience asking better questions and showing effort in the questions you ask. Taking the time to work through questions, type them up in a clear and concise manner, and when stuck, asking them here will actually do well in preparing you for your exams! Also, along the way, take the knowledge you gained and answer questions on this site. Teaching will solidify your knowledge. Good Luck! $\endgroup$ – Lmidh Nov 12 at 13:27
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Just to give a simple example with a single formula, let

$$a_n={(1+(-1)^n)\sqrt2\over n}$$

The sequence is

$$0,\sqrt2,0,{\sqrt2\over2},0,{\sqrt2\over3},0,{\sqrt2\over4},\ldots$$

which clearly has both a rational and an irrational subsequence, each converging to the same limit, namely $0$.

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  • $\begingroup$ This is awesome! Thanks $\endgroup$ – Shreya Singh Nov 12 at 12:58
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Consider $(a_n)$ given by

$a_{2n}= \frac{\sqrt{2}}{2n}$ and $a_{2n-1}= \frac{1}{2n-1}.$

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  • $\begingroup$ So that implies if we have a sequence of real numbers and on that, we define a sequence of n^th rational numbers and another of irrational numbers, then they are both subsequences of the original sequence of real numbers? $\endgroup$ – Shreya Singh Nov 12 at 11:52
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Consider the sequence $$1, 1.4, 1.41, 1.414, 1.4142\ldots$$ that converges to $\sqrt 2$, and the sequence $$\sqrt 2,\sqrt 2,\sqrt 2,\sqrt 2\ldots$$ that also converges to $\sqrt 2$.

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