Question: 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?

Answer: 
> 
> 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:
$$\frac{a\_1 + a\_2 + a\_3 + \cdots + a\_n}{n} \geq \sqrt[n]{a\_1\cdot a\_2 \cdot a\_3 \cdot \cdots \cdot a\_n} $$
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 $\sqrt 2$.