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

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 1

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: $$\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} \geq \sqrt[n]{a_1\cdot a_2 \cdot a_3 \cdot \cdots \cdot a_n} $$

Answer 2

Consider $(a_n)$ given by

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

Answer 3

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 $0$.

Answer 4

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 $\sqrt 2$.