Sequences of Rationals and Irrationals 
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*Let $(x_n)$ be a sequence that converges to the irrational number $x$. Must it be the case that $x_1, x_2, \dots$ are all irrational?

*Let $(y_n)$ be a sequences that converges to the rational number $y$. Must $y_1, y_2, \dots$ all be rational?
This was one of my midterm questions yesterday and I just wanted to clarify my responses. For (1), I said NO and as a counterexample, gave the sequence
$$
(x_n) = (3, 3.1, 3.14, 3.141, 3.1415, \dots)
$$
that converges to $\pi$ (note that each $x_j \in \mathbb{Q}$ since it is a finite decimal expansion). For (2), I said YES but was not sure how to prove it.
Could anyone verify these responses and if I'm correct about (2), offer a proof for why it must be true.
 A: Lemma 1: For any real number $x$, there exists a sequence of rational numbers $(x_n)$ such that $\lim (x_n) = x$.
Proof: take e.g. $x_n$ to be the decimal expansion of $x$ truncated to $n$ digits.
Lemma 2: For any real number $x$, there exists a sequence of irrational numbers $(y_n)$ such that $\lim (y_n) = x$.
Proof: The idea is to start with rationals and “tweak” them to be irrational without changing the limit. This is a common technique in analysis: take a sequence that converges where you want but doesn't have the property you want, and tweak it into another sequence with the same limit and with that desired property. Take $(x_n)$ from lemma $n$ and define $y_n = x_n + \frac{\sqrt 2}{n}$. Since $\sqrt 2$ is irrational, every $y_n$ is irrational. Since $\lim (x_n) = x$ and $\lim \big(\!\frac{\sqrt 2}{n}\!\big) = 0$, we have $\lim (y_n) = x$.
The property “for any $x$, there is a sequence that converges to $x$ and takes values in the set $S$” is known as “$S$ is dense”. These lemmas (which are a little stronger than the properties in your exercise) show respectively that $\mathbb Q$ and $\mathbb R \setminus \mathbb Q$ are dense in $\mathbb R$.
A: For (1), observe how $\mathbb{R}$ is usually constructed - as the set of all finite limit points of sequences in $\mathbb{Q}$. So not only is (1) wrong, you have that for every $x \in \mathbb{R}$ there is a sequence $(a_n)$ contained in $\mathbb{Q}$ such that $\lim_{n\to\infty}a_n = x$.
For (2) - if that was true, then only sequences of rational numbers could converge to zero. Yet, rather obviously, $\lim_{n\to\infty} \frac{\alpha}{n} = 0$ for every $\alpha \in \mathbb{R}$.
A: Both should be no. For (2), just divide by $\pi$ of your first example.
A: Here's a set of numbers that approach $1$ and yet are all irrational:
$(1.31415926\dots,1.03141592\dots,1.00314159\dots,1.00031415\dots,\dots)$
