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  1. Let $(x_n)$ be a sequence of rational numbers that converges to an irrational number $x$. Must it be the case that $x_1, x_2, \dots$ are all irrational?

  2. Let $(y_n)$ be a sequence that converges to a rational number $y$. Define such a sequence where $y_1, y_2, \dots$ are all irrational.

Now these are the 2 questions in my textbook i noticed when posting this that someone had asked the answer to 1) and a similar but less difficult version of 2.

A couple things i noted the defined answer for 1 he used $\pi$ and simply added more digits to it for each part of the sequence. so firstly id like to ask would that really be a full mark answer on a analysis final? secondly and more importantly i was wondering if there was a way to define this where the final number wasn't transcendental.

For 1) I used $ (1+1/n)^n $ s.t. $n \in \mathbb{N} $. For every $n$ this sequence is rational but for $\lim_{ n \to \infty }(1+1/n)^n = e $.

My Question: (i) Is this true and if so is there a definable sequence that holds but doesn't define a transcendental number?

EDIT: Golden ratio from Fibonacci sequence.

For 2) I picked $ 2^{1/n} $ s.t. $ n>1$ and $n \in \mathbb{N}$. I believe every value of this sequence is irrational but the limit should be 1.

My Question: (ii) does the above work? and if possible could I have another example of a sequence where every value is irrational but the limit is rational?

EDIT: $\frac{\sqrt 2}{n}$.

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    $\begingroup$ If $(x_n)$ converges to $x$ is the answer to part (i) (and all the $x_n$ are non-zero), then $(x_n/x)$ is a solution to part (ii). $\endgroup$ Commented Sep 14, 2016 at 2:54
  • $\begingroup$ thats very clever thanks! $\endgroup$ Commented Sep 14, 2016 at 3:00

2 Answers 2

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You basically have both answers right, just need a couple of tweaks.

Question:(i) Is this true and if so is there a definable sequence that holds but doesnt define a transcendental number?

It is true that $a_n = (1 + \frac{1}{n})^n$ is a sequence of rationals whose $\lim_{n \to \infty} a_n = e$ is irrational.

For a sequence of rationals that converges to an algebraic (non transcendental) irrational, take for example the partial sums of the Taylor series expansion of $\sqrt{1+x}$ at $x=1$ which converge to $\sqrt{2}$.

Question:(ii) does the above work? and if possible could i have anther example of a sequence where every value is irrational but the limit is rational?

The above works, except you may want to define $a_n = 2^\frac{1}{n+1}$ so that $a_1$ is irrational as well.

For other such sequences, pick any $r \in \mathbb{R} \setminus \mathbb{Q}$ and let $a_n = \frac{r}{n}$.

[ EDIT ] I see that you just edited the question to add $\frac{\sqrt{2}}{n}$ as an example of the latter.

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It doesn't have to be $\pi$ or $e$, you could do the digit by digit approximation and converge to say $\sqrt{2}$ which is algebraic and not transcendental.

I would say that that answer of digit by digit is basically full mark; you could probably include that $|x - x_n| < 10^{-n}$ which means that the sequence is Cauchy to be more specific.

Your second sequence of $\{2^{1/n}\} \to 1$ is also correct. See if you can prove that each number in the sequence is irrational by mimicking the proof that $\sqrt{2}$ is irrational.

Here's another sequence

$$a_n = 1 + \frac{\sqrt{2}}{10^n}$$

We see that $\lim_{n \to \infty} a_n = 1$ and each $a_n$ is irrational because you can create a common denominator and note that the numerator will be irrational and the denominator and integer.

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