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Consider the metric space $\langle \mathbb I,d\rangle$ where $\mathbb I$ is the set of all irrational numbers, and $d$ is the usual distance metric. For each $n\in\mathbb Z^+$, let $x_n =\frac{n + \sqrt 2}{n-\sqrt 2}$. Then:

a. Prove that $\langle x_n\rangle$ is a sequence in $\mathbb I$.

b. As a sequence in $\langle \mathbb I,d\rangle$, prove that $\langle x_n\rangle$ does not converge.

Please help us answer this problem...

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Hello @Bok and welcome to Math StackExchange! It's best if you let us know what you've tried so far so we can help you. What exactly are you stuck on? –  Edward Hughes Dec 6 '12 at 14:29
    
I notice you originally wrote "Q" and said "irrational," but from context it's clear that you meant "irrational." Just be alert in the future that Q is usually used for exactly the opposite of what you meant the rationals. –  rschwieb Dec 6 '12 at 14:31
    
a) For each n, times the numerator and denominator of x_n by (n + sqrt(2)) and see what happens. –  Adam Rubinson Dec 6 '12 at 14:32
    
b) Hint: What is the approximate value of x_n when n is very large e.g. 100000 ? –  Adam Rubinson Dec 6 '12 at 14:34
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1 Answer 1

$\def\Q{\mathbb Q}\def\l#1{\left\langle#1\right\rangle}\def\I{\mathbb I}$To show that $x_n \in \I$ for each $n$, recall that $\sqrt 2 \not\in\Q$. If $p,q \in \mathbb Z$, $q \ne 0$, we have \begin{align*} x_n = \frac pq &\iff q(n + \sqrt 2) = p(n-\sqrt 2)\\ &\iff \sqrt 2(q + p) = n(p-q)\\ &\iff \sqrt 2 = n \cdot \frac{p-q}{p+q} \in \Q \end{align*} As $\sqrt 2 \not\in \Q$ we have $x_n \not\in \Q$ for each $n$.

To see that $\l{x_n}$ does not converge in $\I$, note that $$ x_n = 1 + \frac{2\sqrt 2}{n-\sqrt 2} $$ So we have $x_n \to 1$ in $\mathbb R$. If $\l{x_n}$ would converge in $\I$, to $y$ say, we would have $x_n \to y$ in $\mathbb R$ also, hence (by uniqueness of limits), $y = 1$. But $1$ is rational. So $\l{x_n}$ cannot converge in $\I$.

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