I am trying to prove that $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$ is not a complete space (with the standard metric). For that purpose, I am looking for a Cauchy sequence in $\mathbb{Q}[\sqrt{2}]$ which is not convergent in $\mathbb{Q}[\sqrt{2}]$. Can somebody help me? Thanks!
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1can you think of a sequence that converges to an element of $\mathbb R\setminus\mathbb Q[\sqrt2]$? – J. W. Tanner Jan 11 '21 at 18:42
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2For example, try $$3, 3.1, 3.14,3.141,3.1415, \dots$$ – Maximilian Janisch Jan 11 '21 at 18:46
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I think the sequence $x_{n+1}=\frac{2x_n}{3}+\frac1{x_n}$ with $x_n = 1$ converges to $\sqrt{3}$, right? – JN_2605 Jan 11 '21 at 18:47
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That's correct, @JN_2605 – J. W. Tanner Jan 11 '21 at 18:48
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1Now you just need to prove that $\sqrt3\notin\Bbb Q[\sqrt2]$. – Arthur Jan 11 '21 at 18:49
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What is the limit of the sequence $\left(1+\frac{\sqrt{2}}{n}\right)^n$? – rtybase Jan 11 '21 at 18:53
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@Arthur good point! Luckily this is adressed on this site :) https://math.stackexchange.com/q/2692648/631742 – Maximilian Janisch Jan 11 '21 at 18:53
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Consider $1, 2, \dfrac32, \dfrac53, \dfrac85, \dfrac{13}8, ...$
J. W. Tanner
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1To someone not familiar with the theory of field extensions, it is not entirely obvious that $\sqrt5\notin \Bbb Q[\sqrt2]$. – Arthur Jan 11 '21 at 18:48
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if $(a+b\sqrt2)^2=a^2+2b^2+2ab\sqrt2=5$ then $2ab=0$, so $a=0$ or $b=0$; from there it follows that $\sqrt5\not\in\mathbb Q[\sqrt2]$ – J. W. Tanner Jan 11 '21 at 18:55
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Note that $\mathbb{Q}$ is dense in $\mathbb{R}$. Hence, there exists a sequence in $\mathbb{Q}$ which converges to $\sqrt{3}$. This sequence is also contained in $\mathbb{Q}[\sqrt{2}]$ (and it's Cauchy, because it converges) but clearly $\sqrt{3}$ is not.
Thomas Bakx
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"clearly $\sqrt{3}$ is not" I don't think it's that clear to someone who isn't familiar with the theory of field extensions. – Arthur Jan 11 '21 at 18:50
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1As already pointed out in another comment, there is an elementary proof by contradiction. I don't even know what a field extension is. – Thomas Bakx Jan 11 '21 at 18:58
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Not particularly elementary, but useful and also a continuation of my comment ...
Let's take an integer $k$ that is not a perfect square. Then $\sqrt{k}$ is algebraic and irrational (root of the polynomial $x^2 -k$). It turns out that $e^{\sqrt{k}}$ is transcendental (i.e. is not algebraic, i.e. not a root of any polynomial with rational coefficients), the proof is easy, given Lindemann–Weierstrass theorem
... by the second formulation of the theorem, if $\alpha$ is a non-zero algebraic number, then $\{0, \alpha\}$ is a set of distinct algebraic numbers, and so the set $\{e^0, e^{\alpha}\} = \{1, e^{\alpha}\}$ is linearly independent over the algebraic numbers and in particular $e^\alpha$ cannot be algebraic and so it is transcendental.
Any element of $\mathbb{Q}[\sqrt{k}]= \{a+b\sqrt{k}:a,b∈Q\}$ is algebraic. From $x=a+b\sqrt{k} \Rightarrow \frac{(x-a)^2}{b^2}-k$ is the polynomial, with rational coefficients. For $b=0$ the polynomial is $x-a$.
$\mathbb{Q}[\sqrt{k}]$is not complete, because
- $1+\frac{\sqrt{k}}{n}\in\mathbb{Q}[\sqrt{k}]$ for $\forall n\in\mathbb{N}\setminus\{0\}$
- $\left(1+\frac{\sqrt{k}}{n}\right)^n\in\mathbb{Q}[\sqrt{k}]$, easy to check using Binomial_theorem
- But $\lim\limits_{n\to\infty}\left(1+\frac{\sqrt{k}}{n}\right)^n=e^{\sqrt{k}}$ transcendental, thus $e^{\sqrt{k}} \notin \mathbb{Q}[\sqrt{k}]$.
rtybase
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