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Show that between any two numbers in $\mathbb{Z[\sqrt2]}$, there is another number in $\mathbb{Z[\sqrt2]}$.

{$(-1+\sqrt2)^1,(-1+\sqrt2)^2,...,(-1+\sqrt2)^n$} represents an infinite sequence of numbers in $\mathbb{Z[\sqrt2]}$ that approaches 0 from the right.

Therefore, all elements of this sequence lie within the interval, (0,$\sqrt2$] which shows that there exists a number in $\mathbb{Z[\sqrt2]}$ between 0 and $\sqrt2$.

I have been able to extend this logice to show that there exists a number in $\mathbb{Z[\sqrt2]}$ that lies between 1 and $\sqrt2$.

Can the logic between extend to apply to the general case?

S. A. Lloyd
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    What have you tried? – Thomas Andrews Oct 23 '22 at 05:22
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    I apologize for submitting an incomplete question. This being my first question and attempt to use MathJax, I misunderstood the preview in the edit screen, and submitted the question to see that my use of MathJax was producing the expressions as I expected. I have since provided my initial thoughts toward a solution. – S. A. Lloyd Oct 23 '22 at 05:29
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    Well if you can show that you can find an element of $\mathbb Z[\sqrt 2]$ of as small a size as you wanted, can you see how this would solve your problem? Next, can you see how your construction of numbers limiting to $0$ helps this? – Arkady Oct 23 '22 at 05:39
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    After your edit and clarification, it is clear you have put genuine effort toward solving your problem, and have written a quality question. My downvote is now an upvote. Welcome to MSE. : ) – Christian E. Ramirez Oct 23 '22 at 06:02
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    The convergence of that sequence shows that there exists an element of $\Bbb{Z}[\sqrt2]$ in any interval of the form $(0,\varepsilon)$ where $\varepsilon>0$. – Jyrki Lahtonen Oct 25 '22 at 04:27
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    Related. Could be a duplicate, but my accepted answer does not really address the current question at all. Rather, at the time I felt that proving $0$ to be a one-sided accumulation point will imply all the other claims in a straightforward manner. Here that question/answer takes the role of the starting point. Tempora mutantur. – Jyrki Lahtonen Oct 25 '22 at 05:20

1 Answers1

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Let $a,b \in \mathbb Z[\sqrt2]$ where $a > b$.

Let $c = a - b$ where $c < 0$.

Let $D = \{b+(-1+\sqrt2)^1,b+(-1+\sqrt2)^2,\dots,b+(-1+\sqrt2)^n\}$ represents an infinite sequece in $\mathbb Z[\sqrt2]$ that approaches $b$ from the right.

Since $\lim_{n\to\infty}{d_n - b} = 0$, $\exists$ $d \in D$ such $d - b < c$.

Therefore, $\exists$ $d \in D$ such that $b < d < a$.

In conclusion, there exists a number in $\mathbb Z[\sqrt2]$ that lies between any two numbers in $\mathbb Z[\sqrt2]$.

S. A. Lloyd
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    I don’t think this is quite right but very close. Please look through and verify once more. – Arkady Oct 24 '22 at 02:50
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    Try the sequence $(b+(\sqrt2-1)^n)_{n\ge0}$ instead. – Jyrki Lahtonen Oct 25 '22 at 04:30
  • My initial proof was flawed. Rather than using the sequence to prove that there existed a postive value less than the difference of a and b, the sequence can be used directly to prove that there exists a value that lies between b and a. Thanx, Arkady and Jykri for your suggestions. – S. A. Lloyd Oct 25 '22 at 05:12
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    A piece of advice : the notations for the sequences are not very "simple": there is one that seems to me to be appropriate here, it is $(1,\sqrt2-1,(\sqrt2-1)^2,...,(\sqrt2-1)^n,...)$. For example, Jacques Dixmier writes them down in his famous book for undergraduates. – Stéphane Jaouen Oct 25 '22 at 05:47