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Prove the following.

(1) $\mathbb{Z}[\sqrt{2}]$ is a dense subset of $\mathbb{R}$.

(2) $1+\sqrt{2}$ is the smallest unit $>1$, that is, no unit $\alpha\in\mathbb{Z}[\sqrt{2}]$ can satisfy $1<\alpha<1+\sqrt{2}$.

(2) Every unit $u\in\mathbb{Z}[\sqrt{2}]$ is a power $u=(1+\sqrt{2})^{m}$, for some $m$.

I'm not sure how to solve (1) and (2). For part (3) it is easy to show that every power is a unit (just use that norm is multiplicative), but I'm not sure that the powers consist of every unit...

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What's your question? –  Don Larynx Sep 27 '13 at 22:32

2 Answers 2

Hints:

Prove that any subgroup of (the additive) group of the reals is either dense or cyclic, and now prove $\;\Bbb Z[\sqrt2]\le\Bbb R\;$ ...and perhaps the following can help, too:

$$|-1+\sqrt2|\le\frac12\implies |-1+\sqrt2|^n\xrightarrow[n\to\infty]{}\;?\ldots$$

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Thanks, I've figured out part (2) and (3) now, but could someone give me help on (1)? –  Rodriguez M. Sep 27 '13 at 22:44
    
@RodriguezM. (1) is the part that this answer is addressing... –  anon Sep 27 '13 at 22:47
    
Highlights: suppose $\;0\neq H\le\Bbb R\;$ . Then either $\;H\;$ has a minimal positive element or it hasn't such. In the second one $\;H\;$ is dense, and in the first one it is cyclic...can you take it from here? –  DonAntonio Sep 27 '13 at 22:48
    
Yes, the dense part was wrong. About finitely generated I think it is unnecessary. –  DonAntonio Sep 27 '13 at 22:52

The numbers of the form $a+b\sqrt{n}$ where $a, b$ for $n\equiv 2,3\pmod 4$ and $a+b, a-b$ for $n\equiv 1\pmod 4$ are integers, form an integer system on the hypercomplex plane, in much the same way as Gaussian and Eisenstein integers form units on the complex plane. The usual complex-number tricks, like conjugate and modulus work, but the outcomes are different. You have both positive and negative modulus, and you have things like unit hyperbolae, rather than circles. But putting $j=-j$ to get conjugates are the same, and $j^2=+1$ here, rather than $i^2=-1$.

The argument is unaffected if $j^2=+n$, save writing $j\sqrt{n}$ everywhere.

The modulus-square $(a+jb)(a-jb)$ gives an integer (+ or -), and therefore if some prime $p \mid (a+jb)(a-jb)$, then $p$ will divide the modulus-square of every multiple of $a+jb$.

From here, if there are several units of the form $a+jb$, such that $\alpha\alpha^{-1} = (a+jb)(a-jb)=1$, and $\beta\beta^{-1}=(A+jB)(A-jB)=1$, etc, then there must be some sort of number of the form $1 \le \alpha^n \beta \lt \alpha$

The lattice is sparse on the hypercomplex plane, so it's relatively easy to search along the unit hyperbola between $1$ and $\alpha$, but one might note that if $\alpha$ and $\beta$ are not both powers of the same number, then there are an infinitely dense number of units on the unit hyperbola. Since the hypercomplex plane is sparsely populated, one can therefore assume some minimal unit greater than 1 exists.

For the case where $j^2=2$, this unit is $1+\sqrt{2}$, in general, Pell's equation provides either the unit or its square or cube. For example, Pell's equation produces the cube of the unit for $\mathbb{Z}[\sqrt{21}]$.

One can demonstrate the density of hypercomplex numbers, by noting that every hypercomplex number system can be written in a kind of base-notation, with various digits.

For example, the system $\mathbb{Z}[\sqrt{2}]$ can be written in a base $1+\sqrt{2}$, where the digits are $1$ and $q=\sqrt{2}$. The two exclusion rules are that 1+q = 1.0, and 1+1.0 = q.0. It is possible to show that every number of the form $\pm a\pm b\sqrt{2}$ reduces to a correctly ordered number in this base. Since there is a member of the set between any other member and any non-member, the set is infinitely dense.

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@YACP The difference is that we're trying to emphisise that it's not -1, but +1. Mathematically there's no difference. People are likely to read $j^2=-1$ and that a typo might exist. –  wendy.krieger Sep 28 '13 at 11:27

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