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We can read from various sources that $\mathbb{Q}(\sqrt{2})$ has class number one, and that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain. However, it also has a group of units: $\mathbb{Z}[\sqrt{2}]^\times =(1 + \sqrt{2})^\mathbb{Z}$. Let's try an example:

  • $79 = (9 \times 9) - 2 \times (1 \times 1) = (9 + \sqrt{2})\times(9 - \sqrt{2})$

  • $41 = (7 \times 7) - 2 \times (2 \times 2) = (7 + 2 \sqrt{2})\times(7 - 2 \sqrt{2}) $

This is just a quick way to get two relatively prime numbers in the ring of integers $\mathcal{O}_K$. Can we get a continued fraction for the ratio of the two of them?

$$ \frac{7 + 2 \sqrt{2}}{9 + \;\,\sqrt{2}} = \frac{59}{79}+ \frac{25}{79}\sqrt{2}\in \mathbb{Q}(\sqrt{2})$$

So now I have to qualify a bit what I mean by Eucliean algorithm. The norm $N\big(a + b \sqrt{2}\big) = a^2 - 2b^2$. Then I guess we have

$$ y = m x + b $$

with $m \in \mathbb{Z}[\sqrt{2}]$ and $N(b) < N(x)$. And then we iterate the Euclidean algorithm until we obtain the GCD. Hopefully that is right?

I guess we'd have to flip them:

$$ \frac{9 + 1 \times \sqrt{2}}{7 + 2 \times \sqrt{2}} = \frac{59}{41} + \frac{9}{41} \sqrt{2} \approx 1 \in \mathbb{Q}(\sqrt{2})$$

This is a really crummy approimation. Let's try a "subtractive" Euclidean algorithm...

  • $(9 + \sqrt{2}) - (7 + 2 \sqrt{2}) = 2 - \sqrt{2} < 1 $ looks good.
  • $(9 + \sqrt{2}) + (7 + 2 \sqrt{2}) = 16 + 3 \sqrt{2}$ Looks worse. $16 \times 16 - 3 \times 2 \times 2 > 100 $.

And two other possibilities:

  • $(9 + \sqrt{2}) - \sqrt{2} \times(7 + 2 \sqrt{2}) = 5 - 6 \sqrt{2}$
  • $(9 + \sqrt{2}) + \sqrt{2} \times (7 + 2 \sqrt{2}) = 13 + 8 \sqrt{2}$

and I've started to conclude the quotient should be $m=1$. The first step could read:

$$ 9 + \sqrt{2} = 1 \times \big(7 + 2 \times \sqrt{2}\big) + \big( 2 - \sqrt{2}\big)$$

I have my doubts. We could have chosen another basis $\mathbb{Z}[\sqrt{2}] = 1 \mathbb{Z}\oplus (\sqrt{2}-1)\mathbb{Z}$ as a $\mathbb{Z}$-module.

cactus314
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  • For a start, why not rationalize the denominator of $(7+2\sqrt 2)/(9+\sqrt 2)$ by multiplying its numerator and denominator by $9-\sqrt 2;$? – DanielWainfleet Aug 28 '18 at 15:54
  • Your concept looks correct - you should be able to just run the Euclidean algorithm to do it. (It's not immediately clear to me how you find your $m$, mind you, but I think that's only a sign that I need coffee...) – Steven Stadnicki Aug 28 '18 at 23:28
  • @StevenStadnicki Right. what is the Euclidean algorithm here? one certainly exists. i don't think it's straightforward. just my opinion. – cactus314 Aug 29 '18 at 01:30
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    Did you look at https://math.stackexchange.com/questions/2524792/gcd-of-two-elements-in-mathbb-z-left-frac1-sqrt-112-right (which is over in the 'related' column)? It suggests looking at an 'approximate' division of the two items, which in your case is $(9+\sqrt{2})/(7+2\sqrt{2})=59/41-(11/41)\sqrt{2}$ as a sort of guiding principle. More generally, the set of multiples of any number in $\mathbb{Z}[\sqrt{2}]$ forms a lattice and it shouldn't be too hard to find the 'closest' of those multiples to your target number (though the distance in question isn't exactly 'Euclidean' in the other sense). – Steven Stadnicki Aug 29 '18 at 05:19
  • math.SE suggests this one is related https://math.stackexchange.com/q/1856408/4997 – cactus314 Aug 29 '18 at 13:28

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