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Consider $[0,1]$ with addition modulo one is an abelian group.

Subgroup generated by $\sqrt2$ is dense in $[0,1]$. I think this generator set is $\{\sqrt2 n|n\in \mathbb Z\}$

I know that any closed subgroup of $[0,1]$ which is infinite is equal to $[0,1]$.

But I cannot show for any interval centered at any element in $[0,1]$ intersects with $\{\sqrt2 n|n\in \mathbb Z\}$

I have seen the trick using taking power of $\sqrt2-1$ but this is an additive group and we need to approach it in group way not rings.

Jale'de jaled
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  • If that's the generator set, I can promise you it isn't dense. For example, there is an open set containing $1$ that doesn't contain any element of the generator set. I can only assume the original question meant $\mathbb Q(\sqrt2)$ – Rushabh Mehta Jan 23 '21 at 16:05
  • Why not? Can you eleborate, and Can you give hint about the case when $Q(\sqrt2)$ – Jale'de jaled Jan 23 '21 at 16:37
  • See here or here for a proof of your statement that the subgroup of $\Bbb R/\Bbb Z$ generated by $\sqrt{2}$ is dense (which is true, of course). –  Jan 23 '21 at 17:02
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    @DonThousand: Addition is supposed to be modulo 1. – Vercassivelaunos Jan 23 '21 at 17:03

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