I've looked for help with this question but I have not found anything, I hope this is not a duplicate.

Define the set $A=\{\mid x+y\sqrt{2}\mid \ : x,y\in \mathbb{Z}\ \mbox{and} \mid x+y\sqrt{2}\mid\gt0 \}$, prove that the infimum of this set is zero.

This is what I've thought: Proving that $[y\sqrt{2}]$ (where $[\cdot ]$ is the floor function) can be close to zero as we want would prove the proposition, so I suppose that this doesn't hold and I want to arrive to the contradiction that for some $y_0$: $[y_0\sqrt{2}]=0$, this would contradict the fact that $\sqrt{2}$ is irrational.

Can someone give me a hand? Thanks! (This problem arised in the context of integer lattices)

  • 2
    $\begingroup$ I would suggest using a combination of the fractional part of integer multiples of $\sqrt 2$ and the pigeonhole principle. If you can find two fractional parts which are close together, can you find a number of the form you are looking for which is close to zero? $\endgroup$ – Mark Bennet Aug 6 '14 at 18:25
  • $\begingroup$ You're right, I've fixed it, thanks. I'm thinking based on the comment from Mark. $\endgroup$ – Daniel Aug 6 '14 at 18:45

Hint: Let $x=-1 + \sqrt 2 \in A$. Then $0<x<1$ and $x^n \in A$ $\forall n \in \mathbb N$.

  • $\begingroup$ Thanks! This is a really simple argument, and easily generalizable. My question is... it doesn't matter whether $\sqrt{2}$ is irrational or not? Because the same argument works if this number is rational but non integer. $\endgroup$ – Daniel Aug 6 '14 at 18:53
  • $\begingroup$ @Devilathor: It actually doesn't: For example if you had $1/2$ instead of $\sqrt{2}$, then $(1/2)^2$ wouldn't be in the set. You need some kind of algebraic integrality condition (in fact I think you need a square root of an integer to apply the same exact argument). $\endgroup$ – Najib Idrissi Aug 6 '14 at 19:01
  • $\begingroup$ @Najib: Oh yes, that's true, so we've got to make sure that $x^n$ lies in the set for all $n$, now that you mention it... By that reason this argument cannot be generalizable to numbers of the form $x+yr$ with $r$ irrational, am I right? If i'm not wrong, the statement is true in that case too. $\endgroup$ – Daniel Aug 6 '14 at 19:09
  • $\begingroup$ @Devilathor: As I said you need some kind of condition. For example if $r^2 = ur+v$ where $u,v \in \mathbb{Z}$, then you can use the same argument. $\endgroup$ – Najib Idrissi Aug 6 '14 at 19:19

Here's an alternative solution, using one of my favorite questions: Characterizing Dense Subgroups of the Reals.

Let $B=\left\{x+y\sqrt{2}:x,y\in\mathbb{Z}\right\}$. Then $B$ is a subgroup of $\mathbb{R}$. By the referred question, $B$ is either dense or of the form $B=\mathbb{Z}a$ for some $a\in\mathbb{R}$. If the latter were true, we would have $1\in B=\mathbb{Z}a$, so $a\in\mathbb{Q}$, but also $\sqrt{2}\in B=\mathbb{Z}a\subseteq\mathbb{Q}$, an absurd.

Thus, $B$ is dense in $\mathbb{R}$. Since $B=-B$, we can approximate $0$ by positive elements of $B$, that is, by elements of $A=\left\{|x|:x\in B,x\neq 0\right\}=\left\{x:x\in B,x>0\right\}$, so $\inf A=0$.

  • $\begingroup$ Thanks a lot, this is a very important answer because it's easily generalized to the set $B_r=\{x+yr:x,y\in \mathbb{Z}\}$ where $r$ is irrational. $\endgroup$ – Daniel Aug 6 '14 at 19:28

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