$x+y\sqrt{2}$ infimum ($x,y\in \mathbb{Z}$)

Question

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)

Answer 1

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

Answer 2

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 $\inf A=0$.