- #1
pantboio
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I have the following assignment:consider the map $$|\cdot|:\mathbb{Z}\longrightarrow \mathbb{N},\qquad |a+ib|:=a^2+b^2$$1) Prove that $|\alpha|<|\beta|$ iff $|\alpha|\leq |\beta|-1$ and $|\alpha|<1$ iff $\alpha=0$2) Let $\alpha,\beta\in\mathbb{Z},\beta\neq 0$. Prove that the map $f:\mathbb{Z}\longrightarrow\mathbb{Z}, f(\gamma):=\alpha-\gamma\beta$ is the composition of a dilatation by the factor $\sqrt{|\beta|}$, a rotation (angle?) and a translation.3) Deduce that there exists $\gamma\in\mathbb{Z}$ such that $|f(\gamma)|$ is strictly smaller than $|\beta|$.$\textbf{Hint:}$ compare the size of a cell of the lattice $f(\mathbb{Z})$ and the size of the set of points whose distance to $0$ is $\leq\sqrt{|\beta|}$.What i did: point 1) is a trivial consequence of the fact that the norm takes integer non negative values. For point 2), I use complex multiplication of numbers which is: multiply absolute values and add angles. For point 3), I'm actually waiting for a miracle... I suppose i should prove that there exists a cell in $f(\mathbb{Z})$ intersecting the open ball centered at the origin with radius $\sqrt{|\beta|}$, but i have no idea how to write down this. Only thing i noticed is that $f$ acts with a rotation, which does not affect distance from the origin, so that the only changes in $|\gamma|$ come from dilatation and by adding $\alpha$.Could someone put me on the right direction? Thanks in advance
