Norms in $\mathbb{Q}[i]$ So I ran through this problem in my study, and need a bit of clarification.
Theorem: Let $r+si \in \mathbb{Q}[i]$. Then there is an element $a+bi \in \mathbb{Z}[i]$
such that 
$N((r+si)-(a+bi)) <1$.
Proof:
Pick integer $a$ and $b$ such that $|r-a| \leq 1/2$ and $|s-b| \leq 1/2$.
Now we have
$N((r+si)-(a+bi)) = N((r-a)+(s-b)i)$. Then:
$=(r-a)^2+(s-b)^2 \leq (1/2)^2 + (1/2)^2 = 1/2 <1$. 
End proof.
So I understand that they choose integers $a$ and $b$ closest to $r$ and $s$, so that the real numbers $r$ and $s$ can only have a maximum difference of $1/2$ to its closest integer in $\mathbb{Z}[i]$. And the normalization function is $a^2+b^2$. What I don't understand is what I am necessarily looking at. Or even the point of this proof, as it's not clarified. To me, choosing the values closest to $r$ and $s$ seems a bit arbitrary, and this seems more like an exercise of simple arithmetic than a proof. Like this is more of something like;
we have $4x$, prove that there is some x that $4x<5$. 
Also the exercise for eisenstein numbers is left for exercise, and I imagine its pretty much the same, choosing the same values of $a$ and $b$, the only difference is the norm function for $\mathbb{Q}[w]$.
 A: Can you picture the grid of points $\Bbb Z[i]$ in the complex plane? This problem shows that for any point $a+bi$ of $\Bbb Q[i]$, the circle about that point of radius $1$--or really, any radius greater than $\frac1{\sqrt2}$--will have at least one point $m+ni$ of $\Bbb Z[i]$ inside it. We shouldn't choose an integer $m$ too far from $a$, for then it wouldn't matter how close $n$ was to $b$. Likewise, we shouldn't choose $n$ too far from $b$. Since $a,b\in\Bbb Q$ are arbitrary, then the best we can do is choose $m,n$ to be within $\frac12$ away from $a,b$ (respectively). Fortunately, that is enough.
A: The point of this exercise is that it is an important step in giving the Gaussian integers the structure of a Euclidean domain. For that you need to have some division procedure that, given any $\def\Z{\Bbb Z}x,y\in\Z[i]$ with $y\neq0$, will produce a quotient $q$ and remainder $r$, both in $\Z[i]$, such that $x=qy+r$ and $N(r)<N(y)$ for some function $N:\Z[i]\to\Bbb N$. The most obvious choice, and one that works here, is to take for $N$ the norm function (square of the absolute value as a complex number). But you need some idea of getting your quotient $q$ (then $r=x-qy$ will follow). The idea is to compute the exact quotient $\frac xy$ in $\Bbb C$, which will actually lie in $\Bbb Q[i]$ but usually not in $\Z[i]$, and to take for $q$ an element of $\Z[i]$ that is sufficiently close to$~\frac xy$. Since $|r|=|x-qy|=|y|\cdot|\frac xy-q|$, one needs, in order to get $N(r)<N(y)$, to ensure that $|\frac xy-q|<1$ or equivalently $N(\frac xy-q)$. So you need to solve exactly the statement of the theorem, with $r+si=\frac xy$ and $a+bi=q$ (this $r\in\Z$ should not be confused with the remainder). The idea of rounding the real and imaginary parts of $\frac xy$ to the nearest integer works fine for the Gaussian integers, and actually gives the best possible quotient (that is, with smallest remainder in the sense of the norm $N(r)$), though possibly an ex aequo with another possibility. Note that the situation is more subtle than in $\Z$, where Euclidean division can round the quotient always down, and is usually taken to do that (although one doesn't have to do this); doing such an operation in $\Z[i]$ to real and imaginary parts would however not be good enough to ensure the condition $N(r)<N(y)$ that is required for a Euclidean algorithm.
The situation for the Eisenstein integers is similar, but not quite the same. One does not have integer real and imaginary parts, but rather integer $a,b$ in an expression $a+bw$ (where $w$ is a 3rd root of unity). Taking $s+tw=\frac xy$ with $s,t\in\Bbb Q$ and rounding both to the nearest integer will ensure an approximation at distance strictly less than $1$, but it won't necessarily be the best approximation. In practice it is good enough, and going for the best possible quotient would require considerable extra effort, with in the end little gain.
