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Consider the ring of integral quaternions, $\mathbb{I}$, with norm-like function $N(a+bi+cj+dk) = a^2+b^2+c^2+d^2$.

Let $z,w \in \mathbb{I}$, with $w \neq 0$. Prove that $\exists q, r \in \mathbb{I}$ such that $z = qw + r$, with $N(r) < N(w)$.

We already know such things like: $N$ is multiplicative, there are precisely 24 units in $\mathbb{I}$ every element in $\mathbb{I}$ is associate to an element with all coefficients integers.

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    I tried extending the proof that division works in $\mathbb{Z}[i]$ by looking at the "pretend" $z/w$ and finding two integers $e, f$ which are within 1/2 of each component of $z/w$, and using $q = e+fi$. However, in the quaternion case this means we get the remainder has norm at most 2, and there are plenty of integral quaternions with that small of a norm...2011-11-08

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Your comment is basically correct; I think you just made a calculation error. If each component of $x = z/w$ (of course, we have to pick a side for the division here), is rounded to the nearest integer to produce a quaternion $x_0$ with integer coefficients, then the norm between $x$ and $x_0$ will be no more than $(1/2)^2 * 4 = 1$ (not 2 as you said in the comment). We are done, unless the distance is exactly 1, but I think you can figure out what happens in that case.

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    +1 It seems to me that the covering radius of the Hurwitz order $\mathbb{I}$ is $1/\sqrt2$, i.e. the same as with the lattice of the Gaussian integers in the complex plane.2011-11-08