Proving Dyadic Squares Interaction Rules

[Mr Davis 97]

Homework Statement

A rational number p/q is dyadic if q is a power of 2, q = 2k for some nonnegative
integer k. For example, 0, 3/8, 3/1, −3/256, are dyadic rationals, but 1/3, 5/12
are not. A dyadic interval is [a, b] where a = p/2k and b = (p + 1)/2k. For
example, [.75, 1] is a dyadic interval but [1, π], [0, 2], and [.25, .75] are not. A
dyadic cube is the product of dyadic intervals having equal length. The set of
dyadic rationals may be denoted as Q2 and the dyadic lattice as Qm2
.
Prove that any two dyadic squares (i.e., planar dyadic cubes) of the same
size are either identical, intersect along a common edge, intersect at a
common vertex, or do not intersect at all.

Homework Equations

The Attempt at a Solution

After constructing many dyadic squares, I can intuitively see why this is true, but I don't know what a proof would really entail. How do I show in general that these are the 4 and only 4 ways dyadic squares could interact?

EDIT:
Actually, I might have a rough solution. The length of any dyadic interval is $1/2^k$, hence, if any two dyadic squares have the same length, then it must be the case that they have the same value of k. Hence, any dyadic square is determined by an integer coordinate pair $(a,b)$ where $[a/2^k, (a+1)/2^k] \times [b/2^k, (b+1)/2^k]$ is an arbitrary dyadic square. Since $a$ and $b$ are integers, the possible dyadic squares of the same fixed length can be represented as occupying an integer lattice where all squares are unit squares in the lattice, which clearly shows that any two dyadic squares of the same size are either identical, intersect along a common edge, intersect at a common vertex, or do not intersect at all.

 

[haruspex]

Homework Statement

A rational number p/q is dyadic if q is a power of 2, q = 2k for some nonnegative
integer k. For example, 0, 3/8, 3/1, −3/256, are dyadic rationals, but 1/3, 5/12
are not. A dyadic interval is [a, b] where a = p/2k and b = (p + 1)/2k. For
example, [.75, 1] is a dyadic interval but [1, π], [0, 2], and [.25, .75] are not. A
dyadic cube is the product of dyadic intervals having equal length. The set of
dyadic rationals may be denoted as Q2 and the dyadic lattice as Qm2
.
Prove that any two dyadic squares (i.e., planar dyadic cubes) of the same
size are either identical, intersect along a common edge, intersect at a
common vertex, or do not intersect at all.

Homework Equations

The Attempt at a Solution

After constructing many dyadic squares, I can intuitively see why this is true, but I don't know what a proof would really entail. How do I show in general that these are the 4 and only 4 ways dyadic squares could interact?

EDIT:
Actually, I might have a rough solution. The length of any dyadic interval is $1/2^k$, hence, if any two dyadic squares have the same length, then it must be the case that they have the same value of k. Hence, any dyadic square is determined by an integer coordinate pair $(a,b)$ where $[a/2^k, (a+1)/2^k] \times [b/2^k, (b+1)/2^k]$ is an arbitrary dyadic square. Since $a$ and $b$ are integers, the possible dyadic squares of the same fixed length can be represented as occupying an integer lattice where all squares are unit squares in the lattice, which clearly shows that any two dyadic squares of the same size are either identical, intersect along a common edge, intersect at a common vertex, or do not intersect at all.

Yes, I think that's good enough. A pedant might ask for a more rigorous proof of the last part.