🚀 go-pugleaf

In the referenced article, gauthier@ug.cs.dal.ca (Paul Gauthier) writes:
>
>Is it possible to specify uniquely, an arbitrary point in 3D space using 
>only 2 integers? IE: You point to any point in space (using a conventional
>three-tuple (x, y, z)). Is it possible for me to respond with 2 integers
>which corrispond to that point and no other? You can even use real numbers
>in your three-tuple, but my response has to be two integers. Also, the
>3D space is infinite in all directions, so the values of each component
>of your three-tuple are completely unconstrained. Is it possible for me
>to use some co-ordinate system and label the point you give me uniquely
>with 2 integers?
>

No - in fact more is true: given any large finite number N you can't even
specify a point on the real line using N integers in the way you suggest. The
reason is that the real line (and 3-dimensional Euclidean space) is bigger than
the integers: in the jargon we say that the integers are "countable" but the
reals are "uncountable". So there can never be what you hope for: what
mathematicians call a "bijection" between \Z^{2} and \R^{3}.

However if you are allowed to discretise your space ie model \R^{3} with a
countable lattice which may (without loss of generality) be viewed as \Z^{3}
then you can specify any point on this lattice using just one integer. There
are lots of ways to do this: one simple (certainly not the best) way would be
to map the three integers (x,y,z) [bijectively] to the positive integer

	2^{|x|}*3^{|y|}*5^{|z|}*7^{1+sgn(x)}*11^{1+sgn(y)}*13^{1+sgn(z)}

(where sgn(x)= +1 if x>0, 0 if x=0 and -1 if x<0)

You could even use this approach if you are allowed to model \R^{3} by \Q^{3},
which is a dense subset, since the rationals are countable (ie there are as
many rationals as natural numbers but more reals). But none of this is very
helpful for considering functions which might arise physically (I assume - if
I'm wrong I'd be most interested to see a counterexample to my assertion!). If
you want to do any mathematics involving "continuous" or "differentiable"
functions then this is not going to provide a way to do it. In short three
dimensions means just that - you can't cut it down to two!

Simon