Suppose that I drew an infinite number of disjoint closed curves in the plane (such as circles, squares, etc.). Suppose that I then tell you that there is one curve for each positive real number.

You would not have too much trouble believing my assertions at this point. For example, I could have drawn all circles with center at the origin. They are all disjoint, and for each positive real number x, there is a corresponding circle - namely, the circle of radius x.

But suppose that I also tell you that all the curves I drew were figure eights. Can you believe my assertions now?

(A figure eight is a curve in the plane obtained from the basic "8" shape by any combination of translation, rotation, expansion, or shrinking.)

(In reply to re: still thinking by Charlie)

Charlie wrote:
"Even though there are many more irrationals than rationals, you can always make the separation between rationals as small as you like--there is no "next rational". If your 8 at rational a/b were any finite size, c (as measured by the distance of its outer part from its center), presumably smaller than 1, you could take 1/(100*ceil(c)) and add it to a/b, and you'd get another rational number (or 1/ceil(c), etc.)."

My idea is that the rationals aren't a problem but the irrationals are.

I'm not sure I understand your formula above since for a small value of c, I assume that ceil(c) = 1 which added to a/b gives (a+b)/b.   Maybe you meant (a/b)+(1/100)*c  which is clearly very close to a/b.   But plotting the 8 for that value would be at [100a+bc]/[100b], or x=100a+bc,  y=100b  which is very far from x=a, y=b.      So for the rationals, close numbers map to (x,y) coordinates that are potentially far away from each other.

But I agree, I can't figure out a way to map the irrationals that guarantees no overlap with other irrationals, unless someone knows more than I do about the "magnitude" of the infinity of irrational numbers, eg Aleph null, vs Aleph one etc.

Posted by Larry on 2005-02-16 03:56:09