You have an ink stamp that is so amazingly precise that, when inked and pressed down on the plane, it makes every circle whose radius is an irrational number (centered at the center of the stamp) black.

Is it possible to use the stamp three times and make every point in the plane black?

If it is possible, where would you center the three stamps?

(In reply to by Benjamin J. Ladd)

I think you're reverting to the idea that an ink line will have depth and area, in which case this problem makes no sense to even address.

"For any given point on a sheet of paper, this unique stamp will ink the entire surface."

No, if i center the stamp at exactly (0, 0) on a Cartesian plane, I know that the point (0, 3) is a rational distance (exactly three units) away, and will not be inked by that single stamp.

Actually, looking at a single radius from a single point, the distances must actually alternate rational and irrational points. You can prove that there exists an irrational number between any two unequal rational numbers, and that there exists a rational number between any two unequal irrational numbers.

To prove the first one, assume two rational numbers a and b. Let x = a + (b-a)(√2)/2. Thus, x is irrational, and a < x < b.

To prove the second, assume two irrational numbers x and y. Conceptually, if you write x and y out in decimal form, if they are not equal, the first digit that is different must exist at some countable location in the number, where you can just pick a rational (terminating) number that fits between them.
More formally:
if (y-x < 1), n = ceil(-log(y-x))
if (y-x ≤ 1), n = 0
then, let a = x(10^(n+1)) and b = y(10^(n+1))
b - a ≥ 10, so:
a < ceil(a) < floor(b) < b
a = x/(10^(n+1)) < ceil(x)/(10^(n+1)) < floor(y)/(10^(n+1)) = b
Then, let z = floor(b)/(10^(n+1)), so z is rational, and
x < z < y.

In any case, the answer is that a single stamp will cover an entire plane.

Posted by DJ on 2003-12-01 23:34:51