Dear all, Can the plane be n-coloured, such that there are no mono- chromatic continuous curves of nonzero length? For n = 4, there is the 'Millar colouring': Colour a point (x,y): * red if x and y are both rational; * green if x is irrational, y is rational; * yellow if x is rational, y is irrational; * blue if x and y are both irrational. It is obvious that no monochromatic continuous curves can exist, as both the ordinate and abcissa are fixed (as if they varied by a nonzero amount, they would cross an infinity of rational and irrational points). With a little effort, it is possible to define a 3-colouring with similar properties: * red if x and y are both irrational; * red if x is a dyadic rational and y is irrational; * green if x is a non-dyadic rational and y is irrational; * blue if x is irrational and y is rational; * blue if x is a non-dyadic rational and y is rational; * green if x is a dyadic rational and y is rational; (A dyadic rational has a terminating binary expansion; a non-dyadic rational has a recurring binary expansion; an irrational has an aperiodic binary expansion.) It is clear that no red and blue curves exist, as they can only be horizontal lines, and are terminated by the infinity of green points on the lines. Also, no green lines exist, as dyadic and non-dyadic rationals are separated by an infinity of irrational numbers. For 2-colourings, this is not so easy. The 'reduced Millar colouring', where a point is red if x and y are both either rational or irrational, and blue otherwise, does not satisfy the property. For example, the curve y = x is exclusively red. A variant, the 'Elliott colouring' was proposed, with the following rules: * red if x and y are both irrational; * blue if x is irrational, y is rational; * blue if x is a dyadic rational, y is irrational; * red if x is a dyadic rational, y is rational. However, Appendix I shows how the reals can be continuously deformed to map the ordinary rationals to dyadic rationals, thereby rendering the Elliott colouring equivalent to the reduced Millar colouring. So, I leave the n = 2 case as an 'exercise for the reader'. Sincerely, Adam P. Goucher ***** Appendix I ***** Consider the following sequences, which can be derived from the Stern-Brocot tree: [0] [0,1/2] [0,1/3,1/2,2/3] [0,1/4,1/3,2/5,1/2,3/5,2/3,1/4] ... For each rational in the interval [0,1), there is a constant k such that for all n > k, the rational appears in the nth sequence. A similar sequence of sequences can be created for the dyadic rationals in the interval [0,1): [0] [0,1/2] [0,1/4,1/2,3/4] [0,1/8,1/4,3/8,1/2,5/8,3/4,7/8] ... This gives a natural bijection between the rationals in [0,1) to the dyadic rationals in [0,1), which can be extended to a bijection between all rationals and all dyadic rationals. It also extends to a mapping from the reals to the reals, which operates as follows: 1. Express the real number, x, as its integer part followed by binary expansion: x = n . a_1 a_2 a_3 a_4 a_5 ... 2. Let L_0 = n, H_0 = n+1. 3. Iterate for all natural numbers i: (a) If a_i = 0, let L_i = L_(i-1), H_i = mediant(L_(i-1),H_(i-1)) (b) If a_i = 1, let H_i = H_(i-1), L_i = mediant(L_(i-1),H_(i-1)) 4. The sequences L_i and H_i converge to a real number, y, from below and above, respectively. ***** End of appendix I *****