Artur
Posts:
72
Registered:
12/13/04


Re: A rational X irrational game
Posted:
Mar 19, 2004 2:35 PM


David C. Ullrich <ullrich@math.okstate.edu> wrote in message news:<th5m501fv6f645mm5tej3va2llt079rt35@4ax.com>... > On 19 Mar 2004 07:30:09 0800, artur@opendf.com.br (Artur) wrote: > > >Hello > >I've been thinking about the following problem (or puzzle) but hasn't > >come to a conclusion yet. Maybe someone can give a hint. > > > >A and B are playing the following game: On the real line, A chooses a > >closed interval I1 of length 0<L1<=1. Then, B chooses a closed > >interval I2, of length 0<L2<=1/2, contained in I1. Then A, in turn, > >chooses a closed interval I3, of length 0<L3<=1/3, contained in I2, > >and so on. We know there's one, and only one, real number x that > >belongs to all of the intervals I1, I2, I3.... If x is rational, A > >wins the game, and if x is irrational, then B wins. We are asked to > >find an strategy that B should follow so that he will certainly win > >the game no matter how B chooses his intervals. > > > >Since the rationals are countable and the irrationals are not, it > >really seems that there is such a strategy that assures B will ever > >win, but I couldn't find it so far. > > Hint: B doesn't have to force a win on the first move. On > the first move B prevents one way he can lose, on the > next move he prevents another method of losing... > Ummm... based on your hint, this may work: Once the interval I1 has been choosen by A, B enumerates the rationals in I1. Let {x_1, x_2....} be such enumeration. Then, B chooses I2 in such a way that x_1 is not in I2 (this is always possible, because the length of I2, though positive, is allowed to be arbitrarily small). In the next move, A chooses an I3 contained in I2. And, in his turn, B chooses an I4 that doesn't contain x_2 . Carrying on with this method of choices, B ensures the game will generate a sequence {Im} of closed intervals such that no x_n belongs to all of the Im's. Therefore, the element x, common to all Im's, is not covered by the enumeration {x_1, x_2....}, which implies x must be irrational. And B wins! Artur

