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Topic: A rational X irrational game
Replies: 10   Last Post: Mar 20, 2004 5:52 PM

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Artur

Posts: 72
Registered: 12/13/04
Re: A rational X irrational game
Posted: Mar 19, 2004 2:35 PM
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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




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