The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
W. Dale Hall

Posts: 129
Registered: 12/6/04
Re: A rational X irrational game
Posted: Mar 19, 2004 7:57 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Keith A. Lewis wrote:
> (Artur) writes in article <> dated 19 Mar 2004 07:30:09 -0800:

>>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 the players agree and cooperate to converge on a single number, you can
> pin it down like that. But they can't be guaranteed to do that (in fact
> they probably won't if they are competitive).

How can the final intersection of these nested intervals, no matter
how they are chosen, ever contain more than a single number? The nth
interval is shorter than 1/n in length. If there are two numbers in the
intersection, they differ by some amount, and there is a N so that the
reciprocal 1/N is smaller than that difference. That means that from
step N on, at most one of those numbers is in the intersection. The
intersection must be nonempty, since these are closed intervals of R,
and the real line has the property that a nested sequence of nonempty
compact intervals has nonempty intersection.

>>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.

> Within any In of length Ln > 0, there will be both rational and irrational
> numbers. In fact, the cardinality of the respective sets of remaining
> rational and irrational numbers after move n is the same as it was before
> move 1. The game never ends.

It ends in the limit. Take 1/2 second for step 1, 1/4 second for
step 2, and so forth with 1/2^N second for step N. The game is over
in one second, albeit the moves are somewhat rapid towards the end.

>>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.

> --Keith Lewis klewis {at}
> The above may not (yet) represent the opinions of my employer.


Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.