1

This is actually a question I find really hard to answer.any hints are appreciated. By the way feel free to edit the tags as i really do not know which category is this question is in.

Two genius mathematician have guessed two numbers which their difference is 1. Each of them asks the question "Have you guessed my number yet" repeatedly after the other one.

Prove that there will be a question which's answer would be "yes"

Minuano
  • 617
  • "Which their subtraction is 1"? – Batman May 22 '14 at 20:35
  • oh sorry!I meant if you subtract them the answer would be 1.I guess their is extra right? sorry for my crappy english.I'm trying to get used to it. – Minuano May 22 '14 at 20:36
  • Can the numbers be complex numbers? :) –  May 22 '14 at 20:38
  • I still don't quite understand the game. If you have numbers $x$ and $y$ such that $|x-y|=1$, then you know that $x=y+1$ or $x=y-1$. So the person with number $y$ guesses $y+1$ then $y-1$ and the person with number $x$ guesses $x+1$ then $x-1$, and you have to get the right number within 2 rounds. – Batman May 22 '14 at 20:39
  • @Batman but the problem is,how do you exactly guess the numbers?There are plenty of numbers right? :D – Minuano May 22 '14 at 20:41
  • 2
    The numbers are positive integers, right? – Andrés E. Caicedo May 22 '14 at 20:41
  • In your example no one asks "have you guessed my number yet?" – mvw May 22 '14 at 20:42

1 Answers1

4

Suppose I know the numbers are positive integers.

If my number is $1$, and I am asked whether I know the other, I say "yes" because it is $2$. Else I say "no". If my colleague also says "no" the pair cannot be $(1,2)$ or $(2,1)$.

On the next round we confirm or exclude either $(2,3)$ or $(3,2)$ etc.

If I have $n$ it is impossible to exclude both $(n, n-1)$ and $(n, n+1)$ so the process must come to an end.

Mark Bennet
  • 100,194