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Finite number of steps (own)
Posted: Wed Jan 05, 2011 8:35 pm
by nayel
Start with any real number other than $0$ and $1$. At each step, replace the number $x$ by either $1/x$ or $1-x$. Prove that, if we can get from $a$ to $b$ in a finite number of steps, the maximum number of steps required is two.
Example:
Re: Finite number of steps (own)
Posted: Wed Jan 05, 2011 9:37 pm
by Avik Roy
This problem is an interesting one. Though I think the maximum number of steps needed is 3.
Re: Finite number of steps (own)
Posted: Wed Jan 05, 2011 10:51 pm
by anumoshsad
Nice problem and nice solution.
I also think the number of steps needed is 3.
Let $f(x)=\dfrac{1}{x}$ and $g(x)=1-x$.So $g\circ f= 1- \dfrac{1}{x},f\circ g=\dfrac{1}{1-x},f\circ g\circ f=g\circ f\circ g=\dfrac{x}{x-1}$.
And the fact that the set $\{id,f,g,g\circ f, f\circ g,f\circ g\circ f\}$ is closed under the composition of functions proves it all.
Re: Finite number of steps (own)
Posted: Wed Jan 05, 2011 11:50 pm
by nayel
Yes, I'm sorry, the number of steps required should be 3. Nice solutions!
Try this slightly modified version I just posted:
viewtopic.php?f=23&t=285
Re: Finite number of steps (own)
Posted: Fri Jan 07, 2011 6:59 pm
by abir91
সুন্দর সমস্যা এবং সমাধান। কিন্তু maximum এর জায়গায় minimum হবে মনে হইতেছে।