On Mon, Dec 5, 2011 at 3:31 PM, saurabh singh <saurab...@gmail.com> wrote:
> I was wondering can we design a machine(Even hypothetical) that can find > a *perfect square root *of any integer thats given to it. > My logic why we can't is since there are uncountably infinite real > numbers, there will be uncountably infinite numbers requiring infinite > states on a turing machine.But since there are only finite number of > states,we cant make such a machine.And since we cant make a turing machine > for calculating the square root we cant make any computing machine for the > same. > I am not sure about my logic though.Thats why i have this doubt. > > Just a thought, If you are saying that there are infinite real numbers then it will require infinite number of states on turing machine. So, the same explanation holds for every arithmetic operation. If you talk about addition then also there are infinite number of numbers so there must be infinite number of states and so not possible to have such a machine according to your argument but we do have such machines. My point is that you are wrong somewhere that since there are infinite real numbers so we must have infinite number of states in turing machine. -- > Saurabh Singh > B.Tech (Computer Science) > MNNIT ALLAHABAD > > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Aamir Khan | 3rd Year | Computer Science & Engineering | IIT Roorkee -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
